Ch19 ISM

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1867 1867

The Second Law of Thermodynamics

Conceptual Problems

1

1 •• Modern automobile gasoline engines have efficiencies of about 25%.Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but

About what percentage of the heat of combustion is not used for work but released as heat? (

released as heat? (aa) 25%, () 25%, (bb) 50%, () 50%, (cc) 75%, () 75%, (d d ) 100%, () 100%, (ee) You cannot tell from) You cannot tell from the data given.

the data given.

Determine the Concept

Determine the ConceptThe efficiency of a heat The efficiency of a heat engine is the ratio of the work engine is the ratio of the work done per cycle

done per cycleW W to the heat absorbed from the high-temperature reservoir to the heat absorbed from the high-temperature reservoir QQhh..

The percentage of the heat of combustion (heat absorbed from the The percentage of the heat of combustion (heat absorbed from the high-temperature reservoir) is the ratio of

temperature reservoir) is the ratio of QQcctoto QQhh. We can use the relationship. We can use the relationship

between

betweenW, QW, Qhh, and, andQQcc((W W

==

QQ_h_h

−−

QQ_cc) to find) to findQQcc//QQhh..

Use the definition of efficiency and Use the definition of efficiency and the relationship between

the relationship between W, QW, Qhh, and, and Q Qccto obtain:to obtain: hh cc h h cc h h h h 1 1 Q Q Q Q Q Q Q Q Q Q Q Q W W

−−

==

−−

==

ε ε Solving for

Solving for QQcc// QQhh yields:yields:

ε ε

−−

==

11 h h cc Q Q Q Q Substitute for

Substitute for ε ε to obtain:to obtain: ₁₁ ₀₀_..₂₅₂₅ ₀₀_..₇₅₇₅

h h cc

==

−−

==

Q Q Q Q and

and

( ( ))

cc is correct.is correct.

2

2 •• If a heat engine does 100 kJ of work per cycle while releasing 400 kJIf a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? (

of heat, what is its efficiency? (aa) 20%, () 20%, (bb) 25%, () 25%, (cc) 80%, () 80%, (d d ) 400%, () 400%, (ee) You) You cannot tell from the data given.

cannot tell from the data given.

done per cycleW W to the heat absorbed from the high-temperature reservoir to the heat absorbed from the high-temperature reservoir QQhh. We. We

can use the relationship between

can use the relationship betweenW, QW, Qhh, and, and QQcc((W W

==

QQ_h_h

−−

QQ_cc) to express the) to express the

efficiency of the heat engine

efficiency of the heat engine in terms of in terms of QQccandandW W ..

the relationship between W, QW, Qhh, and, and Q Qccto obtain:to obtain: _W_W Q Q Q Q W W W W Q Q W W cc cc h h ₁₁ 1 1

++

==

++

==

ε ε

(2)

Substitute for

Substitute for QQccandand W W to obtain:to obtain:

2 2 .. 0 0 kJ kJ 100 100 kJ kJ 400 400 1 1 1 1

==

++

==

and

( ( ))

aa is correct.is correct.

3

3 •• If the heat absorbed by a heat engine is 600 kJ per cycle, and itIf the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (

releases 480 kJ of heat each cycle, what is its efficiency? (aa) 20%, () 20%, (bb) 80%,) 80%, ((cc) 100%, () 100%, (d d ) You cannot tell from the data given.) You cannot tell from the data given.

==

QQ_h_h

−−

efficiency of the heat engine in terms of

efficiency of the heat engine in terms of QQccandandQQhh..

−−

==

−−

==

ε ε Substitute for

Substitute for QQccandand QQhhto obtain:to obtain:

2 2 .. 0 0 kJ kJ 600 600 kJ kJ 480 480 1 1

−−

==

and

( ( ))

4

4 •• Explain what distinguishes a refrigerator from aExplain what distinguishes a refrigerator from a ″″heat pump.heat pump.″″

Determine the Concept The job of a refrigerator is to move heat from its coldThe job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its

direction that is opposite its

″″

naturalnatural

″″

direction of flow, analogous to the use of adirection of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter.

in the summer and warm them in the winter.

5

5 •• [SSM][SSM] An air conditioner’s COP is mathematically identical to thatAn air conditioner’s COP is mathematically identical to that of a refrigerator, that is,

of a refrigerator, that is, cc A

ACC rreeff

C

COOPP

=

CCOOPP

==

QQ

W

W . However a heat pump’s COP is. However a heat pump’s COP is

defined differently, as defined differently, as hh hp hp COP COP

==

QQ W

W . Explain clearly. Explain clearlywhywhythe two COPs arethe two COPs are

defined differently.

defined differently. Hint Hint :: Think of the end use of the three different devices.Think of the end use of the three different devices. Determine the Concept

Determine the ConceptThe COP is defined so as to be a measure of theThe COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior,

quantity is the heat drawn from the already colder interior, QQcc. For a heat pump,. For a heat pump,

the ideas is to focus on

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Substitute for

Substitute for QQccandand W W to obtain:to obtain:

2 2 .. 0 0 kJ kJ 100 100 kJ kJ 400 400 1 1 1 1

==

++

==

and

( ( ))

3

3 •• If the heat absorbed by a heat engine is 600 kJ per cycle, and itIf the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (

releases 480 kJ of heat each cycle, what is its efficiency? (aa) 20%, () 20%, (bb) 80%,) 80%, ((cc) 100%, () 100%, (d d ) You cannot tell from the data given.) You cannot tell from the data given.

==

QQ_h_h

−−

efficiency of the heat engine in terms of

efficiency of the heat engine in terms of QQccandandQQhh..

−−

==

−−

==

ε ε Substitute for

Substitute for QQccandand QQhhto obtain:to obtain:

2 2 .. 0 0 kJ kJ 600 600 kJ kJ 480 480 1 1

−−

==

and

( ( ))

4

4 •• Explain what distinguishes a refrigerator from aExplain what distinguishes a refrigerator from a ″″heat pump.heat pump.″″

Determine the Concept The job of a refrigerator is to move heat from its coldThe job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its

direction that is opposite its

″″

naturalnatural

″″

direction of flow, analogous to the use of adirection of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter.

in the summer and warm them in the winter.

