lectures12 to 25
Transcript of lectures12 to 25
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Lecture 12: Perfect Differentials
If A is a state function, then dA is said to be aperfect differential. A necessary and sufficientcondition for a function of two variables A(x,y) tohave a perfect differential is that
xy
A
yx
A
∂∂∂=
∂∂∂ 22
dyyxgdxyxfdyy
Adx
x
AdA
xy
),(),( +=
∂∂+
∂∂=
The necessary and sufficient condition is
yxx
g
y
f
∂∂=
∂∂
Example: A= x2 sin(xy)
dA = {2x sin(xy) +x2ycos(xy)}dx + x3cos(xy)dy
∑f/∑y = 3x2 cos(xy) - x3y sin(xy) = ∑g/∑x
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Thermodynamic Perfect Differentials
dU = TdS - PdV
ïVS S
P
V
T
∂∂−=
∂∂
dH = TdS + VdP
ïPS S
V
P
T
∂∂=
∂∂
Figure 16. TheMaxwell square
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Goal of Thermodynamic Manipulations: to expressany quantity in terms of V, T, P, S, n, a, kT, and CV.
PT
V
V
∂∂= 1α
Thermal expansion
Ideal gas: TP
nR
V
11 ==α
TT P
V
V
∂∂−= 1κ
Isothermal compressibility
Ideal gas: PP
nRT
VT
112
=
−−=κ
VV T
UC
∂∂=
Heat Capacity, constant V
Ideal gas: CV = 3/2nR
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Lecture 13: Thermodynamic Calculations
Partial derivatives of implicit functions
Consider a function A(x,y,z)
dzz
Ady
y
Adx
x
AdA
yxzxzy ,,,
∂∂+
∂∂+
∂∂=
x, y, and z are said to be implicit functions of eachother; i.e., z = z(x,y).
In addition, x, y, z, may be functions of some othervariable, t.
Examples: A = U, H, A, Gx, y, z, t = P, V, T, S.
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Properties of implicit functions
zA
zA
x
yy
x
,
,
1
∂∂
=
∂∂
zy
zx
zA
x
A
y
A
y
x
,
,
,
∂∂
∂∂
−=
∂∂
zA
zA
zA
t
y
t
x
y
x
,
,
,
∂∂
∂∂
=
∂∂
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Origin of the minus sign:
dzz
Ady
y
Adx
x
AdA
yxzxzy ,,,
∂∂+
∂∂+
∂∂=
Set dA=0:
dzz
Ady
y
Adx
x
A
yxzxzy ,,,
0
∂∂+
∂∂+
∂∂=
Set dz=0 and divide through by dx:
zAzxzy x
y
y
A
x
A
,,,
0
∂∂
∂∂+
∂∂=
( )
zx
zy
zAy
A
xA
x
y
,
,
,
∂∂
∂∂
−=
∂∂
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Manipulations of the Maxwell Square
Z
Z
X
YY
X
∂∂
=
∂∂ 1
Y
X
Z
X
ZY
Z
Y
X
∂∂
∂∂
−=
∂∂
Z
Z
Z
W
YW
X
Y
X
∂∂
∂∂
=
∂∂
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Example 1. Internal Pressure
dTCdVdU VT += π
TT V
U
∂∂=π
dU = TdS - PdV
PV
ST
TT −
∂∂=π
PT
PT
V
−
∂∂=
P
P
VT
VT
T
P −
∂∂
∂∂
−=
PT
T
−=κα
Ideal gas: pT = TP/T - P =0
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Example 2. The Joule-Thompson Coefficient
A pump drives a high pressure gas through a porousplug adiabatiaclly. The work done on the gas onthe high pressure side is PiVi. The work doneby the gas on the low pressure side is PfVf.
Figure17.TheJoule-Thomaseffect.
