mahalakshmi - vidyarthiplus

N.Saravanan, AP/Chemistry, Mahalakshmi Engineering College,Trichy-621213.

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI - 621 213

Sub. Code: CY 6151 Semester : I

Subject : ENGINEERING CHEMISTRY-I Unit –II: CHEMICAL THERMODYNAMICS

2 Marks Questions and Answers

01. Define the terms System , Surroundings and Boundary.

System: The Part of the universe.

Surroundings: Remaining part of the universe.

Boundary: Separating the system and surroundings.

02. What are the types of systems?

Isolated system, closed system, opens system.

03. What do you mean by an isolated system, closed system and open system?

Isolated system: which cannot exchange both energy and matter? Ex: universe.

Closed system: only heat exchange

Open system: only mass exchange.

04. What do you mean by a state of a system?

Exchange takes place at boundary between system and surroundings.

State: that condition in which all variables are fixed and unvarying.

05. Define the isothermal, adiabatic , isobaric, isochoric change.

Isothermal change: one at constant temperature. dT=0

Adiabatic change: No heat is exchanged into or out of a system i.e, insulated.

Isobaric change: one at constant pressure. dP=0

Isochoric change: one at constant volume. dV=0

06. State different form of definitions of second law of thermodynamics.

Work can always be converted into heat, but heat cannot be completely converted into work ,

only a fraction of heat can be converted into work and the rest remains unavailable and

unconverted.

Clausius statement: It is impossible to construct a machine, which can transfer heat from a cold

body to a hot body, unless some external work is done on the machine.

Kelvin statement: It is impossible to take heat from a hot body and convert it completely into work

by a cyclic process without transferring a part of heat to a cold body.

07. Define the statement of Entropy. Give its mathematical expression.

It is a measure of degree of disorder or randomness in a molecular system

Ex: solidliquid, entropy increases. Liquid solid, entropy decreases.

ΔS = revq

T qrev = heat change and T= Temperature.

08. What are the properties of a system?

Two properties. 1. Intensive properties 2. Extensive properties.

1. Intensive properties: Depends on only the nature of the substance present a system.

Ex:Temperature , Pressure, Concentration.

2. Extensive properties: Depends on the amount of the substance present a system.

Ex:Mass, Volume, E, S, G.

Prepared by:

N.Saravanan., AP/Chemistry.

www.Vidyarthiplus.com



09. Define clausius inequality.

The amount of heat transferred to the system divided by its temperature is equal to or less

than zero is known as clausius inequality. 0Q

T

δQ differential heat transfer at the system boundary during a cycle.

T absolute temperature. integration over a cycle.

10. Derive the mathematical form of Gibbs- Helmholtz equation.

P

GG H T

T

11. What are the criteria of spontaneity?

A process to occur in nature. Ex: Falls, Heat flows , Gas expands.

A spontaneous change is one-way.

ΔG = -ve , the process is spontaneous.

ΔG = 0 , the process is in equlibrium

ΔG = -ve , the process is non-spontaneous.

12. Derive the mathematical form of Clausius - clapeyron equation.

( )B A

dP q

dT T V V

13. What are the applications of Clausius - clapeyron equation?

a) Calculation of latent heat of vapourisation.

b) Calculation of boiling point.

c) Calculation of vapour pressure at another temperature.

d) Calculation of molar elevation constant.

14. Derive the mathematical form of van’t hoff isotherm.

ln lnc d

C Deq a b

A B

P PG RT K RT

P P

15. Derive the mathematical form of van’t hoff equation.

2 2 1

1 1 2

log2.303

K T TH

K R TT

16. What is the significance of Van’t hoff equation?

a) Equilibrium constant (K) and ΔH of a reaction can be calculated using Van’t hoff equation.

b) To find equilibrium constant at another temperature is known.

c) ΔH0 = 2.303R x slope

d) The equilibrium constant increases with temperature.

17. Define work function and free energy.

Work function: Helmholtz free energy (A)

The part of internal energy (E) , which is isothermally available is called

work function. A=E-TS.