5

5 •• [SSM][SSM] An air conditioner’s COP is mathematically identical to thatAn air conditioner’s COP is mathematically identical to that of a refrigerator, that is,

of a refrigerator, that is, cc A

ACC rreeff

C

COOPP

=

CCOOPP

==

QQ

W

W . However a heat pump’s COP is. However a heat pump’s COP is

defined differently, as defined differently, as hh hp hp COP COP

==

QQ W

W . Explain clearly. Explain clearlywhywhythe two COPs arethe two COPs are

defined differently.

defined differently. Hint Hint :: Think of the end use of the three different devices.Think of the end use of the three different devices. Determine the Concept

Determine the ConceptThe COP is defined so as to be a measure of theThe COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior,

quantity is the heat drawn from the already colder interior, QQcc. For a heat pump,. For a heat pump,

the ideas is to focus on

(4)

6

6 •• Explain why you cannot cool your kitchen by leaving your refrigerator Explain why you cannot cool your kitchen by leaving your refrigerator door open on a hot day. (Why does turning on a room air conditioner cool down door open on a hot day. (Why does turning on a room air conditioner cool down the room, but opening a refrigerator door does not?)

the room, but opening a refrigerator door does not?)

Determine the Concept As described by the second law of thermodynamics,As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the refrigerator or air conditioner. The heating coils on a refrigerator are inside the room and so the refrigerator

room and so the refrigerator actually heats the room in actually heats the room in which it is located. which it is located. TheThe heating coils on an air conditioner are outside one’s living space, so the waste heating coils on an air conditioner are outside one’s living space, so the waste heat is vented to the outside.

heat is vented to the outside.

7

7 •• Why do steam-power-plant designers try to increase the Why do steam-power-plant designers try to increase the temperaturetemperature of the steam as much as possible?

of the steam as much as possible?

Determine the Concept Increasing the temperature of the steam increases theIncreasing the temperature of the steam increases the Carnot efficiency, and generally increases the

Carnot efficiency, and generally increases the efficiency of efficiency of any heat engine.any heat engine.

8

8 •• To increase the efficiency of a Carnot engine, you shouldTo increase the efficiency of a Carnot engine, you should

((aa) decrease the temperature of the hot reservoir, () decrease the temperature of the hot reservoir, (bb) increase the temperature of ) increase the temperature of the cold reservoir, (

the cold reservoir, (cc) increase the temperature of the hot reservoir, () increase the temperature of the hot reservoir, (d d ) change the) change the ratio of maximum volume to minimum volume.

ratio of maximum volume to minimum volume.

Determine the ConceptBecause the efficiency of a Carnot cycle engine is givenBecause the efficiency of a Carnot cycle engine is given by by h h cc C C 11 T T T T

−−

==

ε

ε , you should increase the temperature of the hot reservoir., you should increase the temperature of the hot reservoir.

( ( ))

c c isis

correct. correct.

9

9 •••• [SSM][SSM] Explain why the following statement is true: To Explain why the following statement is true: To increase theincrease the efficiency of a Carnot engine, you should make the difference between the two efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle

Carnot cycle refrigerator refrigerator , you should make the difference between the two, you should make the difference between the two operating temperatures as small as possible.

operating temperatures as small as possible.

Determine the ConceptA Carnot-cycle refrigerator is more efficient when theA Carnot-cycle refrigerator is more efficient when the temperatures are close together because it is easier to extract heat from an already temperatures are close together because it is easier to extract heat from an already cold interior if the temperature of the exterior is close to

cold interior if the temperature of the exterior is close to the temperature of thethe temperature of the interior of the refrigerator. A Carnot-cycle heat engine

interior of the refrigerator. A Carnot-cycle heat engine is more efficient when theis more efficient when the temperature difference is large because then more work is done by the engine for temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir.

each unit of heat absorbed from the hot reservoir.

10

10 •••• A Carnot engine operates between a cold temperature reservoir of A Carnot engine operates between a cold temperature reservoir of 27

27ooC and a high temperature reservoir of 127C and a high temperature reservoir of 127

°°

C. Its efficiency is (C. Its efficiency is (aa) 21%,) 21%, ((bb) 25%, () 25%, (cc) 75%, () 75%, (d d ) 79%.) 79%.

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Determine the ConceptThe efficiency of a Carnot cycle engine is given byThe efficiency of a Carnot cycle engine is given by

h h cc C C 11 T T T T

−−

==

ε

ε wherewhereT T ccandand T T hh (in kelvins) are the temperatures of the cold (in kelvins) are the temperatures of the cold and hotand hot

reservoirs, respectively. reservoirs, respectively.

Substituting numerical values for Substituting numerical values for T T cc

and

and T T hh yields:yields: ₄₀₀₄₀₀_K_K 00..2525

K K 300 300 1 1 C C

==

−−

==

( ( ))

bb is correct.is correct. 11

11 •••• The Carnot engine in Problem 10 is run in reverse as a refrigerator. ItsThe Carnot engine in Problem 10 is run in reverse as a refrigerator. Its COP is (

COP is (aa) 0.33, () 0.33, (bb) 1.3, () 1.3, (cc) 3.0 () 3.0 (d d ) 4.7.) 4.7.

Determine the ConceptThe coefficient of performance of a Carnot cycle engineThe coefficient of performance of a Carnot cycle engine run in reverse as refrigerator is given by

run in reverse as refrigerator is given by

W W Q Q_cc ref ref COP

COP

==

. We can use the relationship. We can use the relationship between

betweenW, QW, Qcc, and, and QQhhto eliminateto eliminate W W from this expression and then use thefrom this expression and then use the

relationship, applicable only to a device operating in a Carnot cycle, relationship, applicable only to a device operating in a Carnot cycle,

h h cc h h cc T T T T Q Q Q Q

==

toto express the refrigerator’s COP in terms of

express the refrigerator’s COP in terms of T T ccandand T T hh..

