Chapter 5: Entropy and the Second and Third Laws of
Thermodynamics
Problem numbers in italics indicate that the solution is included in the
Student’s Solutions Manual.
Questions on Concepts
Q5.1) Which of the following processes is spontaneous set? a) The reversible isothermal
expansion of an ideal gas. b) The vaporization of superheated water at 102º C and 1 bar.
c) The constant pressure melting of ice at its normal freezing point by the addition of an
infinitesimal quantity of heat. d) The adiabatic expansion of a gas into a vacuum.
a) is not spontaneous because the system and surroundings are in equilibrium. b)
is spontaneous because the equilibrium phase under the stated conditions is a gas.
c)is not spontaneous because the process is reversible. d) is spontaneous because
at equilibrium, the density of a gas is uniform throughout its container.
Q5.2) Why are Sfusion and Svaporization always positive?
This is the case because Hfusion and Hvaporization are always positive. In each of
these transitions, attractive forces must be overcome.
Q5.3) Why is the efficiency of a Carnot heat engine the upper bound to the efficiency of
an internal combustion engine?
This is the case because the maximum work that can be done on the surroundings
in the expansion of a gas is in a reversible process.
Q5.4) The amplitude of a pendulum consisting of a mass on a long wire is initially
adjusted to have a very small value. The amplitude is found to decrease slowly with time.
Is this process reversible? Would the process be reversible if the amplitude did not
decrease with time?
No, because dissipative forces are acting on the system. If the amplitude did not
decrease, no dispersive forces act on the system and the oscillation is reversible in
the limit of very small amplitudes. If the amplitude did not decrease with time, no
dissipative forces act on the system, and because the amplitude is very small, the
process is reversible.
Q5.5) A process involving an ideal gas is carried out in which the temperature changes
at constant volume. For a fixed value of T, the mass of the gas is doubled. The process
is repeated with the same initial mass and T is doubled. For which of these processes is
S greater? Why?
5-1
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics

S is greater if the mass is doubled, because S increases linearly with the
amount of material. By contrast, S only increases as the logarithm of the
temperature. This increase is much slower than a linear increase.
Q5.6) Under what conditions does the equality S 
H
hold?
T
dq
 reversible
 reversible  H . This is the case for
, the equality holds if dq
T
a reversible process at constant pressure.
Because S 
Q5.7) Under what conditions is S < 0 for a spontaneous process?
For a spontaneous process, S + S surroundings < 0. The inequality S < 0 is
satisfied only if S surroundings > 0 and S surroundings  S .
Tf
Q5.8) Is the equation S 

Ti
V
f
Tf 
CV

dT   dV  CV ln
 V f  Vi  valid for an
T

Ti 
Vi
ideal gas?
No. Because  and  are not independent of V for an ideal gas, they can’t be
taken out of the integral.
Q5.9) Without using equations, explain why S for a liquid or solid is dominated by the
temperature dependence of S as both P and T change.
It is useful to think of S = S(V,T). Because V changes very little with P for a
liquid or solid, entropy changes are dominated by changes in T rather than P for
liquids and solids.
Q5.10) You are told that S = 0 for a process in which the system is coupled to its
surroundings. Can you conclude that the process is reversible? Justify your answer.
No. The criterion for reversibility is S + Ssurroundings = 0. To decide if this
criterion is satisfied, Ssurroundings must be known.
Problems
P5.1) Beginning with Equation (5.5), use Equation (5.6) to eliminate Vc and Vd to arrive
at the result wcycle  nR Thot  Tcold  ln
Vb
.
Va
5-2
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
wab   nRThot ln
Vb
Va
wbc  nCV ,m  Tcold  Thot 
wcd  nRTcold ln
Vd
Vc
wda  nCV ,m  Thot  Tcold    wbc
wtotal   nRThot ln
Vb
V
 nRTcold ln d
Va
Vc
ThotVc 1  TcoldVb 1 and TcoldVa 1  ThotVd 1
Solving the last two equations for Thot and equating the results,
Vb 1
Va 1
V
V
V
V

T
. Therefore, b  a or c  b and
cold
 1
 1
Vc
Vd
Vc Vd
Vd Va
Thot  Tcold
wtotal  nRThot ln
Vb
V
V
 nRTcold ln d  nR Thot  Tcold  ln b
Va
Vc
Va
P5.2) Consider the reversible Carnot cycle shown in Figure 5.2 with 1 mol of an ideal
gas with CV = 3/2R as the working substance. The initial isothermal expansion occurs at
the hot reservoir temperature of Thot = 600ºC from an initial volume of 3.50 L (Va) to a
volume of 10.0 L (Vb). The system then undergoes an adiabatic expansion until the
temperature falls to Tcold = 150ºC. The system then undergoes an isothermal compression
and a subsequent adiabatic compression until the initial state described by Ta = 600ºC and
Va = 3.50 L is reached.
a) Calculate Vc and Vd.
b) Calculate w for each step in the cycle and for the total cycle.
c) Calculate  and the amount of heat that is extracted from the hot reservoir to do 1.00 kJ
of work in the surroundings.
a) We first calculate Vc and Vd.
1
Tc  Vc 
 