Conservation of energy (with q=0)gives
Uf = Ui + PiVi - PfVf
Uf + PfVf = Ui + PiVi
Hf = Hi
dPdPP
TdT
H
µ=
∂∂=
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P
T
P
T
H C
T
H
P
H
P
T µµ −=
∂∂
∂∂
−=
∂∂=
dH = TdS + VdP
VP
ST
P
H
TTT +
∂∂=
∂∂=µ
VT
VT
P
+
∂∂−=
= -aVT + V
=V(1 -aT)
aTinv=1
Ideal gas: mT = (1 - (1/T)T) = 0
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Lecture 14: Thermodynamic Calculations 2
Example 3. CP - CV
dH = dqP,rev + VdP
= CPdT + VdP
= TdS + VdP
VV
PP T
STC
T
STC
∂∂=
∂∂= ,
S = S(T,P)
dPP
SdT
T
SdS
TP
∂∂+
∂∂=
dPT
VdT
T
C
P
P
∂∂−=
dPVT
dTCP α−=
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dPTVdTCTdS P α−=
VP
V T
PTVC
T
ST
∂∂−=
∂∂ α
T
PPV
P
VT
V
TVCC
∂∂
∂∂
+= α
TPV
TVCC
κα 2
−=
Ideal gas:
nRT
VP
T
TVPTVCC
TVP ====− 2
2
κα
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Application to a real gas:
2mm V
a
bV
RTP −
−=
bV
RT
V
aP
mm −=+
2
( ) RTbVV
aP m
m
=−
+
2
( ) RdTdVV
abV
V
aP m
mm
m
=
−−
+
322
P
m
m T
V
V
∂∂
= 1α
2
)(21
m
m
m
m
RV
bVa
bV
TV −−
−
=α
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We can simplify this result by taking a seriesexpansion:
Let mVb /1 =ε (dimensionless)
and 22 / mVa=ε (atm-1)
a ~ 0.1 to 10, b ~ 0.01 to 0.001
For P = 1 atm and T = 298,42
243
1 1010,1010 −−−− −≈−≈ εε
1
211
)1(2
1
−
−−
−= εε
εα
R
VT m
1
21
2
1
11−
−
−≈ ε
ε RT
V
Tm
1
21
21
1−
−+≈ εεRT
V
Tm
+−≈ 21
21
1 εεRT
V
Tm
+−≈
mm RTV
a
V
b
T
21
1
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Substituting Vm =RT/P,
+−≈22
21
1
TR
aP
RT
bP
Tα
Tinv = 1/a fl 22
2
TR
aP
RT
bP
inv
=
\ Tinv º 2a/bR
For example, Tinv(H2) º 227 K
Tinv(CO2) º 2079 K
We can also determine the Joule-Thompsoncoefficient:
)1( −=−= TC
V
C PP
T αµµ
−+−≈ 12
122TR
aP
RT
bP
C
V
P
−≈
−= bRT
a
Cb
RT
a
RTC
PV
mPP
212
.
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Application to the virial equation:
++= 2
)()(1
mmm V
TC
V
TBRTPV
dTdT
dC
V
RT
dT
dB
V
RTdT
V
C
V
BRdV
V
C
V
BRTPdV
mmmmm
mmm
++
+++
+−= 2232 1
2
dTdT
dC
VdT
dB
V
RT
V
C
V
BRdV
V
C
V
BRTP
mmmmm
mm
++
++=
++ 1
12
232
++
++++
=
∂∂
32
22
2
11
1
mm
mmmmm
P
m
m
V
C
V
BRTP
dT
dC
V
T
dT
dB
V
T
V
C
V
B
V
RT
T
T
V
V
++
++++
=
2
22
21
11
mmm
mmmm
m
V
C
V
B
PV
RT
dT
dC
V
T
dT
dB
V
T
V
C
V
B
PV
RT
Tα
−−
++++≈
222
211
1
mmmmmm V
C
V
B
dT
dC
V
T
dT
dB
V
T
V
C
V
B
Tα
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17
Lecture 15: Introduction to the Second Law
Significance of the First Law
Implications:
1. Allows us to calculate the energy change for anyprocess (Thermochemistry)
2. Allows us to calculate q and w for a given path
3. Rules out perpetual motion machines of thefisrst kind
Limitations:
1. Tells us nothing about the feasibility of aprocess. DU<0 and DH<0 are not validindicators.
2. Tells us nothing about equilibrium
3. Does not treat q or w as state variables
4. Says nothing about perpetual motion of thesecond kind (qØw machines)
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The heart of the problem is our understanding of q.Heat is a chaotic form of energy, and is not 100%useable.
1. We would like to have a measure of chaos.
2. We would like to have a measure of theusefulness of heat.
The quantity of interest is the entropy, S.
1. S is a state variable. It is the extensivecounterpart to T that appears in the Eulerrelation.
2. S has an absolute value, S=0 for perfect order.
3. An incremental change of heat, dq, leads tomore disorder at low T than at high T. Transferof dq from a hot body to a cold body leads tomore useful work at high T than at low T.Hence, we expect that S, as measure of disorderand as a measure of the usefulness of heat,varies inversely with T.
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Lecture 16. Entropy
We would like to discover what it would take tomake dq a perfect differential. We will use anideal gas to test out ideas.
Let’s start with a problem that we alreadyunderstand.Let’s demonstrate that dV is a perfect differential.
V = V(T,P) = nRT/P
dV = f(T,P)dT + g(T,P)dP
P
nR
T
Vf
P
=
∂∂=
2P
nRT
P
Vg
T
−=
∂∂=
PT T
g
P
nR
P
f
∂∂=−=
∂∂
2
Q.E.D. (but hardly surprising)
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Now let’s examine dwrev.
dPP
nRTnRdTPdVdwrev +−=−=
f(T,V) = -nR
g(T,V) = nRT/P
0=
∂∂
TP
f
P
nR
T
g
P
=
∂∂
\ dwrev is not a perfect differential.
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Now let’s examine dqrev. For an ideal gas:
dPP
nRTnRdTdTCdwdUdq Vrevrev −+=−=
f(T,V) = CV + nR
g(T,V) = -nRT/P
0=
∂∂
TP
f
P
nR
T
g
P
−=
∂∂
\ dqrev is not a perfect differential.
What will it take to make dqrev exact?
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We can “fix up” g(T,P) by dividing it by T.
Define: dS = dqrev/T
dPP
nRdT
T
nRCdS V −
+=
P
dPnR
T
dTCdS P −= (1)
PT T
g
P
f
∂∂==
∂∂
0
\ dS is a perfect differential.