The coefficient of performance of a The coefficient of performance of a refrigerator is given by:

refrigerator is given by: _W_W

Q Q_cc ref ref COP COP

==

or, because or, because W W

==

QQ_h_h

−−

QQ_cc,, cc h h cc ref ref COP COP Q Q Q Q Q Q

−−

==

Dividing the numerator and Dividing the numerator and denominator of this fraction by denominator of this fraction by QQcc

yields: yields: 11 1 1 COP COP cc h h ref ref

−−

==

Q Q Q Q

For a device operating in a Carnot For a device operating in a Carnot cycle: cycle: h h cc h h cc T T T T Q Q Q Q

==

Substitute in the expression for Substitute in the expression for COP

COPref ref to obtain:to obtain:

1 1 1 1 COP COP cc h h C C ref, ref,

−−

==

T T T T

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Substitute numerical values and

evaluate COPref, C: 3.0

1 K 300 K 400 1 COP_ref,_C

=

−

=

( )

c is correct.

12 •• On a humid day, water vapor condenses on a cold surface. During condensation, the entropy of the water (a) increases, (b) remains constant, (c) decreases, (d ) may decrease or remain unchanged. Explain your answer.

Determine the Concept When water vapor condenses, its entropy decreases (the liquid state is a more ordered state than is the vapor state) and the entropy of the universe increases. ( a) is correct.

13 •• An ideal gas is taken reversibly from an initial state Pi,V i, T ito the

final state Pf ,V f ,T f . Two possible paths are (A) an isothermal expansion followed

by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. For these two paths, (a)

Δ

E int A>

Δ

E int B, (b)

Δ

S A>

Δ

S B,

(c)

Δ

S A<

Δ

S B, (d ) None of the above.

Determine the ConceptThe two paths

are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem. P V i V V _f f P i P i f B B A A i T f T

(a) Because E int is a state function and the initial and final states are the same for

the two paths,

Δ

E _int,_A

=

Δ

E _int,_B.

(b) and (c) S , like E int , is a state function and its change when the system moves

from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs. Thus

Δ

S _A

=

Δ

S _B.

(d ) ( d is correct.)

14 •• Figure 19-12 shows a thermodynamic cycle for an ideal gas on anST

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Determine the Concept The processes A

→

B and C

→

D are adiabatic and the processes B

→

C and D

→

A are isothermal. Therefore, the cycle is the Carnot cycle shown in the adjacent PV

diagram. P V A B C D

15 •• Figure 19-13 shows a thermodynamic cycle for an ideal gas on anSV

diagram. Identify the type of engine represented by this diagram.

Determine the Concept Note that A

→

B is an adiabatic expansion, B

→

C is a constant-volume process in which the entropy decreases, C

→

D is an adiabatic compression and D

→

A is a constant-volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). The points A, B, C, and D in Figure 19-13 correspond to points c, d, a, and b, respectively, in Figure 19-3.

16 •• Sketch anST diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.)

Determine the Concept The Otto cycle consists of four quasi-static steps. Refer to Figure 19-3. Therea

→

b is an adiabatic compression,b

→

cis a constant volume heating, c

→

d is an adiabatic expansion and d

→

a is a constant-volume cooling. So, from a to b, S is constant and T increases, from b to c, heat is added to the system and both S and T increase, from c

→

d S is constant while T decreases, and from d to a bothS andT decrease.

To determine how S depends on T

along b

→

candd

→

a,consider the entropy change of the gas from point

b to an arbitrary point on the path

b

→

cwhere the entropy and

temperature of the gas areS andT,

respectively:

T Q S

=

Δ

where, because heat is entering the system, Qis positive.

BecauseW on= 0 for this

constant-volume process:

(

T T b

)

C T C Q Q E

=

Δ

=

−

Δ

_int _in _V _V

Substituting for Qyields:

₍

₎

⎟

⎠

⎞

⎜

⎝

⎛ −

=

−

=

Δ

T T C T T T C S b ₁ b V V

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On pathb

→

c the entropy is given by:

⎟

⎠

⎞

⎜

⎝

⎛ −

+

=

Δ

+

=

T T C S S S S b b b V 1

The first and second derivatives,

dT

dS and d 2S dT 2, give the slope and concavity of the path. Calculate these derivatives assumingC Vis

constant. (For an ideal gasC Vis a

positive constant.): 2 V T T C dT dS _b

=

3 V 2 2 2 T T C dT S d _b

−

=

These results tell us that, along path b

→

c,the slope of the path is positive and the slope decreases asT increases. The concavity of the path is negative for allT.

Following the same procedure on

pathd

→

agives:

_⎠

⎟

⎞

⎜

⎝

⎛ −

+

=

T T C S S _d _V 1 d 2 V T T C dT dS _d

=

3 V 2 2 2 T T C dT S d _d

−

=

These results tell us that, along path d

→

a,the slope of the path is positive and the slope decreases asT increases. The concavity of the path is negative for allT.

An ST diagram for the Otto cycle is shown to the right.

S

T

a _b

c d

17 •• [SSM] Sketch anSV diagram of the Carnot cycle for an ideal gas.

Determine the Concept Referring to Figure 19-8, process 1

→

2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2

→

3 is adiabatic, so S is constant as V increases. Process 3

→

4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4

→

1 is adiabatic, that is, isentropic, and S is constant while V decreases.

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During the isothermal expansion (from point 1 to point 2) the work done by

the gas equals the heat added to the gas. The change in entropy of the gas from point 1 (where the temperature is

T 1) to an arbitrary point on the curve is

given by: 1 T Q S

=

Δ

For an isothermal expansion, the work done by the gas, and thus the heat added to the gas, are given by:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

1 1ln V V nRT W Q

Substituting for Qyields:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

Δ

1 ln V V nR S Since S

=

S ₁

+

Δ

S , we have:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

=

1 1 ln V V nR S S

The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet

program. This graph establishes the curvature of the 1

→

2 and 3

→

4 paths for the SV graph.

V S

An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right.

V

1

2 ₃

4

18 •• Sketch anSV diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.)

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Determine the ConceptThe Otto cycle is shown in Figure 19-3. Processa

→

b

takes place adiabatically and so bothQ= 0 and

Δ

S = 0 along this path. Process

b

→

ctakes place at constant volume.Qin, however, is positive and so, while

Δ

V = 0 along this path,Q > 0 and, therefore

Δ

S > 0. Processc

→

d also takes place adiabatically and so, again, bothQ = 0 and

Δ

S = 0 along this path. Finally, process

d

→

a is a constant-volume process, this time with heat leaving the system and

Δ

S < 0. A sketch of theSV diagram for the Otto cycle follows:

a b

c d

S

19 •• Figure 19-14 shows a thermodynamic cycle for an ideal gas on anSP

diagram. Make a sketch of this cycle on aPV diagram.

