Answers to Selected Problems 
Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 
Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 
Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 
3.1
(a) 268.8 K and 4.62510^{5}
Pa
(b) at 90^{o} (horizontal) 282.6 K and 4.86310^{5}
Pa
(c) at 180^{o} (vertical downward) 296.4 K and
5.110^{5}
Pa
3.2
(a) Q/t
= 3.5110^{8}
J/hr [contractor pays]
(b) Q/t
= 3.2310^{8}
J/hr [contractor collects fee]
3.5
(a) L_{A} = 0.4352 m, L_{B}
=
0.0218 m, T_{A }= T_{B }= 311 K, P_{A
}=
P_{B
}=
7 bar
(b) Same as (a)
(c) There is no real solution given the constraints,
we must specify more about the expansion (or compression) of A or B. For
example, if A expands adiabatically, P_{A }= P_{B
}=
6.68 bar, T_{A }= 263.6 K, T_{B }= 881.3
K, L_{A }= 0.3917 m, and L_{B }= 0.0655 m.
3.12
(a) After 6 s, T = 349.5 K, P = 1.13910^{5}
Pa in the bulge.
(b) After 3 s, T = 454.6 K, P = 6.2610^{3}
Pa in the large tank.
4.1
For the minimum condition T_{A}(final)
= 132 K and for the maximum condition T_{A}(final) =617
K, these extrema require using the object with the lowest mass x heat capacity
product.
4.2
W = 9.807y in J/kg evaporated, for typical
values of ambient dry bulb and wet bulb temperatures of 300 K and 280 K
respectively, y = 16.8 km!
4.3
(a) _{B}
= 0.166 mol/s,_{C}
= 0.833 mol/s
(b) T_{D }= 299.1K
(c) W = 26.4 J/s
(d) S=
13.3 J/K
(e) Same as (d).
(f) =
4000 J/s
4.4 W =  1.1110^{4}
J, T_{f }= 288.7 K, P_{f }= 0.54 bar
4.6
(a) 21.6 s
(b) 391.8 K and 2.2 bar at 10 s
(c) S_{gas}=
4.9310^{3}
J/K = S_{universe}
(d) S(gas
in tank) = 813.6 J/K,
S(gas
vented) = +2020 J/K,
S(surroundings)
= +1203 J/K
(e) W_{min} = +2.8310^{5
}J
4.7
(a) =
2400 W, T = 327.7 K, and Q = 22,200 J after 10 s
(b) T = 1750 K at t = 57.7 s with all stored
work consumed.
4.9
(a) W_{net} =  3.1710^{6}
J
4.12
(a) T_{2} = 227.9 K, T_{1}
= 392.9 K (2) after venting (1) before venting
(b) T_{2} = 211.1 K, T_{1}=
363.9 K
(c) W = 2547 J/mol
(d) W_{max} = 1.3810^{5}
J/kg
4.13
(a) S
= 1.3810^{3}
J/K hr
(b) S
= 2 J/K hr
4.16
(a) Case (1) 1.9810^{2}
kWhr; (2) 2.7810^{2}
kWhr; (3) 2.1610^{2 }kWhr;
(4) 2.7810^{2
}kWhr
(b) Case (1) 420 K, 42 J/K; (2) 300 K, 0 J/K; (3) 300
K, 73.9 J/K; (4) 300 K, 0 J/K
4.18 W = 5.2310^{3} J = Q
4.22
(a) P = 6.3510^{4}
Pa
(b) T = 272.8 K
(c) P = 3.6810^{4}
Pa with 2 pumps
(d) P = 1.9110^{4}
Pa and T
(e) Power (minimum) = 20.9 W
4.24
(a) W_{max}= 198.76 kJ/kg (Carnot + expansion
work)
(b) W_{p }= ideal pump work (estimated
to include PE for lifting water and for the compression of gas space in
storage tanks)
4.30
(a) _{
}=
6.3810^{6} J/s
(b)=
0.90 = (actual power)/(maximum power)
(c) no, the maximum power outputs are the same
5.2
(a)
(b)
5.4
(a)
(b)
(c) U
(d)
5.7
(a)
(b)
5.10 T_{f} = 1087 K
5.12
5.21 V_{c} = RT
= (C_{p}/C_{v}) RT » 330 m/s
(use mass units for R with molecular weight) distance at 2 s »
660 m
5.22
(a)
5.23
(b) No, the ratio as shown in Problem 5.17 is expressed
in terms of PVT properties and an isentropic derivative which in
turn requires nonPVT property information specifically related
to the temperature dependence of energy stored in U.
5.24
(a)
(b)
(c)
(d)
(e)
5.25
(a)
(b)
(c) Yes, using the GibbsDuhem relationship:
6.2
thus,
at equilibrium
6.3

from chemical equilibrium criteria, intersections define equilibrium states.
6.4
(a)



