CHM 3410 – Problem Set 3
Due date: Monday, September 21st (Note: Our first exam is Friday, September 25th).
Do all of the following problems. Show your work.
"Entropy, in spite of one hundred years of history, is not very well understood yet and so as such you will win every
time you use entropy in an argument." - John von Neumann
1) The Joule-Thompson coefficient for nitrogen (N2) at T = 300.0 K and p = 1.000 atm is JT = + 0.27 K/atm.
a) In a particular experiment nitrogen gas undergoes a Joule-Thompson expansion from an initial pressure
of 130. atm and an initial temperature T = 300.0 K to a final pressure of 1.0 atm. Estimate the temperature change
for the gas.
b) Is your estimated temperature change in part a likely to be larger, smaller, or about the same magnitude
as the actual temperature change that will occur if the above experiment is carried out (Hint - Look at Fig 2.32 in
Atkins.)
2) Consider one mole of an ideal monatomic gas (recall that for a monatomic gas C V,m = 3R/2 = 12.472 J/mol-K,
Cp,m = 5R/2 = 20.786 J/mol.K). The gas is initially at a pressure p = 4.000 atm and a temperature T = 400.0 K. Find
Ssyst, Ssurr, and Suniv for the following processes:
a) An isothermal reversible compression of the gas to a final pressure p f = 10.000 atm.
b) An reversible cooling of the gas from T i = 400.0 K to Tf = 300.0 K, at a constant pressure p = 4.00 atm.
c) An adiabatic reversible compression of the gas to a final pressure p f = 10.000 atm.
d) An adiabatic irreversible expansion of the gas against a constant pressure pex = pf = 1.000 atm.
Note that these are the same processes you considered in problem 2 of problem set 2. You can use any
information you previously obtained for these processes (or any information in the problem set 2 solutions) to do this
problem.
3) A "supercooled" liquid is a liquid that has been cooled to below its normal freezing point yet which is still in the
liquid state. Supercooled liquids will spontaneously freeze to form a solid.
Consider the following irreversible process involving 1.000 mol of supercooled water.
(pi = 1.00 atm, Ti = - 10.0 C)  (pf = 1.00 atm, Tf = - 10.0 C)
Find Ssyst, Ssurr, and Suniv for the above process. Note that Cp,m (H2O(s)) = 38.09 J/mol .K. HINT: Recall that
S can only be calculated for reversible pathways, and so you will need to find such a pathway to do this problem.
Y
Also do the following from Atkins:
Exercises
2.30 a Assume 1.00 mol of gas. Recall that, as derived in class, dU = T dV + CV dT (Eq
2.42). You need to know how the expansion is carried out to find q and w. Assume the expansion is isothermal and
irreversible against an external pressure equal to the final pressure of the gas.
Problems
2.24 This is actually a simple problem.
2.34 This one is not quite so simple. Start with the equation for internal energy, which is
(U/V)T = T (p/T)V - p , and transform it into the "partner" equation. HINT: You need to somehow get a term
involving H in your final result. Recall H = U + pV, which can be used to do this.
3.14 You only need to find the value for molar entropy at T = 298.15 K. Below the
lowest temperature in the data (14.14 K) you may assume the Debye approximation is valid, that is, that C p,m = aT3,
where the value for a is found from the data at T = 14.14 K.
EXTRA CREDIT:
Problem 2.40. This is actually an easy problem, but I like it because it is an interesting
application of thermodynamics.
Solutions
1) NOTE: I have changed the problem from the original so that it illustrates what I had actually intended, an example
of a Joule-Thompson expansion
a) We may estimate the temperature change using
JT = (T/p)H  (T/p)H
If we solve for T we get
T = JT p = ( 0.27 K/atm) [ (1 - 130) atm ] = - 34.8 K
b) The value for JT used in part a is for p = 1.00 atm and T = 300. K. However, the expansion occurs over
a range of pressures from 1 atm to 130 atm. An examination of Fig. 2.32 of Atkins shows that for nitrogen the value
for JT becomes smaller as pressure increases since increasing pressure brings you closer to the boundary between
cooling and heating, where JT = 0. Our value is therefore larger than it should be in the high pressure region, and so
our estimated change in temperature is larger than the true value that will occur for this process.
