CHAPTER 15 THERMODYNAMICS
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The plunger of a bicycle tire pump is pushed down
rapidly with the end of the pump sealed so that no air escapes. Since the compression
occurs rapidly, there is no time for heat to flow into or out of the system. Therefore, to a
very good approximation, the process may be treated as an adiabatic compression that is
described by Equation 15.4:
W  (3 / 2)nR(Ti  Tf )
The person who pushes the plunger down does work on the system, therefore W is negative.
It follows that the term (Ti  Tf ) must also be negative. Thus, the final temperature Tf must
be greater than the initial temperature Ti. This increase in temperature is evidenced by the
fact that the pump becomes warm to the touch.
Alternate Explanation:
Since the compression occurs rapidly, there is no time for heat to flow into or out of the
system. Therefore, to a very good approximation, the process may be treated as an adiabatic
compression. According to the first law of thermodynamics, the change in the internal
energy is U  Q  W  W , since Q = 0 for adiabatic processes. Since work is done on
the system, W is negative; therefore the change in the internal energy, U, is positive. The
work done by the person pushing the plunger is manifested as an increase in the internal
energy of the air in the pump. The internal energy of an ideal gas is proportional to the
Kelvin temperature. Since the internal energy of the gas increases, the temperature of the air
in the pump must also increase. This increase in temperature is evidenced by the fact that
the pump becomes warm to the touch.
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2.
REASONING AND SOLUTION
The work done in an isobaric process is given by
Equation 15.2: W  P(Vf  Vi ) . According to the first law of thermodynamics, the change
in the internal energy is U  Q  W  Q  P(V f  V i ) .
One hundred joules of heat is added to a gas, and the gas expands at constant pressure
(isobarically). Since the gas expands, the final volume will be greater than the initial
volume. Therefore, the term P(Vf  Vi ) will be positive. Since Q = +100 J, and the term
P(Vf  Vi ) is positive, the change in the internal energy must be less than 100 J. It is not
possible that the internal energy increases by 200 J.
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3.
REASONING AND SOLUTION The internal energy of an ideal gas is proportional to its
Kelvin temperature (see Equation 14.7). In an isothermal process the temperature remains
constant; therefore, the internal energy of an ideal gas remains constant throughout an
isothermal process. Thus, if a gas is compressed isothermally and its internal energy
increases, the gas is not an ideal gas.
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5.
REASONING AND SOLUTION
a. It is possible for the temperature of a substance to rise without heat flowing into
substance. Consider, for example, the adiabatic compression of an ideal gas. Since
process is an adiabatic process, Q = 0. The work done by the external agent increases
internal energy of the gas. Since the internal energy of an ideal gas is proportional to
Kelvin temperature, the temperature of the gas must increase.
the
the
the
the
b. The temperature of a substance does not necessarily have to change because heat flows
into or out of it. Consider, for example, the isothermal expansion of an ideal gas. Since the
internal energy of an ideal gas is proportional to the Kelvin temperature, the internal energy,
U, remains constant during an isothermal process. The first law of thermodynamics gives
U  Q  W  0 , or Q = W. The heat that is added to the gas during the isothermal
expansion is used by the gas to perform the work involved in the expansion. The
temperature of the gas remains unchanged. Similarly, in an isothermal compression, the
work done on the gas as the gas is compressed causes heat to flow out of the gas while the
temperature of the gas remains constant.
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6.
REASONING AND SOLUTION The text drawing shows a pressure-volume graph in
which a gas undergoes a two-step process from A to B and from B to C.
From A to B: The volume V of the gas increases at constant pressure P. According to the
ideal gas law (Equation 14.1), PV  nRT , the temperature T of the gas must increase.
According to Equation 14.7, U  (3 / 2)nRT , if T increases, then U , the change in the
internal energy, must be positive. Since the volume increases at constant pressure ( V
increases), we know from Equation 15.2, W  PV , that the work done is positive. The
first law of thermodynamics (Equation 15.1) states that U  Q  W ; since U and W are
both positive, Q must also be positive.
From B to C The pressure P of the gas increases at constant volume V. According to the
ideal gas law (Equation 14.1), PV  nRT , the temperature T of the gas must increase.
According to Equation 14.7, U  (3 / 2)nRT , if T increases, then U , the change in the
internal energy, must be positive. Since the process occurs isochorically ( V  0 ), and
according to Equation 15.2, W  PV , the work done is zero. The first law of
thermodynamics (Equation 15.1) states that U  Q  W ; since W = 0, Q is also positive
since U is positive.
These results are summarized in the table below:
AB
U
Q
W
+
+
+
B C
+
+
0
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9.
REASONING AND SOLUTION When a solid melts at constant pressure, the volume of
the resulting liquid does not differ much from the volume of the solid. According to the first
law of thermodynamics, U  Q  W  Q  P(V f  V i )  Q . Hence, the heat that must be
added to melt the solid is used primarily to increase the internal energy of the molecules.
The internal energy of the liquid has increased by an amount Q = mLf compared to that of
the solid, where m is the mass of the material and Lf is the latent heat of fusion.
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15. REASONING AND SOLUTION
The efficiency of a Carnot engine is given by
Equation 15.15: efficiency  1  ( TC / TH ) .