6.5
(a),
(b) T = 366.5 K, P_{A,1} = P_{A,2}
= 2 bar, P_{B,2} = 1 bar
N_{B,1}
= 0 N_{B,2} = 1 mole N_{A,1} +
N_{A,2}
= 4 moles
(c) P_{1} = 2 bar, N_{1}
= 5 moles; V_{1} = 2.5 RT/10^{5} =
0.0752 m^{3}_{ }with T = 367 K and no gas on side
2.
6.7
Fraction evaporated 0.227
7.1
if A_{NN} = 0 then A_{VV}
= 0 = A_{NN }/V ^{2}
7.3
T = 55.1 K
7.6
for
= 2.5
x_{A}  x_{B} 
0.1  0.286, 0.614 
0.2 
0.4 (critical pt.)

0.3 
no solution

0.4  0.2, 0.4 (critical pts.) 
0.5 
0.175, 0.325

0.6 
0.085, 0.28

7.11
8.4
At 365.8 K, 16.5 bar
8.5
At T = 319.4 K
P_{r}  K_{H}/K_{S}  Z = PV/RT 
0.1  0.100  0.967 
0.2  0.194  0.932 
0.4  0.369  0.862 
1.0  0.798  0.598 
4.0  0.426  0.547 
10.0  0.296  1.108 
8.9
Using an isenthalpic expansion across an insulated value
to 1 bar, 0.236 kg of dry ice could
be produced per kg of CO_{2} drawn from the cylinder.
For an isentropic expansion, 0.41 kg dry ice/kg of CO_{2} is possible.
8.10
(a) T_{f} = 379.3 K at 48.3 bar
(b) W =  9.110^{7}
J
8.14
T_{B,f}256
to 257 K, P_{B,f}21.5
bar, x_{g} = 0.63 (fraction vapor)
8.15
(a) Power = 508 kW
(b) Power = 516 kW
8.30

= 90.9 kW and
= 68.3 kW
8.31
9.1
(a) 31.36 kJ/mole water added to acid at the start;
61.1 kJ/mole and added to water
at the start
(b) maximum heat load occurs at the start
(c) total Q =  28 kJ/mol acid = 11.2 kJ/mol
of solution same for both cases
9.2
(b) Q =  3.86 10^{5}
J
(c)
(d)
9.3
mole % NH_{3 }= 51.37 %
9.6
W = 326 J
9.7
w = water and s = salt (NaCl)
9.8
(a) W_{min} = 1069 J (PREOS)
(b) W_{min} = 1812 J (ideal gas mixture)
9.9
9.11
x salt (wt fraction)  
0.01  7.2  8.5 
0.05  42.2  43.9 
0.10  90.5  91.1 
0.15  156.6  142.0 
0.20  242.0  197.0 
0.25  355.0  257.0 
9.13
9.14
(a)
(b) At 85^{o}C^{ }(188 K), there are
two phases in equilibrium at x_{EtOH} = 0.07 and
0.35.
9.20
=
675 W
9.21
(a) 1.3210^{5}
W
(b) See Section 14.4 for details, information provided
in part (a) is not sufficient, ideal gas state heat capacities and a mixture
PVTN
EOS are needed.
9.27
10.2
H  H^{o }= PV  RT  a/V
10.3
10.4
10.6
10.7
b =
a =
10.9
(a)
(b)
(c) for T/T_{c} = 1.00001,
10.10
11.4
(a)
(b)
(c) for only positive values of ,
a large number of values
exist as a function of x_{i} that yield
extrema, e.g.
for x_{1} = 0.2
0  3.236 
0.500  1.699 
1.000  1.000 
1.500  0.586 
2.000  0.209 
2.118  0 
11.5
for
for
11.7
(c) fraction of solution precipitated = 0.0067
11.8
for 2suffix Margules T_{c} = w/2
k
11.9
for quasichemical model T_{c} = w/2.23
k
12.3
(a) selected values of are
given below for LiCl
 
 

