Download - UNIVERSITY OF NAIROBI Lecture 2
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UNIVERSITY OF NAIROBI
Lecture 2
1ST Law of Thermodynamics
1ST Law
Thermodynamic Processes
Heat capacities
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Objectives
Explain 1ST Law, its significance and limitations
Explain various thermodynamic processes.
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Introduction
A Donkey is an engine: It eats (= heat input) to get
energy in order to work -Concept of 1ST Law
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RECAL:
Energy may be converted from one form to another e.g., heat to work but TOTAL ENERGY MUST BE CONSERVED
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1ST Law of Thermodynamics
Defines relationship between heat (dQ), work (dW) & internal energy (dU) of a system by mandating conservation of energy.
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dQ = dU + dW
Heat flow into/out of system
Change in internal energy
Work done by (or on) system
1ST Law
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1ST Law Simplified
“We eat in order to work’
Or
“No animal works continously without eating or wearing out”
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dQ = dU + dW
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For mechanical device e.g., gas in cylinder
dW = Fdx = PA dx dW = P(Adx) = PdV
dQ = dU + PdV ………………….(2)
Gas dQ
dx Area A
Movable piston P
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Internal Energy of Ideal Gas
From Kinetic Theory of gases, internal
energy (U) of ideal gas with f degrees of
freedom is:
f = translational & rotational degrees
of freedom. Vibrational ones are
negligible
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f 3 (monatomic), 5 (diatomic) etc
)1.......(..........22 TNKRTUB
ff
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Note
U is a state function i.e., it does not depend on the path taken/process
dUabc = dUadc S.T
U depends only on T
For reversible process,
dU = 0
P
V
T a
B
b
c
d
C
dU 0
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dQ & dW are path dependent & are not state functions S.T dWabc dWadc
C C
dQdW 0&0
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Significance of 1ST Law
It provides means of determining dU
Limitations of 1ST Law Does not define efficiency of systems i.e.,
how much dQ is converted into dW
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Perpetual Machines of 1st kind
A machine that works continously without energy input (dQ) or change in (dU) does not exist.
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Worked Example
A device working reversibly takes in 1000 J of heat, ejects 200 J of heat to a cold reservoir and produces 800 J of work.
Does this device violate 1ST Law ?
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Soln
(i) 1ST Law mandates conservation of Energy
Since system is reversible dU = 0
dW = dQ
W (800) = Qh (1000) - Qc (200)=800J
device doesn't violate 1ST Law since no energy is created nor destroyed.
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Thermodynamics and Laws of nature Rule of Nature: We eat to live
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Laws of Nature Laws of Thermodynamics
1. We eat when hungry
Zeroth Law: Heat flows from hot to Cold
2. We eat in order to work
1st Law: dQ = dU + dW
3. (i) If you eat, you must shit
(ii) You live once
2nd Law: (i) Efficiency of systems <
100% (ii) Nature is irreversible,
dS 0
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Lecture Evaluation
State 1ST Law giving its significance and limitations
Explain perpetual machines of 1ST Kind
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Thermodynamic processes
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Process = passage of a system from an initial to a final state of equilibrium as a result of transfers of heat, work or Internal Energy
An ideal gas can undergo 4 main Processes:-
isothermal (heat transfer at const temp)
isobaric (const P)
isochoric (const V) and
adiabatic (no heat input)
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Other Processes
o Reversible process → the system is
close to equilibrium at all times. NB.
there are no truly reversible processes
in nature.
o Cyclic process → the final and initial
state are the same.
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These processes can be represented on P-V diagram
Isothermal (dT=0)
V
P
a
Adiabatic (dQ=0)
c Isobaric
(dP = 0)
Isochoric
(dV = 0)
d b
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Applications of 1ST Law
Used to find work done in various thermodynamic processes for purposes of designing a (heat engine) with max possible efficiency
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(a) Isothermal process (dT = 0)
Work done = area under P-V curve i.e.,
but
Also since
2
1
21
V
V
PdVW
V
P
PV = c
V2 V1
W 2
1
isothermal V
nRTP
1
2ln
2
1
V
VnRT
V
dVnRTW
V
V
2211 VPVP 2
1lnP
PnRTW
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(b) Isochoric Process (dV = 0)
heat input only increases dU
V
P
V1,2
T1
T2
1
2
gasmonoatomicforRdTdUdQ2
3
0
2
1
21
PdVW
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(c) Isobaric process (dP = 0)
1ST Law becomes:
dQ = dU + PdV = dH
V
P
V1
1
2
V2
Where dH = Entalpy = heat involved in chemical
reaction (at const P)
012
2
1
21
VVPPdVW
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Worked Example
An ideal gas is taken through cyclic process ABCA. Determine
(i) Net heat transferred to system during one complete cycle
(ii) Net heat input for reversed cycle ACBA.
