CHAPTER 15
THERMODYNAMICS
CONCEPTUAL QUESTIONS
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1.
SSM
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:
The person who pushes the plunger down does work on the system, therefore
W
is negative.
It follows that the term must also be negative. Thus, the ﬁnal temperature
T
f
must be
greater than the initial temperature
T
i
. 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 , since
Q
= 0 for adiabatic processes. Since work is done on the system,
W
is
negative; therefore the change in the internal energy,
Δ
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: . According to the ﬁrst law of thermodynamics, the change in the internal
energy is .
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 will be positive. Since
Q
= +100 J, and the term 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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4
. REASONING AND SOLUTION
According to the ﬁrst law of thermodynamics
(Equation 15.1), the change in the internal energy is
Δ
U
=
Q
–
W
. The process is isochoric,
which means that the volume is constant. Consequently, no work is done, so
W
= 0. The
process is also adiabatic, which means that no heat enters or leaves the system, so
Q
= 0.
According to the ﬁrst law, then,
Δ
U
=
Q
–
W
= 0. There is no change in the internal energy,
and the internal energy of the material at the end of the process is the same as it was at the
beginning.
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5
. REASONING AND SOLUTION
a. It is possible for the temperature of a substance to rise without heat ﬂowing into
the substance. Consider, for example, the adiabatic compression of an ideal gas. Since the
process is an adiabatic process,
Q
= 0. The work done by the external agent increases the
internal energy of the gas. Since the internal energy of an ideal gas is proportional to the
Kelvin temperature, the temperature of the gas must increase.
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,
Δ
U
, remains constant during an isothermal process. The ﬁrst law of
thermodynamics gives , 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 pressurevolume graph in
which a gas undergoes a twostep 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), , the temperature
T
of the gas must increase. According to
Equation 14.7, , if
T
increases, then , the change in the internal energy, must be positive.
Since the volume increases at constant pressure ( increases), we know from Equation 15.2, ,
that the work done is positive. The ﬁrst law of thermodynamics (Equation 15.1) states that ;
since 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), , the temperature
T
of the gas must increase. According to
Equation 14.7, , if
T
increases, then , the change in the internal energy, must be positive.
Since the process occurs isochorically (), and according to Equation 15.2,
,
the work done is
THERMODYNAMICS
zero. The ﬁrst law of thermodynamics (Equation 15.1) states that ; since
W
= 0,
Q
is also
positive since is positive.
These results are summarized in the table below:
Q
W
+
+
+
+
+
0
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7
. REASONING AND SOLUTION
Since the process is an adiabatic process,
Q
= 0. Since
the gas expands into chamber B under zero external pressure, the work done by the gas is
W
=
P
Δ
V
= 0. According to the ﬁrst law of thermodynamics, the change in the internal
energy is, therefore, zero: . The internal energy of an ideal gas is proportional to the Kelvin
temperature of the gas (Equation 14.7). Since the change in the internal energy of the gas is
zero, the temperature change of the gas is zero. The ﬁnal temperature of the gas is the same
as the initial temperature of the gas.
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8
.
SSM
REASONING AND SOLUTION
A material contracts when it is heated. To
determine the molar speciﬁc heat capacities, we ﬁrst calculate the heat
Q
needed to raise the
temperature of the material by an amount
Δ
T
. From the ﬁrst law of thermodynamics, .
When the heating occurs at constant pressure, the work done is given by Equation 15.2: .
When the volume is constant,
Δ
V
= 0. Therefore, we have:
and
Equation 15.6 indicates that the molar heat capacities will be given by
.
Therefore
and
Since the material contracts when it is heated,
V
f
is less than
V
i
. Therefore the term
is negative. Hence, the numerator of
C
P
is smaller than the numerator of
C
V
.
Therefore,
C
V
is larger than
C
P
.
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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, . 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
=
mL
f
compared to that of the solid, where
m
is the mass of
the material and
L
f
is the latent heat of fusion.
Chapter 15 Conceptual Questions
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10
. REASONING AND SOLUTION
According to Equation 14.7, the Kelvin temperature
T
of
the gas is related to its internal energy
U
by . The change in the internal energy is given by
the ﬁrst law of thermodynamics (Equation 15.1), .
It is desired to heat a gas so that its temperature will be as high as possible. If the
process occurs at constant pressure, so that the volume of the gas increases, work is done by
the gas. The available heat is used to do work
and
to increase the internal energy of the gas.
On the other hand, if the process is carried out at constant volume, the work done is zero,
and all of the heat increases the internal energy of the gas. From Equation 14.7, the internal
energy is directly proportional to the Kelvin temperature of the gas. Since the internal
energy increases by a greater amount when the process occurs at constant volume, the
temperature increase is greatest under conditions of constant volume. Therefore, if it is
desired to heat a gas so that its temperature will be as high as possible, you should heat it
under conditions of constant volume.
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11
. REASONING AND SOLUTION
A hypothetical device takes 10 000 J of heat from a hot
reservoir and 5000 J of heat from a cold reservoir and produces 15 000 J of work.
a. According to the ﬁrst law of thermodynamics, . This is a statement of energy
conservation. The hypothetical device does not violate energy conservation. It does not
create or destroy energy. It converts one form of energy (15 000 J of heat) into another form
of energy (15 000 J of work) with no gain or loss.
b. This hypothetical device does violate the second law of thermodynamics. It
converts all of its input heat (15 000 J) into work (15 000 J). Therefore, the efﬁciency of
this device is 1.0 or 100 %. But Equation 15.15 is a consequence of the second law of
thermodynamics and sets the limits of the maximum possible efﬁciency of any heat device.
Since
T
C
is greater than 0 K, the ratio
must be positive. Furthermore, since
T
C
<
T
H
,
the ratio
must be less than one. Therefore, Equation 15.15 implies that the
efﬁciency of any device must be less than 1 or 100%. Since the efﬁciency of the
hypothetical device is equal to 100%, it violates the second law of thermodynamics.
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12
. REASONING AND SOLUTION
According to the second law of thermodynamics, heat
ﬂows spontaneously from a substance at a higher temperature to a substance at a lower
temperature and does not ﬂow spontaneously in the reverse direction. Therefore, according
to the second law of thermodynamics, work must be done to remove heat from a substance
at a lower temperature and deposit it in a substance at a higher temperature. In other words,
the second law requires that energy in the form of work must be supplied to an air
conditioner in order for it to remove heat from a cool space and deposit the heat in a warm
space. An advertisement for an automobile that claimed the same gas mileage with and
without the air conditioner would be suspect. Since the car would use more energy with the
THERMODYNAMICS
air conditioner on, the car would use more gasoline. Therefore, the mileage should be less
with the air conditioner running.
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13
. REASONING AND SOLUTION
Carnot's principle states that the most efﬁcient engine
operating between two temperatures is a reversible engine. This means that a
reversible
engine operating between the temperatures of 600 and 400 K must be more efﬁcient than an
irreversible
engine operating between the
same two temperatures
. No comparison can be
made with an irreversible engine operating between two temperatures that are different than
600 and
.
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14
. REASONING AND SOLUTION
The efﬁciency of a Carnot engine is given by
Equation 15.15:
. Three reversible engines A, B, and C, use the
same cold reservoir for their exhaust heats. They use different hot reservoirs with the
following temperatures: (A) 1000 K; (B) 1100 K; and (C) 900 K. We can rank these
engines in order of increasing efﬁciency according to the following considerations. The ratio
is inversely proportional to the value of
T
H
. The ratio
will be smallest for
engine B; therefore, the quantity 1 –
will be largest for engine B. Thus, engine B
has the largest efﬁciency. Similarly, the ratio
will be largest for engine C; therefore,
the quantity 1 –
will be smallest for engine C. Thus, engine C has the smallest
efﬁciency. Hence, the engines are, in order of increasing efﬁciency: engine C, engine A, and
engine B.
