i zakon termodinamike du =dq +dw+ dwe dh =du + pdv u c h p ... termodinamika.pdf · jedan mol gasa...

VV T

UC ⎟⎠⎞

⎜⎝⎛=partpart

edwdwdqdU ++=

PP T

HC ⎟⎠⎞

⎜⎝⎛=partpartpdVdUdH +=

dPPHdTCdH

Tp ⎟

⎠⎞

⎜⎝⎛partpart

+=dVPdTCdVVUdTCdU uV

TV +=⎟

⎠⎞

⎜⎝⎛+=partpart

CPm ndash CVm = RCP ndash CV = nR

Izotermski reverzibilni zapreminski rad gasa u IGS

Vpw Δminus= Izotermski revetzibilni zapreminski rad isparavanja p-napon pare

Izotermski procesi

I zakon termodinamike

2

1

1

2 lnlnPPRT

VVnRTw minus=minus=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛minus=minus= 11)(

2

11

1

2112

VCnR

VVVad VVTC

TTTCTTCw

Adijabatski zapreminski rad

1

2

1

2

2

1

1

2

1

2

2

1

PP

TT

VV

TT

VV

TT R

CR

CCR PVV

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγ

⎟⎟⎠

⎞⎜⎜⎝

⎛===

1

2

2

1 VV

PPconstPVPV VP CC

Adijabatski procesi

Jednačina adijabate

RTnUVPUH gasΔ+Δ=Δ+Δ=Δ

2

12

2 TTT

qw minus=

minus=η Efikasnost toplotne mašine

0gege dSTdqdS II zakon termodinamike-

Reverzibilni i ireverzibilni

1

2

1

212 lnln

VVR

TTCSSS V +=minus=Δ

oksistotoki

iirev

sis SSSTHSS

TqS Δ+Δ=Δ

Δminus=Δ==Δ sumν Ireverzibilni

1

2

1

212 lnln

PPR

TTCSSS p minus=minus=Δ

tr

trtr T

HS Δ=Δ Fazni prelazi

VV

V

V TSTCodnosno

TC

TS

⎟⎠⎞

⎜⎝⎛partpart

==⎟⎠⎞

⎜⎝⎛partpart

( ) int int==minus=2

1

2

1

ln12

T

T

T

TVVV TdC

TdTCSSSΔ


1

2

1

ln12

T

T

T

TPPP TdC

TdTCSSSΔ

VT TP

VS

⎟⎠⎞

⎜⎝⎛partpart

=⎟⎠⎞

⎜⎝⎛partpart

PT TV

PS

⎟⎠⎞

⎜⎝⎛partpart

minus=⎟⎠⎞

⎜⎝⎛partpart

int ⎟⎠⎞

⎜⎝⎛partpart

minus=minus=Δ dPTVSSS

P12int ⎟

⎠⎞

⎜⎝⎛partpart

=minus=Δ dVTPSSS

V12

Maksveloverelacije

TV VU

TPTP ⎟

⎠⎞

⎜⎝⎛partpart

minus⎟⎠⎞

⎜⎝⎛partpart

=

sum=

minus=minus=Δn

iiimeš xnRSSS

112 ln Entropija mešanja

TP PH

TVTV ⎟

⎠⎞

⎜⎝⎛partpart

+⎟⎠⎞

⎜⎝⎛partpart

=

Termodinamičke jednačinestanja

WkS ln= k-Bolcmanova konstanta k=RNA=138middot10-23 JK

Helmholcova i Gipsova slobodnaenergija

TV VS

TP

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

PVA

Tminus=⎟

⎠⎞

⎜⎝⎛partpart S

TA

Vminus=⎟

⎠⎞

⎜⎝⎛partpart

dA=minus PdV minus SdT dG = VdP ndash SdT

TP PS

TV

⎟⎟⎠

⎞⎜⎜⎝

⎛minus=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

STG

P

minus=⎟⎠⎞

⎜⎝⎛partpart

VPG

T

=⎟⎠⎞

⎜⎝⎛partpart

pTGTHG ⎟⎠⎞

⎜⎝⎛+=partΔpartΔΔ 2

)(TH

TTG

p

ΔpartΔpart

minus=⎟⎠⎞

⎜⎝⎛

00 ln

PP

RTGG mmm +=

ZADATAK 1

Izračunati rad potreban da ptica mase 120g uzleti do visine od 50m oda) Površine Zemljeb) Površine Meseca (g=16ms-2)

Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2

minus=sdotsdotminus=minus=


ZADATAK 2

Hemijska reakcija se dešava u cilindru preseka 100cm2 Kaorezultat klip se pomerio za 20cm nasuprot pritiska od 1 atm Koliki je rad

Rešenje

JmPaVPw 65202102000101325 36 minus=sdotsdotminus=Δminus= minus

Zadatak 3

U ciklusu 1 mol gasa u idealnom gasnom stanju vrši rad od 4186J Koliko je q

Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4

Rad (u J) koji izvrši gas u IGS pri širenju nasuprot pritisku od 1 atm i od 10 do 30 L je

a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje

JmPaw 5202610)1030(101325 33 minus=sdotminusminus= minus

Zadatak 5

Jedan mol gasa u IGS u početku na pritisku od 1 atm i 300K se reverzibilno zagreva do 400 K pri konstantnoj zapreminiIzračunati krajnji pritisak i promenu unutrašnje energijeToplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int