2) As a general comment, note that we will make extensive use of the results previously found for these processes in
problem set 2, and will not rederive them here.
a) The process is reversible and isothermal, so
Ssyst = if (dq)rev/T = (1/T) if (dq)rev = q/T = ( - 3047. J)/(400.0 K) = - 7.62 J/K.
Ssurr = if (dq)rev,surr/T = (1/T) if (dq)rev,surr = - qsyst/T = ( + 3047. J)/(400.0 K) = + 7.62 J/K.
Suniv = Ssyst + Ssurr = (- 7.62 J/K) + (+ 7.62 J/K) = 0.
b) The process is reversible and carried out at constant pressure, so
Ssyst = if (dq)rev/T
But for the constant pressure cooling of a substance dq = n Cp,m dT, and so (recalling that Cp,m is constant)
Ssyst = if [ n Cp,m/T ] dT = n Cp,m if dT/T = n Cp,m ln(Tf/Ti)
= (1.00 mol) (20.786 J/mol.K) ln(300.0/400.0)
= - 5.98 J/K
Since the process is reversible Suniv = 0 = Ssyst + Ssurr
Ssurr = - Ssyst = + 5.98 J/K
If you want a good mental picture of how the above process could be carried out reversibly consider the following
method. The system is initially in thermal contact with a temperature bath at T = 400.00 K. It is placed in contact
with a temperature bath at T = 399.99 K and allowed to come to equilibrium. It is then placed in contact with a
temperature bath at T = 399.98 K and allowed to come to equilibrium. This continues until the final temperature, T
= 300.00 K, is reached. In the limit of an infinite number of the above steps, each with an infintesimally smaller
change in the temperature for the bath, the process will be reversible.
c) The process is adiabatic, and so qsurr = 0, which means Ssurr = 0.
The process is reversible, so Ssyst = if (dq)rev/T = 0. Therefore Suniv = 0. Life is good.
d) The process is adiabatic, and so qsurr = 0, which means Ssurr = 0.
Unfortunately, the process is irreversible, and so we cannot use the actual pathway for the process to calculate Ssyst.
Instead, we need to find a reversible pathway that connects the initial and final states.
The inital and final states for the process are
pi = 4.00 atm
Ti = 400.0 K
pf = 1.00 atm
Tf = 280.0 K
One possible reversible pathway is the following
Step 1 - An isothermal reversible expansion of the gas from pi = 4.00 atm to pf = 1.00 atm, at T = 400.0 K.
Step 2 - A constant pressure cooling of the gas from T i = 400.0 K to Tf = 280.0 K, at p = 1.00 atm
We have already found expressions for Ssyst for step 1 (problem 2a) and step 2 (problem 2b) and so will simply use
the results.
Ssyst,1 = q/T = - (nRT ln(pf/pi) = - nR ln(pf/pi) = - (1.00 mol) (8.314 J/mol.K) ln (4.00/1.00) = + 11.53 J/K
T
Ssyst,2 = n Cp,m ln(Tf/Ti) = (1.00 mol) (20.786 J/mol.K) ln(280./400.) = - 7.41 J/K
So Ssyst = Ssyst,1 + Ssyst,2 = + 11.53 J/K + (- 7.41 J/K) = + 4.12 J/K
And Suniv = Ssyst + Ssurr = + 4.12 J/K + 0. = 4.12 J/K.
3) To find Ssyst requires that we find a reversible pathway connecting the initial and final states. One such pathway
is as follows.
Step 1 - Reversible heating of liquid water from T = - 10.0 C to T = 0.0 C.
Step 2 - Phase transition from liquid to solid water, which is reversible at T = 0.0 C.
Step 3 - Reversible cooling of solid water from T = 0.0 C to T = - 10.0 C.
