Ch 344 – Problems for Class # 23
Due: Mar. 25, 2015
1. Problem
Starting with the Shomate equation for the constant pressure heat capacity, work out the expressions for ∆H and ∆S for the constant pressure heating or cooling of a substance.
Solution
The Shomate equation is given by
C P = A + Bt + Ct2 + Dt3 + E/t2
where t = T /1000, T being the temperature in degrees Kelvin, and the heat capacity is in units
of J/K mol.
Assuming this form for the constant pressure heat capacity, the change in enthalpy for heating or
cooling is given by
Z
∆H
T2
=
T1
Z
T2
=
T1
C P dT
(
T
A+B
+C
1000
T
1000
2
+D
T
1000
3
)
2
+ (1000) E/T
2
dT
T
T3
T4
(1000)2 E 2
T2
+C
+
D
−
2(1000)
3(1000)2
4 × (1000)3
T
T1
B
C
2
2
3
3
A (T2 − T1 ) +
T − T1 +
T − T1
2(1000) 2
3(1000)2 2
1
D
1
4
4
2
T
−
T
−
(1000)
E
+
−
1
4(1000)3 2
T2
T1
=
=
AT + B
The enthalpy change in this equation is given in units of J/mol. If it is to be expressed in kJ/mol,
the quantity should be divided by 1000, and can be written as
1
1
1
1
1
−
∆H = A (t2 − t1 ) + B t22 − t21 + C t32 − t31 + D t42 − t41 − E
2
3
4
t2
t1
1
In a similar way, the entropy change is given by
Z T2
CP
∆S =
dT
T
T1
Z T2 T2
A
1
T
2
3
+D
+ (1000) E/T
dT
+B
+C
=
T
1000
(1000)2
(1000)3
T1
T
T
T3
(1000)2 E 2
T2
=
A ln(T ) + B
+D
−
+C
(1000)
2(1000)2
3 × (1000)3
2T 2
T1
B
C
= A ln (T2 /T1 ) +
(T2 − T1 ) +
T 2 − T12
(1000)
2(1000)2 2
(1000)2 E
1
D
1
3
3
− 2
+
T − T1 −
3(1000)3 2
2
T22
T1
1
E 1
1
1
2
2
3
3
= A ln (t2 /t1 ) + B (t2 − t1 ) + C t2 − t1 + D t2 − t1 −
− 2
2
3
2 t22
t1
where this quantity is expressed in units of J/K mol.
2. Problem
In class, a system was constructed in which two substance, A and B, were in thermal contact but
the system was insulated from its surroundings. The heat capacities of the two substances, CA
and CB , were assumed to be constants, independent of temperature. The two substances initially
had temperatures, TA and TB .
(a) Apply the first two laws of thermodynamics to show that the temperatures of the two substances will change and end up with the same temperature.
(b) Consider a system where 10 g of aluminum at 500 K are placed in 1 L of water at 300 K.
What is the final temperature of this system?
Solution
(a) The system can change by transferring heat from one substance to the other. Let TAf and
TBf be the final temperatures of the two substances. For any heat transfer between the two
substances, the heat lost by one object will be gained by the other. Since the system is
insulated, the overall heat gained by the system must be zero, which translates to
qA + qB
= CA (TAf − TA ) + CB (TBf − TB )
=
0
This can be solved for TBf to give
CA
(TAf − TA )
CB
The change in entropy for the heat transfer is given by
TBf = TB −
∆S
= CA ln (TAf /TA ) + CB ln (TBf /TB )
CA
= CA ln (TAf /TA ) + CB ln 1 −
(TAf − TA ) /TB
CB
The second law of thermodynamics says that a spontaneous change in the system will be
associated with an increase in entropy. It follows that the system will strive for a maximum
in entropy. The only free variable we have is TAf , so the maximum change in entropy is
associated with a vanishing derivative with respect to this quantity.
