CHM 3410 – Problem Set 3
Due date: Monday, September 20th (NOTE: Our first exam is on Friday, September 24th. It will cover material from
Chapters 1, 2, and 3 of Atkins, and handouts.)
Do all of the following problems. Show your work.
1) A 10.00 g block of iron metal (sp,Fe = 0.449 J/g.K) at T = 400.0 K is placed in contact with a 20.00 g block of
aluminum metal (sp,Al = 0.903 J/g.K) at T = 300.0 K. Heat flows between the two metal blocks until equilibrium is
reached. Find Tf (the final temperature) and S (change in entropy) for the process.
2) Starting with the expression pV = constant (where  = Cp/CV), which applies to an adiabatic reversible expansion
or compression of an ideal gas with constant heat capacity, derive the following relationship (which you will need for
problem 3).
(pi/pf) = (Ti/Tf)/(-1)
(2.1)
3) Consider the following three step cyclic process indicated below. The process is carried out on 1.000 mol of an
ideal monatomic gas (CV,m = 3/2 R = constant). Th = 300.0 K, Tc = 250.0 K, and pA = 1.000 atm.
The three steps in the process are as follows:
step 1 – An isothermal reversible expansion from A to B.
step 2 – A constant volume reversible cooling from B to C
step 3 – An adiabatic reversible compression from C to A
a) What are Ucycle, Hcycle, and Scycle, the change in internal energy, enthalpy, and entropy, for the entire
cycle? (Hint – You should be able to answer this part of the problem without carrying out any calculations.)
b) What are pB amd pC, the pressures at points B and C?
c) What are wcycle and qcycle, the work and heat for the entire cycle?
4) For the process
2 Cu2O(s) + O2(g)  4 CuO(s)
(4.1)
find Hrxn, Srxn, and Grxn at
a) T = 25.0  C
b) T = 100.0 C
The thermochemical data needed to do this problem is in Table 2.8 of the Appendix of Atkins. You may assume that
the constant pressure molar heat capacities for the reactants and product are constant over the temperature range in
the problem.
5) Starting from the relationship
dG = V dp - S dT
(5.1)
show the following
a) (G/p)T = V
b) (G/T)p = - S
c) (V/T)p = - (S/p)T
Also do the following from Atkins:
Exercises
3.2b Calculate the molar entropy of a constant volume sample of argon at T =250. K
given that it is 154.84 J/mol.K at 298 K. (Hint: You may assume argon is an ideal monatomic gas).
Problems
3.14 The compound 1,3,5-trichloro-2,4,6-trifluorobenzene is an intermediate in the
conversion of hexachlorobenzene to hexafluorobenzene, and its thermodynamic properties have been examined by
measuring its heat capacity over a wide temperature range (R. L. Andon and J. F. Martin, J. Chem. Soc. Faraday
Trans. I, 871 (1973).) Some of the data are as follows
T (K)
Cp,m (J/mol.K)
14.14
9.492
16.33
12.70
20.03
18.18
31.15
32.54
44.08
46.86
64.81
66.36
T (K)
Cp,m (J/mol.K)
100.90 140.86 183.59 225.10 262.99 298.06
95.05 121.3 144.4 163.7 180.2 196.4
Find the value for S, the absolute entropy, at T = 298.15 K. (Hint: The above compound is a solid over the
temperature range of the problem. Below 14.14 K you may assume that the Nernst equation applies, that is, that C p,m
= aT3, where a is a constant).
Solutions.
1) It helps to think about what is happening in this problem. Two metal blocks at different initial temperatures are
placed in contact with one another. Heat will flow from the hot block to the cold block. The temperature of the hot
block will decrease, and the temperature of the cold block will increase, until an equal final temperature T f is
reached.