ZADATAK 6 Jedan mol gasa prelazi iz stanja sa pritiskom P0 i zapreminom V0 u stanje sa duplo većom zapreminom i pritiskom Na p-V dijagramu taj proces je predstavljen pravom linijom Koliki rad izvrši gas u tom procesua)PoV0 b) PoV02 c) 3 PoV02d) 2 PoV0 e) 3 PoV0 f) ne znam

P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=

ZADATAK 7U termodinamičkom procesu promena unutrašnje energije

sistema je ∆U = -300 J sistem prima toplotu od 100 J i širi se nasuprot pritiska od 1 bar Kolika je promena zapremine (L)

a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35

=sdotsdot=minusminus

=Δ

minus=Δsdotminus=Δminus=

minus=minusminus=+=Δ

minus

Zadatak 8bull Električni grejač snage 15W greje masu od 12g

vode tokom jednog minuta Ako je početna temperatura vode 35oC kolika će biti krajnja temperatura vode (K)Specifični toplotni kapacitet vode je 1 calstepg

bull RešenjeSnaga grejača je P=ΔUt Energija kojugrejač oslobađa jeΔU=Psdott=15Js sdot60s=900J

KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ

Zadatak 9bull 1 mol vode isparava Kolika je promena entalpije ako je

pritisak 1 bar Promena unutrašnje energije pri isparavanju je 407kJmol

molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==

rangrangminus=ΔΔ+Δ=Δ

Rešenje

Zadatak 10Domaći

bull Koliko energije treba dovesti masi od 135kg vode da bi se zagrejala od 20oC do

bull temperature ključanja Pretpostaviti da je Cvs=4186JgK

ZADATAK 11Izračunati ΔU za reakciju sagorevanja 10 mola propana na 25oC ako je ΔH=-2058 kJ

Rešenje

C3H8(t)+5O2(g)rarr3CO2(g)+4H2O(t)

Δng=3-5=-2

ΔU=ΔH-ΔngRT=-2058000J+2sdot83145sdot298=-2053kJ

ZADATAK 12Izračunati rad širenja pri elektrolizi 50 g vode pri konstantnom pritisku i temperaturi od 25oC

Rešenje2H2O(t)rarr2H2(g)+O2(g)

w=-PspΔV=-Psp(Vk-Vp)asymp-PspVk= nRTP

nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

ZADATAK 13 Rad reverzibilnog izotermskog širenja jednog mola gasa u

idealnom gasnom stanju od zapremine V1 do zapremine V2 je dat izrazom

a) ndashp(V2 -V1) b) ndashp(V1 -V2) c)

d) CV(T2-T1) e) f) ne znam

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14

bull Izračunati napon pare vode na 35oC ako se vodena para širi reverzibilno i izotermski od zapremine 5 cm3 do zapremine 100 cm3 ako se pri tome vrši rad od 534 mJ

Rešenje

w=-p(V2-V1) w=-0534J p=0534Nm95sdot10-6m3=5621Pa

Zadatak 15

Dva mola idealnog gasa podleže izotermalnoj reverzibilnoj ekspanzijiod početne zapremine V1 do krajnje zapremine 10V1 i vrši rad od41860J Ako je početni pritisak 100 bar kolika je početna zapremina(u L) i temperatura (u K)

Rešenje

w=nRTlnV2V1=nRTmiddot23 P1V1=nRT=w23=4186023=18200JV1=18200100middot105=182middot10-3m3=182 LnRT=18200J T= 182002middot8314=1093K

Zadatak 16

Uzorak argona mase 656g zauzima zapreminu od 185dm3

Na 305K a) Izračunati rad koji bi se izvršio kada se gas širi izotermski nasuprot konstantnog spoljašnjeg pritiska od 77kPa dok mu se zapremina ne poveća na 25 dm3b) Izračunati rad koji bi se izvršio kada bi se ista ekspanzija vršila reverzibilno

Rešenjea)

b)

JmPaVPw sp 251910521077 333 minus=sdotsdotsdotminus=Δminus= minus

Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=

=sdotsdotsdotminus=minus=

Zadatak 17

Kada se 3 mola kiseonika zagreje pri konstantnom pritisku od 325atm njihova temperatura poraste od 260 K do 285 KAko je CPm=204JKmol izračunati ΔH q i ΔU

Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203

=sdotsdotminus=ΔminusΔ=Δ

=minussdot=Δsdot==Δ

ZADATAK 18

Toplotni kapacitet gasa u IGS varira sa temperaturom prema izrazu

za 1 mol Izračunati q w ΔU i ΔH za jedan mola gasa kada temperatura raste od 0oC do 100oCa) Na konstantnom pritiskub) Na konstantnoj zapremini