a. Lowering the Kelvin temperature of the cold reservoir by a factor of four makes the ratio
TC / TH one-fourth as great.
b. Raising the Kelvin temperature of the hot reservoir by a factor of four makes the ratio
TC / TH one-fourth as great.
c. Cutting the Kelvin temperature of the cold reservoir in half and doubling the Kelvin
temperature of the hot reservoir makes the ratio TC / TH one-fourth as great.
Therefore, all three possible improvements have the same effect on the efficiency of a
Carnot engine.
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20. REASONING AND SOLUTION A refrigerator is advertised as being easier to "live with"
during the summer, because it puts into your kitchen only the heat that it removes from the
food. The advertisement is describing a refrigerator in which heat is removed from the
interior of the refrigerator and deposited outside the refrigerator without requiring any work.
Since no work is required, the flow must be spontaneous. This violates the second law of
thermodynamics, which states that heat spontaneously flows from a higher-temperature
substance to a lower-temperature substance, and does not flow spontaneously in the reverse
direction. Heat can be made to flow from a cold reservoir to a hot reservoir, but only when
work is done. Both the heat and the work are deposited in the hot reservoir.
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22. REASONING AND SOLUTION The second law of thermodynamics states that the total
entropy of the universe does not change when a reversible process occurs (S universe  0)
and increases when an irreversible process occurs (S universe  0) .
An event happens somewhere in the universe and, as a result, the entropy of an object
changes by –5 J/K. If the event is a reversible process, then the entropy change for the rest
of the universe must be +5 J/K; this results in a total entropy change of zero for the universe.
If the process is irreversible, the only possible choice for the change in the entropy of the
rest of the universe is +10 J/K; this results in a total entropy change of +5 J/K for the
universe. The choices –5 J/K and 0 J/K are not possible choices for the entropy change of
the rest of the universe, because they imply that the total entropy change would be negative.
This would violate the second law of thermodynamics.
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23. REASONING AND SOLUTION When water freezes from a less-ordered liquid to a moreordered solid, its entropy decreases. This decrease in entropy does not violate the second
law of thermodynamics, because it is a decrease for only one part of the universe. In terms
of entropy, the second law indicates that the total change in entropy for the entire universe
must be either zero (reversible process) or greater than zero (irreversible process). In the
case of freezing water, heat must be removed from the water and deposited in the
environment. The entropy of the environment increases as a result. If the freezing occurs
reversibly, the increase in entropy of the environment will exactly match the decrease in
entropy of the water, with the result that Suniverse  S water  S environment  0 . If the
freezing occurs irreversibly, then the increase in entropy of the environment will exceed the
decrease in entropy of the water, with the result that Suniverse  S water  S environment  0.
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PROBLEMS
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1.
REASONING Since the change in the internal energy and the heat released in the
process are given, the first law of thermodynamics (Equation 15.1) can be used to find the
work done. Since we are told how much work is required to make the car go one mile, we
can determine how far the car can travel. When the gasoline burns, its internal energy
decreases and heat flows into the surroundings; therefore, both U and Q are negative.
SSM
SOLUTION According to the first law of thermodynamics, the work that is done when one
gallon of gasoline is burned in the engine is
W  Q  U  1.00  10 8 J – (–1.19  10 8 J) = 0.19  10 8 J
Since 6.0  10 5 J of work is required to make the car go one mile, the car can travel
0.19  10 8 J
1 mile I
F
G
H6.0  10 J J
K
5
32 miles
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3.
REASONING AND SOLUTION
determined from Equation 14.7 as
The change in the internal energy of the gas can be
U  32 nRT  32 (1.00mol )[8.32J/(mol K)](550K  350 K)  2500 J
a. The heat can be found from the first law of thermodynamics:
Q  U  W  2500 J  ( 6200 J)  3700 J
b. Since Q is negative, heat flows out of the gas .
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12. REASONING For segment AB, there is no work, since the volume is constant. For
segment BC the process is isobaric and Equation 15.2 applies. For segment CA, the work
can be obtained as the area under the line CA in the graph.
SOLUTION
a. For segment AB, the process is isochoric, that is, the volume is constant. For a process in
which the volume is constant, no work is done, so W = 0 J .
b. For segment BC, the process is isobaric, that is, the pressure is constant. Here, the
volume is increasing, so the gas is expanding against the outside environment. As a result,
the gas does work, which is positive according to our convention. Using Equation 15.2 and
the data in the drawing, we obtain
c h
c
7 .0  10 Pa hc
5.0  10
W  P V f  Vi
5
3
hc
m 3  2 .0  10 3 m 3
h
2 .1  10 3 J
c. For segment CA, the volume of the gas is decreasing, so the gas is being compressed and
work is being done on it. Therefore, the work is negative, according to our convention. The
magnitude of the work is the area under the segment CA. We estimate that this area is 15 of
the squares in the graphical grid. The area of each square is
5
–3
(1.0  10 Pa)(1.0  10
3
2
m ) = 1.0  10 J
The work, then, is
W = – 15 (1.0  102 J) = 1.5  10 3 J
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20. REASONING AND SOLUTION As Section 14.1 discusses, the number of moles n is
given by the mass m divided by the mass per mole:
n
6. 0 g
m

