CHEM 444 HW#6
Answer key by Jiahao Chen
due October 27, 2004
Grading policy: One point is given for each numbered equation shown in the solution, or for written
statements and/or equations to the same effect. The total score for this problem set is 35 points with a
maximum of 1 bonus point.
1. (20-14) Show that
∆S̄ = C̄V ln
T2
V2
+ R ln
T1
V1
(1)
if one mole of ideal gas is taken from T1 , V1 to T2 , V2 , assuming that C̄V is independent of temperature.
Calculate the value of ∆S̄ if one mole of N2 (g) is expanded from 20.0 dm3 at 273 K to 300 dm3 at 400
K. Take C̄P = 29.4 JK−1 mol−1 . [8 points]
Solution. By definition, the infinitesimal change in entropy is
dS =
¯
dq
T
(2)
where d¯ denotes an inexact differential. But from the first law of thermodynamics,
¯ = dU − dw
¯
dq
(3)
and assuming a fixed pressure, the differential work done on the system is
¯ = −P dV
dw
(4)
Substituting all these relations into the definition of the differential entropy (2) gives
dS =
¯
1
P
dU − dw
= dU + dV
T
T
T
now integrate to get
Z
S2
Z
T2
dS =
S1
T1
1
T
∂U
∂T
Z
T2
dT +
P
T1
P
dV
T
(5)
Note
that formally, a substitution of variables at fixed pressure was done to convert the integral
R
dU into an integral over temperature.
Now recognize that by definition,
∂U
∂T
≡ CP
(6)
P
which was assumed to be constant, and
Z
S2
dS = S2 − S1 ≡ ∆S
S1
1
(7)
CHEM 444
HW#6 Solutions
Furthermore the ideal gas equation can also be used to evaluate
nR
P
=
T
V
(8)
So
Z
∆S
T2
= CP
T1
dT
+ nR
T
Z
V2
V1
dV
V
T2
V2
= CP ln
+ nR ln
T1
V1
(9)
Finally dividing throughout by the number of moles n gives the molar quantities
V2
T2
+ R ln
T1
V1
∆S̄ = C̄P ln
Common mistake. Limits, limits, limits! Note that integrating the differential entropy should also
be accomplished using a definite integral, i.e.
Z S2
∆S ≡ S2 − S1 =
dS
S1
Common mistake. It is not true in general that the (integrated) change in entropy is
∆S =
qrev
(wrong!)
T
2. (20-15 paraphrased) Consider a two-compartment system in a rigid container, with the two subsystems having the same temperature but different pressures and the wall that separates them being
flexible rather than rigid. Show that in this case,
dS =
dVB
(PB − PA )
T
(10)
Interpret this result with regard to the sign of dVB when PB > PA and when PB < PA . [9 points]
Solution. The entropy of subsystem j ∈ {A, B} is
dSj
=
=
d̄qj
T
dU − d̄w
T
(11)
(12)
from the first law of thermodynamics, where in the specific case of mechanical work,
d̄w
⇒ dSj
= −P dV
dU + P dV
=
T
(13)
Now since entropy is extensive, it must be additive in the sense that
dS
= dSA + dSB =
dUA + PA dVA
dUB + PB dVB
+
T
T
(14)
But since the system is isolated (see note)
dU = dUA + dUB = 0
2
(15)
CHEM 444
HW#6 Solutions
and since the system is held in a rigid container with fixed volume,
dV = dVA + dVB = 0
(16)
so
PA dVA
PB dVA
dVB
−
=
(PB − PA )
T
T
T
Note that the second law of thermodynamics says that for an isolated system the change
in entropy of that system for a spontaneous process must always be positive, i.e.
dS =
(17)
dS > 0
So in the case that PB > PA the second law implies that
PB > PA ⇒ dVB > 0
(18)
so that the subsystem B expands in volume. In the converse case the second law implies
PA > PB ⇒ dVB < 0
(19)
in order to have dS > 0. This implies that in this case, the subsystem B contracts, i.e. the
subsystem A expands.
Note on what constitutes an isolated system. The textbook on p. 826 gives a very misleading
definition of an isolated system. The only defining characteristics of an isolated system are that
neither matter nor energy is allowed to flow in or out of a system. However it is untrue
that an isolated system must necessarily have a fixed volume; the universe is an excellent counterexample! In contrast, the system described in this problem is both isolated and isochoric
(of fixed volume).