Tb  Vb 
1
 T 1
V
; c  c
Vb  Tb 
1
Vc  423 K 1 5  423 K 

 3 

Vb  873 K 
 873 K 
1
Td  Vd 
 
Ta  Va 

3
2
 2.96; Vc  29.6 L
1
 T 1
V
; d  d 
Va  Ta 
1
Vd  423 K 1 5  423 K 

 3 

Va  873 K 
 873 K 

3
2
 2.96; Vd  10.4 L
5-3
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
b) We next calculate w for each step in the cycle, and for the total cycle.
V
10.0 L
wab   nRTa ln b  1 mol  8.314 J mol1K 1  873 K  ln
 7.62  103J
Va
3.50 L
3
wbc  nCV ,m Tc  Tb   1 mol   8.314 J mol1K 1   423 K  873 K   5.61  103J
2
V
10.4 L
wcd   nRTc ln d  1 mol  8.314 J mol1K 1  423 K  ln
 3.68  103 J
Vc
29.6 L
3
wda  nCV ,m Ta  Td   1 mol   8.314 J mol1K 1   873 K  423 K   5.61  103J
2
3
3
wtotal  7.62  10 J  5.61  10 J  3.68  103J + 5.61  103J  3.94  103J
Tcold
423 K
w
 1
 0.515 q   1.94
Thot
873 K

Therefore, 1.94 kJ of heat must be extracted from the surroundings to do 1.00 kJ of work
in the surroundings.
  1
P5.3) Using your results from Problem P5.2, calculate q, U, and H for each step in
the cycle and for the total cycle described in Problem P5.2.
a→b
U = H = 0 because T = 0.
q = –w = 7.62  103J
b→c
U = w = –5.61  103J because q = 0. H = H +nRT = –9.35  103J
c→d
U = H = 0 because T = 0.
q = –w = –3.68  103J
3
d→a
U = w = 5.61  10 J because q = 0. H = H +nRT = 9.35  103J
qtotal = 7.62  103J – 3.68  103J = 3.94  103J = – wtotal
Utotal = Htotal = 0
P5.4) Using your results from Problems P5.2 and P5.3, calculate S, Ssurroundings, and
Stotal for each step in the cycle and for the total cycle described in Problem P5.2.
qreversible 7.62  103J

 8.73 J K 1
T
873 K
 0 because qreversible  0 Stotal = 0
a→b
S  Ssurroundings 
b→c
S  Ssurroundings
c→d
S  Ssurroundings 
S  Ssurroundings
d→a
round-off error.
Stotal = 0
qreversible 3.68 103J

 8.70 J K 1 Stotal = 0
T
423 K
 0 because qreversible  0 . Stotal = 0 to within the
For the cycle, S = Ssurroundings = Stotal = 0 to within the round-off error.
5-4
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
P5.5) Calculate S if the temperature of 1 mol of an ideal gas with CV = 3/2R is
increased from 150 to 350 K under conditions of (a) constant pressure and (b) constant
volume.
a) at constant pressure
Tf
350 K
3 
S  nCP ,m ln  1 mol   + 1  8.314 J mol1K 1  ln
= 17.6 J K 1
2 
Ti
150 K
b) at constant volume
Tf
3
350 K
S  nCV ,m ln  1 mol   8.314 J mol1K 1  ln
= 10.6 J K 1
Ti
2
150 K
P5.6) One mole of N2 at 20.5ºC and 6.00 bar undergoes a transformation to the state
described by 145ºC and 2.75 bar. Calculate S if
2
3
C P ,m
3 T
5 T
8 T
.

30.81

11.87

10

2.3968

10

1.0176

10
J mol 1 K 1
K
K2
K3
S  nR ln
Pf
Pi
Tf
 n
Ti
C P ,m
dT
T
 1 mol  8.314 J mol1K 1  ln
2.75 bar
6.00 bar
2
3


3 T
5  T 
8  T 
30.81

11.87

10

2.3968

10

1.0176

10

 
  
418.15 
K
K
K  T

 
d J K 1
T
K
293.65
418
2

T
T
5 T 
 6.48 J K  30.81ln  0.1187  1.1984 10
J K 1
2
K
K
K  293.5

1
 6.48 J K 1  10.89 J K 1  1.48 J K 1  1.06 J K 1  0.16 J K 1  16.8 J K 1
P5.7) One mole of an ideal gas with CV = 3/2R undergoes the transformations described
in the following list from an initial state described by T = 300 K and P = 1.00 bar.
Calculate q, w, U, H, and S for each process.
a) The gas is heated to 450 K at a constant external pressure of 1.00 bar.
b) The gas is heated to 450 K at a constant volume corresponding to the initial volume.
c) The gas undergoes a reversible isothermal expansion at 300 K until the pressure is half
of its initial value.
a) The gas is heated to 450 K at a constant pressure of 1.00 bar.
5-5
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
nRTi 1 mol  8.314 J mol1K 1  300 K

 2.49  102 m3
Pi
105 Pa
Tf
450 K
V f  Vi 
 2.49  102 m3 = 3.74  102 m3
Ti
300 K
Vi 


w   Pexternal V  105 Pa  3.74  102 m3  2.49  102 m3  1.25  103 J
1
1
3  8.314 J mol K
 150 K = 1.87  103J
2
3
q  H  U  w  1.87 10 J + 1.25 103J = 3.12 103J
T
450 K
3

S  nCP ,m ln f  1 mol   + 1  8.314 J mol1K 1  ln
= 8.43 J K 1
Ti
300 K
2

b) The gas is heated to 450 K at a constant volume corresponding to the initial volume.
U  nCV ,m T  1 mol 
w = 0 because V = 0.
3  8.314 J mol1K 1
 150 K = 1.87  103J
2
5  8.314 J mol1K 1
H  nCP ,m T  1 mol 
 150 K = 3.12  103J
2
T
450 K
3
S  nCV ,m ln f  1 mol     8.314 J mol1K 1  ln
= 5.06 J K 1
Ti
300 K
2
c) The gas undergoes a reversible isothermal expansion at 300 K until the pressure is half
of its initial value.
U  q  nCV ,m T  1 mol 
U = H = 0 because T = 0.
Vf
wreversible   q   nRT ln
 1 mol  8.314 J mol 1K 1  300 K  ln 2  1.73 103J
Vi
S 
qreversible 1.73  103J