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15
. REASONING AND SOLUTION
The efﬁciency of a Carnot engine is given by
Equation 15.15:
.
a. Lowering the Kelvin temperature of the cold reservoir by a factor of four makes
the ratio
onefourth as great.
b. Raising the Kelvin temperature of the hot reservoir by a factor of four makes the
ratio
onefourth 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
onefourth as great.
Therefore, all three possible improvements have the same effect on the efﬁciency of a
Carnot engine.
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16
.
SSM
REASONING AND SOLUTION
A refrigerator is kept in a garage that is not
heated in the cold winter or airconditioned in the hot summer. In order to make ice cubes,
the refrigerator uses electrical energy to provide the work to remove heat from the interior of
Chapter 15 Conceptual Questions
the freezer and deposit the heat outside of the refrigerator. In the summer, the "hot" reservoir
will be at a higher temperature than it is in the winter. Therefore, more work will be required
to remove heat from the interior of the freezer in the summer, and the refrigerator will use
more electrical energy. Hence, it will cost more for the refrigerator to make a kilogram of ice
cubes in the summer.
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17
. REASONING AND SOLUTION
The coefﬁcient of performance of a heat pump is given
by Equation 15.17: coefﬁcient of performance =
. From the conservation of
energy,
. Thus, the ratio
can be written
. The job of
a heat pump is to remove heat from a cold reservoir, and deliver it to a hot reservoir;
therefore, the ratio
must be nonzero and positive. Hence, the coefﬁcient of
performance,
, must always be greater than one.
18.
REASONING AND SOLUTION
In a refrigerator, the interior of the unit is the cold
reservoir, while the warmer exterior of the room is the hot reservoir. An air conditioner is
like a refrigerator, except that the room being cooled is the cold reservoir, and the outdoor
environment is the hot reservoir. Therefore, an air conditioner cools the inside of the house,
while a refrigerator warms the interior of the house.
19
. REASONING AND SOLUTION
Heat pumps
can
deliver more energy into your house
than they consume in operating. A heat pump consumes an amount of energy
,
which it
uses to make an amount of heat
ﬂow from the cold outdoors into the warm house. The
amount of energy the heat pump delivers to the house is
. This is greater
than the amount of energy,
, consumed by the heat pump.
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 highertemperature
substance to a lowertemperature 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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21
. REASONING AND SOLUTION
On a summer day, a window air conditioner cycles on
and off, according to how the temperature within the room changes. When the unit is on it
THERMODYNAMICS
will be depositing heat, along with the work required to remove the heat, to the outside.
Therefore, the outside of the unit will be hotter when the unit is on. Hence, you would be
more likely to fry an egg on the outside part of the unit when the unit is on.
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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 and increases
when an irreversible process occurs .
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.
23
. REASONING AND SOLUTION
When water freezes from a lessordered liquid to a more
ordered 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 . 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
24
. REASONING AND SOLUTION
Since we can interpret the increase of entropy as an
increase in disorder, the more disordered system will have the greater entropy.
a. The popcorn that results from the kernels is more disorderly than a handful of popcorn
kernels; therefore, the popcorn that results from the kernels has the greater entropy.
b. A salad has more disorder after it has been tossed; therefore, the tossed salad has the
greater entropy.
c. A messy apartment is more disorderly than a neat apartment; therefore, a messy
apartment has the greater entropy.
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25
. REASONING AND SOLUTION
A glass of water contains a teaspoon of dissolved sugar.
After a while, the water evaporates, leaving behind sugar crystals. The entropy of the sugar
crystals is less than the entropy of the dissolved sugar, because the sugar crystals are in a
more ordered state. However, think about the water. The entropy of the gaseous water
vapor is greater than the entropy of the liquid water, because the molecules in the vapor are
Chapter 15 Conceptual Questions
in a less ordered state. Since the increase in the entropy of the water is greater than the
decrease in entropy of the sugar, the net change in entropy of the universe is positive. The
process, therefore, does not violate the entropy version of the second law of
thermodynamics.
THERMODYNAMICS
26
. REASONING AND SOLUTION
Since we can interpret the increase of entropy as an
increase in disorder, the more disordered state will have the greater entropy. The ﬁnished
building is the most ordered state; therefore it has the smallest entropy. The burnedout shell
of a building is the most disordered state; therefore it has the largest amount of entropy.
The states can be ranked in order of decreasing entropy (largest ﬁrst) as follows:
(3) the burnedout shell of a building, (1) the unused building material, and (2) the building.
Chapter 15 Conceptual Questions
CHAPTER 15
THERMODYNAMICS
PROBLEMS
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1.
SSM
REASONING
Energy in the form of work leaves the system, while energy in the
form of heat enters. More energy leaves than enters, so we expect the internal energy of the
system to decrease, that is, we expect the change
Δ
U
in the internal energy to be negative.
The ﬁrst law of thermodynamics will conﬁrm our expectation. As far as the environment is
concerned, we note that when the system loses energy, the environment gains it, and when
the system gains energy the environment loses it. Therefore, the change in the internal
energy of the environment must be opposite to that of the system.
SOLUTION
a. The system gains heat so
Q
is positive, according to our convention. The system does
work, so
W
is also positive, according to our convention. Applying the ﬁrst law of
thermodynamics from Equation 15.1, we ﬁnd for the system that
As expected, this value is negative, indicating a decrease.
b. The change in the internal energy of the environment is opposite to that of the system, so
that
.
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2.
REASONING
The change
Δ
U
in the individual’s internal energy is given by the first law of
thermodynamics as
(Equation 15.1), where
Q
is the heat and
W
is the work.
Note that
Q
is negative since the individual gives off heat and that
W
is positive since the
individual does work. We will apply the first law twice, once to the individual walking and
again to the individual jogging. By taking advantage of the fact that
Δ
U
is the same in each
case, we will be able to obtain the heat given off during the walking.
SOLUTION
Applying the first law of thermodynamics to walking and jogging gives
Since
, we have
THERMODYNAMICS
The value for
is negative, indicating that the individual gives off heat. The
magnitude of this heat is
.
3.
REASONING
Since the student does work,
W
is positive, according to our convention.
Since his internal energy decreases, the change
Δ
U
in the internal energy is negative. The
ﬁrst law of thermodynamics will allow us to determine the heat
Q
.
SOLUTION
a. The work is
W
= +1.6
×
10
4
J .
b. The change in internal energy is
Δ
U
= –4.2
×
10
4
J .
c. Applying the ﬁrst law of thermodynamics from Equation 15.1, we ﬁnd that
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4.
REASONING
According to the discussion in Section 14.3, the internal energy
U
of a
monatomic ideal gas is given by
(Equation 14.7), where
n
is the number of
moles,
R
is the universal gas constant, and
T
is the Kelvin temperature. When the
temperature changes to a final value of
T
f
from an initial value of
T
i
, the internal energy
changes by an amount
Solving this equation for the final temperature yields
We are given
n
and
T
i
, but must determine
Δ
U
. The change
Δ
U
in the internal energy of the gas is related to
the heat
Q
and the work
W
by the first law of thermodynamics,
Δ
U
=
Q
−
W
(Equation 15.1). Using these two relations will allow us to find the final temperature of the
gas.
SOLUTION
Substituting
Δ
U
=
Q
−
W
into the expression for the final temperature gives
Chapter 15 Problems
Note that the heat is positive (
Q
= +2438 J) since the system (the gas) gains heat, and the
work is negative (
W
=
−
962 J), since it is done on the system.
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5.
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
Δ
U
and
Q
are negative.
SOLUTION
According to the ﬁrst law of thermodynamics, the work that is done when one
gallon of gasoline is burned in the engine is
Since
of work is required to make the car go one mile, the car can travel
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6.