Free energy: Gibbs free energy.

It is known as isothermally available energy present in a system. G=H-TS

18. Derive the mathematical form of Maxwell’s relations.

(i). s v

T P

V S

(ii).

s P

T V

P S

(iii). T V

S P

V T

(iv).

T P

S V

P T




PART – B QUESTIONS AND ANSWERS

01. Explain the terminology in thermodynamics.

System: The Part of the universe.

Surroundings: Remaining part of the universe.

Boundary: Separating the system and surroundings.

Types of systems

Isolated system, closed system, opens system.

Isolated system: which cannot exchange both energy and matter? Ex: universe.

Closed system: only heat exchange

Open system: only mass exchange.

State of a system: Exchange takes place at boundary between system and surroundings.

State: that condition in which all variables are fixed and unvarying.

Isothermal change: one at constant temperature. dT=0

Adiabatic change: No heat is exchanged into or out of a system i.e, insulated.

Isobaric change: one at constant pressure. dP=0

Isochoric change: one at constant volume. dV=0

02. Explain the entropy change for an ideal gas.

According to I law of thermodynamics dE = q-w ------------- (1)

Where, q = heat change and w = work done.

In reversible isothermal expansion, there is no change of internal energy dE = 0.

Equation (1) becomes, qrev –w = 0 ; qrev = w ---------------(2)

The work done in an expansion of n moles of a gas from volume V1 to V2 at constant

temperature T is given by

W = nRT 2

1

lnV

V-------------------- (3)

Equation (2) becomes

qrev = nRT 2

1

lnV

V ----------------(4)

but ΔS = revq

T and qrev = T ΔS

Equation (4) becomes, T ΔS = nRT 2

1

lnV

V

ΔS = 1

T nRT 2

1

lnV

V ; ΔS = nR 2

1

lnV

V

-------------------------(5)

Equation (5) is known as an entropy change in an isothermal expansion of an ideal gas.

ΔS = 2.303 nR 2

1

logV

V




03. Explain the entropy change for a reversible and irreversible process.

Entropy change for a reversible process:

a) Consider an isothermal and reversible expansion of an ideal gas.

b) If the system absorbs q amount of heat from the surroundings at temperature T, the

increase in entropy of the system is given by system

qS

T

c) The heat loose, the entropy of the surroundings decreases surroundings

qS

T

Total System surroundings

Total

S S S

q qS

T T

TotalS = 0

In a reversible isothermal process, there is no net change in entropy.

Entropy change for a irreversible process:

A system maintained at high temperature T1 and

Its surroundings maintained high temperature T2.

If q amount of heat passes irreversibly from the system to surroundings, then

Decrease in entropy of the system 1

System

qS

T

Increase in entropy of the surroundings 2

Surroundings

qS

T

Net change in entropy Total System SurroundingsS S S

1 2

2 1

1 2

1 2

1 1

q q

T T

qT T

T Tq

TT

Since T1>T2, T1-T2 is positive. Hence ΔSTotal = positive value.

In an irreversible process, the entropy of the system increases.

All the spontaneous process are irreversible and entropy of the system always

increases.

ΔSTotal>0




04. Explain the clausius inequality with proof.

The amount of heat transferred to the system divided by its temperature is equal to or less

than zero is known as clausius inequality. 0Q

T

δQ differential heat transfer at the system boundary during a cycle.

T absolute temperature. integration over a cycle.

Proof:

Consider two heat engines.

One is reversible carnot engine.

Another one is irreversibleirreversible engine.

Both engines absorb the same amount of heat QH at a temperature TH

Both engines reject the heat to a heat sink at a temperature TL

According to I law of thermodynamics,

Wrev = QH-QL rev

Wirrev.=QH-QL irrev.

since the reversible engine is more efficient than the irreversible engine.

Wrev = QH-QL rev > Wirrev.=QH-QL irrev.

For reversible heat engine – Carnot.