We can write this result equivalently as
dU = TdS - PdV
For an ideal gas, setting dU=CVdT and rearranging,
V
nRdV
T
dTC
T
PdV
T
dTCdS VV +=+= (2)
DS is independent of path.
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From Eq. (1): ∫∫ −=∆2
1
2
1
)(P
P
T
T
P P
dPnR
T
dTTCS
−= ∫
1
2ln)(2
1P
PnR
T
dTTC
T
T
P
From eq. (2): ∫∫ +=∆2
1
2
1
)(V
V
T
T
V V
dVnR
T
dTTCS
+= ∫
1
2ln)(2
1V
VnR
T
dTTC
T
T
V
Contributions to S result from increases in bothtemperature and volume and a decrease in pressure.
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Isolating different contributions to DS:
1. Constant pressure (reversible):
1
2ln)(2
1T
TC
T
dTTCS P
T
T
P ==∆ ∫
The second equality assumes a constant heatcapacity.
2. Constant volume (reversible):
1
2ln)(2
1T
TC
T
dTTCS V
T
T
V ==∆ ∫
3. Constant temperature (reversible):
1
2
1
2 lnlnP
PnR
V
VnRS −==∆
4. Adiabatic process:
DSad, rev = 0
DSad, irrev depends on the process
5. Phase transition:
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25
DStrans = DHtrans/Ttrans
Troutons’ Rule: DSvap,m º 85 J mol-1 K-1
fl DHvap,m º 85 Tb.p.J mol-1
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Lecture 17: Calculating the Entropy
Example: Heating of I2 at constant pressure, revisited
Cp,m = a + bT + c/T2
( )
−−−+
=∆ 2
12
212
1
2 112
lnTT
cTTb
T
TaS
Phase a b csolid 40.12 0.04979 0liquid 80.33 0 0vapor 37.40a 0.00059 -0.71e+5
aNote: CP at 300 K is 4.42R
Melting point = 386.8 KBoiling point = 458.4 K
DHfus,m = 15.52 kJ mol-1 fl DSfus,m = 40.12 J mol-1
K-1
DHvap,m = 41.80 kJ mol-1 fl DSvap,m = 91.18 J mol-1
K-1
(Trouton’s rule predicts 85 J mol-1 K-1)
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Calculate the entropy change accompanying theheating of one mole of I2 from 100 to 500 K at 1 atm.
DS = 40.12 ln(386.8/100)+ 0.04979 (386.8-100) + 40.12+ 80.33 ln(458.4/386.8) + 91.18
+ 37.40 ln(500/458.4)+0.00059(500-458.4)
-0.5x0.71x105(500-2-458.4-2)
= 216.21 J mol-1 K-1
For comparison, note that Table 2.6 gives DSf,m =116.135 J mol-1 K-1 at 298 K.
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Third Law of Thermodyanamics
If the entropy of every element in its most stablestate at T=0 is taken as zero, then every substancehas a positive entropy which at T=0 may becomezero, and which does become zero for all perfectcrystalline substances, including compounds.
Law of Dulong and Petit: CV = 3R for all atomiccrystals. (Atkins, p288)
Data taken from Atkins Table 2.2 and McQuarie,Statistical Mechanics, p 203:
Metal CV/R(constant term)
QD (K)
Al 2.49 390Cu 2.72 315Pb 2.66 88
This “law” clearly violates the Third Law, because,if CV is a constant,
T
dTC
T
T
V∫2
1
is ill-behaved as T1Ø0.
Experimentally it is observed that0lim
0=
→ VT
C
This discovery was crucial in the foundation ofquantum mechanics.
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29
Debye Heat Capacity:
Figure 18. The Debye Heat Capacity
The Debye theory treats a crystal as having acontinuous distribution of frequencies, n, with amaximum cutoff frequency, nD. This modelpredicts that CV is given by
( )∫Θ
−
Θ=
T
x
x
DV
D
dxe
exTRC
/
02
43
19
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where x = hn/kT and the Debye temperature isgiven by
k
h DD
ν=Θ
Limiting behavior:RCV
T3lim =
∞→
34
0 5
12lim
Θ=
→D
VT
TRC
π
This is the famous T3 temperature law for the heatcapacity and entropy.
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Lecture 18: The Second Law: Examples
How to deal with irreversible processes:
2 moles of an ideal atomic gas occupy a volume of 25liters at a temperature of 300K. Calculate the entropychange of the system and its surroundings in each ofthe following isothermal processes.
A. Suppose that one wall of the vessel is actually apiston that is held in place by a pin. The pressurebehind the piston is 10 atm. The pin is removedand the piston is allowed to push on the gasirreversibly until it comes to rest on its own.
B. Suppose that the gas is compressed reversibly tothe same final volume as in the previous question.
For later reference, note that P1 = nRT1/V1 = 1.97 atm
Let’s treat the reversible case first.
dqrev = -dwrev = PdV = nRT dV/V
dS = dqrev/T = nR dV/V
DS = nR ln (V2/V1)
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32
T = 300 K, V1 = 25 L, P2=10 atm, V2 = nRT/P2 =4.92 L
DS = -1.63R
dqrev,surr = - dqrev,sys fl DSsurr = -DSsys
DStot = DSsys + DSsurr = 0
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Now let’s treat the irreversible case.