REASONING AND SOLUTION
a. For the weight lifter
Δ
U
=
Q
−
W
=
−
mL
v
−
W
=
−
(0.150 kg)(2.42
×
10
6
J/kg)
−
1.40
×
10
5
J =
b. Since 1 nutritional calorie = 4186 J, the number of nutritional calories is
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THERMODYNAMICS
7.
REASONING
We will apply the ﬁrst law of thermodynamics as given in Equation 15.1
(
Δ
U
=
Q
–
W
) to the overall process. First, however, we add the changes in the internal
energy to obtain the overall change
Δ
U
and add the work values to get the overall work
W
.
SOLUTION
In both steps the internal energy increases, so overall we have
Δ
U
= 228 J + 115 J = +343 J. In both steps the work is negative according to our
convention, since it is done on the system. Overall, then, we have
W
= –166 J – 177 J = –343 J. Using the ﬁrst law of thermodynamics from Equation 15.1,
we ﬁnd
Since the heat is zero, the overall process is adiabatic .
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8.
REASONING
When a gas expands under isobaric conditions, its pressure remains
constant. The work
W
done by the expanding gas is
W
=
P
(
V
f
−
V
i
), Equation 15.2, where
P
is the pressure and
V
f
and
V
i
are the final and initial volumes. Since all the variables in this
relation are known, we can solve for the final volume.
SOLUTION
Solving
W
=
P
(
V
f
−
V
i
) for the final volume gives
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9.
SSM
REASONING
According to Equation 15.2,
W
=
P
Δ
V
, the average pressure
of
the expanding gas is equal to
, where the work
W
done by the gas on the bullet
can be found from the workenergy theorem (Equation 6.3). Assuming that the barrel of the
gun is cylindrical with radius
r
, the volume of the barrel is equal to its length
L
multiplied
by the area (
π
r
2
) of its cross section. Thus, the change in volume of the expanding gas is
.
SOLUTION
The work done by the gas on the bullet is given by Equation 6.3 as
The average pressure of the expanding gas is, therefore,
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Chapter 15 Problems
10.
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. 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
(1.0
×
10
5
Pa)(1.0
×
10
–3
m
3
) = 1.0
×
10
2
J
The work, then, is
W = – 15 (1.0
×
10
2
J) =
______________________________________________________________________________
11.
REASONING
AND
SOLUTION
a. Starting at point
A
, the work done during the ﬁrst (vertical) straightline segment is
W
1
=
P
1
Δ
V
1
=
P
1
(0 m
3
) = 0 J
For the second (horizontal) straightline segment, the work is
W
2
=
P
2
Δ
V
2
= 10(1.0
×
10
4
Pa)6(2.0
×
10
–3
m
3
) = 1200 J
For the third (vertical) straightline segment the work is
W
3
=
P
3
Δ
V
3
=
P
3
(0 m
3
) = 0 J
For the fourth (horizontal) straightline segment the work is
W
4
=
P
4
Δ
V
4
= 15(1.0
×
10
4
Pa)6(2.0
×
10
–3
m
3
) = 1800 J
The total work done is
THERMODYNAMICS
W
=
W
1
+
W
2
+
W
3
+
W
4
=
b. Since the total work is positive, work is done
.
______________________________________________________________________________
12.
REASONING
The pressure
P
of the gas remains constant while its volume increases by an
amount
Δ
V
. Therefore, the work
W
done by the expanding gas is given by
(Equation 15.2).
Δ
V
is known, so if we can obtain a value for
W
, we can use this expression
to calculate the pressure. To determine
W
, we turn to the first law of thermodynamics
(Equation 15.1), where
Q
is the heat and
Δ
U
is the change in the internal
energy.
Δ
U
is given, so to use the first law to determine
W
we need information about
Q
.
According to Equation 12.4, the heat needed to raise the temperature of a mass
m
of material
by an amount
Δ
T
is
where
c
is the material’s specific heat capacity.
SOLUTION
According to Equation 15.2, the pressure
P
of the expanding gas can be
determined from the work
W
and the change
Δ
V
in volume of the gas according to
Using the first law of thermodynamics, we can write the work as
(Equation 15.1). With this substitution, the expression for the pressure becomes
(1)
Using Equation 12.4, we can write the heat as
, which can then be substituted
into Equation (1). Thus,
13.
SSM
REASONING
The work done in an isobaric process is given by Equation 15.2,
; therefore, the pressure is equal to
. In order to use this expression,
Chapter 15 Problems
we must ﬁrst determine a numerical value for the work done; this can be calculated using
the ﬁrst law of thermodynamics (Equation 15.1),
.
SOLUTION
Solving Equation 15.1 for the work
W
, we ﬁnd
Therefore, the pressure is
The change in volume
Δ
V
, which is the ﬁnal volume minus the initial volume, is negative
because the ﬁnal volume is 0.010 m
3
less
than the initial volume.
______________________________________________________________________________
14.
REASONING
AND
SOLUTION
According to the ﬁrst law of thermodynamics, the change
in internal energy is
Δ
U
=
Q
–
W
. The work can be obtained from the area under the graph.
There are sixty squares of area under the graph, so the positive work of expansion is
Since
Q
= 2700 J, the change in internal energy is
______________________________________________________________________________
15.
SSM
WWW
REASONING AND SOLUTION
The ﬁrst law of thermodynamics states
that
. The work
W
involved in an isobaric process is, according to Equation
15.2,
. Combining these two expressions leads to
. Solving for
Q
gives
(1)
Since this is an expansion,
, so
. From the ideal gas law,
, we
have
. Since
, it follows that
. The internal energy of an
ideal gas is directly proportional to its Kelvin temperature
T
. Therefore, since
, it
follows that
. Since both terms on the right hand side of Equation (1) are positive,
the left hand side of Equation (1) must also be positive. Thus,
Q
is positive. By the
convention described in the text, this means that
THERMODYNAMICS
______________________________________________________________________________
16.
REASONING
AND
SOLUTION
The rod's volume increases by an amount
Δ
V
=
β
V
0
Δ
T
,
according to Equation 12.3. The work done by the expanding aluminum is, from
Equation 15.2,
W
=
P
Δ
V
=
P
β
V
0
Δ
T
= (1.01
×
10
5
Pa)(69
×
10
–6
/C°)(1.4
×
10
–3
m
3
)(3.0
×
10
2
C°) =
______________________________________________________________________________
17.
REASONING
AND
SOLUTION
Since the pan is open, the process takes place at constant
(atmospheric) pressure
P
0
. The work involved in an isobaric process is given by Equation
15.2:
W
=
P
0
Δ
V
. The change in volume of the liquid as it is heated is given according to
Equation 12.3 as
Δ
V
=
β
V
0
Δ
T
, where
β
is the coefﬁcient of volume expansion. Table 12.1
gives
for water. The heat absorbed by the water is given by
Equation 12.4 as
Q
=
cm
Δ
T
, where
is the speciﬁc heat capacity of
liquid water according to Table 12.2. Therefore,
where
is the density of the water (see Table 11.1). Thus, we ﬁnd
______________________________________________________________________________
18.
REASONING
We can use the first law of thermodynamics,
Δ
U
=
Q
−
W
(Equation 15.1) to
find the work
W
. The heat is
Q
=
−
4700 J, where the minus sign denotes that the system (the
gas) loses heat. The internal energy
U
of a monatomic ideal gas is given by
(Equation 14.7), where
n
is the number of moles,
R
is the universal gas constant, and
T
is the
Kelvin temperature. If the temperature remains constant during the process, the internal
energy does not change, so
Δ
U
= 0 J.
SOLUTION
The work done during the isothermal process is
Chapter 15 Problems
The negative sign indicates that work is done on the system.