d∆S
dTAf
CA
−CA /(CB TB )
+ CB
TAf
1 − (CA /CB ) (TAf − TA ) /TB
CA
CB CA
=
−
TAf
CB TB − CA (TAf − TA )
= 0
=
2
Solving this for TAf is an exercise in algebra. There results
TAf =
CA TA + CB TB
CA + CB
This can be inserted into the equation for TBf to give
CA CA TA + CB TB
TBf = TB −
− TA
CB
CA + CB
1
CA
=
(CA + CB ) TB −
(CA TA + CB TB − CA TA − CB TA )
CA + CB
CB
1
=
{CA TB + CB TB − CA TB + CA TA }
CA + CB
CA TA + CB TB
=
CA + CB
This is precisely the same as TAf . The two substances come to the same temperature spontaneously.
(b) The heat capacity of aluminum is∗ 0.897 J/g K and that for water † at 30◦ C is 4.1784 J/g
K. The density of water at this temperature is 0.99565 g/cm3 . A 10 g sample of aluminum
therefore has a heat capacity of 8.97 J/K and 1 L of water has a heat capacity of 4160 J/K.
Using these values and the initial temperatures, the final temperature of the system will be
Tf =
(8.97 J/K)(500 K) + (4160 J/K)(300 K)
= 300.4 K
(8.97 J/K) + (4160 J/K)
The temperature of the water will hardly rise at all.
3. Problem
Show that
∂S
∂V
=
T
α
κT
Solution
The Maxwell relation associated with the helmholtz free energy is
∂S
∂P
=
∂V T
∂T V
The cyclic rule gives
Rearranging this then leads to
∗ CRC
† Ibid.,
∂S
∂V
∂P
∂T
V
∂T
∂V
P
∂V
∂P
= −1
T
1
(∂T /∂V )P (∂V /∂P )T
(∂V /∂T )P
= −
(∂V /∂P )T
(1/V ) (∂V /∂T )P
=
−(1/V ) (∂V /∂P )T
α
=
κT
= −
T
Handbook of Chemistry and Physics, 88th Ed., Ed. D. R. Lide, (CRC Press, Boca Raton, 2008), p. 4-135.
p. 6-4.
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4. Problem
For Extra Credit, evaluate the change in entropy for the isothermal volume change for a real gas.
Solution
To start, the entropy change for this process is given by the integral
Z V2 ∂S
∆S =
dV
∂V T
V1
Z V2
α
=
dV
κ
T
V1
In the last problem set, the quantities in the integrand for a real gas were worked out. They are:
2P (dB2 /dT − B2 /T )
1
1+
α =
T
RΞ (1 + Ξ)
1
2B2 P
κT =
1−
P
RT Ξ (1 + Ξ)
where
1/2
4B2 P
Ξ= 1+
RT
It is actually more convenient to evaluate these integrals in terms of the accompanying pressure
change (although integrals in V can be evaluated in closed form). Since this is an isothermal
process, the volumes imply definite pressures, and the desired integral can be written as
Z P2
α ∂V
∆S =
dP
∂P T
P1 κT
Z P2
= −
V α dP
P1
The volume is expressed in terms of the pressure as
V =
nRT
{1 + Ξ}
2P
Therefore,
∆S
=
=
=
=
P2
1
2P (dB2 /dT − B2 /T )
nRT
(1 + Ξ)
1+
dP
−
2P
T
RΞ (1 + Ξ)
P1
Z
nR P2 1
Ξ
2 (dB2 /dT − B2 /T )
−
+ +
dP
2 P1
P
P
RΞ
P
nR
1−Ξ
2 (dB2 /dT − B2 /T ) RT Ξ 2
−
ln P + 2Ξ + ln
+
2
1+Ξ
R
2B2 P1
P2
1−Ξ
T dB2 /dT
nR
ln P + ln
+
+ 1 Ξ −
2
1+Ξ
B2
P1
Z
The pressures at the endpoints are evaluated at the desired volumes according to
n
n2 B2
P = RT
+
V
V2
These expression are readily programmed into an Excel spreadsheet or some other mathematical
program.
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