= 5.76 J K 1
T
300 K
P5.8) Calculate Ssurroundings and Stotal for each of the processes described in Problem
P5.7. Which of the processes is a spontaneous process? The state of the surroundings for
each part is as follows:
a) 450 K, 1 bar
b) 450 K, 1 bar
c) 300 K, 0.500 bar
a) The gas is heated to 450 K at a constant pressure of 1.00 bar.
q
3.12  103J
S surroundings 

 6.93 J K 1
Tsurroundings
450 K
Stotal  S  Ssurroundings  8.43 J K1  6.93 J K1  1.50 J K 1
The process is spontaneous.
5-6
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
b) The gas is heated to 450 K at a constant volume corresponding to the initial volume.
q
1.87  103J
S surroundings 

 4.16 J K 1
Tsurroundings
450 K
Stotal  S  Ssurroundings  5.06 J K1  4.16 J K1  0.90 J K1
The process is spontaneous.
c) The gas undergoes a reversible isothermal expansion at 300 K until the pressure is half
of its initial value.
q 1.73 103J
Ssurroundings  
 5.76 J K 1
T
300 K
Stotal  S  Ssurroundings  5.76 J K1  5.76 J K1  0
There is no natural direction of change in this process because it is reversible.
P5.9) At the transition temperature of 95.4ºC, the enthalpy of transition from rhombic to
monoclinic sulfur is 0.38 kJ mol–1.
a) Calculate the entropy of transition under these conditions.
b) At its melting point, 119ºC, the enthalpy of fusion of monoclinic sulfur is
1.23 kJ mol–1. Calculate the entropy of fusion.
c) The values given in parts (a) and (b) are for 1 mol of sulfur; however, in crystalline
and liquid sulfur, the molecule is present as S8. Convert the values of the enthalpy and
entropy of fusion in parts (a) and (b) to those appropriate for S8.
a) Stransition 
b) S fusion 
H transition
0.38 kJ mol1

 1.03 J K 1 mol1
Ttransition
 273.15+95.4 K
H fusion
T fusion

1.23 kJ mol1
 3.14 J K 1 mol1
 273.15  119 K
c) Each of the S in parts (a) and (b) should be multiplied by 8.
Stransition = 8.24 J K–1 mol–1
S fusion = 25.12 J K–1 mol–1
P5.10)
a) Calculate S if 1 mol of liquid water is heated from 0º to 100ºC under constant
pressure if CP,m = 75.291 J K–1 mol–1.
5-7
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
b) The melting point of water at the pressure of interest is 0ºC and the enthalpy of fusion
is 6.0095 kJ mol–1. The boiling point is 100ºC and the enthalpy of vaporization is 40.6563
kJ mol–1. Calculate S for the transformation H2O(s, 0ºC) → H2O(g, 100ºC).
a) The heat input is the same for a reversible and an irreversible process.
dq  dqreversible  nC P ,m dT
S  n 
CP ,m
T
dT  nCP ,m ln
Tf
Ti
 1 mol  753 J mol 1 K 1 ln
373.15 K
273.15 K
 23.49 J K 1
b)
S fusion 
H fusion
T fusion
Svaporization 

6009 J
 22.00 J K 1
273.15 K
H vaporization
Tvaporization

40656 J
 108.95 J K 1
373.15 K
Stotal  S fusion  Svaporization  Sheating   22.00  108.95  23.49 J K 1
 154.4 J K 1
P5.11) One mole of an ideal gas with CV,m = 5/2R undergoes the transformations
described in the following list from an initial state described by T = 250 K and P = 1.00
bar. Calculate q, w, U, H, and S for each process.
a) The gas undergoes a reversible adiabatic expansion until the final pressure is half its
initial value.
b) The gas undergoes an adiabatic expansion against a constant external pressure of 0.500
bar until the final pressure is half its initial value.
c) The gas undergoes an expansion against a constant external pressure of zero bar until
the final pressure is equal to half of its initial value.
a) The gas undergoes a reversible adiabatic expansion until the final pressure is half its
initial value.
q = 0 because the process is adiabatic.
5-8
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
1
Vf 
 
Ti  Vi 
Tf
1
 Tf 
 
 Ti 
 Pi

 Pf
1



1

P 
 Tf 
;     i 
P 
 Ti 
 f 
P
;
 i
Ti  Pf
Tf



1

7
5
7
5
1
2

 1.00 bar 

   2.00  7  0.820
Ti  0.500 bar 
T f  0.820  250 K  205 K
Tf
5  8.314 J mol1K 1
U  w  nCV ,m T  1 mol 
  205 K  250 K   935 J
2
7  8.314 J mol1K 1
  205 K  250 K   1.31 103J
2
S = 0 because qreversible = 0.
H  nCP ,m T  1 mol 
b) The gas undergoes an adiabatic expansion against a constant external pressure of 0.500
bar until the final pressure is half its initial value.
q = 0 because the process is adiabatic.
T T 
nCv,m T f  Ti   nRPexternal  f  i 
P P
i 
 f