______________________________________________________________________________
19.
SSM
REASONING
According to the ﬁrst law of thermodynamics (Equation 15.1),
. For a monatomic ideal gas (Equation 14.7),
. Therefore, for the
process in question, the change in the internal energy is
. Combining the last
expression for
with Equation 15.1 yields
This expression can be solved for
.
SOLUTION
a. The heat is
, since it is absorbed by the system. The work is
,
since it is done
by
the system. Solving the above expression for
and substituting the
values for the data given in the problem statement, we have
b. Since
is negative,
must be greater than
; this change
represents a
in temperature.
Alternatively, one could deduce that the temperature decreases from the following physical
argument. Since the system loses more energy in doing work than it gains in the form of
heat, the internal energy of the system decreases. Since the internal energy of an ideal gas
depends only on the temperature, a decrease in the internal energy must correspond to a
decrease in the temperature.
______________________________________________________________________________
20.
REASONING
Since the gas is expanding adiabatically, the work done is given by
Equation 15.4 as
. Once the work is known, we can use the ﬁrst law of
thermodynamics to ﬁnd the change in the internal energy of the gas.
SOLUTION
a. The work done by the expanding gas is
THERMODYNAMICS
b. Since the process is adiabatic,
Q
= 0, and the change in the internal energy is
______________________________________________________________________________
21.
REASONING
During an adiabatic process, no heat flows into or out of the gas (
Q
= 0 J).
For an ideas gas, the final pressure and volume (
P
f
and
V
f
) are related to the initial pressure
and volume (
P
i
and
V
i
) by
where
γ
is the ratio of the specific
heat capacities at constant pressure and constant volume (
γ
=
in this problem). We will
use this relation to find
V
f
/
V
i
.
SOLUTION
Solving
for
V
f
/
V
i
and noting that the pressure doubles
(
P
f
/
P
i
= 2.0) during the compression, we have
______________________________________________________________________________
22.
REASONING
According to the first law of thermodynamics
(Equation 15.1),
where
Δ
U
is the change in the internal energy,
Q
is the heat, and
W
is the work. This
expression may be solved for the heat.
Δ
U
can be evaluated by remembering that the
internal energy of a monatomic ideal gas is
(Equation 14.7), where
n
is the
number of moles,
R
= 8.31 J/(mol
⋅
K) is the universal gas constant, and
T
is the Kelvin
temperature. Since heat is being added isothermally, the temperature remains constant and
so does the internal energy of the gas. Therefore,
Δ
U
= 0 J. To evaluate
W
we use
(Equation 15.3), where
V
f
and
V
i
are the final and initial volumes of the
gas, respectively.
SOLUTION
According to the first law of thermodynamics, as given in Equation 15.1, the
heat added to the gas is
Chapter 15 Problems
Using the fact that
Δ
U
= 0 J for an ideal gas undergoing an isothermal process and the fact
t h a t
(Equation 15.3),
we can rewrite the
expression for the
heat as follows:
Since the volume of the gas doubles, we know that
V
f
= 2
V
i
. Thus, it follows that
23.
SSM
REASONING
When the expansion is isothermal, the work done can be calculated
from Equation (15.3):
. When the expansion is adiabatic, the work done
can be calculated from Equation 15.4:
.
Since the gas does the same amount of work whether it expands adiabatically or
isothermally, we can equate the right hand sides of these two equations. We also note that
since the initial temperature is the same for both cases, the temperature
T
in the isothermal
expansion is the same as the initial temperature
for the adiabatic expansion. We then have
or
SOLUTION
Solving for the ratio of the volumes gives
______________________________________________________________________________
24.
REASONING
a. The work done by the gas is equal to the area under the pressureversusvolume curve. We
will measure this area by using the graph given with the problem.
THERMODYNAMICS
Volume, m
3
2.00
4.00
6.00
8.00
0
10.0
12.0
2.00
4.00
6.00
0
Pressure (
×
10
5
Pa)
A
B
b. Since the gas is an ideal gas, it obeys the ideal gas law,
PV
=
nRT
(Equation 14.1). This
implies that
. All the variables except for
T
B
in this relation are
known. Therefore, we can use this expression to find the temperature at point B.
c. The heat
Q
that has been added to or removed from the gas can be obtained from the first
law of thermodynamics,
Q
=
Δ
U + W
(Equation 15.1), where
Δ
U
is the change in the
internal energy of the gas and
W
is the work done by the gas. The work
W
is known from
part (a) of the problem. The change
Δ
U
in the internal energy of the gas can be obtained
from Equation 14.7,
where
n
is the number of moles,
R
is
the universal gas constant, and
T
B
and
T
A
are the final and initial Kelvin temperatures. We
do not know
n,
but we can use the ideal gas law (
PV
=
nRT
) to replace
nRT
B
by
P
B
V
B
and
to replace
nRT
A
by
P
A
V
A
.
SOLUTION
a. From the drawing we see that the area under the curve is 5.00 “squares,” where each
square has an area of
Therefore, the work
W
done
by the gas is
b. In the Reasoning section, we have seen that
. Solving this relation
for the temperature
T
B
at point B, using the fact that
P
A
=
P
B
(see the graph), and taking the
values for
V
B
and
V
A
from the graph, we have that
Chapter 15 Problems
c. From the Reasoning section we know that the heat
Q
that has been added to or removed
from the gas is given by
Q
=
Δ
U + W
. The change
Δ
U
in the internal energy of the gas is
. Thus, the heat can be expressed as
We now use the ideal gas law (
PV
=
nRT
) to replace
nRT
B
by
P
B
V
B
and
nRT
A
by
P
A
V
A
.
The result is
Taking the values for
P
B
,
V
B
,
P
A
, and
V
A
from the graph and using the result from part a that
W
= 2.00
×
10
6
J, we find that the heat is
______________________________________________________________________________
25.
REASONING
AND
SOLUTION
a. Since the curved line between A and C is an isotherm, the initial and ﬁnal temperatures
are the same. Since the internal energy of an ideal monatomic gas is
U
= (3/2)
nRT
, the
initial and ﬁnal energies are also the same, and the change in the internal energy is
Δ
U
= 0.
The ﬁrst law of thermodynamics, then, indicates that for the process A
→
B
→
C, we have
The heat is equal to the work. Determining the work from the area beneath the straight line
segments AB and BC, we ﬁnd that
b. The minus sign is included because the gas is compressed, so that work is done on the
gas. Since the answer for
Q
is negative, we conclude that heat ﬂows out of the gas .
______________________________________________________________________________
26.
REASONING
AND
SOLUTION
Step A
→
B
The internal energy of a monatomic ideal gas is
U
= (3/2)
nRT
. Thus, the change is
THERMODYNAMICS
The work for this constant pressure step is
W
=
P
Δ
V
. But the ideal gas law applies, so
The ﬁrst law of thermodynamics indicates that the heat is
Step B
→
C
The internal energy of a monatomic ideal gas is
U
= (3/2)
nRT
. Thus, the change is
The volume is constant in this step, so the work done by the gas is
.
The ﬁrst law of thermodynamics indicates that the heat is
Step C
→
D
The internal energy of a monatomic ideal gas is
U
= (3/2)
nRT
. Thus, the change is
The work for this constant pressure step is
W
=
P
Δ
V
. But the ideal gas law applies, so
The ﬁrst law of thermodynamics indicates that the heat is
Step D
→
A
The internal energy of a monatomic ideal gas is
U
= (3/2)
nRT
. Thus, the change is
Chapter 15 Problems
The volume is constant in this step, so the work done by the gas is
The ﬁrst law of thermodynamics indicates that the heat is
______________________________________________________________________________
27.