Here, the cyclic integral of the heat transfer divided by the temperature.

.

.

0

L revH

H H

L rev H

H H

QQQ

T T T

Q Q

T T

Q

T

For irreversible heat engine:

Same value of heat transfer from QH. but heat rejection QL is more in irreversible engine than the

reversible one.

QL irrev> QL rev

for any reversible or irreversible heat engine , we obtain the clausius inequality

0

Q

T




05. Explain Gibbs – Helmholtz equation (or) Relation between ΔG and ΔH (or) relation between

free energy and work function.

From Gibbs free energy,

G = H-TS ; H = E+PV; G = E+PV-TS

For an infinitesimal change,

dG = dE + PdV+VdP-TdS-SdT---------------------------------------(1)

But according to I & II law of thermodynamics ,

dE = TdS- PdV------------------------------------------------------------(2)

subs eqn (2) in eqn (1), we get

dG = TdS- PdV + PdV+VdP-TdS-SdT

dG = VdP- SdT--------------------------------------------------------(3)

At constant pressure dP = 0 ; eqn (3) becomes

dG = - SdT -----------------------------------------------------------------(4)

P

GS

T

---------------------------------------------------------------(5)

Subs eqn (5) in G = H-TS, we get

G = H +P

GT

T

(Or) -----------(6)

Equation (6) is Gibbs – Helmholtz equation.

For any two states of the system, the equation (4) may be written as

1 1

2 2

dG =-S dT Initialstate

dG =-S dT Finalstate

To get the change ,

2 1 2 1

2 1 2 1

dG -dG =-S dT- -S dT

d G -G =- S -S dT

d ΔG =-ΔSdT--------------------------(7)

At constant pressure the eqn.(7) becomes,

(8)

P

GS

T

According to free energy , ΔG = ΔH – TΔS

(9)G H

ST

Subs eqn (9) in eqn (8),

G H

T

=

P

G

T

G H

=T

P

G

T

Eqn(10) is another form of Gibbs – Helmholtz equation.

G- P

GT

T

= H

G

(10)= H+T

P

G

T




Significance:

(i). It relates ΔG and ΔH with temperature at constant pressure.

(ii). It gives the nature of a reaction. ΔG value = -ve, the reaction is spontaneous.

ΔG value = 0, the reaction is in equlibrium.

Applications:

1. Calculate ΔH for the cell reaction: It is calculated as follows.

ΔH = 0

0

P

EnF E T

T

2. Calculate the emf of a cell: It is calculated as follows.

00

P

H EE T

nF T

3. Calculation of entropy change: It is calculated as follows. ΔG = ΔH – TΔS

4. It is used to calculate ΔH value from free energy change of two different temperatures.

06. The free energy change accompanying a given process is -85.77 kJper mole at 250C and –

83.68 kJ at 350C. calculate the change in enthalpy for the process at 300C.

Gibbs – Helmholtz equation is G

= H+T

P

G

T

ΔG1=

-85.77 kJ

ΔG2=

-83.68 kJ ΔG = ΔG2 - ΔG1 = 2.09.

T1 = 250C = 25+273 = 298 K

T2 = 350C = 35+273 = 308 K

2.09 /10 0.209

P

G

T

0 1 2 298 30830 303

2 2

85.77 83.68

2

84.725

84.725 303 0.209

84.725 63.327

148.052

P

T TG C K

kJ

GH G T

T

X

H kJ




07. Derive the clausius – clapeyron Equation.

A system contains two phases like A and B.