Common mistake. Again, please do not write (integrated) changes as being related to differential
quantities without an integral! In particular,
∆S 6= dS =
d̄q
T
Common mistake. For some reason, it was very popular to mix up cause and effect in the interpretation of the sign of dS. It is not true that PB > PA nessarily implies dS > 0! You are
tacitly assuming dVB > 0 in order to get that result! Conversely, dS is always positive for a
spontaneous process in an isolated system. This is simply a restatement of the second law
of thermodynamics, which insists that some volume change must occur to re-establish equilibrium.
3. (20-32) Derive the equation dH = T dS + V dP . Show that
∆S̄ = C̄P ln
T2
P2
− R ln
T1
P1
(20)
for the change of one mole of an ideal gas from T1 , P1 to T2 , P2 , assuming that C̄P is independent of
temperature. [10 + 1 points]
Solution. The enthalpy is defined from the Legendre transform of the internal energy by eliminating
the dependence on volume [1 bonus point for showing the Legedre transformation]:
∂U
V
(21)
H (S, P ) = U (S, V ) −
∂V S
⇒ H = U + PV
(22)
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CHEM 444
HW#6 Solutions
Differentiating the above equation gives
(23)
dH = dU + P dV + V dP
Also, the first law of thermodynamics states that
dU = d̄q + d̄w
(24)
which can be eliminated for exact differentials through the definition of the differential entropy
dS =
d̄q
⇒ d̄q = T dS
T
(25)
and the differential (mechanical) work
d̄w = −P dV
(26)
Combining all these differential forms gives the desired result
dH = T dS − P dV + P dV + V dP = T dS + V dP
To solve for the desired change in entropy, first make dS the subject of the equation:
dS =
dH
V
− dP
T
T
(27)
which we can simplify by noting that by changing variables and recalling the original definition
of the enthalpy, that by definition
∂U
∂H
dT =
dT = CP dT
(28)
dH =
∂T P
∂T P
and also from the ideal gas equation
V
nR
=
T
P
(29)
and hence
dP
dT
− nR
T
P
which can be converted into molar quantities by dividing throughout by n:
dS = CP
dS̄ = C̄P
dT
dP
−R
T
P
(30)
Integrating with the assumption C̄P 6= C̄P (T ) gives
Z
∆S
T2
= C̄P
T1
= C̄P ln
dT
−R
T
Z
P2
P1
dP
P
(31)
T2
P2
− R ln
T1
P1
Common mistake. Despite what the question suggests, the proper way is to get the molar change in
entropy is to divide throughout by the number of moles n at some point in the derivation, rather
than setting n = 1 mol and then inserting the overbars¯ad hoc. All extensive quantities such as
S and V (which several people forgot to convert to molar volume V̄ !) then acquire overbars as
their molar counterparts are intensive.
Note on Cp . While the derivation on pp. 783-4 of the textbook is correct for ideal gases only, it is not
difficult to show that the result is also true in general, as is shown in (28).
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CHEM 444
HW#6 Solutions
4. (21-2) The molar heat capacity of H2 O (l) has an approximately constant value of C̄P = 75.4 J ·
K−1 mol−1 from 0o C to 100o C. Calculate ∆S if two moles of H2 O (l) are heated from 10o C to 90o C at
constant pressure. [4 points]
Solution. The result from the previous problem can be used immediately, but also noting that the
heating process is isobaric, i.e. P2 = P1 , so
∆S̄
= C̄P ln
T2
P1
T2
− R ln
= C̄P ln
T1
P1
T1
(32)
and therefore
∆S
T2
T1
(273 + 90) K
(2 mol) 75.4 J · K−1 mol−1 ln
(273 + 10) K
= nC̄P ln
(33)
=
(34)
=
37.5 J · K−1
(35)
5. (21-6) Show that because C̄P − C̄V = R for an ideal gas,
∆S̄P = ∆S̄V + R ln
T2
T1
Check numerically that your answers to Problems (21-4) and (21-5) differ by R ln 2 = 0.693R = 5.76 J ·
K−1 mol−1 . [4 points]
Solution. Note that from equations (21.5) and (21.9) (p. 854-5),
Z
∆S̄P
=
T2
∆S̄V +
T1
Z T2
=
=
C̄P − C̄V
dT
T
R
dT
T1 T
T2
∆S̄V + R ln
T1
∆S̄V +
(36)
(37)
(38)
And since the processes in in Problem (21-5) and Problem (21-4) have the same T1 and T2 , the
difference is simply
∆S̄P − ∆S̄V = R ln
T2
600 K
= R ln
= R ln 2 = 0.693R = 5.76 J · K−1 mol−1
T1
300 K
(39)
Note. You weren’t asked to derive the relation C̄P − C̄V = R. Also if you were lazy and just looked up
the answer at the back of the book (or the solutions manual) for the last part, I gave you credit
if you cited the source.
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