SSM
REASONING
According to Equation 15.5,
. The ideal gas law states
that
for both the initial and ﬁnal conditions. Thus, we have
Since the ratio of the temperatures is known, the last expression can be solved for the ﬁnal
pressure
.
SOLUTION
Since
, and
, we ﬁnd that
______________________________________________________________________________
28.
REASONING
AND
SOLUTION
The
threestep process is shown on the
P

V
diagram at the right. From the ﬁrst law of
thermodynamics,
Q
=
Δ
U
+
W
(1)
However, the ideal gas is back in its initial
state at the end of the threestep process, so
that
Δ
U
= 0 overall. With this value for
Δ
U
, Equation (1) becomes
Q
=
W
, and we
conclude that
Using Equations 15.3 for the isothermal work and Equation 15.2 for the isobaric work, and
remembering that there is no work done in an isochoric process, we ﬁnd that
THERMODYNAMICS
In this result,
T
= 438 K,
P
is the pressure for step
, and 2
PV
0
= 2(
P
3
V
0
)/3 = 2
nRT
/3.
In addition, we know that
n
= 1 mol. Therefore,
Since this answer is positive,
by the gas.
______________________________________________________________________________
29.
SSM
REASONING AND SOLUTION
a. The ﬁnal temperature of the adiabatic process is given by solving Equation 15.4 for
T
f
.
b. According to Equation 15.5 for the adiabatic expansion of an ideal gas,
.
Therefore,
From the ideal gas law,
; therefore, the ratio of the pressures is given by
Combining the previous two equations gives
Solving for
V
f
we obtain
Chapter 15 Problems
Therefore,
______________________________________________________________________________
30.
REASONING
When the temperature of a gas changes as a result of heat
Q
being added, the
change
Δ
T
in temperature is related to the amount of heat according to
where
C
is the molar specific heat capacity, and
n
is the
number of moles. The heat
Q
V
added under conditions of constant volume is
where
C
V
is the specific heat capacity at constant volume and is given by
and
R
is the universal gas constant. The heat
Q
P
added under
conditions of constant pressure is
where
C
P
is the specific heat capacity at
constant pressure and is given by
It is given that
Q
V
=
Q
P
, and
this fact will allow us to find the change in temperature of the gas whose pressure remains
constant.
SOLUTION
Setting
Q
V
=
Q
P
, gives
Algebraically eliminating
n
and solving for
Δ
T
P,
we obtain
______________________________________________________________________________
31.
SSM
REASONING AND SOLUTION
According to the ﬁrst law of thermodynamics
(Equation 15.1),
.
Since the internal energy of this gas is doubled by
the addition of heat, the initial and ﬁnal internal energies are
U
and 2
U,
respectively.
Therefore,
THERMODYNAMICS
Equation 15.1 for this situation then becomes
. Solving for
Q
gives
(1)
The initial internal energy of the gas can be calculated from Equation 14.7:
a. If the process is carried out isochorically (i.e., at constant volume), then W = 0, and the
heat required to double the internal energy is
b. If the process is carried out isobarically (i.e., at constant pressure), then
, and
Equation (1) above becomes
(2)
From the ideal gas law,
, we have that
, and Equation (2) becomes
(3)
The internal energy of an ideal gas is directly proportional to its Kelvin temperature. Since
the internal energy of the gas is doubled, the ﬁnal Kelvin temperature will be twice the
initial Kelvin temperature, or
Δ
T
= 350 K. Substituting values into Equation (3) gives
______________________________________________________________________________
32.
REASONING AND SOLUTION
a. The amount of heat needed to raise the temperature of the gas at constant volume is given
by Equations 15.6 and 15.8,
Q
=
n
C
V
Δ
T
. Solving for
Δ
T
yields
b. The change in the internal energy of the gas is given by the ﬁrst law of thermodynamics
with
W
= 0, since the gas is heated at constant volume:
Chapter 15 Problems
c. The change in pressure can be obtained from the ideal gas law,
______________________________________________________________________________
33.
REASONING
AND
SOLUTION
The heat added at constant pressure is
Q
=
C
p
n
Δ
T
= (5
R
/2)
n
Δ
T
The work done during the process is
W
=
P
Δ
V
. The ideal gas law requires that
Δ
V
=
nR
Δ
T
/
P
, so
W
=
nR
Δ
T
. The required ratio is then
Q
/
W
=
______________________________________________________________________________
34.
REASONING
The heat
Q
P
that needs to be added under constantpressure conditions is
given by
, according to Equation 15.6, where
C
P
is the molar specific heat
capacity at constant pressure,
n
is the number of moles of the gas, and
Δ
T
is the change in
the temperature. We have values for
n
and
Δ
T
, but not for
C
P
. For an ideal gas of any type,
however, Equation 15.10 applies, so that
, where
C
V
is the molar specific heat
capacity at constant volume and
R
= 8.31 J/(mol
⋅
K) is the universal gas constant. This result
is useful, because we can evaluate
C
V
from the heat added under constantvolume
conditions.
SOLUTION
According to Equation 15.6, we know that
. Substituting
(Equation 15.10) into this expression, we obtain
(1)
Using Equation 15.6 again, this time for constantvolume conditions, we have
,
where
Q
V
is the heat added. Substituting this result into Equation (1) gives
THERMODYNAMICS
35.
SSM
REASONING AND SOLUTION
The amount of heat required to change the
temperature of the gas is given by Equation 15.6, where
C
P
is given by Equation 15.7.
____________________________________________________________________________________________
36.
REASONING
AND
SOLUTION
The total heat generated by the students is
Q
= (200)(130 W)(3000 s) = 7.8
×
10
7
J
For the isochoric process.
Q
=
C
v
n
Δ
T = (5
R
/2)
n
Δ
T
The number of moles of air in the room is found from the ideal gas law to be
Now
______________________________________________________________________________
37.
REASONING
The power rating
of the heater is equal to the heat
Q
supplied to the gas
divided by the time
t
the heater is on,
(Equation 6.10b). Therefore,
. The
heat required to change the temperature of a gas under conditions of constant pressure is
given by
where
C
P
is the molar speciﬁc heat capacity at
constant pressure,
n
is the number of moles, and
Δ
T
=
T
f
−
T
i
is the change in temperature.
For a monatomic ideal gas, the speciﬁc heat capacity at constant pressure is
Equation (15.7), where
R
is the universal gas constant. We do not know
n,
T
f
and
T
i
, but we
can use the ideal gas law,
PV
=
nRT
, (Equation 14.1) to replace
nRT
f
by
P
f
V
f
and to replace
nRT
i
by
P
i
V
i
, quantities that we do know.
SOLUTION
Substituting
into
and using the fact that
give
Chapter 15 Problems
Replacing
RnT
f
by
P
f
V
f
and
RnT
i
by
P
i
V
i
and remembering that
P
i
=
P
f
, we ﬁnd
Since the volume of the gas increases by 25.0%,
V
f
= 1.250
V
i
. The time that the heater is on
is
______________________________________________________________________________
38.
REASONING
AND
SOLUTION
The change in volume is
Δ
V
= –
sA
, where
s
is the
distance through which the piston drops and
A
is the piston area. The minus sign is included
because the volume decreases. Thus,
The ideal gas law states that
Δ
V
=
nR
Δ
T
/
P
. But
. Thus,
. Using these expressions for
Δ
V
and
Δ
T
, we ﬁnd that
______________________________________________________________________________
39.
REASONING AND SOLUTION
Let
P
,
V
, and
T
represent the initial values of pressure,
volume, and temperature. The ﬁrst process is isochoric, so
Q
1
=
C
V
n
Δ
T
1
= (3
R
/2)
n
Δ
T
1
The ideal gas law for this process gives
Δ
T
1
= 2
PV
/(
nR
), so
Q
1
= 3
PV
.