nRPexternal
T f  nCv,m 

Pf



nRPexternal 
  Ti  nCv,m 

Pi




RPexternal
 Cv,m 
Pi
T f  Ti 
 C  RPexternal
 v,m
Pf

T f  214 K


8.314 J mol1K 1  0.500 bar 
1 1
2.5

8.314
J
mol
K




1.00 bar
  250 K 

8.314 J mol1K 1  0.500 bar 


1 1
 2.5 8.314 J mol K +


0.500 bar



5  8.314 J mol1K 1
  214 K  250 K   748 J
2
7  8.314 J mol1K 1
H  nCP ,m T  1 mol 
  214 K  250 K   1.05  103 J
2
Pf
Tf
S  nR ln  nCP ,m ln
Pi
Ti
U  w  nCV ,m T  1 mol 
0.500 bar
7  8.314 J mol 1K 1
214 K
+ 1 mol 
 ln
1.00 bar
2
250 K
1
1
1
 5.76 J K  4.52 J K  1.24 J K
 1 mol  8.314 J mol 1K 1  ln
c) The gas undergoes an expansion against a constant external pressure of zero bar until
the final pressure is equal to half of its initial value.
5-9
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
T and w = 0. Therefore U =H = 0 and q = U – w = 0.
Pf
0.500 bar
S   nR ln
 8.314 J mol1K 1  ln
= 5.76 J K 1
Pi
1.00 bar
P5.12) The standard entropy of Pb(s) at 298.15 K is 64.80 J K–1 mol–1. Assume that the
2
CP ,m  Pb, s 
T
5 T
 22.13  0.01172  1.00 10
. The
heat capacity of Pb(s) is given by
J mol 1 K 1
K
K2
melting point is 327.4ºC and the heat of fusion under these conditions is 4770 J mol–1.
C  Pb, l 
T
 32.51  0.00301 .
Assume that the heat capacity of Pb(l) is given by P ,m1
1
J K mol
K
a) Calculate the standard entropy of Pb(l) at 500ºC.
b) Calculate H for the transformation Pb(s, 25ºC) → Pb(l, 500ºC).
a) S m  Pb, l , 773 K   S m  Pb, S , 298.15 K 
H fusion 773.15 CP ,m
C P ,m
 
d T / K  
 
d T / K 
T / K
T fusion
T / K
298.15 
600.55 
600.55
 64.80 J mol 1 K 1
22.13  0.01172 T / K   1.00  105 T / K 
d T / K 

T
/
K


298.15
600.55

4770 J mol1
600.55 K
773.15
32.51  0.00301T / K 
 
d T / K 
T
/
K


600.55

 64.80 J mol 1 K 1  20.40 J mol 1 K 1  7.94 J mol1 K 1  7.69 J mol 1 K 1
 100.8 J mol 1 K 1
600.55
b) H total 

CPsolid
, m d T / K  +H fusion +
298.15
773.15

CPliquid
, m d T / K 
600.55
1
1
 8918 J mol  4770 J mol  5254 J mol 1
 18.94  103 J mol 1
P5.13) Between 0ºC and 100ºC, the heat capacity of Hg(l) is given by
CP ,m  Hg, l 
T
 30.093  4.944 103 . Calculate H and S if 1 mol of Hg(l) is raised in
1
1
J K mol
K
temperature from 0º to 100ºC at constant P.
5-10
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
Tf
H m   CP ,m d T / K 
Ti




 30.093 T f  Ti  2.472  10 3 T f2  Ti 2 J mol 1
 2.84  103 J mol 1
Tf
Tf
CP , m
 T / K d T / K   30.093ln T
S m 
i
Ti


 4.944  103 T f  Ti  8.90 J K 1 mol 1
P5.14) One mole of a van der Waals gas at 27ºC is expanded isothermally and
reversibly from an initial volume of 0.020 m3 to a final volume of 0.060 m3. For the van
a
 U 
6
–2
der Waals gas, 
  2 . Assume that a = 0.556 Pa m mol , and that
 V T Vm
–6
b = 64.0 × 10 m3 mol–1. Calculate q, w, U, H, and S for the process.
Vf
U 
 U 
  V  dV
Vi
T
for a van der Waals gas
 1
1
U  a 

V
 m ,i Vm , f



as shown in Example Problem 3.5
1
1


U  0.556 Pa m 6 mol 2 

1 mol  18.5 J
3
3 
 0.020 m 0.060 m 
Vf
Vf
Vf
dV
dV
w    PdV   RT 
a 2
V b
V
Vi
Vi m
Vi
  RT ln
V
 b
 1 1
 a  
Vi  b   V f Vi 
f
0.059936
0.019936
1
1


 0.556 Pa m 6 mol2 

 2.73 103 J
3
3 
0.060
m
0.020
m


 8.314 J mol 1 K 1  300.15 K ln
5-11
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
Pi 
RT
a 8.314 J mol1 K 1  300.15 K 0.556 Pa m6 mol
 2

 1.237 105 Pa
2
3
3
Vi  b Vi
0.019936 m
 0.020 m 
Pf 
RT
a 8.314 J mol1 K 1  300.15 K 0.566 Pa m6 mol 2
 2