Because S is a state variable, DSsys has the same valueas in the reversible case, so long as the final state isthe same. But DSsurr is different.
Suppose the outside world has a constant pressure.Then DHsurr equals the heat loss of the system,regardless if the process is reversible or not. That is:
DSsurr = -qrev/Tsurr = DHsurr/Tsurr = -qirrev/Tsurr
Note: 1) The minus sign comes from the fact thatheat leaves the system and enters the surroundings.2) This principle works because the surroundingheat bath is so large that its temperature does notchange when q is added or removed.
qirrev = -wirrev = PexDV
DSsurr = -PexDV/T = 8.157R (verify this!)
DStot = DSsys + DSsurr = 6.54 R
The total entropy change for a spontaneous processis greater than zero.
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Let’s consider next the equivalent adiabaticprocesses.
For the reversible case, DSsys = DSsurr = 0
For the irreversible case, DSsurr = 0 because no heatenters the surroundings. But for DSsys we must findan equivalent reversible path. (Recall the hour exam.)
-Pex (V2 - V1) = CV (T2 - T1)
PexV2 = P2V2 = nRT2
P2V1 + CVT1 = CVT2 + nRT2 = 5/2 nRT2
P2V1 + CVT1 = (P2/P1)(P1V1) + CVT1
= (5.078 + 3/2)nRT1 = 5/2 nRT2
T2 = 1399 K, V2 = 12.95 L (verify this!)
DSsys = CV ln (T2/T1) + nR ln (V2/V1)
= 3R ln (2.361) + 2R ln (0.518)= 1.262 R = DStot
Stot Again, DStot> 0 for an irreversible process.
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35
The bottom line in these examples:
1. Because DS is a state quantity, DSsys is the samefor reversible and irreversible processes, providedthat the final state is the same for both cases. If it isnot (as in the adiabatic example above), then for theirreversible process it is necessary to construct areversible path and calculate DSsys along that path.DSsys=0 for reversible adiabatic processes but not forirreversible ones.
2. DSsurr is given by -q/Tsurr, both for reversible andirreversible processes. It follows that DSsurr=0 for alladiabatic processes.
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36
Another example: Thermal equilibration of twoobjects.
Suppose we have two identical copper blocks, oneheated to 500 K and the other at 300 K. Supposethey are brought into contact and allowed to reach acommon final temperature, T.
1. Irreversible heat transfer:
CP(500 - T) = CP(T - 300)
\ T = 400
DS = CP ln(T/500) + CP ln(T/300)
= PP CC 0645.0500300
400ln
2
=
⋅
Again, we find that the total entropy increases for aspontaneous (irreversible) process.
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2. Reversible heat transfer. (Suppose the copperblocks are used as the heat source and sink for areversible engine.)
DS = CP ln(T/500) + CP ln(T/300) = 0
Solve for T:
ln(T/500) = -ln(T/300) = ln(300/T)
T2 = 300x500
T = 387 K
The geometric mean is less than the arithmetic mean.
Work extracted = 2(400 - 387)CP
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It is instructive to examine this problem in the limitof very large heat sources.
Irreversible transfer of heat q from a hot reservoir atThot to a cold reservoir at Tcold via a thin wire:
Figure 19. Irreversibleheat transfer.
DShot = -q/Thot
DScold = q/Tcold
DStotal = q[1/Tcold -/Thot] > 0
w = 0
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Reversible transfer of heat q from a hot reservoir atThot to a cold reservoir at Tcold via a gas and a piston:
Figure 20.Reversibleheat transfer.
1) Isothermalexpansion;2) Adiabaticexpansion3) Isothermalcompression
1. Transfer of q from Thot to the gas by anisothermal expansion:
DShot = -q/Thot = -DSgas,1, w = -q
2. Adiabatic expansion of the gas, cooling it to Tcold
DSgas,2 = 0, w = -Cv (Thot - Tcold)
3. Transfer of q from the gas to Tcold to by anisothermal compression:
DScold,3 = q/Tcold = -DSgas,3, w=qDSgas = -q[1/Tcold -1/Thot], DStotal = 0
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40
Lecture 19. Theorems Associated with theSecond Law
We will use the following formulations of theSecond Law to prove various theorems. In thenext lecture we will show that they areequivalent.
1. Entropy formulation: The entropy of an isolatedsystem increases in the course of a spontaneouschange: DStot > 0.
2. Kelvin-Planck statement: It is impossible for asystem to undergo a cyclic process for which the onlyeffects are the flow of heat into the system from aheat reservoir (a Areversible heat source@) and theperformance of an equivalent amount of work by thesystem on its surroundings (a Areversible worksource@).
3. Clausius statement: It is impossible for a system toundergo a cyclic process for which the only effectsare the flow of heat into the system from a coldreservoir and flow of an equal amount of heat out ofthe system into a hot reservoir.
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1. DStot = 0 for a reversible process. Decompose thepath into an adiabatic and an isothermal component.DSsysªDS = 0 along the adiabatic path and DS = q/Talong the isothermal path. DSsurr = 0 along theadiabatic path and DSsys = -q/T along the isothermalpath. \ DStot = DS + DSsurr = 0.