The second process is isobaric, so
Q
2
=
C
P
n
Δ
T
2
= (5
R
/2)
n
Δ
T
2
THERMODYNAMICS
The ideal gas law for this process gives
Δ
T
2
= 3
PV
/
nR
, so
Q
2
= (15/2)
PV
. The total heat is
Q
=
Q
1
+
Q
2
= (21/2)
PV
.
But at conditions of standard temperature and pressure (see Section 14.2),
P
= 1.01
×
10
5
Pa
and
V
= 22.4 liters = 22.4
×
10
–3
m
3
, so
______________________________________________________________________________
40.
REASONING
According to Equation 15.11, the efﬁciency of a heat engine is
,
where
is the magnitude of the work and
is the magnitude of the input heat. Thus,
the magnitude of the work is
. We can apply this result before and after the tune
up to compute the extra work produced.
SOLUTION
Using Equation 15.11, we ﬁnd the work before and after the tuneup as
follows:
Subtracting the “before” equation from the “after” equation gives
______________________________________________________________________________
41.
SSM
REASONING AND SOLUTION
The efﬁciency of a heat engine is deﬁned by
Equation 15.11 as
,
where
is the magnitude of the work done and
is
the magnitude of the heat input. The principle of energy conservation requires that
, where
is the magnitude of the heat rejected to the cold reservoir
(Equation 15.12). Combining Equations 15.11 and 15.12 gives
______________________________________________________________________________
42.
REASONING
According to the definition of efficiency given in Equation 15.11, an engine
with an efficiency
e
does work of magnitude
, where
is the magnitude of
the input heat. We will apply this expression to each engine and utilize the fact that in each
Chapter 15 Problems
case the same work is done. We expect to find that engine 2 requires less input heat to do
the same amount of work, because it has the greater efficiency.
SOLUTION
Applying Equation 15.11 to each engine gives
Since the work is the same for each engine, we have
It follows, then, that
which is less than the input heat for engine 1, as expected.
43.
REASONING AND SOLUTION
a. The efﬁciency is
, where
is the magnitude of the work and
is the
magnitude of the input heat. It follows that
b. The magnitude of the rejected heat is
______________________________________________________________________________
44.
REASONING
The efficiency
e
of an engine can be expressed as (see Equation 15.13)
, where
is the magnitude of the heat delivered to the cold reservoir
and
is the magnitude of the heat supplied to the engine from the hot reservoir. Solving
this equation for
gives
. We will use this expression twice, once for
the improved engine and once for the original engine. Taking the ratio of these expressions
will give us the answer that we seek.
THERMODYNAMICS
SOLUTION
Taking the ratio of the heat rejected to the cold reservoir by the improved
engine to that for the original engine gives
But the input heat to both engines is the same, so
. Thus, the ratio
becomes
______________________________________________________________________________
45.
SSM
WWW
REASONING
The efﬁciency of either engine is given by Equation 15.13,
. Since engine A receives three times more input heat, produces ﬁve
times more work, and rejects two times more heat than engine B, it follows that
,
, and
. As required by the principle of energy
conservation for engine A (Equation 15.12),
Thus,
(1)
Since engine B also obeys the principle of energy conservation (Equation 15.12),
(2)
Substituting
from Equation (2) into Equation (1) yields
Solving for
gives
Chapter 15 Problems
Therefore, Equation (2) predicts for engine B that
SOLUTION
a. Substituting
and
into Equation 15.13 for engine A, we
have
b. Substituting
into Equation 15.13 for engine B, we have
______________________________________________________________________________
46.
REASONING
AND
SOLUTION
The efﬁciency is given by
e
= 1
−
(
T
C
/
T
H
) = 1
−
[(200 K)/(500 K)] = 0.6 =
The work is
= (0.6)(5000 J) =
______________________________________________________________________________
47.
REASONING
We will use the subscript “27” to denote the engine whose efficiency is
27.0% (
e
27
= 0.270) and the subscript “32” to denote the engine whose efficiency is 32.0%
(
e
32
= 0.320). In general, the efficiency
e
Carnot
of a Carnot engine depends on the Kelvin
temperatures,
T
C
and
T
H
, of its cold and hot reservoirs through the relation (see
Equation 15.15)
e
Carnot
= 1
−
(
T
C
/
T
H
). Solving this equation for the temperature
T
C, 32
of the
engine whose efficiency is
e
32
gives
T
C, 32
= (1
−
e
32
)
T
H, 32
. We are given
e
32
, but do not
know the temperature
T
H, 32
. However, we are told that this temperature is the same as that
of the hot reservoir of the engine whose efficiency is
e
27
, so
T
H, 32
=
T
H, 27
. The temperature
T
H, 27
can be determined since we know the efficiency and cold reservoir temperature of this
engine.
THERMODYNAMICS
SOLUTION
The temperature of the cold reservoir for engine whose efficiency is
e
32
is
T
C, 32
= (1
−
e
32
)
T
H, 32
. Since
T
H, 32
=
T
H, 27
, we have that
T
C, 32
= (1
−
e
32
)
T
H, 27
(1)
The efficiency
e
27
is given by Equation 15.15 as
e
27
= 1
−
(
T
C, 27
/
T
H, 27
). Solving this
equation for the temperature
T
H, 27
of the hot reservoir and substituting the result into
Equation 1 yields
______________________________________________________________________________
48.
REASONING
The efficiency
e
Carnot
of a Carnot engine is
(Equation 15.15), where
T
C
and
T
H
are, respectively, the Kelvin temperatures of the cold and
hot reservoirs. After the changes are made to the temperatures, this same equation still
applies, except that the variables must be labeled to denote the new values. We will use a
“prime” for this purpose. From the original efficiency and the information given about the
changes made to the temperatures, we will be able to obtain the new temperature ratio and,
hence, the new efficiency.
SOLUTION
After the reservoir temperatures are changed, the engine has an efficiency that,
according to Equation 15.15, is
where the “prime” denotes the new engine. Using unprimed symbols to denote the original
engine, we know that
and
. With these substitutions, the efficiency of
the new engine becomes
(1)
To obtain the original ratio
, we use Equation 15.15:
Chapter 15 Problems
Substituting this original temperature ratio into Equation (1) gives
49.
SSM
REASONING
The efﬁciency
e
of a Carnot engine is given by Equation 15.15,
, where, according to Equation 15.14,
. Since the
efﬁciency is given along with
and
, Equation 15.15 can be used to calculate
.
Once
is known, the ratio
is thus known, and Equation 15.14 can be used to
calculate
.
SOLUTION
a. Solving Equation 15.15 for
gives
b. Solving Equation 15.14 for
gives
______________________________________________________________________________
50.
REASONING
AND
SOLUTION
In order to ﬁnd out how many kilograms of ice in the tub
are melted, we must determine
, the amount of exhaust heat delivered to the cold
reservoir. Since the hot reservoir consists of boiling water (
T
H
= 373.0 K) and the cold
reservoir consists of ice and water (
T
C
= 273.0 K), the efﬁciency of this engine is
The work done by the engine is
THERMODYNAMICS
= (0.2681)(6800 J) = 1823 J
Therefore, the amount of heat delivered to the cold reservoir is
= 6800 J – 1823 J = 4977 J
Using the deﬁnition of the latent heat of fusion (melting)
L
f
, we ﬁnd that the amount of ice
that melts is
______________________________________________________________________________
51.
REASONING
AND
SOLUTION
The efﬁciency of the engine is
e
= 1
−
(
T
C
/
T
H
) so
(i) Increase
T
H
by 40 K;
e
= 1
−
[(350 K)/(690 K)] = 0.493
(ii) Decrease
T
C
by 40 K;
e
= 1
−
[(310 K)/(650 K)] = 0.523
The greatest improvement is made by
the temperature of the cold reservoir.