0.1  0.9651  0.9653 
0.01  0.9032  0.9044 
0.1  0.7850  0.7927 
1.0  0.7636  0.7758 
5.0  1.9586  2.0222 
10.0  7.9948  11.0827 
(b)
given that s = 9 mol/kg for pure LiCl in H_{2}O.
Using the Meissner model for
predictions for both pure and mixed electrolyte.
K_{sp}(9)^{2
}(~6)^{2}
= 2916 for pure LiCl at 25^{o}C
Iterating
s*=2.4 and
3H for mixed LiCl and HCl at 25^{o}C
(c) Difficult to say without more information on how
much Na_{2}SO_{4
}influences for
LiCl. For example is the effect on
large enough to compensate for the Cl^{} common ion effect?
12.8
Some representative values for
molality (NaCl
or LiCl)(CuCl2)
0.1
0.96
0.55
1.0
0.30
0.18
10.0
0.096
0.055
At T = 300 ^{o}C 22.3
so multiply above values by (22.3
573)/(78.5)(298) = 0.74
12.10
I = 0.546 or m = I / 3 =
0.182 molal
13.2
Using Joback’s method with T_{b} = 225.5K
(exp. value)
T_{c}
= 364 K, P_{c} = 46.7 bar, and =
0.1606

= 2.83 + 0.269 T = 3.1510^{4}
T
^{2}
+ 4.210^{8}
T
^{3}
13.3
T_{c} = 663.5 K, P_{c} =
32.21 bar,
V_{c}
= 429.5 cm^{3}/mol, and T_{b}
=
450.38 K
Using Eq. (1318) for vapor pressure:
13.4
For pure caffeine, T_{b}640
K, T_{c} = 872 K, V_{c} = 488.5 cm^{3}/mol,
and P_{c} = 41.46 bar
Using the PR EOS for density estimate for 5 mol % caffeine
CO_{2} mixture at 80 bar, 310 K =
1151 kg/m^{3}. The vapor pressure of liquid caffeine is approximated
by Eq. (1318)
14.2
Using the PR EOS,
14.6
(a)
(b)
(c) Using the RKEOS and a suitable mixing rule,
a_{mix}
and b_{mix} parameters can be calculated and used to calculate
all PVTN properties. With an idealgas state
C_{p}
all derived properties needed for the cycle calculation can be obtained
using a departure function approach. Then,
14.8
(a)
(b)
(about 150 times smaller!)
14.12
(a) and (b)
so CTI’s claim of 1 kWhr
production violates the 2^{nd} Law limit
(c) All heat transfer and work production and utilization
steps have some irreversibility in practical systems.
14.14
(a)
(b) no fundamental laws or concepts are violated
(c)
about 17% more area for heat exchange is required for the GH process
(d)
NB: If an electric motor drive is used for the feed pump in the conventional Rankine system, then the net output would be reduced possibly accounting for the discrepancy.
15.1
15.3
x (Naphthalene) = 0.196
15.6
(c) q (critical quality) =
(e) O_{2}: q = 0.54 and H_{2}: q = 0.36
15.12
x_{KCl} = 0.413 and
T
= 624 K
15.13
15.15
Anesthetic pressure of CCl_{4}
4.910^{3} bar.
15.19
(a)
(b)
(c)
(i) ,
solution is regular
(ii),
less structural ordering in mixture versus pure state
(iii),
more structural ordering in mixture versus pure state, implies
16.9
(a) y_{A} = 0.764 y_{B}
=
0.236
(b) not in equilibrium, estimate y_{B} assuming
equilibrium
16.12
16.14
16.16
using
expansion valve
using
expansion turbine
Consider net energy, entropy, and work flows using 1^{st}
and 2^{nd} Law concepts to show process is not feasible as described.
16.20
(a) P = 21.9 bar
(b) T = 51.5 K
(c)
16.22
T, K C_{p}
(effective),
J/gK
293
4.69
313
7.20
333
8.29
353
6.36
373
3.64
17.3
(a) ;
at A: L_{1}L_{2}HG and at
B:
L_{1}HIG
(b) L_{1} = 0 and M_{1}
= 0 use Pexplicit EOS to determine required A_{ijk}
and
A_{ij}
derivatives
(c)
(d)
_{
in this case}
17.6
d ln P/d(1/T) =
18.1
18.4
_{J/kg
mol}
18.5
T_{i} = 269 K = 4^{o}C
(winter!)
18.6
T_{bath} = 317 K,
efficiency = 0.065
18.8
z = depth = 764 m
18.9
(tube
bottom) = 0.501, if L
4/7
18.11
18.14
y_{tritium} = 0.065 and y_{deuterium}
= 0.538 at rim
18.16
h = 252 m, for equilibrium ocean, fresh water
cannot rise above a level 252 m below sea surface. For the wellmixed case,
process becomes feasible if Z = depth > 9850 m (32,000 ft.).
19.1
19.2
19.5
19.7
x_{butanol} = 0.436
19.8
(a) 0.046 J/m^{2}
(b) 296 K
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