Determine the ConceptProcess A

→

B

is at constant entropy; that is, it is an adiabatic process in which the pressure increases. Process B

→

C is one in which P is constant and S decreases; heat is exhausted from the system and the volume decreases. Process C

→

D is an adiabatic compression. Process D

→

A returns the system to its original state at constant pressure. The cycle is shown in the adjacent PV diagram.

P V A B C D

20 •• One afternoon, the mother of one of your friends walks into his room and finds a mess. She asks your friend how the room came to be in such a state, and your friend replies, ″Well, it is the natural destiny of any closed system to degenerate toward greater and greater levels of entropy. That’s all, Mom.″ Her reply is a sharp ″ Nevertheless, you’d better clean your room!″ Your friend retorts,

″But that can’t happen. It would violate the second law of thermodynamics.″

Critique your friend’s response. Is his mother correct to ground him for not cleaning his room, or is cleaning the room really impossible?

Determine the Concept The son is out of line, here, but besides that, he’s also wrong. While it is true that systems tend to degenerate to greater levels of

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disorder, it is not true that order cannot be brought forth from disorder. What is required is an agent doing work – for example, your friend – on the system in order to reduce the level of chaos and bring about order. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the task. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease.

Estimation and Approximation

21 • Estimate the change in COP of your electric food freezer when it is removed from your kitchen to its new location in your basement, which is 8°C cooler than your kitchen.

Picture the ProblemWe can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen. If we further assume that the freezer operates in a Carnot cycle, then we can use the proportionQ_h Q_c

=

T _h T _c to

express the ratio of the coefficients of performance in terms of the temperatures in the kitchen, basement, and freezer.

The ratio of the coefficients of performance in the basement and

kitchen is given by:

kit c, kit c, basement c, basement c, kit basement COP COP W Q W Q

=

Because W

=

Q_h

−

Q_cfor a heat engine or refrigerator: kit c, kit h, kit c, basement c, basement h, basement c, kit basement COP COP Q Q Q Q Q Q

−

=

Divide the numerators and

denominators by Qc,basement andQc,kit

and simplify to obtain:

1 1 1 1 1 1 COP COP basement c, basement h, kit c, kit h, kit c, kit h, basement c, basement h, kit basement

−

=

−

=

Q Q Q Q Q Q Q Q

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If we assume that the freezer unit operates in a Carnot cycle, then

c h c h T T Q Q

=

and our expression for the ratio of the COPs becomes:

1 1 COP COP basement c, basement h, kit c, kit h, kit basement

−

=

T T T T

Assuming that the temperature in your kitchen is 20

°

C and that the temperature of the interior of your freezer is

−

5

°

C, substitute numerical values and evaluate the ratio of the coefficients of performance: 47 . 1 1 K 268 K 285 1 K 268 K 93 2 COP COP kit basement

=

−

=

or an increase of 47% in the performance of the freezer!

22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room.

Picture the ProblemThe probability that all the molecules in your bedroom are located in the (open) closet is given by

N V V p

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

1

2 _where_N_{is the number of air}

molecules in your bedroom andV 1 andV 2are the volumes of your bedroom and

closet, respectively. We can use the ideal-gas law to find the number of molecules

N . We’ll assume that the volume of your room is about 50 m3 and that the temperature of the air is 20

°

C.

If the original volume of the air in your bedroom isV 1, the probability pof finding the N molecules,

normally in your bedroom, confined to your closet whose volume isV 2 is

given by: N V V p

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

1 2 or, because ₁₀1 ₁ 2 V V

=

, N p

⎟

⎠

⎞

⎜

⎝

⎛

=

10 1 (1)

Use the ideal-gas law to express N :

kT PV N

=

Substitute numerical values and evaluate N :

(

)

(

)

(

)

(

)

molecules 10 252 . 1 K 293 J/K 10 381 . 1 m 50 kPa 325 . 101 27 23 3

×

=

×

=

₋ N

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Substitute for N in equation (1) and evaluate p: 27 27 27 27 10 10 252 . 1 10 252 . 1 10 252 . 1 10 10 10 1 10 1 − × − × ×

≈

=

⎟

⎠

⎞

⎜

⎝

⎛

=

p

23 •• [SSM] Estimate the maximum efficiency of an automobile engine

that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume γ =1.4. (The Otto cycle is discussed in Section 19-1. )

Picture the Problem The maximum efficiency of an automobile engine is given

by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio.

Express the Carnot efficiency of an engine operating between the

temperatures T candT h: h c C 1 T T

−

=

ε

Relate the temperaturesT cand T h to

the volumesV candV h for a

quasi-static adiabatic compression from V c

to V h: 1 h h 1 c c − −

₌

γ γ V T V T

⇒

1 c h 1 c 1 h h c − − −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

γ γ γ V V V V T T Substitute for h c T T to obtain: 1 c h C 1 −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

=

γ ε V V

Express the compression ratio r :

h c V V r

=

Substituting for r yields:

1 C 1 1

−

₋

=

_γ ε r

Substitute numerical values for r and

γ (1.4 for diatomic gases) and

evaluateε C:

( )

8.0 56% 1

1 ₁_.₄ ₁

C

=

−

≈

24 •• You are working as an appliance salesperson during the summer. One day, your physics professor comes into your store to buy a new refrigerator. Wanting to buy the most efficient refrigerator possible, she asks you about the efficiencies of the available models. She decides to return the next day to buy the most efficient refrigerator. To make the sale, you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator, and (b) and the highest rate possible for the heat to be released by the refrigerator if the refrigerator uses 600 W of electrical power.

(14)

Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0

°

C (273 K) and the room temperature to be about 30

°

C (303 K), then the refrigerator must be able to maintain a temperature difference of about 30 K. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir andQh and Qc to find an

upper limit on the COP of a household refrigerator. In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to

time to estimate the rate at which heat is being drawn from the refrigerator compartment.