)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const

kJJTnRHUJKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273

==minus=ΔminusΔ=Δminus=sdotsdotminus=Δminus=Δminus=

=minus+minus==Δ

+==Δ= int int

qkJHUkJHw ==Δ=Δ=Δ= 9149140

ZADATAK 19

Rad reverzibilnog adijabatskog širenja jednog mola gasa u idealnom gasnom stanju od zapremine V1 pri temperaturi T1 do zapremine V2pri temperaturi T2 je dat izrazom


d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC

ZADATAK 20Jedan mol idealnog gasa na 300 K se širi adijabatski i reverzibilno

od 20 bar do 1 bar Koja je temperatura u krajnjem stanju gasa pretpostavljajući da je CV=(32) R

a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==

ZADATAK 21Izračunati krajnji pritisak argona (bar) posle reverzibilnog i adijabatskogširenja pri kome se zapremina poveća dva puta Početni pritisak jeiznosio 100 kPa a γAr=53

Rešenje

barkPakPaPPVVP 31031100

21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

Iz jednačine adijabate se dobija

ZADATAK 22Izračunati rad i promenu unutrašnje energije pri adijabatskom širenju 02 mol Ar od 05 do 10L Početna temperatura je izosila 25oC a molarni toplotni kapacitet Ar na konstantnoj zapremini iznosi 1248 JK-1mol-1

Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12

minus=Δ=minus=minussdotsdot=

=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći

Dva mola idealnog gasa za koji je Cvm=5R2 je reverzibilno zagrevan do 356K na konstantnoj zapremini Početni pritisak i temperatura su bili P1=111kPa i T1=277K Izračunati krajnji pritisak ΔU q i w

Rešenje

Jedan mol idealnog monoatomskog gasa u početku na 10 atm pritisku i temperaturi od 0oC širi se izotermski nasuprot pritiska od 1 atm Uslovi su takvi da je konačna zapremina 10 puta veća od početne krajnji pritisak je jednak spoljašnjem pritisku

(a) Izračunati početnu i krajnju zapreminu(b) Izračunati q w ΔU za proces

Zadatak 24

Kako je V1=224 L to je V2=224 LUkupni proces je izotermski pa je ΔU=0Izvršeni rad je -PΔV=-1013105Nm2 x(00224-000224)m3=w=-20422J q=20422J

ZADATAK 25Na vrlo niskoj temperaturi toplotni kapacitet čvrstih supstancija se može uzeti da je proporcionalan sa T3 pa se može pisati da je C=aT3Kolika je promena entalpije takve supstancije pri zagrevanju od 0 dotemperature T (koja je bliska 0)

Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26

bull Gas se pokorava jednačini stanja

a) Odrediti reverzibilni rad koji se vrši pri zagrevanju gasa od T1 do T2 pri konstantnom pritisku

b) Odrediti reverzibilni izotermski rad pri širenju od V1 do V2

pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222

TTpTTRVVpwpTRTpVpTRTpV

ααααminus+minus=minus=

+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27

Uzorak od 5mol ugljendioksida pri zapremini od 15dm3 na 280K podleže adijabatskoj ekspanziji nasuprot konstantnog pritiskaod 785kPa dok mu se zapremina poveća za faktor 4 Izračunatiq w ΔT ΔU i ΔH (CP=3711JKmol)Rešenje

JKmolKJmolJHnRTUnPVUPVUH

JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333

minus=minussdotsdot+minus=ΔΔ+Δ=Δ+Δ=Δ+Δ=Δ

minus=+=Δminus=Δminussdot

minus==ΔΔsdot=minus=

sdotminussdotsdotsdotminus=Δminus== minus

Zadatak 28Jedan mol idealnog monoatomskog gasa na početnim P1=2 atm i T1=273K je preveden na pritisak P2=4atm reverzibilnim putem definisanim sa PV=const Izračunati V1 V2 i T2 ΔU ΔH q i w

RešenjeIz jedn id g stanja je V1=112 L Pošto je PV=const to je V2=224 L Kombinovanjem PV=const sa PV=RT dobija se TV2=const pa je T2=4T1=1092KΔU=CVΔT=(3R2)819=1021kJ ΔH=CPΔT=17022kJ Da bi se dobilo w treba odrediti w=intPdV Iz početnih uslovaPV=const=2112=0178atmL pa je w=0178intVdV=0089(V2

2-V12)

=0089middot375=333Latm ili w=823cal=34045j q=ΔU-w=1624cal=1361752J

Zadatak 29

Jedan mol idealnog gasa sa CV=2093Jmol step u početku pristandardnim uslovima prolazi kroz sledeći reverzibilni ciklus A izohorsko zagrevanje do dvostruke vrednosti početne temperature od stanja 1 do 2 B adijabatsko širenje od 2 do 3 do početne temperatureC izotermalnu kompresiju od 3 do 1 Izračunati q w ΔU i ΔH zakorake A B i C Rešenje

Korak A V=const T=27315K P=1bar w=0 ΔU=qv =CV ΔT= 2093middot(2T1-T1)=2093middot27315=5717Jii) Korak B q=0 ΔU=w=CV ΔT=2093(T1-2T1)=-5717JmolΔH=-7988Jmoliii) Korak C ΔU=0 T2T3=2 V3V2=(T2T3)CVR V2=V1w=RT1lnV3V1=RT1ln(T2T3) CVR=CVT1sdot2303log2=2093middot27315middot2303middotlog2=396346Jmol q=-396346 Jmol