______________________________________________________________________________
52.
REASONING
AND
SOLUTION
The amount of work delivered by the engines can be
determined from Equation 15.12,
. Solving for
for each engine gives:
The total work delivered by the two engines is
But we know that
, so that
(1)
Since these are Carnot engines,
Similarly, noting that
and that
T
H2
=
T
C1
, we have
Chapter 15 Problems
Substituting into Equation (1) gives
______________________________________________________________________________
53.
SSM
REASONING
The maximum efﬁciency
e
at which the power plant can operate is
given by Equation 15.15,
. The power output is given; it can be used to
ﬁnd the magnitude
of the work output for a 24 hour period. With the efﬁciency and
known, Equation 15.11,
, can be used to ﬁnd
, the magnitude of the input
heat. The magnitude
of the exhaust heat can then be found from Equation 15.12,
.
SOLUTION
a. The maximum efﬁciency is
b. Since the power output of the power plant is
P
= 84 000 kW, the required heat input
for a 24 hour period is
Therefore, solving Equation 15.12 for
, we have
______________________________________________________________________________
54.
REASONING
AND
SOLUTION
The temperature of the gasoline engine input is
T
1
= 904 K, the exhaust temperature is
T
2
= 412 K, and the air temperature is
T
3
= 300 K.
The efﬁciency of the engine/exhaust is
e
1
= 1
−
(
T
2
/
T
1
) = 0.544
The efﬁciency of the second engine is
e
2
= 1
−
(
T
3
/
T
2
) = 0.272
THERMODYNAMICS
The magnitude of the work done by each segment is
and
since
Now examine
to ﬁnd the ratio of the total work produced by both engines
to that produced by the ﬁrst engine alone.
But,
, so that
. Therefore,
______________________________________________________________________________
55.
SSM
REASONING AND SOLUTION
The efﬁciency
e
of the power plant is three
fourths its Carnot efﬁciency so, according to Equation 15.15,
The power output of the plant is
watts. According to Equation 15.11,
. Therefore, at 33% efﬁciency, the magnitude of the heat
input per unit time is
From the principle of conservation of energy, the heat output per unit time must be
The rejected heat is carried away by the ﬂowing water and, according to Equation 12.4,
. Therefore,
Chapter 15 Problems
Solving the last equation for
Δ
t
, we have
______________________________________________________________________________
56.
REASONING
The expansion from point a to
point b and the compression from point c to point d
occur isothermally, and we will apply the first law
of thermodynamics to these parts of the cycle in
order to obtain expressions for the input and
rejected heats, magnitudes
and
,
respectively. In order to simplify the resulting
expression for
, we will then use the fact
that the expansion from point b to point c and the
compression from point d to point a are adiabatic.
SOLUTION
According to the first law of thermodynamics, the change in internal energy
Δ
U
is given by
(Equation 15.1), where
Q
is the heat and
W
is the work. Since
the internal energy of an ideal gas is proportional to the Kelvin temperature and the
temperature is constant for an isothermal process, it follows that
for such a case.
The work of isothermal expansion or compression for an ideal gas is
(Equation 15.3), where
n
is the number of moles,
R
is the universal gas constant,
T
is the
Kelvin temperature,
V
f
is the final volume of the gas, and
V
i
is the initial volume. We have,
then, that
Applying this result for
Q
to the isothermal expansion (temperature =
T
H
) from point a to
point b and the isothermal compression (temperature =
T
C
) from point c to point d, we have
THERMODYNAMICS
where
and
for the isotherm at
T
H
and
and
for the isotherm
at
T
C
. In this problem, we are interested in the magnitude of the heats. For
Q
H
, this poses
no problem, since
,
is positive, and we have
(1)
However, for
Q
C
, we need to be careful, because
and
is negative. Thus,
we write for the magnitude of
Q
C
that
(2)
According to Equations (1) and (2), the ratio of the magnitudes of the rejected and input
heats is
(3)
We now consider the adiabatic parts of the Carnot cycle. For the adiabatic expansion or
compression of an ideal gas the initial pressure and volume (
P
i
and
V
i
) are related to the
final pressure and volume (
P
f
and
V
f
) according to
(15.5)
where
γ
is the ratio of the specific heats at constant pressure and constant volume. It is also
true that
(Equation 14.1), according to the ideal gas law. Substituting this
expression for the pressure into Equation 15.5 gives
Chapter 15 Problems
Applying this result to the adiabatic expansion from point b to point c and to the adiabatic
compression from point d to point a, we obtain
Dividing the first of these equations by the second shows that
With this result, Equation (3) becomes
57.
REASONING
The coefficient of performance COP is defined as
(Equation 15.16), where
is the magnitude of the heat removed from the cold reservoir
and
is the magnitude of the work done on the refrigerator. The work is related to the
magnitude
of the heat deposited into the hot reservoir and
by the conservation of
energy,
. Thus, the coefficient of performance can be written as (after some
algebraic manipulations)
The maximum coefficient of performance occurs when the refrigerator is a Carnot
refrigerator. For a Carnot refrigerator, the ratio
is equal to the ratio
of the
Kelvin temperatures of the hot and cold reservoirs,
(Equation 15.14).
SOLUTION
Substituting
into the expression above for the COP gives
THERMODYNAMICS
______________________________________________________________________________
58.
REASONING AND SOLUTION
We know that
Therefore,
______________________________________________________________________________
59.
SSM
WWW
REASONING AND SOLUTION
Equation 15.14 holds for a Carnot air
conditioner as well as a Carnot engine. Therefore, solving Equation 15.14 for
, we
have
______________________________________________________________________________
60.
REASONING
Since the refrigerator is a Carnot device, we know that
(Equation 15.14). We have values for
T
H
(the temperature of the kitchen) and
(the
magnitude of the heat removed from the food). Thus, we can use this expression to
determine
T
C
(the temperature inside the refrigerator), provided that a value can be obtained
for
(the magnitude of the heat that the refrigerator deposits into the kitchen). Energy
conservation dictates that
(Equation 15.12), where
is the magnitude of
the work that the appliance uses and is known.
SOLUTION
From Equation 15.14, it follows that
Substituting
(Equation 15.12) into this result for
T
C
gives
Chapter 15 Problems
61.
REASONING
AND
SOLUTION
a. The work is
b. The coefﬁcient of performance (COP) is
______________________________________________________________________________
62.
REASONING
The coefﬁcient of performance of an air conditioner is
, according
to Equation 15.16, where
is the magnitude of the heat removed from the house and
is the magnitude of the work required for the removal. In addition, we know that the ﬁrst
law of thermodynamics (energy conservation) applies, so that
, according to
Equation 15.12. In this equation
is the magnitude of the heat discarded outside. While
we have no direct information about
and
, we do know that the air conditioner is a
Carnot device. This means that Equation 15.14 applies:
. Thus, the
given temperatures will allow us to calculate the coefﬁcient of performance.
SOLUTION
Using Equation 15.16 for the deﬁnition of the coefﬁcient of performance and
Equation 15.12 for the fact that
, we have
Equation 15.14 applies, so that
. With this substitution, we ﬁnd
THERMODYNAMICS
______________________________________________________________________________
63.
REASONING
The conservation of energy applies to the air conditioner, so that
, where
is the amount of heat put into the room by the unit,
is the
amount of heat removed from the room by the unit, and
is the amount of work needed
to operate the unit. Therefore, a net heat of
is added to and heats up the
room. To ﬁnd the temperature rise of the room, we will use the COP to determine
and
then use the given molar speciﬁc heat capacity.
SOLUTION
Let COP denote the coefﬁcient of performance. By deﬁnition (Equation
15.16),
, so that
The temperature rise in the room can be found as follows:
.
Solving for
Δ
T
gives
______________________________________________________________________________
64.