 4.146 104 Pa
2
3
Vf  b Vf
0.059936 m
 0.060 m3 
H  U    PV   U  Pf V f  PV
i i


 18.5 J  4.146  104 Pa  0.060 m3  1.237  105 Pa  0.020 m3  32.1 J
q  U  w  2.73  103 J
S 
qreversible 2.73  103 J

 9.10 J K 1
T
300.13 K
P5.15) The heat capacity of -quartz is given by
CP ,m  -quartz, s 
2
T
5 T

11.30

10
. The coefficient of thermal
J K 1 mol1
K
K2
expansion is given by  = 0.3530 × 10–4 K–1 and Vm = 22.6 cm3 mol–1. Calculate Sm for
the transformation -quartz (25ºC, 1 atm) → -quartz (225ºC, 1000 atm).
 46.94  34.31  103
From Equations (5.23) and (5.24)
Tf
S   CP
Ti

dT
 V  Pf  Pi
T

498 K


  46.94 ln
 34.31  103   498  298  5.65  10 5  4982  2982  J K 1 mol1
298 K


3
1m
1.0125×105 Pa
3
1
4
1
 22.6 cm mol  6
 0.3530  10 K  999 atm 
10 cm3
atm
1
1
1
1
1
1
 22.11 J K mol  0.0807 J K mol  21.88 J K mol


P5.16) Calculate Ssurroundings and Stotal for the processes described in parts (a) and (b)
of Problem P5.11 Which of the processes is a spontaneous process? The state of the
surroundings for each part is as follows:
a) 250 K, 0.500 bar
b) 300 K, 0.500 bar
Ssurroundings = 0 for a) and b) because q = 0.
a) Stotal = S + Ssurroundings = 0 + 0 = 0. The process is not spontaneous.
b) Stotal = S + Ssurroundings = 1.24 J K–1 + 0 = 1.24 J K–1. The process is spontaneous.
5-12
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
P5.17) Calculate S, Stotal, and Ssurroundings when the volume of 85.0 g of CO initially
at 298 K and 1.00 bar increases by a factor of three in (a) an adiabatic reversible
expansion, (b) an expansion against Pexternal = 0, and (c) an isothermal reversible
expansion. Take CP,m to be constant at the value 29.14 J mol–1 K–1 and assume ideal gas
behavior. State whether each process is spontaneous.
a) an adiabatic reversible expansion
Ssurroundings = 0 because q = 0.S = 0 because the process is reversible.
Stotal = S + Ssurroundings = 0. The process is not spontaneous.
b) an expansion against Pexternal = 0
T and w = 0. Therefore U = q = 0.
Vf
85.0 g
S  nR ln

 8.314 J mol1K 1  ln 3
1
Vi 28.01 g mol
 3.03 mol  9.13 J mol1K 1 = 27.7 J K 1
Stotal = S + Ssurroundings = 27.7 J K–1 + 0 = 27.17 J K–1. The process is spontaneous.
c) an isothermal reversible expansion
T = 0. Therefore U = 0.
Vf
w   q   nRT ln
 3.03 mol  8.314 J mol 1K 1  298 K  ln 3  8.25 103J
Vi
qreversible 8.25 103J

= 27.7 J K 1
T
298 K
 q 8.25 103J
Ssurroundings 

 27.7 J K 1
T
298 K
Stotal = S + Ssurroundings = 27.7 J K–1 – 27.7 J K–1 = 0. The system and surroundings
are at equilibrium.
S 
P5.18) One mole of an ideal gas with CV,m = 3/2R is transformed from an initial state
T = 600 K and P = 1.00 bar to a final state T = 250 K and P = 4.50 bar. Calculate U,
H, and S for this process.
3  8.314 J mol1K 1
  250 K  600 K   4.36  103J
2
5  8.314 J mol1K 1
H  nCP ,m T  1 mol 
  250 K  600 K   7.27  103J
2
U  nCV ,m T  1 mol 
5-13
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
S   nR ln
Pf
Pi
 nCP ,m ln
Tf
Ti
4.500 bar
5  8.314 J mol 1K 1
250 K
+ 1 mol 
 ln
1.00 bar
2
600 K
1
1
1
 12.5 J K  18.2 J K  30.7 J K
 1 mol  8.314 J mol 1K 1  ln
P5.19) An ideal gas sample containing 2.50 moles for which CV,m = 3/2R undergoes the
following reversible cyclical process from an initial state characterized by T = 450 K and
P = 1.00 bar:
a) It is expanded reversibly and adiabatically until the volume doubles.
b) It is reversibly heated at constant volume until T increases to 450 K.
c) The pressure is increased in an isothermal reversible compression until P = 1.00 bar.
Calculate q, w, U, H, S, Ssurroundings, and Stotal for each step in the cycle, and for the
total cycle. The temperature of the surroundings is 300 K.
a) The temperature at the end of the adiabatic reversible expansion is calculated.
1
5
2
Vf 
1

     2  3   2  3  0.630
Ti  Vi 
T f  0.630  450 K = 283 K
Tf
The initial and final volume and the final pressure are calculated.
nRTi 2.50 mol  8.314 J mol 1K 1  450K
Vi 

= 9.25  102 m3
Pi
105 Pa
V f  2Vi  0.185 m 3
Pf  Pi
Vi T f
1 283 K
 1.00 bar  
= 0.314 bar
V f Ti
2 450 K
q = 0 because the expansion is adiabatic.
3  8.314 J mol1K 1
U  w  nCV ,m T  2.50 mol 
  283 K  450 K   5.21  103J
2
5  8.314 J mol1K 1
H  nCP ,m T  2.50 mol 
  283 K  450 K   8.68  103J
2
S = 0 because qreversible = 0. Ssurroundings = 0 because q = 0. Stotal = 0.
b) w = 0 because V = 0.
5-14
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
3  8.314 J mol1K 1
U  q  nCV ,m T  2.50 mol 
  450 K  283 K   5.21  103J
2
5  8.314 J mol1K 1
H  nCP ,m T  2.50 mol 
  450 K  283 K   8.68  103J
2
The pressure at the end of the process is calculated in order to calculate S.
Tf
450 K
Pf  Pi
 0.314 bar 
= 0.500 bar
Ti
283 K
Pf
Tf
S   nR ln  nCP ,m ln
Pi
Ti
0.500 bar
5  8.314 J mol 1K 1
450 K
 2.50 mol 
 ln
0.314 bar
2
283 K
1
1
1
 9.63 J K  24.10 J K  14.5 J K
 2.50 mol  8.314 J mol 1K 1  ln
S surroundings 
q
Tsurroundings
5.21  103J