30 Jedan mola gasa u idealnom gasnom stanju prolazi kroz termodinamički ciklus koji se sastoji od reverzibilnih promena (koraka) A B i C i stanja 1 2 i 3 i koji je prikazan na slici Popuni Tablice 1 i 2 za dati ciklus

21

3

A

BC

273 546

224

448V[dm3]

T[K]

Stanje P Pa V m3 T K

1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546

Korak Ime procesa q J w J ΔU JABC

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorskiizotermskiizobarski

340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0

31 Jedan mol monoatomskog gasa u idealnom gasnom stanju prolazi kroz ciklus koji se sastoji iz tri procesa što je prikazano na slici Ispuniti tablice 1 i 2

P atm

V L

1

2

3

A B (q=0)

C1

2

224

Stanje P Pa V m3sdot10-3 T K

1 101325 2242 2026503 101325

Proces Tip procesa q J w J ΔU J

A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći

Jedan mola gasa u idealnom gasnom stanju prolazi kroz termodinamički ciklus koji se sastoji od

a) izotermske kompresije od 2 atm i 10L do 20 atm i 1L

b) izobarske ekspanzije kojom se gas vraća do zapremine od 10 L a temperatura menja od T1 do T2

c) izohorskog hlađenja do početnog stanja Nacrtati grafik i popuniti tablicu vrednostima q w i

ΔU za procese i ciklus

Stepen korisnog dejstva mašine koja hladnjaku preda jednu trećinu količine toplote uzete od grejača je

a) 025 b) 035 c) 067

d) 05 e) ne znam

Izračunati entropiju topljenja (S) u Jmol K za KCl čija je tačka topljenja 7700C Promena entalpije topljenja 268 kJmol

a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34

Za sledeću reakciju na 250CCuO(č)+H2(g)rarrCu(č)+H2O(g)

vrednosti standardnih entropija su S0

CuOč=4263 JKmol S0H2g=13068 JKmol S0

Cuč=3315 JKmol i S0

H2Og=18883 JKmol Odrediti da li će se reakcija odigravati spontano sa aspekta sistema

RešenjeStandardna promena entropije u reakciji je

Za gornju reakciju promena standardne entropije je

Pošto je promena entropije za sistem pozitivna to je reakcija spontana sa aspekta sistema

)tan()( 000 tireakSproduktiSS minus=Δ

( )KmolJ

KmolJSSSSS gHcCuOgOHcCu

6748

6813063428318815330)(

0)(

0)(

0)(

022

=

=minusminus+=minusminus+=Δ

Zadatak 35

36Grafit i dijamant su dve alotropske modifikacije ugljenika Izračunati ∆Ssis ∆Sok i ∆Stot za hemijsku reakciju u kojoj grafit i gasoviti vodonik grade metan

C(graf)+2H2(g)rarrCH4(g) ∆Ho298m=-7481kJmol

574 130684 18626 Somf (JKmol)

Rešenje∆Ssis=18626-2x130684-574=-80848JKmol∆Sok=74810298=251JKmol∆Stot=-80848+251=17019JKmol

Koja od sledećih reakcija je praćena najpozitivnijom promenom entropijea) 2 CO(g) + O2(g)rarr 2 CO2(g)

b) N2(g) + O2(g) rarr 2 NO(g)

c) 2 CH4(g) + O2(g) rarr 2 CH3OH(t)

d) 2 H2O2(t) + N2H4(t) rarr N2(g) + 4 H2O(g)

e) C(č grafit) + H2O(g) rarr CO(g) + H2(g)

Zadatak 37

Zadatak 38Za reakciju CHCl3(t)+Cl2(g)rarrCCl4(t)+HCl(g)na 250C standardne entropije suS0

298(CHCl3(t))=20302 JKmolS0

298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0

298(HCl(g))=18691 JKmola toplotni kapaciteti su

C0p(CHCl3(t))=11548 JKmol

C0p(Cl2(g))=3436 JKmol

C0p(CCl4(t))=13263 JKmol

C0p(HCl(g))=2884 JKmol

Odrediti standardnu promenu entropije reakcije na 50oC

Rešenje

sumsum minus=Δj

jTji

iTiT tireakSnproduktiSnS )tan()( 0

0

0

KmolJKmolJS 8324)0722322039118653214(0298 minus=minusminus+=Δ

sumsum minus=Δi

iPii

iPiP tireakCnproduktiCnC )tan()( 0

0

0

( ) KmolJKmolJCP 63113634481158428631320 =minusminus+=Δ

KmolJdTT

S 8923298323ln6311832463118324

323

298

0323 minus=+minus=+minus=Δ int

Zadatak 39

Koliki je porast entropije kada zagrevamo 1 mol hloroforma CHCl3 od 240 do 330K ako je Cp=(9147+75sdot10-2T) JmolK

Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40

Izračunati promenu entropije kada se idealan gas čiji je CVm=5R2 komprimuje do jedne trećine svoje početne zapreminei istovremeno zagreje do tri puta veće temperature od početne

Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41

U sistemu se odigrava proces u kome se entropija menja za551JK Za vreme procesa 15 kJ toplote dodato je sistemu na 350K Da li je proces termodinamićki reverzibilan

Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928

Proces nije termodinamički reverzibilan-ireverzibilan je

Zadatak 42Uzorak bakra (M=63546gmol) mase 275kg i toplotnog kapaciteta 2444JKmol se hladi na konstantnom pritisku od 330 K do 275 K Izračunati a)energiju koja se mora razmeniti u vidu toplote i b) promenu entropije sistemaRešenje

a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝

⎛ sdot=Δ==Δ minusminus

minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint

Zadatak 43Uzorak azota mase 35 g na 230 K i 211 atm širi se izotermalno do pritiska od 43atm Izračunati promenu entropije gasa

Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ

Zadatak 44Uzorak idealnog gasa u početku na 270K 120 atm i 110Lkomprimuje se izotermalno Do koje zapremine treba da se komprimuje da bi se entropija smanjila za 30JK

Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus

Zadatak 45Jedan mol čvrstog bakra se širi izotermski od 200 bar do 1 bar Izračunati promenu entropije sistema za proces u kome je

a gustina ρ=896sdot103kgm-315106611 minusminussdot=⎟⎠⎞

⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus

minusminusminusminus sdot=minus⎟⎟

⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus

=minusminus=minus=⎟⎠⎞

⎜⎝⎛partpart

minus=Δ intint αα

Zadatak 46Jedan mol CO2(g) na 273 K se hladi do CO2(t) na 1944 K Hlađenje se vrši reverzibilno i ireverzibilno stavljanjem uzorka u tečni vodonik na 1396 K Izračunati promenu entropije za proces ako je standardna entalpija isparavanja 231752 kJmol-1 na 1944 K i ako je Cp=3222+(2218sdot10-3)T+(-2347sdot10-6)T2

RešenjeMože se razmatrati proces u dva koraka Prvi je hlađenje gasa na konstantnom pritisku do tačke ključanja i drugi prevođenje u tečno stanje

KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1

=Δminus=minusminus=

=minus

+minussdotminus

+minussdot+⎟⎠⎞

⎜⎝⎛=

=minus

+⎥⎦⎤

⎢⎣⎡ sdotminus+sdot+=

Δminus+=Δ

minusminus

minusminusintint

Stoga je ukupna promena entropije za reverzibilan proces

0=Δ+Δ=Δ oksisiz SSS

Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ

Ako se promena vrši na irevrzibilan način toplota preneta u okolini je

kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int

Zadatak 47Promena Gipsove energije za neki izobarski proces možeda se prikaže izrazom

Izračuneti promenu entropije i entalpije za taj proces=minusminus=Δ )(342786 KTJG

RešenjePoznato je da je

( )

JTTSTGH

JTTT

GSp

786342342786

342342786

minus=+minusminus=Δ+Δ=Δ

=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ

Zadatak 48Kada se dva mola gasa na 330 K i 35 atm izotermski komprimuje entropija opadne za 250JK Izračunati krajnji pritisak gasa i promenu Gipsove energije za kompresijuRešenje

PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞

⎜⎝⎛ Δminus=⎟

⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2

=minussdotminus=Δminus=Δ

=sdotsdot

sdotsdot==Δ

Zadatak 49Kada pritisak 35 g uzorka tečnosti raste izotermalno od1atm do 3000 atm Gibsova energija raste za 12 kJ Izračunati gustinu tečnosti (u gcm3)

33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50

Jedan mola idealnog monoatomskog gasa se prevodi iz početnog stanja (224 L 273 K1atm S=20 calK) u krajnje stanje ( 124L 2 atm 303K) Izračunati ΔU ΔH ΔS i ΔG za ovu termodinamičku promenu

Rešenje

bull Označićemo stanja gasabull Početno Krajnje

V=224L=224middot10-3m3 V=124L=124middot10-3m T=273 K T=303 K

P= 1 atm=1013middot105Pa P=2atm=2026middot105Pa S= 20 calKmol=8372JKmol

Pošto je gas u idealnom gasnom stanju to jeΔU=CvΔT=(3R2)x30=37413J ΔH=CpΔT=(5R2)x30=62355J

bull Promena entropije i Gipsove funkcije će se odrediti razmatrajući promenu kroz dva stupnja

bull 1 00224m3 273 K 1013middot105Pa rarr 002486 m3 303 K 1013middot105Pa ΔS1=CplnT2T1=(5R2)ln111=2167JK (Skrajnje=2169JK)

ΔG1=ΔH 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=-2544651 Jbull 2 002486 m3 303 K 1013middot105Pararr 00124 m3

303 K 2026middot105PaΔS2=RlnP1P2=8314x23log05=-5763JKΔG2=RTlnP2P1=174614JUkupne promene su ΔS=2167-5763=-3596JK ΔG=-2544651+174614=-798511J