(a) Using its definition, express the

COP of a household refrigerator: _W

Q_c

COP

=

(1)

Apply conservation of energy to

the refrigerator to obtain: c h

Q Q

W

+

=

⇒

W

=

Q_h

−

Q_c

Substitute for W and simplify to obtain: 1 1 COP c h c h c

−

=

−

=

Q Q Q Q Q

Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to Q and_h Qc:

c h c h T T Q Q

=

Substitute for c h Q Q to obtain: 1 1 COP c h max

−

=

T T

evaluate COPmax: 9.1

1 K 273 K 303 1 COP_max

=

−

=

(b) Solve equation (1) for Qc: Q_c

=

W

(

COP

)

(2)

Differentiate equation (2) with

respect to time to obtain:

(

)

_dt

dW dt

dQ

COP

c

=

Substitute numerical values and evaluate dt dQ_c :

( )(

9.1 600J/s

)

5.5kW c

=

dt dQ

(15)

25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K, the average temperature of the surface of Earth is about 290 K. The solar constant (the intensity of sunlight reaching Earth’s atmosphere) is about

1.37 kW/m2. (a) Estimate the total power of the sunlight hitting Earth. (b) Estimate the net rate at which Earth’s entropy is increasing due to this solar radiation.

Picture the ProblemWe can use the definition of intensity to find the total power of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun.

(a) Using its definition, express the intensity of the Sun’s radiation on Earth in terms of the power P

delivered to Earth and Earth’s cross sectional area A:

A P I

=

Solve for Pand substitute for Ato obtain: 2 R I IA P

=

π

where R is the radius of Earth. Substitute numerical values and

evaluateP:

(

)(

)

W 10 75 . 1 W 10 746 . 1 m 10 37 . 6 kW/m 37 . 1 17 17 2 6 2

×

=

×

=

×

=

π P

(b) Express the rate at which Earth’s entropyS Earth changes due to the

flow of solar radiation: Earth

Earth T P dt dS

=

Substitute numerical values and evaluate dt dS _Earth : s J/K 10 02 . 6 K 290 W 10 746 . 1 14 17 Earth

⋅

×

=

×

=

dt dS

26 •• A 1.0-L box contains N molecules of an ideal gas, and the positions of the molecules are observed 100 times per second. Calculate the average time it should take before we observe all N molecules in the left half of the box if N is equal to (a) 10, (b) 100, (c) 1000, and (d ) 1.0 mole. (e) The best vacuums that have been created to date have pressures of about 10 –12 torr. If a vacuum chamber has the same volume as the box, how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe, which is about 1010years.

Picture the ProblemIf you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100% chance of it being on one side or the other. With two molecules, there are four

(16)

possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both being on a particular side, or a 50% chance of them both being on either side.

Extending this logic, the probability of N molecules all being on one side of the box isP= 2/2 N , which means that, if the molecules shuffle 100 times a second,

the time it would take them to cover all the combinations and all get on one side or the other is

( )

100 2

2 N

t

=

. In (e) we can apply the ideal gas law to find the number of molecules in 1.0 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K.

(a) Evaluatet for N = 10 molecules:

(

100s

)

5.12s 5s 2 2 1 10

≈

=

₋ t

(b) Evaluatet for N = 100 molecules:

(

)

y 10 2 s 10 3.156 y 1 s 10 34 . 6 s 100 2 2 20 7 27 1 100

×

≈

×

=

₋ t

(c) Evaluatet for N = 1000 molecules:

(

1

)

1000 s 100 2 2 −

=

t

To evaluate 2 let1000 10 x

=

21000 and take the logarithm of both sides of the equation to obtain:

(

1000

)

ln2

=

xln10

⇒

x

=

301 Substitute to obtain:

(

)

y 10 2 s 10 3.156 y 1 s 10 5 . 0 s 100 2 10 291 7 299 1 301

×

≈

×

=

₋ t (d ) Evaluatet for N = 1.0 mol =6.022

×

1023 molecules:

₍

1

₎

10 022 . 6 s 100 2 2 23 − ×

=

t To evaluate 26.022×1023 let 23 10 022 . 6 2

10 x

=

× _{and take the logarithm}

of both sides of the equation to obtain:

)

ln2 ln10 10 022 . 6

×

23

=

x

⇒

x

≈

1023

(17)

Substituting for x yields:

(

)

y 10 s 10 3.156 y 1 s 100 2 10 23 23 10 7 1 10

≈

⎟

⎠

⎞

⎜

⎝

⎛

×

≈

₋ t

(e) Solve the ideal gas law for the

number of molecules N in the gas: _kT

PV N

=

Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N :

)

(

)( )

(

)

(

)

molecules 10 22 . 3 K 300 J/K 10 381 . 1 L 0 . 1 Pa/torr 32 . 133 torr 10 7 23 12

×

=

×

=

− ₋ N Evaluate t for N = 3.22

×

107 molecules:

₍

1

₎

10 22 . 3 s 100 2 2 7 − ×

=

t To evaluate 23.22×107 let 7 10 22 . 3 2

10 x

=

× and take the logarithm of both sides of the equation to obtain:

)

ln2 ln10 10 22 . 3

×

7

=

x

⇒

x

≈

107

Substituting for x yields:

(

)

y 10 s 10 3.156 y 1 s 100 2 10 7 7 10 7 1 10

≈

×

=

₋ t

Express the ratio of this waiting time to the lifetime of the universe t universe:

7 7 10 10 10 universe 10 y 10 y 10

≈

=

t t or universe 107 10 t t

≈

Heat Engines and Refrigerators

27 • [SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work

during each cycle. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle?

Picture the Problem(a) The efficiency of the engine is defined to be

h

Q W

=

whereW is the work done per cycle andQh is the heat absorbed from

the hot reservoir during each cycle. (b) Because, from conservation of energy,

c

h W Q

(18)

released to the cold reservoir during each cycle. (a) Qhabsorbed from the hot reservoir

during each cycle is given by: 0.200 500J

J 100 h

=

ε W Q (b) Use Q_h

=

W

+

Q_cto obtain: ₅₀₀_J ₁₀₀_J ₄₀₀_J h c

=

Q

−

W

=

−

=

Q

28 • A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does 0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle?