Free energy changes of two phases GA & GB

The system is in equilibrium GA = GB -------------------------------(1)

eqn (1) becomes, GA + d GA = GB + d GB -----------------------(2)

We know that, G = H – TS ; H = E + PV ;

G = E + PV – TS --------------------------------------(3)

dG = dE + PdV + VdP– TdS –SdT---------------------------------(4)

But dE = dq – dw ; dq = dE + dw = dE + PdV

dq = dE + PdV--------------------------------------------------------------(5)

for reversible equation;

dqdS

T

; dq = TdS-------------------(6)

subs eqn (6) in eqn (5)

TdS = dE + PdV

dE = TdS – PdV------------------------------------------------------------(7)


dG = TdS – PdV + PdV + VdP– TdS –SdT

dG = VdP– SdT-----------------------------------------------------------(8)

dGA = VAdP– SAdT------------------------------------------------------(9)

DgB= VBdP– SBdT -----------------------------------------------------(10)

Where VA, VB are molar volumes of phase A & B

SA & SB are molar entropies.

GA = GB ; dGA = dGB

VAdP– SAdT = VBdP– SBdT

SBdT– SAdT = VBdP– VAdP

dT (SB - SA) = dP(VB-VA )

(15)S - SB A dP

=dTV -V B A

When one mole of entropy change. So ΔS= SA - SB

(16)

dP=

dTV -V B A

S

We know that entropy change ΔS = q

T

Eqn (17) is known as Clausius – clapeyron equation.

Applications of Clausius – clapeyron equation.

(1). Calculation of latent heat of vapourisation.

(2). Calculation of Boiling point.

(3). Calculation of vapour pressure at another temperature.

(4). Calculation of molar elevation constant.

dP q=

dT T(V -V )B A

(17)

dPdT V -V B AT

q




08. Derive Maxwell’s relations.

We know that , (1). dE = TdS – PdV (2). dH = TdS + VdP (3). dA = -SdT – PdV (4). dG = -SdT + VdP (1). According to internal energy ; The combined form of I and II law is , dE = TdS – PdV ---------------------------------------(1) If V is constant then dV = 0 , eqn (1) becomes

(2)V

ET

S

If S is constant then dS = 0 , eqn (1) becomes

(3)S

EP

V

Diff. eqn (2) w.r.t V at constant S

2

(4)S

E T

S V V

Diff. eqn (3) w.r.t S at constant V

2

(5)V

E P

V S S

Compare eqns. (4) & (5) Eqn (6) is one form of maxwell’s relations. (2). According to enthalpy ; dH = TdS + VdP---------------------------------------(7) If P is constant then dP = 0 , eqn (7) becomes

(8)P

HT

S

If S is constant then dS = 0 , eqn (7) becomes

(9)S

HV

P

Diff. eqn (8) w.r.t P at constant S

2

(10)S

H T

S P P

Diff. eqn (9) w.r.t S at constant P

2

(11)P

H V

S P S

Compare eqns. (10) & (11)

Eqn (12) is Second form of maxwell’s relations.

(6)S V

T P

V S

(12)S P

T V

P S




03. According to work function A; dA = -SdT – PdV---------------------------------------(13) If V is constant then dV = 0 , eqn (13) becomes

(14)V

AS

T

If T is constant then dT = 0 , eqn (13) becomes

(15)T

AP

V

Diff. eqn (14) w.r.t V at constant T

2

(16)T

A S

T V V

Diff. eqn (15) w.r.t T at constant V

2

(17)V

A P

V T T


[[[[ [[

Eqn (18) is Third form of maxwell’s relations. 04. According to Gibbs free energy (G); dG = -SdT + VdP-----------------------------------------(19) If P is constant then dP = 0 , eqn (19) becomes

(20)P

GS

T

If T is constant then dT = 0 , eqn (19) becomes

(21)T

GV

P

Diff. eqn (20) w.r.t P at constant T

2

(22)T

G S

T P P

Diff. eqn (21) w.r.t T at constant P

2

(23)P

G V

P T T


[[[[ [[

Eqn (24) is Fourth form of maxwell’s relations. The four types of Maxwell relations are as follows;

S V

T P

V S

S P

T V

P S

T V

S P

V T

T P

S V

P T

(18)T V

S P

V T

(24)T P

S V

P T




09. Derive van’t hoff isotherm.

It gives the quantitative relationship between the free energy change and equilibrium constant.