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛minus=minus= 11)(

2

11

1

2112

VCnR

VVVad VVTC

TTTCTTCw

Adijabatski zapreminski rad

1

2

1

2

2

1

1

2

1

2

2

1

PP

TT

VV

TT

VV

TT R

CR

CCR PVV

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

γγ

⎟⎟⎠

⎞⎜⎜⎝

⎛===

1

2

2

1 VV

PPconstPVPV VP CC

Adijabatski procesi

Jednačina adijabate

RTnUVPUH gasΔ+Δ=Δ+Δ=Δ

2

12

2 TTT

qw minus=




1

2

1

212 lnln

VVR

TTCSSS V +=minus=Δ

oksistotoki

iirev

sis SSSTHSS

TqS Δ+Δ=Δ


1

2

1

212 lnln

PPR


tr

trtr T


VV

V

V TSTCodnosno

TC

TS

⎟⎠⎞

⎜⎝⎛partpart

==⎟⎠⎞

⎜⎝⎛partpart


1

2

1

ln12

T

T

T

TVVV TdC

TdTCSSSΔ


1

2

1

ln12

T

T

T

TPPP TdC

TdTCSSSΔ

VT TP

VS

⎟⎠⎞

⎜⎝⎛partpart

=⎟⎠⎞

⎜⎝⎛partpart

PT TV

PS

⎟⎠⎞

⎜⎝⎛partpart

minus=⎟⎠⎞

⎜⎝⎛partpart

int ⎟⎠⎞

⎜⎝⎛partpart


P12int ⎟

⎠⎞

⎜⎝⎛partpart

=minus=Δ dVTPSSS

V12

Maksveloverelacije

TV VU

TPTP ⎟

⎠⎞

⎜⎝⎛partpart

minus⎟⎠⎞

⎜⎝⎛partpart

=

sum=

minus=minus=Δn

iiimeš xnRSSS


TP PH

TVTV ⎟

⎠⎞

⎜⎝⎛partpart

+⎟⎠⎞

⎜⎝⎛partpart

=




TV VS

TP

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

PVA

Tminus=⎟

⎠⎞

⎜⎝⎛partpart S

TA

Vminus=⎟

⎠⎞

⎜⎝⎛partpart


TP PS

TV

⎟⎟⎠

⎞⎜⎜⎝

⎛minus=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

STG

P

minus=⎟⎠⎞

⎜⎝⎛partpart

VPG

T

=⎟⎠⎞

⎜⎝⎛partpart

pTGTHG ⎟⎠⎞


)(TH

TTG

p

ΔpartΔpart

minus=⎟⎠⎞

⎜⎝⎛

00 ln

PP

RTGG mmm +=

ZADATAK 1


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









2

12

2 TTT

qw minus=




1

2

1

212 lnln

VVR

TTCSSS V +=minus=Δ

oksistotoki

iirev

sis SSSTHSS

TqS Δ+Δ=Δ


1

2

1

212 lnln

PPR


tr

trtr T


VV

V

V TSTCodnosno

TC

TS

⎟⎠⎞

⎜⎝⎛partpart

==⎟⎠⎞

⎜⎝⎛partpart


1

2

1

ln12

T

T

T

TVVV TdC

TdTCSSSΔ


1

2

1

ln12

T

T

T

TPPP TdC

TdTCSSSΔ

VT TP

VS

⎟⎠⎞

⎜⎝⎛partpart

=⎟⎠⎞

⎜⎝⎛partpart

PT TV

PS

⎟⎠⎞

⎜⎝⎛partpart

minus=⎟⎠⎞

⎜⎝⎛partpart

int ⎟⎠⎞

⎜⎝⎛partpart


P12int ⎟

⎠⎞

⎜⎝⎛partpart

=minus=Δ dVTPSSS

V12

Maksveloverelacije

TV VU

TPTP ⎟

⎠⎞

⎜⎝⎛partpart

minus⎟⎠⎞

⎜⎝⎛partpart

=

sum=

minus=minus=Δn

iiimeš xnRSSS


TP PH

TVTV ⎟

⎠⎞

⎜⎝⎛partpart

+⎟⎠⎞

⎜⎝⎛partpart

=




TV VS

TP

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

PVA

Tminus=⎟

⎠⎞

⎜⎝⎛partpart S

TA

Vminus=⎟

⎠⎞

⎜⎝⎛partpart


TP PS

TV

⎟⎟⎠

⎞⎜⎜⎝

⎛minus=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

STG

P

minus=⎟⎠⎞

⎜⎝⎛partpart

VPG

T

=⎟⎠⎞

⎜⎝⎛partpart

pTGTHG ⎟⎠⎞


)(TH

TTG

p

ΔpartΔpart

minus=⎟⎠⎞

⎜⎝⎛

00 ln

PP

RTGG mmm +=

ZADATAK 1


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









VV

V

V TSTCodnosno

TC

TS

⎟⎠⎞

⎜⎝⎛partpart

==⎟⎠⎞

⎜⎝⎛partpart


1

2

1

ln12

T

T

T

TVVV TdC

TdTCSSSΔ


1

2

1

ln12

T

T

T

TPPP TdC

TdTCSSSΔ

VT TP

VS

⎟⎠⎞

⎜⎝⎛partpart

=⎟⎠⎞

⎜⎝⎛partpart

PT TV

PS

⎟⎠⎞

⎜⎝⎛partpart

minus=⎟⎠⎞

⎜⎝⎛partpart

int ⎟⎠⎞

⎜⎝⎛partpart


P12int ⎟

⎠⎞

⎜⎝⎛partpart

=minus=Δ dVTPSSS

V12

Maksveloverelacije

TV VU

TPTP ⎟

⎠⎞

⎜⎝⎛partpart

minus⎟⎠⎞

⎜⎝⎛partpart

=

sum=

minus=minus=Δn

iiimeš xnRSSS


TP PH

TVTV ⎟

⎠⎞

⎜⎝⎛partpart

+⎟⎠⎞

⎜⎝⎛partpart

=




TV VS

TP

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

PVA

Tminus=⎟

⎠⎞

⎜⎝⎛partpart S

TA

Vminus=⎟

⎠⎞

⎜⎝⎛partpart


TP PS

TV

⎟⎟⎠

⎞⎜⎜⎝

⎛minus=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

STG

P

minus=⎟⎠⎞

⎜⎝⎛partpart

VPG

T

=⎟⎠⎞

⎜⎝⎛partpart

pTGTHG ⎟⎠⎞


)(TH

TTG

p

ΔpartΔpart

minus=⎟⎠⎞

⎜⎝⎛

00 ln

PP

RTGG mmm +=

ZADATAK 1


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









TV VU

TPTP ⎟

⎠⎞

⎜⎝⎛partpart

minus⎟⎠⎞

⎜⎝⎛partpart

=

sum=

minus=minus=Δn

iiimeš xnRSSS


TP PH

TVTV ⎟

⎠⎞

⎜⎝⎛partpart

+⎟⎠⎞

⎜⎝⎛partpart

=




TV VS

TP

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

PVA

Tminus=⎟

⎠⎞

⎜⎝⎛partpart S

TA

Vminus=⎟

⎠⎞

⎜⎝⎛partpart


TP PS

TV

⎟⎟⎠

⎞⎜⎜⎝

⎛minus=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

STG

P

minus=⎟⎠⎞

⎜⎝⎛partpart

VPG

T

=⎟⎠⎞

⎜⎝⎛partpart

pTGTHG ⎟⎠⎞


)(TH

TTG

p

ΔpartΔpart