Figure 21.Decompositionof a bath intoadiabats andisotherms.
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2. Clausius inequality:
TdqdS /≥
Proof:0≥totdS
0≥+ surrdSdS
TdqdSsurr /−=
0≥−T
dqdS
TdqdS /≥∴
TqdScyc /∫≥∆But DScyc = 0 because S is a state variable.
0/ ≤∴ ∫ Tqd
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An important consequence of the Clausius inequalityis that the entropy change along an irreversible path
is positive.
Figure 22. The entropy of anirreversible process.
Let AØB be an irreversible and isolated process.\ qAØB=0.
Let BØA be a reversible process that completes thecycle.
The Clausius inequality fl ∫ <A
B T
dq0
But q along path BØA is a reversible. This impliesthat along this path,
∫ <A
B
rev
T
dq0
\DS(BØA) < 0
\DS(AØB) > 0
Q.E.D
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The efficiency of an engine is defined by
hot
cold
hot
coldhot
hot
coldhot
absorbed q
q
q
q
q
work−=
−=+== 1ε
Figure 23. Athermodynamic engine.The First Law givesqhot + qcold + w =0 .
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3. All reversible engines have the same efficiency.
Figure 24. Proof of the theorem.
Suppose engine 1 absorbs q1,hot and produces q1,cold
and w1. Similarly, engine 2 has parameters q2,hot,q2,cold, and w2. Suppose that e2 > e1. Run engine 2 inreverse (i.e., as a refrigerator), such that its input is-q1,cold and w2. Its greater efficiency implies that lesswork, w2 < -w1, is needed to remove q1,cold and lessheat, q2,hot, is generated.
Net result: q1,hot - |q2,hot| is converted into |w1| - w2.This violates the Kelvin formulation.
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4. The efficiency of a reversible engine is given by
hot
cold
hotengine T
T
q
w−== 1ε
Proof: Use a Carnot Cycle
1
2
2
1 lnlnV
VnR
T
TCS V +=∆
Path 1:A
B
A
B
hot
hotVBA V
VnR
V
VnR
T
TCS lnlnln =+=∆ →
Recall:
=
A
Bhot V
VnRTq ln1
Path 2:B
C
hot
coldVCB V
VnR
T
TCS lnln +=∆ →
B
C
hot
coldVCB
V
V
T
T
nR
C
nR
Slnln +=∆ →
0ln =
=
B
CnR
C
hot
cold
V
V
T
TV
This result is expected because dqrev = 0 along anadiabat.
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Path 3:A
B
C
D
cold
coldVDC V
VnR
V
VnR
T
TCS lnlnln −=+=∆ →
Recall:
=
C
Dcold V
VnRTq ln3
Path 4: 0=∆ → ADS
For the entire cycle: DS=0
Heat budget:cold
hot
cold
hot
T
T
q
q−=
Work done in one cycle: w = -(qhot + qcold)
Engine efficiency:hot
cold
hot
coldhot
hot T
T
q
q
w−=
+== 1ε
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Actually, we don’t need to use the Carnot cycle (orany specified system) to prove this result. It allcomes directly from the First and Second Laws.
First Law: qh + qc + w = 0 (1)
Second Law: qh/Th + qc/Tc = 0 (2)
Substituting Eq. (2) into Eq. (1) gives immediately:
h
c
h T
T
q
w −=−1
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49
5. Derivation of the Clausius inequality using theefficiency of a Carnot engine.
Suppose a cycle contains an irreversible part.Construct a Carnot engine that includes that path. It’sefficiency will be less than ideal.
hot
coldhotirrhot
irrcold
irrhot
irrhot
irr
T
TT
q
q
w −<+=−
hot
coldirrhot
irrcold
T
T
q
q −<
0<+hot
irrhot
cold
irrcold
T
q
T
q
Take the sum of all contributions from the mini-Carnot cycles used to describe an arbitrary cyclicpath:
0<∫ T
dqirr
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50
Lecture 20: Formulations of the Second Law
1. Entropy formulation: The entropy of an isolatedsystem increases in the course of a spontaneouschange: DStot > 0.
2. Kelvin-Planck statement: It is impossible for asystem to undergo a cyclic process for which the onlyeffects are the flow of heat into the system from aheat reservoir (a Areversible heat source@) and theperformance of an equivalent amount of work by thesystem on its surroundings (a Areversible worksource@).
3. Clausius statement: It is impossible for a system toundergo a cyclic process for which the only effectsare the flow of heat into the system from a coldreservoir and flow of an equal amount of heat out ofthe system into a hot reservoir.
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51
Equivalence of the various formulations of theSecond Law.
First we will show that the Kelvin and Clausiusformulations are equivalent.
Proof that Kelvin fl Clausius:Suppose this statement is false. Set up a conventionalengine that converts qhotØqcold + w. Then hook up ananti-Clausius refrigerator to our engine and transferqcold back into the hot reservoir. Net result: qhot -
|qcold| is converted into |w|, in contradiction to theKelvin formulation.
Figure25. Proof thatKelvin fl Clausius.