Consider the following reaction.

(1)aA bB cC dD

We know that G = H – TS ; H = E + PV.

G = E + PV – TS -----------------------> (2)

dG = dE + PdV+VdP-TdS-SdT--------> (3)

but according to I and II law;

dE = TdS- PdV -------------------------------(4)


dG = TdS- PdV + PdV+VdP-TdS-SdT

dG = VdP-SdT---------------------------------(5)

at constant Temperature ,dT=0;

T

dG =VdP

free energy change for 1 mole

dG = VdP

. (6)dP

dG RTP

On integrating the eqn (6);

dP

dG RTP

---------------(7)

Where G0 = integration constant.

Free energies like A,B,C,D

Pressures PA,PB,PC,PD

Free energies GA,GB,GC,GD

ΔG = GProducts – G Reactants

DC A

0 0 0 0D D B BC C A A

0 0 0 0D B D BC A C A

c d0 DC

a bBA

ΔG= G +G - G +GB -----------(8)

ΔG= G +cRTlnP +G +dRTlnP - G +aRTlnP +G +bRTlnP

ΔG= G +G - G +G +RT clnP +dlnP - alnP +blnP ---(9)

P .PΔG=ΔG +RTln --------(10)

P .P

When equilibrium ΔG =0, eqn (10) becomes

00 ln eqG RT K

0 ln (11)eqG RT K

Subs eqn (11) into eqn(10)

c dDC

a bBA

lnP .P

ΔG= +RTlnP .P

eqRT K -------(12)

The above equation is known as van’t hoff isotherm.

G = G0 +RT lnP

c d

C D

a b

A B

ln -P .P

-ΔG= RTlnP .P

eqRT K




10. Derive the van’t hoff isochore of van’t hoff equation.

This equation is obtained by combining of van’t hoff isotherm and Gibbs – Helmholtz equation.

According to van’t hoff isotherm, 0 ln pG RT K ------------(1)

Diff w.r.t T at constant P,

0

PP

0

2 PP

0

0 2 P

dlnKRlnK -RT (2)

dT

(2) ,

dlnK-RTlnK -RT (3)

dT

dlnKΔG -RT (4)

dT

P

P

P

G

T

Eqn XT

GT

T

GT

T

Gibbs – Helmholtz equation;

0

0 0 (5)G = H +T

P

G

T

Subs eqn (5) into eqn(4)

0

0

0

0

0

0

0

0

2 P

0

2 P

2 P

2 P

P2

(8)

dlnK-RT (6)

dT

dlnK-RT

dT

dlnK-RT

dT

dlnKRT

dT

dlnK

dT RT

H +T

T H

H

H

H

PP

PP

G GT

T T

G GT

T T

Integrating the above eqn T1 , T2 and 1 2

p pK ,K

22

p

1p

1

TK

P 2KT

2 1p p

2 1

ΔH dTd lnK = (9)

R T

ΔH 1 1lnK lnK

R T T




2 1

1 2

1 2

T T

T T

ΔH 1 1

R T T

ΔH

R

2 1

1 2

2 1 T T

T T

ΔHlnK lnK

R

----------------(8)

The above equation is known as van’t hoff equation.

11. The equilibrium constant for the reaction 2 2 3N +3H 2NH is 1.64x10-4 atm and 0.144x10-4

atm at 4000C and 5000C respectively. Calculate the heat of reaction in terms of calories(R=1.987cal).

T1 = 4000C = 400+273 = 673 K

T2 = 5000C = 500+273 = 773 K

K1= 1.64x10-4 atm

K2= 0.144x10-4 atm

R=1.987cal

2 1

1 2

2

1

T T

T T

K ΔHlogK 2.303R

4

4

100ΔH1.0564

2.303 1.987 673 773

ΔH 25.14 .

0.144 10 773 6731.64 10 673 773

ΔHlog2.303X1.987

X X X

k cal

XX X

2 1

1 2

2

1

T T

T T

K ΔHlogK 2.303R



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