minus=⎟⎠⎞

⎜⎝⎛

00 ln

PP

RTGG mmm +=

ZADATAK 1


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










TV VS

TP

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

PVA

Tminus=⎟

⎠⎞

⎜⎝⎛partpart S

TA

Vminus=⎟

⎠⎞

⎜⎝⎛partpart


TP PS

TV

⎟⎟⎠

⎞⎜⎜⎝

⎛minus=⎟

⎠⎞

⎜⎝⎛

partpart

partpart

STG

P

minus=⎟⎠⎞

⎜⎝⎛partpart

VPG

T

=⎟⎠⎞

⎜⎝⎛partpart

pTGTHG ⎟⎠⎞


)(TH

TTG

p

ΔpartΔpart

minus=⎟⎠⎞

⎜⎝⎛

00 ln

PP

RTGG mmm +=

ZADATAK 1


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









ZADATAK 1


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









ZADATAK 2


Rešenje


Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 3


Rešenje

w=-4186J ΔU=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 5


Rešenje

JqU

JKKmolJmoldTCnqw

PaKKPa

TTPP

T

Tv

11247

1124710031485110

135100300400101325

2

1

1

212

==Δ

=sdotsdotsdot===

=sdot=sdot=

int


P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










P

VVo 2Vo

Po

2Po

1

2Rešenje

ooooo VPVPPw

23

2)2(

=+

=



a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











a) 2 b) 3 c) 4d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVmJVPw

JJJwwqU

410104100000

400

400101

400100300

3333

35


=Δ



minus




KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje












KKT

CCK

CgJgJ

CU

UCCgJCgJgJ

mUU

k

opko

ovs

s

svs

ovss

07326921715308

925292173592179217

186475

18647512900

=+=

=+=Δ+====

Δ=Δ

ΔΔ

====Δ

=Δ

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











molkJmolmPamolJH

molmPa

KKmolJP

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

35

=sdotsdot+=Δ

=sdot

==


Rešenje

Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 10Domaći




Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










Rešenje


Δng=3-5=-2





nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje












nRTPsp

sp minus=minus

kJJ

KKmolJmolggmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=





1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje













1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC

Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 14


Rešenje


Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 15


Rešenje


Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 16



Rešenjea)

b)


Jdm

dmKKmolJmolgg

VVnRTw

7152518

21ln305314840

566ln 3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









ZADATAK 18



)(400101720)( KTKJCp +=

Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Rešenje

a) P=const

V=const


JqH

dTdTCHq

P

PP

114831411448312314946483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273


=minus+minus==Δ

+==Δ= int int


ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









ZADATAK 19



d) CV(T1-T2) e) f)

1

2lnVV

RTminus

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

minus⎟⎠

⎞⎜⎝

⎛1

2

11

VCR

V VV

TC⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡minus⎟⎟

⎠

⎞⎜⎜⎝

⎛1

2

12

RC

V

V

VVTC



a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRTR

RRCCPPR

TTC VPP

590201ln

300ln

25

25lnln

22

1

2

1

2

==

=+==


Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










Rešenje


21 3

5

212

12 ==⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ



Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










Rešenje

JUqJKmolJKmolw

KKLLT

VVT

VVT

VmV CR

CnR

2750275)2988187(481220

81872980150

11

6660

2

11

2

12


=sdot⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minusminus

ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









ZADATAK 23Domaći


Rešenje



Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











Zadatak 24



Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










Rešenje

4

4

0

3 aTdTaTHT

==Δ int

Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 26




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Rešenjea)

b)