 17.4 J K 1
300 K
Stotal  S  S surroundings  14.5 J K 1  17.4 J K 1  2.90 J K 1
c) H = U = 0 because T = 0.
Vf
P
w   q   nRT ln
  nRT ln i
Vi
Pf
1
= 6.48 103J
2
Pf
1.00 bar
S   nR ln
 2.50 mol  8.314 J mol1K 1  ln
 14.5 J K 1
Pi
0.500 bar
 2.50 mol  8.314 J mol 1K 1  450K  ln
S surroundings 
q
Tsurroundings

6.48  103 J
= 21.6 J K 1
300 K
Stotal  S surroundings  S  7.1 J K 1
For the entire cycle,
wtotal = –5.21  103J + 0 + 6.48  103J = 1.27  103J
qtotal = 0 + 5.21  103J – 6.48  103J = –1.27  103J
Utotal = –5.21  103J + 5.21  103J + 0 = 0
Htotal = –8.68  103J + 8.68  103J + 0 = 0
Stotal = 0 + 14.5 J K–1 – 14.5 J K–1 = 0
Ssurroungings=S total = 0 – 17.4 J K–1 + 21.6 J K–1 = 4.20 J K–1
P5.20) One mole of H2O(l) is compressed from a state described by
5-15
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
P = 1.00 bar and T = 298 K to a state described by P = 800 bar and T = 450 K. In
addition,  = 2.07 × 10–4 K–1 and the density can be assumed to be constant at the value
997 kg m–3. Calculate S for this transformation, assuming that  = 0.
From Equation (5.24),
Tf
S 

Ti
Pf

Tf
CP
dT   V  dP  nCP ,m ln  nVm,i  Pf  Pi
T
Ti
Pi

450 K
298 K
3
18.02  10 kg mol1
 1mol 
 2.07  104 K 1  799 bar  105 Pa bar 1
3
997 kg m
 1mol  75.3 J mol1K 1  ln
 31.0 J K 1  0.299 J K 1  30.7 J K 1
P5.21) A 25.0 g mass of ice at 273 K is added to 150.0 g of H2O(l) at 360 K at constant
pressure. Is the final state of the system ice or liquid water? Calculate S for the process.
Is the process spontaneous?
Assume initially that the final state is water. If this is not the case, the calculated
temperature will be below 273 K. Calculate S for the ice and water separately, and add
them to get the overall S for the process.




H 2O
ice
nice H ice
 nH 2O CPH,2mO T f  Ti H 2O  0
fusion  nice C P , m T f  Ti
Tf 
niceCPH,2mOTi ice  nH 2O CPH,2mOTi H 2O  nice H ice
fusion
niceCPH,2mO  nH 2O CPH,2mO

25.0 g ice
150 g H 2 O
 75.291 J K 1mol1  273 K+
 75.291 J K 1mol1  360 K
1
18.02 g ice mol
18.02 g H 2 O mol1
25.0 g ice
 6008 J mol1
1
18.02 g ice mol
25.0 g ice
150 g H 2 O
 75.291 J K 1mol1 
 75.291 J K 1mol1
1
1
18.02 g ice mol
18.02 g H 2 O mol

T f  336 K
S is calculated for the ice. It consists of melting the ice at 273 K and heating the
resulting water to 336 K.
5-16
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
S  n

H fusion
T fusion
 nCP ,m ln
Tf
T fusion
25.0 g ice
6008 J mol 1
25.0 g ice
336 K

+
 75.291 J K 1mol 1 ln
1
1
18.02 g ice mol
273 K
18.02 g ice mol
273 K
 30.5 J K 1  21.7 J K 1  52.2 J K 1

S is calculated for the water. It consists of cooling the water from 360 K to 336 K.
Tf
150.0 g
336 K
S  nCP ,m ln

 75.291 J K 1mol 1  ln
=  43.2 J K 1
1
T fusion 18.02 g mol
360K
Stotal = 0 + 52.2 J K–1 – 43.2 J K–1 = 9.0 J K–1. The process is spontaneous.
P5.22) 15.0 g of steam at 373 K is added to 250.0 g of H2O(l) at 298 K at a constant
pressure of 1 bar. Is the final state of the system steam or liquid water? Calculate S for
the process.
Assume that the final state is liquid water. If this is not the case, the calculated
temperature will be greater than 373 K.