We have previously (problem 2b) found expressions for the entropy change for step 1 and 3 (note that these assume
Cp,m is constant). As discussed in class, the entropy change for a phase transition at the normal transition point is
Spt = Hpt/Tpt. So
S1 = n Cp,m(H2O()) ln(Tf/Ti) = (1.00 mol) (75.291 J/mol.K) ln(273.15 K/263.15 K) = + 2.808 J/K
S2 = - n Hfus/Tfus = - (1.00 mol) (6008. J/mol) / (273.15 K) = - 22.00 J/K
S3 = n Cp,m(H2O(s)) ln(Tf/Ti) = (1.00 mol) (38.09 J/mol.K) ln(263.15 K/273.15 K) = - 1.421 J/K
and so Ssyst = S1 + S2 + S3 = 2.808 J/K + (- 22.00 J/K) + (- 1.421 J/K) = - 20.61 J/K
To find the entropy change for the surroundings we need to know what Hfus is at T = 263.15 K.
(Remember, however, that our process is freezing, which is the opposite of fusion).
Hfus(263.15 K) = Hfus(273.15 K) + 273.15263.15 Cp,m dT = Cp,m T
where Cp,m = Cp,m(H2O()) - Cp,m(H2O(s)) = (75.291 J/mol.K) - (38.09 J/mol.K) = 37.20 J/mol.K
So Hfus(263.15 K) = 6008. J/mol + (37.20 J/mol .K) ( - 10.0 K) = 5636. J/mol
And so Ssurr = - qsyst/T = + Hfus(263.15 K) / 263.15 K = (5636. J/mol) / (263.15 K) = 21.42 J/mol .K
Finally, Suniv = Ssyst + Ssurr = - 20.61 J/K + 21.42 J/K = + 0.81 J/K
Note that Suniv > 0, as expected, since the freezing of a supercooled liquid is an irreversible process.
Exercise 2.30a
Recall that dU = T dV + CV dT. For an isothermal irreversible expansion dT = 0, and so
dU = T dV
Integrating both sides gives
U = if T dV
But T = a/Vm2 = an2/V2, so
U = if (an2/V2) dV = - an2 [ (1/Vf) - (1/Vi) ] For N2 , a = 1.352 L2.atm/mol2
So
U = - (1.352 L2.atm/mol2) (1.00 mol) 2 [ (1/24.8 L) - (1/1.0 L) ]
= 1.297 L.atm 101.3 J = 131. J
L.atm
To find the value for work we must first find the final pressure.
p = nRT - an2 = (1.00 mol) (0.082057 L.atm/mol.K) (298. K) - (1.352 L2.atm/mol2) (1.00 mol)2
(V - nb)
V2
[ (24.8 L) - (1.00 mol) (0.0387 L/mol) ]
(24.8 L) 2
= 0.9876 atm - 0.0022 atm = 0.9854 atm
and so w = - if pex dV = - pf if dV = - pf (Vf - Vi) = - (0.9854 atm) [(24.8 - 1.0) L] = - 23.45 L.atm 101.3 J = - 2376.
J
L.atm
Finally, since U = q + w, q = U - w = 131. J - (- 2376. J) = 2507. J
Problem 2.24
We begin with the relationship
Cp - CV = T (p/T)V (V/T)p
But (p/T)V (T/V)p (V/p)T = - 1, and so (p/T)V = - (V/T)p (p/V)T = - (V/T)p
(V/p)T
Substituting this result gives
Cp - CV = - T (V/T)p (V/T)p = - T (V/T)p2
(V/p)T
(V/p)T
(Note: The book has this relationship wrong).
For an ideal gas, V = nRT/p, and so
(V/T)p = nR/p
(V/p)T = - nRT/p2
and so Cp - CV = - T (nR/p)2 = nR , as expected.