Picture the Problem(a) The efficiency of the engine is defined to be

h Q W

=

ε where W is the work done per cycle andQhis the heat absorbed from

the hot reservoir during each cycle. (b) We can apply conservation of energy to the engine to obtain Q_h

=

W

+

Q_c and solve this equation for the heatQcreleased

to the cold reservoir during each cycle. (a) The efficiency of the heat engine

is given by: 400J 30% J 120 h

=

Q W ε

(b) Apply conservation of energy to the engine to obtain:

c

h W Q

Q

=

+

⇒

Q_c

=

Q_h

−

W

Substitute numerical values and evaluateQc: J 280 J 120 J 400 c

=

−

=

Q

29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine.

Picture the ProblemWe can use its definition to find the efficiency of the engine and the definition of power to find its power output.

(a) The efficiency of the heat engine

is given by: _h c h c h h Q 1 Q Q Q Q Q W

−

=

−

=

ε

Substitute numerical values and evaluateε : % 40 J 100 J 60 1

−

=

ε

(19)

(b) The power outputPof this engine is the rate at which it does work: dt dQ Q dt d dt dW P

=

ε _h

=

ε h

evaluateP:

(

)

_0.500_s 80W J 100 0.40

_⎟⎟

=

⎠

⎞

⎜⎜

⎝

⎛

=

P

30 • A refrigerator absorbs 5.0 kJ of heat from a cold reservoir and releases 8.0 kJ to a hot reservoir. (a) Find the coefficient of performance of the

refrigerator. (b) The refrigerator is reversible. If it is run backward as a heat engine between the same two reservoirs, what is its efficiency?

Picture the ProblemWe can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.

(a) The COP of a refrigerator is

defined to be: W

Q_c

COP

=

Apply conservation of energy to relate the work done per cycle to

Qh andQc:

c

h Q

Q

W

=

−

Substitute for W to obtain:

c h c COP Q Q Q

−

=

evaluate COP: 8.0kJ 5.0kJ 1.7 kJ 5.0 COP

=

−

=

(b) The efficiency of a heat pump

is defined to be: Q_h

W

=

ε

Apply conservation of energy to the

heat pump to obtain: _h

c h c h ₁ Q Q Q Q Q

−

=

−

=

ε

Substitute numerical values and evaluateε : % 38 kJ 8.0 kJ 0 . 5 1

−

=

31 •• [SSM] The working substance of an engine is 1.00 mol of a

monatomic ideal gas. The cycle begins atP1 =1.00 atm andV 1 =24.6 L. The gas

is heated at constant volume to P2 =2.00 atm. It then expands at constant pressure

(20)

pressure is again 1.00 atm. It is then compressed at constant pressure to its

original state. All the steps are quasi-static and reversible. (a) Show this cycle on a PV diagram. For each step of the cycle, find the work done by the gas, the heat absorbed by the gas, and the change in the internal energy of the gas. (b) Find the efficiency of the cycle.

Picture the ProblemTo find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from Q

=

C _V

Δ

T and Q

=

C _P

Δ

T . We can use the 1st law of thermodynamics to find the change in internal energy for each step of the cycle. Finally, we can find the efficiency of the cycle from the work done each cycle and the heat thatenters the system each cycle.

(a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T 1:

(

)(

)

(

)

K 300 K mol atm L 10 8.206 mol 1.00 L 24.6 atm 1.00 2 1 1 1

=

⎟

⎠

⎞

⎜

⎝

⎛

⋅

×

=

− nR V P T

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

K T

T ₂

=

2 ₁

=

600

The volume doubles while the pressure remains constant between

states 2 and 3. Hence:

K T

T ₃

=

2 ₂

=

1200

The pressure is halved while the volume remains constant

between states 3 and 4. Hence:

K T

T 1₂ ₃ 600

(21)

For path 1 2: 0 Δ ₁₂ 12

=

P V

=

W and

(

600K 300K

)

3.74kJ K mol J 8.314 Δ Δ ₂3 12 2 3 12 V 12

⎟

−

=

⎠

⎞

⎜

⎝

⎛

⋅

=

C T R T Q

The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1stlaw of thermodynamics: on in int Δ E

=

Q

+

W Because W ₁₂

=

0: _Δ ₃_.₇₄_kJ 12 12 int,

=

Q

=

E For path 2 3:

(

)(

)

4.99kJ atm L J 101.325 L 24.6 L 49.2 atm 2.00 Δ 23 23 on

⎟

=

−

⎠

⎞

⎜

⎝

⎛

⋅

−

=

−

=

−

=

W P V W

(

1200K 600K

)

12.5kJ K mol J 8.314 Δ Δ ₂5 23 2 5 23 P 23

⎟

−

=

⎠

⎞

⎜

⎝

⎛

⋅

=

C T R T Q

Apply Δ E_int

=

Q_in

+

W _onto obtain: _Δ ₁₂_.₅_kJ ₄_.₉₉_kJ ₇_.₅_kJ

23 int,

=

−

=

E For path 3 4: 0 34 34

=

P

Δ

V

=

W and

(

600K 1200K

)

7.48kJ K mol J 8.314 Δ Δ Δ 2 34 ₂3 3 34 V 34 int, 34

⎟

−

=

−

⎠

⎞

⎜

⎝

⎛

⋅

=

E C T R T Q

Apply Δ E_int

=

Q_in

+

W _onto obtain: _Δ ₇_.₄₈_kJ ₀ ₇_.₄₈_kJ

34 int,

=

−

+

=

−

E For path 4 1:

(

)(

)

2.49kJ atm L J 101.325 L 2 . 9 4 L 24.6 atm 1.00 Δ ₄₁ 41 on

⎟

=

⎠

⎞

⎜

⎝

⎛

⋅

−

=

−

=

−

=

W P V W and

(

300K 600K

)

6.24kJ K mol J 8.314 Δ Δ 2 41 ₂5 5 41 P 41

⎟

−

=

−

⎠

⎞

⎜

⎝

⎛

⋅

=

C T R T Q

(22)

Apply Δ E _int

=

Q_in

+

W _onto obtain: _Δ ₆_.₂₄_kJ ₂_.₄₉_kJ ₃_.₇₅_kJ

41

int,

=

−

+

=

−

E

For easy reference, the results of the preceding calculations are summarized in the following table:

Process W , kJon Q , kJin Δ Eint

(

=

Qin

+

W on

)

, kJ

1

→

2 0 3.74 3.74

2

→

3

−

4.99 12.5 7.5

3

→

4 0

−

7.48

−

7.48

4

→

1 2.49

−

6.24

−

3.75

(b) The efficiency of the cycle is given by:

(

)

23 12 41 23 in by Q Q W W Q W

+

−

+

−

=

ε

Substitute numerical values and evaluateε

:

% 15 kJ 5 . 12 kJ 3.74 kJ 2.49 kJ 4.99

≈

+

−

=

Remarks: Note that the work done per cycle is the area bounded by the

rectangular path. Note also that, as expected because the system returns to its initial state, the sum of the changes in the internal energy for the cycle is zero.