The first thing to do is to find T f. Since the heat leaving the hot block is equal in magnitude to the heat
entering the cold block, we may say
qh + qc = 0 = Ti,hTf mh sh dT + Ti,cTf mc sc dT
where “h” and “c” are used to label quantities associated with the hot and cold block, m is mass, s is specific heat,
and Ti is the initial block temperature. Since we assume the specific heat is constant over the temperature range in
the problem, we can pull m and s outside the integrals, to get
0 = mh sh Ti,hTf dT + mc sc Ti,cTf dT
= mh sh (Tf – Ti,h) + mc sc (Tf – Ti,c)
= mhshTf – mhshTi,h + mcscTf – mcscTi,c
Putting all of the terms containing Tf on one side of the equation and all of the other terms on the other side gives
mhshTf + mcscTf = mhshTi,h + mcscTi,c
(mhsh + mcsc) Tf = mhshTi,h + mcscTi,c
Tf = mhshTi,h + mcscTi,c = [ (10.00)(0.449)(400.0) + (20.00)(0.903)(300.0) ] = 319.9 K
(mhsh + mcsc)
[ (10.00)(0.449) + (20.00)(0.903) ]
Now, we may find S by saying
S = Sh + Sc
where the terms on the right are the change in entropy for the hot and cold block. Since the process is spontaneous
we expect S > 0. Since dS = (dq)rev/T = (mspdT)/T
Sh = Ti,hTf mhsh dT/T = mhsh Ti,hTf dT/T = mhsh ln(Tf/Ti,h)
Sc = Ti,cTf mcsc dT/T = mcsc Ti,cTf dT/T = mcsc ln(Tf/Ti,c)
So
Sh = (10.00 g) (0.449 J/g.K) ln (319.9/400.0) = - 1.003 J/K
Sc = (20.00 g) (0.903 J/g.K) ln (319.9/300.0) = + 1.160 J/K
S = ( - 1.003 J/K) + 1.160 J/K = + 0.157 J/K
2)
piVi = pfVf
But V = nRT/p. If we substitute for Vi and Vf, we get
pi (nRTi/pi) = pf (nRTf/pf)
pi1- Ti = pf1- Tf
If we now collect pressure terms on one side and temperature terms on the other side, we get
(pi/pf)(1-) = (Tf/Ti)
If we take both sides to the 1/(1-) power, we get
(pi/pf) = (Tf/Ti)/(1-)
Flipping Ti and Tf introduces a negative sign in the exponent, giving the final result
(pi/pf) = (Ti/Tf)/(-1)
3)
a) Since U, H, and S are all state functions, and the process is cyclic, it follows immediately that
Ucycle = Hcycle = Scycle = 0
b) It is easiest to first find pC, and then find pB. The process C  A is adiabatic and reversible, and the
molar heat capacity is constant.
CV,m = 3/2 R so Cp,m = 5/2 R and so  = (Cp,m)/(CV,m) = 5/3
/(-1) = (5/3)/(2/3) = 5/2
Using equn 2.1
pC/pA = (Tc/Th)/(-1)
pC = pA (Tc/Th)/(-1) = (1.000 atm) (250.0/300.0) 5/2 = 0.6339 atm.
To find pB we note that process B  C is a constant volume reversible cooling. Since for an ideal gas at constant
volume p ~ T, it follows
pB/pC = (Th/Tc)
pB = pC (Th/Tc) = (0.6339 atm) (300.0/250.0) = 0.7607 atm
c) We now have sufficient information to find wcycle. Once we do this, we know from the first law
Ucycle = wcycle + qcycle = 0 , qcycle = - wcycle
Since we have previously discussed the individual steps in the cycle we will not rederive the expressions for work
but will simply use the previous results
For A  B (isothermal reversible expansion)
wAB = nRT ln(pB/pA) = (1.000 mol) (8.314 J/mol.K) (300.0 K) ln(0.7607/1.000) = - 682. J
For B  C (constant volume reversible cooling)
wBC = 0
For C  A (adiabatic reversible compression)
qCA = 0
wCA = UCA = nCV,m(Th – Tc) = (1.000 mol)(12.472 J/mol.K)(300.0 K – 250.0 K) = 624. J
So wcycle = - 682. J + 0. J + 624. J = - 58. J
qcycle = - wcycle = + 58. J
Note, by the way, that the magnitude of wcycle is just equal to the area inside of the ABC curve in the plot of p vs V.