[ ])()()()()()(

121212

111222



+=+=

)()(ln

)()(ln

)( 2

1

1

22

1

2

1TVTVRT

TVTVRTdV

TVRTpdVw

V

V

V

V αα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 27



JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










JwqUKTKmolJmol

JCwTTCwJw

mPaVPwq

m

VV

4551)524(3148553532)(

53532524)31481137(5

535353532

10)15154(105780 333







2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











2-V12)


Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 29




21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










21

3

A

BC

273 546

224

448V[dm3]

T[K]


1 224sdot10-3 2732 224sdot10-3 5463 448sdot10-3 546


Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105


340458 0 340458-314650 0314650

-56743 226912 -34045887678 -87738 0


P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -175584783 -4785 0

273546

41379

32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









32 Domaći







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje


















( )KmolJ


6748

6813063428318815330)(

0)(

0)(

0)(

022

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











574 130684 18626 Somf (JKmol)







Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje














Zadatak 37



298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











298(Cl2(g))=22307 JKmolS0

298(CCl4(t))=21453 JKmolS0







Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Rešenje

sumsum minus=Δj

jTji


0

0


sumsum minus=Δi

iPii


0

0


KmolJdTT

S 8923298323ln6311832463118324

323

298


Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 39


Rešenje

[ ] [ ] molKJTT

TdTCSSS p

883575613291057ln4791

TdTT)1075(9147

330240

2330240

330

240

2-330

240240330

=+=sdot+=

=sdot+==minus=Δ

minus

intint

Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 40


Rešenje

molKnJnRnRRn

nRRnVVnR

TTnCS

VVTTRC

mV

mv

7133ln233ln3ln

25

31ln3ln

25lnln

33

25

1

2

1

2

1212

==minussdot

=+sdot=+=Δ

===

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdotΔ==Δ

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

111

3

minus=sdotminus=

=minus⎟⎟⎠

⎞⎜⎜⎝


minus

KJKJKmolJmol

TdTnC

TdTC

TqS

T

Tmp

T

T

pT

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====Δ intintint


Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










Rešenje

KJKmolJmolg

gppnRS 516)

34121ln()3148(

0142835ln

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛==Δ


Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










Rešenje

molKKmolLatm

LatmRTPVn 5960

27008201121

=sdot

sdot==

LLmolJKmolJKLnRSVV

VV

nRS

VV

nRS

6546011)3148)(5960(03exp()11()exp(

)exp(ln

11112

1

2

1

2

=sdot=minus=Δ=

=Δ

=Δ

minusminusminus



⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











⎜⎝⎛partpart

= KTV

V P

α

Rešenje

KJbarkgmkgmolmolK

PPVdPVdPTVsistemS

P

PP

10342)2001(109681054663110661

)()(

333

1315

12

2

1

minusminus


⎠

⎞⎜⎜⎝

⎛sdotsdot

sdotsdotsdotminus


⎜⎝⎛partpart




KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje











KJSKJ

dTTTT

HTdTCS

ok

T

T klj

ispPsis

4713146131211192512

4194223175))273()4194((

2)104723()2734194)(101822(

2734194ln2232

4194223175)104723()101822(2232

226

3

4194

273

6302

1


=minus

+minussdotminus


⎜⎝⎛=

=minus

+⎥⎦⎤


Δminus+=Δ

minusminus

minusminusintint



Reverzibilan proces

KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









KJS

KJKJ

TqS

tot

ok

71731461311863

18639613

26010

=minus=Δ

===Δ


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((2

101822)2734194(2232 336

223

4194

273

0

=

=⎥⎦

⎤⎢⎣

⎡minusminus

sdotminusminus

sdot+minusminus=

=Δ+=

minusminus

int




( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje












( )

JTTSTGH

JTTT

GSp

786342342786

342342786


=minusminuspartpart

minus=⎟⎠⎞

⎜⎝⎛partΔpart

minus=Δ


PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










PaKmolJmol

KJPaP

nRSPP

nRS

PP

PPnRS

652

122

1

2

1

10594131482

25exp1054553

expexpln

sdot=⎟⎠

⎞⎜⎝

⎛sdotminus

minussdot=

⎟⎠⎞


⎠⎞

⎜⎝⎛ Δ==Δ

JKJKSTG

JPPnRTG

825025330

882461054553

105941ln33031482ln 5

6

1

2


=sdotsdot

sdotsdot==Δ


33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje










33

53

886088612000

10013129991035

cmgmkgJ

PakgGPm

VmPVG

==

sdotsdotsdotsdot=

ΔΔ

==Δ=Δminus

ρ

ρρ

Zadatak 50


Rešenje









Zadatak 50


Rešenje









Rešenje









i zakon termodinamike du =dq +dw+ dwe dh =du + pdv u c h p ... termodinamika.pdf · jedan mol gasa...

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