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52
Proof that Clausius fl Kelvin:
Suppose this statement is false. Set up a conventionalrefrigerator to convert qhot≠qcold + w. Then hook upan anti-Kelvin refrigerator to our refrigerator and useit to generate the work needed to run it by directtransfer of heat out of the cold reservoir. Net result:an amount of heat equal to qcold + w is transferredfrom the cold reservoir to the hot reservoir, incontradiction to the Clausius formulation.
Figure26. Proof thatClausius fl Kelvin.
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Next we will show that the Entropy Formulation isequivalent to the Clausius Formulation. We will usethe Clausius inequality as a description of theEntropy Formulation.
Clausius inequality fl Kelvin Formulation:
Suppose the Kelvin formulation were false.Use an engine operating at temperature T0 towithdraw heat q and convert it into work w.
With respect to the engine, q > 0 and w < 0. (Thisfollows because DU = q + w = 0 for a cycle.)
q > 0 and T0 constant fl ∫ > 00T
dq
But this contradicts the Clausius inequalityfl the Kelvin formulation must be true.
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Kelvin Formulation fl Clausius inequality:
Suppose that DStot < 0 for a spontaneous process.Then the Clausius inequality becomes
∫ > 0T
dq
For an engine connected to an isothermal bath atT=T0, it follows that q > 0 over the entire cycle.q + w =0 fl w < 0 for the cycle. This is contrary tothe Kelvin formulation, proving the theorem.
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55
Lecture 21: Thermodynamic Engines
1. Reversible engines:
Figure 27.Thermodynamics of areversible engine.
First Law: -dqhot = dqcold + dw
Second Law: 0=+cold
cold
hot
hot
T
dq
T
dq
hothot
coldcold dq
T
Tdq −=
dwdqT
Tdq hot
hot
coldhot −=
dwT
Tdq
hot
coldhot −=
−1
hot
cold
hotengine T
T
dq
dw −=−= 1ε
1lim0
=∞→
→ engine
TT
hot
cold
ε
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56
2. Reversible heat pumps:
Figure 28.Thermodynamics of areversible heat pump adrefrigetator.
First Law: dqhot = -dqcold - dw
Second Law: 0=+cold
cold
hot
hot
T
dq
T
dq
hothot
coldcold dq
T
Tdq −=
dwdqT
Tdq hot
hot
coldhot −=
dwT
Tdq
hot
coldhot −=
−1
coldhot
hothotpumpheat TT
T
dw
dq
−=−=_ε
∞=→ pumpheat
TT hotcold_lim ε
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3. Reversible refrigerators:
First Law: dqhot = -dqcold - dw
Second Law: 0=+cold
cold
hot
hot
T
dq
T
dq
coldcold
hothot dq
T
Tdq −=
dwdqT
Tdq cold
cold
hotcold +=
dwT
Tdq
cold
hotcold =
−1
coldhot
coldcoldorrefrigerat TT
T
dw
dq
−==ε
∞=→ orrefrigerat
TT coldhot
εlim
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58
Lecture 22: Microscopic Basis of TheSecond and Third Laws
S is a measure of the number of microscopic states Wthat are consistent with an observed macroscopicstate (i.e., for a given U, V, n1, n2,…).
Analogy with a pair of dice:
Macroscopicstate
Microscopic state W
2 {1,1} 13 {1,2},{2,1} 24 {1,3},{2,2},{2,1} 35 {1,4},{2,3},{3,2},{4,1} 46 {1,5},{2,4},{3,3},{4,2},{5,1} 57 {1,6},{2,5},{3,4}{4,3},{5,2},{6,1} 68 {2,6}{3,5},{4,4},{5,3},{6,2} 59 {3,6},{4,5},{5,4},{6,3} 4
10 {4,6},{5,5},{6,4} 311 {5,6},{6,5} 212 {6,6} 1
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Molecular example: m energy levels
∑=
=m
iiiENU
1
Microscopic configurations:
{N11, N21,…, Nm1},
{N12, N22,…, Nm2},…,
{N1W, N2W,…, NmW}
Boltzmann’s definition of the entropy:
S = k lnW
k = R/NA = 1.381x10-23 J K-1
S is dimensionless if T has the units of energy.
As TØ0, all molecules drop to E1, so that
W=1
k lnW =0
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60
Exception: 2 (or more) molecular orientations at theground energy level.
Suppose there are N molecules with 2 equivalentorientations. (g=2)
W = 2N
S(T=0) = k ln{2N} = kN ln 2
= kNA(N/NA) ln2
= nR ln 2
Rationalization for using a logarithm is that it makesS extensive.
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61
Example: isothermal expansion
g ~ V/V0
N
V
VkS
=
0
ln
NN
V
Vk
V
VkS
−
=∆
0
1
0
2 lnln
=
−
=∆
1
2
0
1
0
2 lnlnlnV
VNk
V
VNk
V
VNkS
=∆
1
2lnV
VnRS
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Nernst’s Heat Theorem: For any physical or chemicaltransformation,
0lim0
=∆→
ST
R(T) Ø P(T)
Æ ∞
R(0) Ø P(0)
( ) 0)(0
=−+∆ ∫ T
dTCCTS
T
PR
( )T
dTCCTS
T
RP∫ −=∆0
)(
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Lecture 23: Thermodynamic Potentials
Legendre Transforms
The conventional way to describe a curve is by afunction, where for every X we specify a Y:
Y = Y(X)
But we could also map out the curve by drawing thetangent at every point and tabulating the slopes,
dX
dYP =
and intercepts,
PX
Y =−−
0
ψ
PXY −=ψ
for every point on the curve. Elimination of X and Ygives a new function,
y = y(P).