4)
a)
Hrxn = [4 Hf(CuO(s))] – [2 Hf(Cu2O(s)) + Hf(O2(g))]
= [4(- 157.3)] – [2(- 168.6) + 0] = - 292.0 kJ/mol
Srxn = [4 S(CuO(s))] – [2 S(Cu2O(s)) + S(O2(g))]
= [4(42.63)] – [2(93.14) + 205.138)] = - 220.90 J/mol.K
Grxn = [4 Gf(CuO(s))] – [2 Gf(Cu2O(s)) + Gf(O2(g))]
= [4(- 129.7)] – [2(- 146.0) + 0] = - 226.8 kJ/mol
Cp = [4 Cp,m(CuO(s))] – [2 Cp,m(Cu2O(s)) + Cp,m(O2(g))]
= [4(42.30)] – [2(63.64) + 29.355] = 12.56 J/mol .K
Note that we can check the above results by remembering that Grxn = Hrxn - TSrxn. If we substitute for Hrxn
and Srxn, we get Grxn = - 292.0 kJ/mol – (298. K) (-0.22090 kJ/mol.K) = - 226.2 kJ/mol, rwasonably close to the
value found directly using the thermochemical tables.
b) Since the constant pressure molar heat capacities are assumed constant over the temperature range of the
problem we may say
Hrxn(373 K) = Hrxn(298 K) + Cp (75. K)
= - 292.0 kJ/mol + (0.01256 kJ/mol.K)(75. K) = - 291.1 kJ/mol
Srxn(373 K) = Srxn(298 K) + Cp ln(373/298)
= - 220.90 J/mol.K + (12.56 J/mol.K) ln(373./298.) = - 218.1 J/mol.K
Grxn = Hrxn - TSrxn = (- 291.1 kJ/mol) – (373. K) (- 0.2181 kJ/mol.K) = - 209.7 kJ/mol
Note that because Hrxn and Srxn are relatively insensitive to changes in temperature, we often find values for
Grxn at some temperature other than 298. K by using the values for Hrxn and Srxn found at T = 298. K, and
assuming they are approximately constant. if we had done this in part b, we would have gotten
Grxn = Hrxn - TSrxn  (- 292.0 kJ/mol) – (373. K) (- 0.2209 kJ/mol.K) = - 209.6 kJ/mol
the same, within roundoff error, to the value found by the more rigorous method! That they are this close is a
coincidence, but the error introduced for a 75. K temperature change would typically be only a few kJ/mol.
5)
a) dG = V dp - S dT
divide by dp
dG/dp = V dp/dp - S dT/dp = V - S dT/dp
Make T constant
(G/p)T = V - S (T/p)T = V
The last term drops out when T = constant
b) dG = V dp - S dT
divide by dT
dG/dT = V dp/dT - S dT/dT = V dp/dT - S
Make p constant
(G/T)p = V (p/T)p - S = - S
The last term drops out when T = constant
c) Take /T)p of the first expression above and /p)T of the second expression above
/T)p [ (G/p)T = V ] = (2G/Tp) = (V/T)p
/p)T [ (G/T)p = - S ] = (2G/pT) = - (S/p)T
For "well behaved" functions the order of taking the derivatives doesn't matter, and so the left sides of the above
expressions are equal. Therefore the right sides must also be equal, and so
(V/T)p = - (S/p)T
Exercise 3.2 b
Since we know the value for entropy at 298. K, we need to find the change in entropy when we change the
temperature of the gas from 298. K to 250. K by a constant volume reversible cooling. Then
S(250. K) = S(298. K) + S
But for a constant volume reversible cooling of a gas with a constant value for the molar heat capacity, we may say
(note we use CV,m instead of Cp,m because volume is constant. Also note CV,m = 3/2 R).
S = CV,m ln(Tf/Ti) = (12.471 J/mol.K) ln(250./298.) = - 2.19 J/mol.K
So
S(250. K) = (154.84 J/mol.K) + (- 2.19 J/mol.K) = 152.63 J/mol.K
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 assume that Cp,m = aT3. 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
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
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/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 below. 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
T (K)
250
300
350