 1.5 mol
Mass per mole 4.0 g / mol
For an isothermal process we have (see Equation 15.3)
ln
FV I 
G
HV J
K
f
i
9600 J
W

nRT
1.5 mol 8.31 J / mol  K
b
b
g
gb370 K g 2.08
Therefore, Vf/Vi = e2.08 = 8.0 .
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22. REASONING An adiabatic process is one for which no heat enters or leaves the system, so
Q = 0 J. The work is given as W = +610 J, where the plus sign denotes that the gas does
work, according to our convention. Knowing the heat and the work, we can use the first law
of thermodynamics to find the change U in internal energy as U = Q – W (Equation 15.1).
Knowing the change in the internal energy, we can find the change in the temperature by
recalling that the internal energy of a monatomic ideal gas is U = 23 nRT, according to
Equation 14.7. As a result, it follows that U =
3
2
nRT.
SOLUTION Using the first law from Equation 15.5 and the change in internal energy from
Equation 14.7, we have
U  Q  W
Therefore, we find
T 
b g
2 Q W
3nR
or
3
2
nRT  Q  W
b gb g 
3b
0.50 mol g8.31 J / b
mol  K g
2 0 J  610 J
98 K
The change in temperature is a decrease.
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30. REASONING AND SOLUTION The heat required for an isobaric process is
cR hn T
F 8.0 g IJb75 K g
8.31 J / b
mol  K gG
H39.9 g / mol K
Q  Cp n T 

5
2
5
2
310 J
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38. REASONING The change in the internal energy of the gas can be found using the first law
of thermodynamics, since the heat added to the gas is known and the work can be calculated
by using Equation 15.2, W = P V. The molar specific heat capacity at constant pressure can
be evaluated by using Equation 15.6 and the ideal gas law.
SOLUTION
a. The change in the internal energy is
U  Q  W  Q  P V
c
hc
h
 31.4 J  1.40  10 4 Pa 8.00  10 4 m 3  3.00  10 4 m 3  24 .4 J
b. According to Equation 15.6, the molar specific heat capacity at constant pressure is
Cp = Q/(n T). The term n T can be expressed in terms of the pressure and change in
volume by using the ideal gas law:
P V = n R T
or
n T = P V/R
Substituting this relation for n T into Cp = Q/(n T), we obtain
Cp 
Q

PV
R
31.4 J
c1.40  10 Pa hc5.00  10
4
4
m3
h
37 .3 J / (mol  K)
R
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49.
REASONING The efficiency e of a Carnot engine is given by Equation 15.15,
e  1  ( TC / TH ) , where, according to Equation 15.14, ( QC / QH )  ( TC / TH ) . Since the
efficiency is given along with TC and QC , Equation 15.15 can be used to calculate TH .
Once TH is known, the ratio TC / TH is thus known, and Equation 15.14 can be used to
calculate QH .
SSM
SOLUTION
a. Solving Equation 15.15 for TH gives
TH 
TC
1– e

378 K
 1260 K
1– 0.700
b. Solving Equation 15.14 for QH gives
Q H  QC
FT I  (5230 J) F
1260 K I
G
G
J
H378 K J
K
HT K
H
1.74  10 4 J
C
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68. REASONING AND SOLUTION The change in entropy is S = Q/T, where
Q = mLs = (4.00 kg)(5.77  105 J/kg) = 2.31  106 J
Thus,
S = Q/T = (2.31  106 J)/(194.7 K) = 1.19  10 4 J/K
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71.
REASONING AND SOLUTION
a. Since the energy that becomes unavailable for doing work is zero for the process, we
have from Equation 15.19, Wunavailable  T0 Suniverse  0 . Therefore, Suniverse  0 and
according to the discussion in Section 15.11, the process is reversible .
SSM
b. Since the process is reversible, we have (see Section 15.11)
Suniverse  Ssystem  Ssurroundings  0
Therefore,
Ssurroundings  Ssystem  –125 J/K
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