H 2O
 nsteam H vaporization
 nsteamCPH,2mO T f  Ti steam  nH 2OCPH,2mO T f  Ti H 2O  0
Tf 
nsteamC
H 2O steam
P ,m i
T
n
n
H 2O H 2O
H 2O P , m i
steam
H 2O
H 2O
steam P , m
H 2O P , m
n
C
C
T
n
H
H 2O
vaporization
C

15.0 g steam
250 g H 2O
 75.291 J K 1mol 1  373 K +
 75.291 J K 1mol 1  298 K
1
18.02 g mol
18.02 g H 2O mol 1
15.0 g steam
 40650 J mol1
18.02 g mol 1
15.0 g steam
250 g H 2 O
 75.291 J K 1mol1 +
 75.291 J K 1mol 1
1
1
18.02 g mol
18.02 g H 2O mol
+
T f  333 K
S is calculated for the steam. It consists of condensing the steam at 373 K and cooling
the resulting water to 333 K.
S   n

H vaporization
Tvaporization
 nCP ,m ln
Tf
Tvaporization
15.0 g steam 40650 J mol 1
15.0 g ice
333 K

+
 75.291 J K 1mol 1  ln
1
1
18.02 g mol
373 K
18.02 g ice mol
373 K
 90.7 J K 1  7.11 J K 1  97.8 J K 1

5-17
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
S is calculated for the water. It consists of heating the water from 298 K to 333 K.
Tf
250.0 g
333 K
S  nCP ,m ln

 75.291 J K 1mol1  ln
=116 J K 1
1
T fusion 18.02 g mol
298 K
Stotal = 0 + 116 J K–1 –97.8 J K–1 = 18.2 J K–1. The process is spontaneous.
P5.23) The maximum theoretical efficiency of an internal combustion engine is
achieved in a reversible Carnot cycle. Assume that the engine is operating in the Otto
cycle and that CV,m = 5/2R for the fuel–air mixture initially at 298 K (the temperature of
the cold reservoir). The mixture is compressed by a factor of 8.0 in the adiabatic
compression step. What is the maximum theoretical efficiency of this engine? How much
would the efficiency increase if the compression ratio could be increased to 30? Do you
see a problem in doing so?
1
1
7
0.4
 Vf 
1 5 1
     
Ti  Vi 
8
8
T f  2.30  298 K = 684 K
T
298
  1  cold  1 
 0.564
Thot
684
Tf
1
1
7
 2.30
0.4
 Vf 
 1  5  1 
          3.89
Ti  Vi 
 30 
 30 
T f  3.89  298 K = 1162 K
T
298 K
  1  cold  1 
 0.744
Thot
1162 K
Tf
It would be difficult to avoid ignition of the fuel-air mixture before the compression was
complete.
P5.24) 1 mol of H2O(l) is supercooled to –2.25ºC at 1 bar pressure. The freezing
temperature of water at this pressure is 0.00ºC. The transformation
H2O(l) → H2O(s) is suddenly observed to occur. By calculating S, Ssurroundings, and
Stotal, verify that this transformation is spontaneous at –2.25ºC. The heat capacities are
given by CP(H2O(l)) = 75.3 J K–1 mol–1 and CP(H2O(s)) = 37.7 J K–1 mol–1, and
Hfusion = 6.008 kJ mol–1 at 0.00ºC. Assume that the surroundings are at –2.25ºC. [Hint:
Consider the two pathways at 1 bar: (a) H2O(l, –2.25ºC) → H2O(s, –2.25ºC) and (b)
H2O(l, –2.25ºC) → H2O(l, 0.00ºC) → H2O(s, 0.00ºC) → H2O(s, –2.25ºC). Because S is a
state function, S must be the same for both pathways.]
For pathway b) H2O(l, –2.25ºC) → H2O(l, 0.00ºC) → H2O(s, 0.00ºC) →
H2O(s, –2.25ºC)
5-18
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
H fusion
273.15 K
270.90 K
n
 nCP ,m  s  ln
270.90 K
273.15 K
273.15 K
273.15 K
6008 J mol 1
 1 mol  75.3 J mol 1K 1  ln
 1 mol 
270.90 K
273.15 K
S  nCP ,m  l  ln
 1 mol  37.7 J mol1K 1  ln
 0.608 J K 1  22.0 J K 1  0.312 J K 1  21.7 J K 1
To calculate Ssurroundings, we first calculate HP = q.
H 2.25 C  H 0.00 C   CP  solid   CP  liquid   T




H  2.25 C   6008 J mol   37.7 J mol
1
270.90 K
273.15 K

K 1  73.15 J mol 1K 1   2.25 K 
1
 5928 J mol 1  q
S surroundings 
q
Tsurroundings