(- nRT/p2)
Problem 2.34
(H/p)T = (H/V)T (V/p)T
But H = U + pV, and so (H/V)T = (U/V)T + p (V/V)T + V (p/V)T = (U/V)T + p + V (p/V)T
So (H/p)T = [ (U/V)T + p + V (p/V)T ] (V/p)T
But (U/V)T = T (p/T)V - p (the thermodynamic relationship given in the problem). Substituting gives
(H/p)T = [ T (p/T)V - p + p + V (p/V)T ] (V/p)T
= [ T (p/T)V + V (p/V)T ] (V/p)T
= T (p/T)V (V/p)T + V (p/V)T (V/p)T = T (p/T)V (V/p)T + V
But (p/T)V (T/V)p (V/p)T = - 1 , and so (p/T)V = - (V/T)p (p/V)T
If we substitute this result, we get
(H/p)T = - T (V/T)p (p/V)T (V/p)T + V
= - T (V/T)p + V
the final result.
Problem 3.14
S(298.15 K) = 0298.15 (Cp,m/T) dT
= 014.14 (Cp,m/T) dT + 14.14298.15 (Cp,m/T) dT
For temperatures below 14.14 K we assume that Cp,m = a T3. The value for a is found using the heat capacity at
14.14 K
(9.492 J/mol.K) = a (14.14 K)3
a = (9.492 J/mol.K) / (14.14 K)3 = 3.36 x 10-3 J/mol.K4
So 014.14 (Cp,m/T) dT = 014.14 (a T3/T) dT = 014.14 (a T2) dT = (a T3)/3
= (3.36 x 10-3 J/mol.K4) (14.14 K)3 / 3 = 3.166 J/mol.K
For the second integral, 14.14298.15 (Cp,m/T) dT, we will find the value for the integral numerically using the trapezoid
rule. For adjacent temperatures we say
T1T2  (T2 - T1) { [ Cp,m(T2)/T2 ] + [ Cp,m(T1)/T1 ] }/2
T (K)
14.14
16.33
20.03
31.15
44.08
64.81
100.90
140.86
183.59
225.10
262.99
298.06
298.15
Cp,m (J/mol.K)
9.492
12.70
18.18
32.54
46.86
66.36
95.05
121.3
144.4
163.7
180.2
196.4
Cp,m/T (J/mol.K2)
0.6713
0.7777
0.9076
1.0446
1.0631
1.0239
0.9420
0.8611
0.7865
0.7272
0.6852
0.6589
0.658
T (K)
2.19
3.70
11.12
12.93
20.73
36.09
39.96
42.73
41.51
37.89
35.07
0.09
S (J/mol.K)
1.587
3.118
10.854
13.626
21.632
35.475
36.026
35.201
31.417
26.758
23.569
0.0593
The data are plotted on the next page. Note that the area under the curve up to any temperature T represents the
value for S at that temperature (though the value calculated here for T = 298.15 K was found as described above).
14.14298.15 (Cp,m/T) dT   S = 239.322 J/mol.K
And so
S(298.15 K) = 014.14 (Cp,m/T) dT + 14.14298.15 (Cp,m/T) dT = (3.166 + 239.322) J/mol.K = 242.49 J/mol.K
(Cp,m)/T (J/mol.K2)
Plot of (Cp,m)/T vs T
1.2
1
0.8
0.6
0.4
0.2
0
0
50
100
150
200
250
300
350
T (K)
Problem 2.40
The heat capacity of liquid water is Cp,m(H2O()) = 75.291 J/mol.K
The heat capacity of a 65. kg person is
Cp = 65000. g
1 mol
18.02 g
0.075291 kJ
mol.K
= 271.6 kJ/K
Since Cp = (dq/dT)p  (q/T), it follows that
T =
q = 10000. kJ = 36.8 K
Cp
271.6 kJ/K
To find the mass of water that needs to be evaporated each day to maintain normal body temperature we need the
value for Hvap at T = 37. C. Since we are making an estimate, and since the enthalpy of vaporization does not
change much with temperature, we will use Hvap = 44.02 kJ/mol, the value at T = 25.0 C (Table 2.3).
So m = 10000. kJ
1 mol
0.01802 kg = 4.09 kg water (so about 4 liters of sweat per day)
44.02 kJ
mol
Note that the power output of an average person is
power = energy = 10000 x 103 J = 116. J/s = 116. watt
time
86400 s
or about the same power as used in a 100 watt light bulb.