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64
Figure 29. Constructionof the LegendreTransform.
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Example:
2
4
1XY =
XP2
1=
Y = P2
222 2 PPPPXY −=−=−=ψ
In general, the Legendre transform is given by
y = Y - PX,
where it is understood that X and Y are eliminated.
Simply replacing Y(X) by Y(P) is unacceptable,because Y(P) no longer describes a unique curve.
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The fundamental relation U = U(S, V, n) contains allthermodynamic information about an equilibriumstate, using only extensive quantities.
A Legendre transform can be used to replace one ormore extensive quantities by the correspondingintensive variable. For example, we may replace Vand S by -P and T:
SV
UP
∂∂=−
VS
UT
∂∂=
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Legendre Transforms in Thermodynamics
Replace V by -P: H = U + PV = H(S, P, n)
Replace S by T: A = U - TS = A(T, V, n)
Replace V by -P and S by T:
G = U + PV - TS = H - TS = G(P, T, n)
Derivative Relations
dU = TdS - PdV
dH = dU + PdV + VdP = TdS +VdP
dA = dU - TdS - SdT = -SdT - PdV
dG = dU + PdV + VdP - TdS - SdT = VdP - SdT
Two new Maxwell relations:
TV V
S
T
P
∂∂=
∂∂
TP P
S
T
V
∂∂−=
∂∂
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68
We can find all the differential formulas and theMaxwell relations in the Maxwell square.
Figure 16. TheMaxwell square
Macroscopic changes at constant T
DA = DU - TDS
DG = DH - TDS
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Lecture 24: Thermodynamic Potentials,Continued
Inequalities and signposts
dS + dSsurr ¥ 0
dS - dq/T ¥ 0
dS ¥ dq/T (Clausius)
TdS ¥ dq
TdS ¥ dU + PdV
Note: This becomes an equality for a reversible process.
At constant V,TdS ¥ dU
At constant U and V, dS ¥0.
dS > 0 fl S is at a local maximum at equilibriumAt constant S and V,
dU < 0 fl U is at a local minimum at equilibriumThese are postulates of thermodynamics.
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Transform V Ø -P
TdS ¥ dq = dH - VdP
At constant P,TdS ¥ dH
At constant H and P, dS ¥0.
dS > 0 fl S is at a local maximum at equilibrium
At constant S and P,
dH < 0 fl H is at a local minimum at equilibrium
By similar reasoning, at constant V and T,
dA < 0 fl A is at a local minimum at equilibrium
At constant P and T,
dG < 0 fl G is at a local minimum at equilibrium
(To prove these results rigorously, we must also showthat d2H, d2A, and d2G are >0.)
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Differential Quantities
dU = dqV
dH = dqP
dA = dwmax,total
dG = dwmax,other
Maximum Work
For an irreversible process,
TdS ¥ dU + PdV = dU - dw
\ dw ¥ dU - TdS
Note: The work is a maximum when it is the mostnegative, i.e. when w is as small as possible.
\ dwmax = dU - TdS = dA + SdT
At constant temperature,
dwmax = dA
wmax = DA
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72
We can distinguish between mechanical (“PV”) workand other types of work (e.g., electrical):
dw = -PdV + dwother ¥ dU - TdS
dwother ¥ dU - TdS + PdV
dwother,max = dU - TdS + PdV = dG + SdT - VdP
At constant T and P,
dwother,max = dG
Dwother,max = DG
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Properties of the Gibbs Free Energy
dG = VdP - SdT
ST
G
P
−=
∂∂
VP
G
T
=
∂∂
1. Pressure dependence of G
∫=−f
i
P
P
if VdPPGPG )()(
Incompressible material:
)()()( ifif PPVPGPG −=−
Ideal gas:
==− ∫
i
f
P
P
if P
PnRTdP
P
nRTPGPG
f
i
ln)()(
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74
2. Temperature dependence
T
HGS
T
G
P
−=−=
∂∂
Gibbs-Helmholtz Equation:
2
1
T
G
T
G
TT
G
T PP
−
∂∂=
∂∂
222 T
H
T
G
T
HG −=−−=
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75
Thermodynamic Calculations
Recipe for evaluating thermodynamic derivatives
1. If the derivative contains any potentials, bringthem one by one to the numerator and eliminatethem using the thermodynamic square.
2. If the derivative contains the entropy, bring it tothe numerator. If possible, use one of theMaxwell relations to eliminate it. If this doesn’twork, put a ∑T under the ∑S. The numeratorwill now be expressible as either CV or CP.
3. Bring the volume to the numerator. Theremaining derivative will now be expressible interms of a and kT.
4. Invoke TVP TVCC κα /2+=
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76
Example of step 1.