5928 J mol 1
 21.9 J K 1
270.90 K
Stotal = 21.9 J K–1 –21.7 J K–1 = 0.2 J K–1 > 0. The process is spontaneous.
P5.25) An air conditioner is a refrigerator with the inside of the house acting as the cold
reservoir and the outside atmosphere acting as the hot reservoir. Assume that an air
conditioner consumes 1.50 × 103 W of electrical power, and that it can be idealized as a
reversible Carnot refrigerator. If the coefficient of performance of this device is 2.50,
how much heat can be extracted from the house in a 24-hour period?
q
r  cold  2.50; qcold  2.50w
w
3600 s
qcold  2.50  1.50  103 J s 1 
 24 hr = 3.24  108 J
hr
P5.26) The interior of a refrigerator is typically held at 277 K and the interior of a
freezer is typically held at 255 K. If the room temperature is 294 K, by what factor is it
more expensive to extract the same amount of heat from the freezer than from the
refrigerator? Assume that the theoretical limit for the performance of a reversible
refrigerator is valid.
From Equation (5.44)
r 
Tcold
Thot  Tcold
for the freezer r 
255 K
 6.5
294 K  255 K
5-19
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
for the refrigerator r 
277 K
 16.3
294 K  277 K
The freezer is more expensive to operate than the refrigerator by the ratio 16.3/6.5 = 2.5.
P5.27) The Chalk Point, Maryland, generating station supplies electrical power to the
Washington, D.C., area. Units 1 and 2 have a gross generating capacity of 710 MW
(megawatt). The steam pressure is 25 × 106 Pa, and the superheater outlet temperature
(Th) is 540ºC. The condensate temperature (Tc) is 30ºC.
a) What is the efficiency of a reversible Carnot engine operating under these conditions?
b) If the efficiency of the boiler is 91.2%, the overall efficiency of the turbine, which
includes the Carnot efficiency and its mechanical efficiency, is 46.7%, and the efficiency
of the generator is 98.4%, what is the efficiency of the total generating unit? (Another 5%
needs to be subtracted off for other plant losses.)
c) One of the coal burning units produces 355 MW. How many metric tons
(1 metric ton = 1 × 106 g) of coal per hour are required to operate this unit at its peak
output if the enthalpy of combustion of coal is 29.0 × 103 kJ kg–1?
a)  
Thot  Tcold
303 K
 1
 0.627
Thot
813 K
b) The efficiency of the generating plant is the product of the individual efficiencies.
 = 0.912  0.467  0.984  0.95 = 0.398
c) The energy output per hour is given by 355 MW 
3600 s
 1.278  1012 J hr 1
hr
The heat required to output this energy is
Eoutput 1.278  1012 J hr 1
q

 3.211  1012 J hr 1

0.398
The number of tons of coal needed is
3.211  1012 J hr 1
 110.7 ton hr 1 .
9
1
29.0  10 J ton
P5.28) The mean solar flux at the Earth’s surface is ~4.00 J cm–2 min–1. In a
nonfocusing solar collector, the temperature can reach a value of 90.0ºC. A heat engine is
operated using the collector as the hot reservoir and a cold reservoir at 298 K. Calculate
the area of the collector needed to produce one horsepower (1 hp = 746 watts). Assume
that the engine operates at the maximum Carnot efficiency.
5-20
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
  1
Thot
298 K
 1
 0.179
Tcold
363 K
The area required for the solar panel is
2.5  105 J min 1

4.00 J cm

2 1
min
1
 6.25 m2
P5.29) A refrigerator is operated by a 0.25-hp (1 hp = 746 watts) motor. If the interior is
to be maintained at –20ºC and the room temperature is 35ºC, what is the maximum heat
leak (in watts) that can be tolerated? Assume that the coefficient of performance is 75%
of the maximum theoretical value.
The coefficient of performance is
r  6.75 
Tcold
 3.45
Thot  Tcold
The maximum heat that can be removed from the cold reservoir is given by
input power  coefficient of performance
746 W
 3.45  640 W =640 J s 1
hp
If the heat leak is greater than this value, the temperature in the refrigerator will rise.
 0.25 hp 
P5.30) An electrical motor is used to operate a Carnot refrigerator with an interior
temperature of 0ºC. Liquid water at 0ºC is placed into the refrigerator and transformed to
ice at 0ºC. If the room temperature is 20ºC, what mass of ice can be produced in one
minute by a 0.25-hp motor that is running continuously? Assume that the refrigerator is
perfectly insulated and operates at the maximum theoretical efficiency.
We need to find the amount of heat per unit time that can be removed from the interior of
the refrigerator.
Tcold   w 
q
 w
 r 


 t  Thot  Tcold  t 
t
273 K
746 W
 0.25 hp 
 2546 W = 2546 J s 1
293 K  273 K
hp
The number of grams of ice that can be frozen in one minute

by this amount of heat extraction is
q t
2546 J s 1
18.02 g 60 s
mice 
t 


 4.5  102 g
1
H fusion
6008 J mol
mol
min
5-21
Chapter 5/Entropy and the Second and Third Laws of Thermodynamics
P5.31) Calculate S for the reaction H2(g) + Cl2(g) → 2HCl(g) at 650 K. Omit terms in
the temperature-dependent heat capacities higher than
T2
.
K2
From Table 2.4,
T
T2
 1.0835  10 4 2 J K 1mol 1
K
K
2
2 T
4 T
CP  Cl2 , g   22.85  6.543  10
 1.2517  10
J K 1mol 1
2
K
K
T
T2
CP  HCl, g   29.81  4.12  103  6.2231  106 2 J K 1mol 1
K
K
2


T
T
CP  2  29.81  4.12 103  6.2231106 2 J K 1mol1 
K
K


CP  H 2 , g   22.66  4.38  102


T
T2
  22.66  4.38 10 2  1.0835 10 4 2 J K 1mol 1 
K
K




T
T2
  22.85  6.54310 2  1.2517 10 4 2 J K 1mol1 
K
K


T
T2
CP  14.11  0.117  2.460 104 2 J K 1mol1
K
K
S  2S298.15  HCl, g   S 298.15  Cl2 , g   S 298.15  H 2 , g 
 2  186.9 J K 1mol1  223.1 J K 1mol 1  130.7 J K 1mol1
 22.0 J K 1mol1
T
C p
ST  S 298.15  
dT 
T
298.15
2


T
4 T
14.11

0.117

2.460

10
J K 1mol 1 
650 
2
K
K
 d T J K 1mol1
 20.0 J K 1mol1   
T
K
298.15
K
1
1
1
1
 20.0 J K mol  10.99 J K mol  41.33 J K 1mol 1  41.03 J K 1mol1
 20.0 J K 1mol 1  10.69 J K 1mol1  30.69 J K 1mol 1
5-22