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V m3
P
B
6 8 10
C A
6
2
8
4
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Soln
(i) From first law, dQ = dU + dW
But dW = PdV = area under P-V dU = 0 in cyclic process.
dQ = PdV = ½(8-2)(4) = 12 Kj (ii) dQ is the same as in (i)
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Activity 2
An Ideal gas is taken from state a to c either through abc or adc. In processes ab and bc, 600 J and 200J are given to the system respectively. Calculate
(i) dU along ab (ii) dU along abc (iii) Total heat added in adc
2
3
8
P104 Pa
b c
a d
5 V10-3 m3
200J
600J
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Solution
From 1ST Law (dU = dQ – dW)
(i) Along ab, dW = 0: dU = dQ = 600 J (ii) Total dQ along a b c = (600 + 200) = 800 J
dW along ab = 0 dW along bc = PdV = (8 104Pa)(3 10-3m3) = 240 J.
Total dW along a b c = 240 J dU along a b c = dQ - dW = (800 - 240) J = 560 J
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(iii) Along a d c, dU = 560 J
(NB. dU is a state function S.T total dU along a b c = dU along a d c.)
dW along a d c
= P1(V2 - V1) = 90 J
NB. along d c, dW = 0
Total dQ in adc
= dU + dW
= (560 + 90)J = 650J. 2
3
8
P
b c
a d
5 V
200J
600J
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(d) Adiabatic process (dQ = 0)
No heat enters or leaves the system BUT gas expands doing work (dW) as dU decreases with decrease in T (path 1-2)
V
P
V2
2
1
V1
Adiabatic
Isothermal, PV =c
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But
To obtain dW, we first need to obtain Eqn for adiabatic process by substituting for dT in terms P and V in eqn (3).
From the ideal gas law (PV = nRT), differentiating we have
RdTVdPPdV
)2.....(..........23 RdTRdTdU
)1.......(10 LawofFormdUPdV st
)3....(....................0 RdTPdV
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Substituting for dT in (3), gives
Integrating we have
VdPPdV
VdPPdVPdV
1
0
cPV
P
dP
V
dV
lnln1
1
R
VdPPdVdT
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Adiabatic curve has steeper slope ( times) than isothermal curve
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)5........(
2
51ln
AdiabaticforEqncPV
wherecPV
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Activity
Show also that for adiabatic process
cTPii
cTVi
1
)(
)(
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Soln
From
Replacing for P from ideal gas Eqn (PV = nRT)
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cPV
cTV
VTVT
V
V
VnRT
VnRT
1
1
22
1
11
1
2
22
11
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Work done in adiabatic Process
W = area under curve
V1 V2 V m-3
P 1
Adiabatic, PV = c P1
P2 2
W
cPV
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since
But Replacing C we get
2
1
2
1
V
V
V
VV
dVCPdVW
V
CP
.2211 CVPVP
1
2211
VPVPW
1
1
1
21
1 V
C
V
CW
1
11
2
22
1
1
V
VP
V
VPW
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Alternatively
Substituting fot T1 and T2 from PV = nRT, we get
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2
1
2
1
1.23
T
T
V
V
nfordTRdUPdVdW
2123
1223 TTRTTR
1
22112211
23
VPVP
R
VP
R
VPRdW
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Activity
Show that W can also be given by
(i)
21
1TT
RW
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22 2
1 1 1
11 11 2 1 1
1 1 1 1
2 1
1( , )
1
1 1 1
1
VV V
V V V
PVW P V T dV dV PV V
V
PVV V
constVPPV
11(i) From
Soln
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Examples of Adiabatic processes
Adiabatic processes take place rapidly e.g.,
(i) Air compressor
Air let out from an air compressor feels cold
(ii) Opening a bottle of carbonated beverage
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Activity
A gas is rapidly compressed from initial
pressure is 105 Pa and temperature of
220C (= 295 K) to a volume that is a
quarter of its original volume (e.g.,
pumping bike’s tire). Find the final
temperature?
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Soln
From
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cVTVT 1
22
1
11
KV
VTT 514
1
2
1
12
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Activity 2.4
During the ascent of a meteorological helium-gas balloon, its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and pressure inside balloon decreases linearly from 1 bar (=105 N/m2) to 0.5 bar.
(a) If initial T is 300K, find the final T ?
(b) How much work is done by the gas?
(c) How much “heat” does the gas absorb?
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Lecture -Evaluation
Explain the following Processes –Isothermal, Isobaric, Isochoric, Adiabatic
State Eqns of state for Isothermal & Adiabatic & give expressions for work done in these processes
List some applications of Adiabatic process
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Heat Capacities of Ideal Gas
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Heat Capacities of Ideal Gas
Heat absorbed by a system (dQ) is measure by its capacity to absorb heat i.e., its heat capacity (C).
The heat capacity of a system is the amount of heat energy required to produce a unit temperature rise in that system
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dT
dQC
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NB. Heat can be added to a system either at const Pressure (P) or constant Volume (V)
For a constant volume process, the specific heat capacity (Cv) is:- (NB dW =0)
)1(..........2
3R
dT
dU
dT
dQC
VV
V
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If dQ is added at constant pressure, then specific heat capacity at const P (CP) is
But from (PV = nRT )
Cp - Cv = R Mayer’s relation
PP
PdT
PdVdU
dT
dQC
)2.........(RCdT
nRdT
dT
dUC VP
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Ratio of heat Capacities
)3..(..............................67.1 V
P
C
C The Gamma relation
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Summary
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dT
PdVdU
dT
QC
RT
UC
V
V2
3
RCT
HC
V
P
P
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Lecture -Evaluation
Explain Cp and Cv
Why is Cp > Cv always ?
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