1−
∂∂=
∂∂
GG P
U
U
P
1−
∂∂−
∂∂=
GG P
VP
P
ST
1−
∂∂
∂∂
+
∂∂
∂∂−
=
P
V
P
S
V
G
P
GP
S
G
P
GT
1−
∂∂−
+
∂∂−
+
∂∂−
+
∂∂−
−
P
V
P
S
V
TS
VP
TS
P
S
TS
VP
TS
T
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77
Example of step 2.
PP
P
P
T
S C
VT
TC
T
V
T
SP
S
P
T α=
∂∂
=
∂∂
∂∂
−=
∂∂
/
Note that at constant pressure, dqp = dH =TdSBut also dqp = CPdT. \ CPdT = TdS.
T
C
T
S P
P
=
∂∂∴
Another example:
V
TC
T
VT
S
V
S P
P
P
P α/=
∂∂
∂∂
=
∂∂
Example of step 3.
ακ T
P
T
V
T
V
P
V
P
T =
∂∂
∂∂
−=
∂∂
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78
Lecture 25: Fugacity
Chemical potential for a mixture of substances:
dG = -SdT + VdP + m1dn1 + m2dn2
At constant T, P, n1 + n2,
dG = m1dn1 + m2dn2 = (m1 - m2) dn1
At equilibrium, dG = 0 fl m1 = m2
In general,
ijnPTii n
G
≠
∂∂=
,,
µ
For a pure substance, G = nGm fl m = Gm
In this lecture we will consider only this case. For anideal gas,
+=0
0
P
PnRTln)PG(T,P)G(T,
+=0
0
P
PRTln)P(T,P)(T, µµ
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79
Choice of a standard state:
1. Specify T
2. For a condensed phase, P0 = 1 bar.
3. For a gas, the standard state is a hypotheticalstate at 1 atm in which the gas behaves ideally.
Definition of the fugacity, f: For a non-ideal gas,
+=0
0
P
fRTln)P(T,P)(T, µµ
Definition of the fugacity coefficient:
f = f/P
1lim0
=→
φP
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80
How to calculate the fugacity: At constant T,
dm = RT dln f
But we also know that
m
T
m
T
VP
G
P=
∂∂
=
∂∂µ
\ dm = VmdP
\ RT d ln f = VmdP
PRTddPVP
fRTd m lnln −=
dPP
RTVm
−=
dPPRT
V
P
fd m
−=
1ln
dPP
Z
P
dP
RT
PVm
−=
−= 11
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81
Integrating over pressure,
PdP
Z
P
f
P
f
P
f P
P
′
′−==
=
−
∫
= 00
1lnlnlnln φ
Interpretation: Use a thermodynamic path to connectwith the standard state.
Figure 30.Calculationof thefugacity.
Ideal gas @ P=1atm, T Ø Ideal gas @ P=0, T
Ø Real gas @ P=0, T Ø Real gas @ P, T
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82
Note that
idealreal ffP
flnlnln −=
PP
Z
P
Z
PP
f11
ln−=−=
∂
∂
where PPP 11
ln−=
∂
∂
and P
Z
P
f =∂
∂ ln
Integrating along the thermodynamic path,
)(ln)(ln Pfpf idealreal −
∫∫ ′′
++′′
=P
real
P
ideal P
PdZ
P
PdZ
0
0
0
( )∫ ∫∫ ′′
−=′′
+′′
−=P PP
P
PdZ
P
PdZ
P
Pd
0 00
1
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83
Example of a real gas
...1 2 +′+′+= PCPBZ
( ) PdPCBP
f P
′′′+′=
∫0
ln
2
2
1PCB ′+′=
2
2
1PCPB
Pef′+′
=
In the limit of PØ0,
ZRT
PVPB
P
f m ==′+=1
Define Pideal = RT/Vm
ZP
Pideal 1=
idealP
P
P
f =
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84
Numerical example: Ar gas (See table 1.4)
T=273 K T=600KB (liter/mol) -0.0217 0.0190B’=B/RT (atm-1) -9.69x10-4 2.42x10-4
f at 1 atm 0.9990 1.0024f at 100 atm 90.31 102.42
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85
Temperature dependence of the fugacity
Let f be the fugacity at temperature T and pressure P,and f* be the fugacity at temperature T and P=0.
oo f
fRT
f
fRT ln
*ln* −=−µµ
f
fR
TT
*ln
* =− µµ
PPPP T
fR
T
fR
T
T
T
T
∂∂−
∂∂=
∂∂−
∂∂ ln*ln)/()/*(
**
µµ
But the Gibb’s-Helmoltz equation gives:
22
*
*
)/()/*(
RT
H
RT
H
T
T
T
T mm
PP
+−=
∂∂−
∂∂ µµ
Therefore,
22
*
*
ln*ln
RT
H
RT
H
T
f
T
f mm
PP
+−=
∂∂−
∂∂
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86
But in the limit of P*=0, f=0, making the 1st termvanish. Therefore,
2
*ln
RT
HH
T
f mm
P
−=
∂∂
Recall that the pressure dependence of the enthalpy isgiven by the Joule-Thompson coefficient:
µmP
T
m CP
H,−=
∂∂
Example of a van der Waals gas:
3
2
2 )(2)(ln
RT
abP
RT
bP
RT
aP
P
f ++−=
Differentiation with respect to T gives H*-H, andfurther differentiation with respect to P gives m.