32 •• The working substance of an engine is 1.00 mol of a diatomic ideal gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10.0 L to a pressure of 1.00 atm and a

volume of 20.0 L, (2) a compression at constant pressure to its original volume of 10.0 L, and (3) heating at constant volume to its original pressure. Find the

efficiency of this cycle.

Picture the Problem The three steps in the process are shown on the PV

diagram. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

V (L) 2 3 1 P_(atm) 2.639 2 1 0 0 10.0 20.0

The pressures and volumes at the end points of the adiabatic expansion are

related according to:

γ γ 2 2 1 1V PV P

=

⇒

₂ 1 2 1 P V V P γ

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

(23)

evaluateP1: ₁₀_.₀_L

(

1.00atm

)

2.639atm

L 0 . 20 1.4 1

⎟

=

⎠

⎞

⎜

⎝

⎛

=

P

Express the efficiency of the cycle:

h

Q W

=

ε (1)

No heat enters or leaves the system during the adiabatic expansion:

0

12

=

Q

Find the heat entering or leaving the system during the isobaric

compression:

(

)(

)

L atm 35.0 L 20.0 L 10.0 atm 1.00 Δ Δ Δ 2 7 23 2 7 23 2 7 23 V 23

⋅

−

=

−

=

C T R T P V Q

Find the heat entering or leaving the system during the

constant-volume process:

(

)(

)

L atm 0 . 41 L 10.0 atm 1.00 atm 2.639 Δ Δ Δ 2 5 31 2 5 31 2 5 31 V 31

⋅

=

−

=

C T R T PV Q

Apply the 1stlaw of thermodynamics to the cycle (

Δ

E _int,_cycle

=

0) to obtain:

L atm 6.0 L atm 41.0 L atm 35.0 0 Δ 31 23 12 in in int on

⋅

=

⋅

+

⋅

−

=

+

=

−

=

−

=

Q Q Q Q Q E W

Substitute numerical values in equation (1) and evaluateε :

% 15 L atm 41 L atm 6.0

=

⋅

=

ε

33 •• An engine using 1.00 mol of an ideal gas initially at a volume of 24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume, (2) cooling at

constant volume to a temperature of 300 K (3) an isothermal compression to its original volume, and (4) heating at constant volume to its original temperature of 400 K. Assume thatC v =21.0 J/K. Sketch the cycle on aPV diagram and find its

efficiency.

Picture the ProblemWe can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

(24)

The PV diagram of the cycle is shown to the right. A, B, C, and D identify the four states of the gas and the numerals 1, 2, 3, and 4 represent the four steps through which the gas is taken. 0 10 20 30 40 50 60 2 1 1.5 0 0.5 400 K 300 K P ( t a m ) V _(L) 1 2 3 4 A B C D

Express the efficiency of the cycle:

4 h, 3 h, 2 h, 1 h, 4 3 2 1 h Q Q Q Q W W W W Q W

+

=

ε

Because steps 2 and 4 are

constant-volume processes, W 2 =W 4 = 0: _h,₁ _h,₂ _h,₃ _h,₄ 3 1 h 0 0 Q Q Q Q W W Q W

+

=

ε

Because the internal energy of the gas increases in step 4 while no work is done, and because the internal energy does not change during step 1 while work is done by the gas, heat enters the system only during these processes: 4 h, 1 h, 3 1 h Q Q W W Q W

+

=

ε (1)

The work done during the isothermal

expansion (1) is given by:

_⎠

⎟⎟

⎞

⎜⎜

⎝

⎛

=

A B 1 ln V V nRT W

The work done during the isothermal

compression (3) is given by:

_⎠

⎟⎟

⎞

⎜⎜

⎝

⎛

=

C D c 3 ln V V nRT W

Because there is no change in the internal energy of the system during step 1, the heat that enters the system during this isothermal expansion is given by:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

A B h 1 1 ln V V nRT W Q

The heat that enters the system during the constant-volume step 4 is given by:

(

_h _c

)

V V 4 C ΔT C T T Q

=

−

(25)

Substituting in equation (1) yields:

(

h c

)

V A B h C D c A B h ln ln ln T T C V V nRT V V nRT V V nRT

−

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

ε Noting the 2 A B

=

V V and 2 1 C D

=

V V

,substitute and simplify to obtain:

( )

(

)

( )

(

)

( )

(

h c

)

V h c h c h V h c h c h V h c h 2 ln 2 ln 2 ln 2 ln 2 ln 2 1 ln 2 ln T T nR C T T T T T nR C T T T T T nR C T T T

−

+

−

=

−

+

−

=

−

+

⎟

⎠

⎞

⎜

⎝

⎛

+

=

ε

Substitute numerical values and evaluateε :

(

)

(

)

(

)

% 1 . 13 K 300 K 400 2 ln K mol J 314 . 8 mol 00 . 1 K J 0 . 21 K 400 K 300 K 400

=

−

⎟

⎠

⎞

⎜

⎝

⎛

⋅

+

−

=

ε

34 •• Figure 19-15 shows the cycle followed by 1.00 mol of an ideal

monatomic gas initially at a volume of 25.0 L. All the processes are quasi-static. Determine (a) the temperature of each numbered state of the cycle, (b) the heat transfer for each part of the cycle, and (c) the efficiency of the cycle.

Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes. Because the change in internal energy is zero for the isothermal process, we can use the expression for the work done on or by a gas during an isothermal process to find the heat flow during such a process. Finally, we can find the efficiency of the cycle from its definition.