REASONING
AND
SOLUTION
The amount of heat removed when the ice freezes is
= [4186 J/(kg
⋅
C°)](1.50 kg)(20.0 C
) + (1.50 kg)(33.5
×
10
4
J/kg) = 6.28
×
10
5
J
Since the coefﬁcient of performance is
, the magnitude of the work done by
the refrigerator is
The magnitude of the heat delivered to the kitchen is
The space heater has a power output
P
of
Chapter 15 Problems
Therefore,
______________________________________________________________________________
65.
SSM
WWW
REASONING
Let the coefﬁcient of performance be represented by the
symbol COP. Then according to Equation 15.16,
. From the statement of
energy conservation for a Carnot refrigerator (Equation 15.12),
. Combining
Equations 15.16 and 15.12 leads to
Replacing the ratio of the heats with the ratio of the Kelvin temperatures, according to
Equation 15.14, leads to
(1)
The heat
that must be removed from the refrigerator when the water is cooled can be
calculated using Equation 12.4,
; therefore,
(2)
SOLUTION
a. Substituting values into Equation (1) gives
b. Substituting values into Equation (2) gives
______________________________________________________________________________
THERMODYNAMICS
66.
REASONING
According to the conservation of energy, the work
W
done by the electrical
energy is
, where
is the magnitude of the heat delivered to the outside
(the hot reservoir) and
is the magnitude of the heat removed from the house (the cold
reservoir). Dividing both sides of this relation by the time
t
, we have
The term
is the magnitude of the work per second that must be done by the electrical
energy, and the terms
and
are, respectively, the magnitude of the heat per
second delivered to the outside and removed from the house. Since the air conditioner is a
Carnot air conditioner, we know that
is equal to the ratio
of the Kelvin
temperatures of the hot and cold reservoirs,
(Equation 15.14). This
expression, along with the one above for
, will allow us to determine the magnitude of
the work per second done by the electrical energy.
SOLUTION
Solving the expression
for
, substituting the result
into the relation
, and recognizing that
= 10 500 J/s, give
In this result we have used the fact that
T
H
= 273.15 + 33.0
C = 306.15 K and
T
C
= 273.15 + 19.0
C = 292.15 K.
______________________________________________________________________________
67.
SSM
REASONING
The efﬁciency of the Carnot engine is, according to Equation 15.15,
Chapter 15 Problems
Therefore, the magnitude of the work delivered by the engine is, according to Equation
15.11,
The heat pump removes an amount of heat
from the cold reservoir. Thus, the amount
of heat
delivered to the hot reservoir of the heat pump is
Therefore,
. According to Equation 15.14,
, so
.
SOLUTION
Solving for
gives
______________________________________________________________________________
68.
REASONING
According to the discussion on Section 15.11, the change
Δ
S
universe
in the
entropy of the universe is the sum of the change in entropy
Δ
S
C
of the cold reservoir and the
change in entropy
Δ
S
H
of the hot reservoir, or
Δ
S
universe
=
Δ
S
C
+
Δ
S
H
. The change in entropy
of each reservoir is given by Equation 15.18 as
Δ
S
= (
Q
/
T
)
R
, where
Q
is the heat removed
from or delivered to the reservoir and
T
is the Kelvin temperature of the reservoir. In
applying this equation we imagine a process in which the heat is lost by the house and
gained by the outside in a reversible fashion.
SOLUTION
Since heat is lost from the hot reservoir (inside the house), the change in
entropy is negative:
Δ
S
H
=
−
Q
H
/
T
H
. Since heat is gained by the cold reservoir (the
outdoors), the change in entropy is positive:
Δ
S
C
= +
Q
C
/
T
C
. Here we are using the symbols
Q
H
and
Q
C
to denote the magnitudes of the heats. The change in the entropy of the universe
is
THERMODYNAMICS
In this calculation we have used the fact that
T
C
= 273
−
15
C = 258 K and
T
H
= 273 + 21
C = 294 K.
______________________________________________________________________________
69.
REASONING AND SOLUTION
Equation 15.19 gives the unavailable work as
(1)
where
. We also know that
. Furthermore, we can apply
Equation 15.18 to the heat lost from the 394K reservoir and the heat gained by the reservoir
at temperature
T
, with the result that
With these substitutions for
T
0
,
W
unavailable
, and
Δ
S
, Equation (1) becomes
______________________________________________________________________________
70.
REASONING
The total entropy change
of the universe is the sum of the entropy
changes of the hot and cold reservoir. For each reservoir, the entropy change is given by
(Equation 15.18). As indicated by the label “R,” this equation applies only to
reversible processes. For the two irreversible engines, therefore, we apply this equation to
an imaginary process that removes the given heat from the hot reservoir reversibly and
rejects the given heat to the cold reservoir reversibly. According to the second law of
thermodynamics stated in terms of entropy (see Section 15.11), the reversible engine is the
one for which
, and the irreversible engine that could exist is the one for
which
. The irreversible engine that could not exist is the one for which
.
SOLUTION
Using Equation 15.18, we write the total entropy change of the universe as the
sum of the entropy changes of the hot (H) and cold (C) reservoirs.
Chapter 15 Problems
In this expression, we have used
for the heat from the hot reservoir because that
reservoir loses heat. We have used
for the heat rejected to the cold reservoir because
that reservoir gains heat. Applying this expression to the three engines gives the following
results:
Since
for Engine II, it is reversible. Since
for Engine I,
it is irreversible and could exist. Since
for Engine III, it is irreversible
and could not exist.
71.
SSM
REASONING AND SOLUTION
a. Since the energy that becomes unavailable for doing work is zero for the process, we
have from Equation 15.19,
. Therefore,
and
according to the discussion in Section 15.11, the process is
.
b. Since the process is reversible, we have (see Section 15.11)
Therefore,
______________________________________________________________________________
72.
REASONING
AND
SOLUTION
a. We know that the hot and cold waters exchange equal amounts of heat, i.e.,
Δ
Q
hw
=
Δ
Q
cw
, so that (
mc
Δ
T
)
hw
= (
mc
Δ
T
)
cw
, or
(1.00 kg)[4186 J/(kg
⋅
C°)](373 K
−
T
f
) = (2.00 kg)[4186 J/(kg
⋅
C°)](
T
f
−
283 K)
THERMODYNAMICS
Solving for
T
f
, we obtain
.
b. Since
Δ
S
=
mc
ln(
T
f
/
T
i
):
Δ
S
hw
=
m
hw
c
ln[(313 K)/(373 K)] =
−
734 J/K
Δ
S
cw
=
m
cw
c
ln[(313 K)/(283 K)] = +844 J/K
Therefore,
Δ
S
universe
=
Δ
S
hw
+
Δ
S
cw
=
c. The energy unavailable for doing work is, therefore,
W
unavailable
=
T
0
Δ
S
universe
= (273 K)(1.10
×
10
2
J/K) =
______________________________________________________________________________
73.
REASONING
The change
Δ
S
universe
in entropy of the universe for this process is the sum
of the entropy changes for (1) the warm water (
Δ
S
water
) as it cools down from its initial
temperature of 85.0
C to its final temperature
T
f
,
(2) the ice (
Δ
S
ice
) as it melts at 0
C, and
(3) the ice water (
Δ
S
ice water
) as it warms up from 0
C to the final temperature
T
f
:
Δ
S
universe
=
Δ
S
water
+
Δ
S
ice
+
Δ
S
ice water
.
To find the final temperature
T
f
, we will follow the procedure outlined in Sections 12.7 and
12.8, where we set the heat lost by the warm water as it cools down equal to the heat gained
by the melting ice and the resulting ice water as it warms up. The heat
Q
that must be
supplied or removed to change the temperature of a substance of mass
m
by an amount
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