(a) Use the ideal-gas law to find the temperature at point 1:

(

)(

)

(

)

K 301 K mol J 8.314 mol 1.00 L 25.0 kPa 100 1 1 1

=

⎟

⎠

⎞

⎜

⎝

⎛

⋅

=

nR V P T

(26)

Use the ideal-gas law to find the temperatures at points 2 and 3:

(

)(

)

(

)

K 601 K mol J 8.314 mol 1.00 L 25.0 kPa 200 2 2 3 2

=

⎟

⎠

⎞

⎜

⎝

⎛

⋅

=

nR V P T T

(b) Find the heat entering the system for the constant-volume process from 1

→

2:

(

601K 301K

)

3.74kJ K mol J 8.314 Δ Δ ₂3 12 2 3 12 V 12

⎟

−

=

⎠

⎞

⎜

⎝

⎛

⋅

=

C T R T Q

Find the heat entering or leaving the system for the isothermal process from 2

→

3:

(

)

(

)

3.46kJ L 0 . 5 2 L 0 . 0 5 ln K 01 6 K mol J 8.314 mol 1.00 ln 2 3 2 23

⎟⎟

=

⎠

⎞

⎜⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

⋅

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

V V nRT Q

Find the heat leaving the system during the isobaric compression from 3

→

1:

(

301K 601K

)

6.24kJ K mol J 8.314 Δ Δ ₂5 31 2 5 31 P 31

⎟

−

=

−

⎠

⎞

⎜

⎝

⎛

⋅

=

C T R T Q

(c) Express the efficiency of the

cycle: _in Q₁₂ Q₂₃ W Q W

+

=

ε (1)

Apply the 1stlaw of thermodynamics to the cycle: kJ 0.96 kJ 6.24 kJ 3.46 kJ .74 3 31 23 12

=

−

+

=

+

=

∑

Q Q Q Q W

because, for the cycle, Δ E_int

=

0. Substitute numerical values in equation

(1) and evaluateε : % 13 kJ 3.46 kJ 3.74 kJ 0.96

=

+

=

ε

35 •• An ideal diatomic gas follows the cycle shown in Figure 19-16. The temperature of state 1 is 200 K. Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle.

Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition;

(27)

using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determineQin.

(a) Use the ideal-gas law for a fixed amount of gas to find the

temperature in state 2 to the temperature in state 1: 2 2 2 1 1 1 T V P T V P

=

⇒

1 2 1 1 1 2 2 1 2 P P T V P V P T T

=

Substitute numerical values and evaluateT 2:

(

)

(

)

(

1.0atm

)

600K atm 3.0 K 200 2

=

T

Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain: 2 3 2 2 2 3 3 2 3 V V T V P V P T T

=

(

)

(

)

(

100L

)

1800K L 300 K 600 3

=

T

Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain: 3 4 3 3 3 4 4 3 4 P P T V P V P T T

=

(

)

(

)

(

3.0atm

)

600K atm 1.0 K 1800 4

=

T

(b) The efficiency of the cycle is:

in Q

W

=

ε (1)

Use the area of the rectangle to

find the work done each cycle:

(

)(

)

L atm 400 atm 1.0 atm 3.0 L 100 L 300 Δ Δ

⋅

=

−

=

P V W

Apply the ideal-gas law to state 1 to find the product of n and R:

(

)( )

atm/K L 0.50 K 200 L 100 atm 1.0 1 1 1

⋅

=

T V P nR

Noting that heat enters the system between states 1 and 2 and states 2

and 3, express Qin:

₍

_T _T

₎

_nR T nR T nR T C T C Q Q Q 23 2 7 12 2 5 23 2 7 12 2 5 23 P 12 V 23 12 in

Δ

+

Δ

=

Δ

+

Δ

=

Δ

+

Δ

=

+

=

(28)

Substitute numerical values and evaluateQin:

(

)

[

(

)

]

2600atm L K atm L 50 . 0 K 600 K 1800 K 200 K 600 ₂7 2 5 in

⎟

=

⋅

⎠

⎞

⎜

⎝

⎛

⋅

−

+

−

=

Q

Substitute numerical values in equation (1) and evaluateε :

% 15 L atm 2600 L atm 400

=

⋅

=

ε

36 ••• Recently, an old design for a heat engine, known as theStirling engine

has been promoted as a means of producing power from solar energy. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume, (3) an isothermal expansion of the gas, and (4) cooling of the gas at constant volume. (a) SketchPV andST diagrams for the Stirling cycle. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero.

Picture the Problem(a) The PV and ST cycles are shown below. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes.

P T T _c h (1) (2) (3) (4) 1 2 3 4 S T T h c T V _{= 0} V _{= 0} Δ Δ

(b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle:

41 34 23 12 cycle Δ Δ Δ Δ ΔS

=

S

+

S

+

S

+

S (1)

Express the entropy change for the isothermal process from state 1 to state 2:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

1 2 12 ln Δ V V nR S

(29)

Similarly, the entropy change for the isothermal process from state 3 to state 4 is:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

3 4 34 ln Δ V V nR S

or, becauseV 2 =V 3 andV 1 =V 4,

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

1 2 2 1 34 ln ln Δ V V nR V V nR S

The change in entropy for a constant-volume process is given by:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

∫

i f V V isochoric ln Δ f i T T nC T dT nC T dQ S T T

For the constant-volume process

from state 2 to state 3:

_⎠

⎟⎟

⎞

⎜⎜

⎝

⎛

=

h c V 23 ln Δ T T C S

For the constant-volume process

from state 4 to state 1:

_⎠

⎟⎟

⎞

⎜⎜

⎝

⎛

−

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

h c V c h V 41 ln ln Δ T T C T T C S

Substituting in equation (1) yields:

0 ln ln ln ln Δ h c V 1 2 h c V 1 2 cycle

⎟⎟

=

⎠

⎞

⎜⎜

⎝

⎛

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

=

T T C V V nR T T C V V nR S

37 •• ″As far as we know, Nature has never evolved a heat engine″ —Steven Vogel, Life’s Devices, Princeton University Press (1988). (a) Calculate the

efficiency of a heat engine operating between body temperature (98.6ºF) and a typical outdoor temperature (70ºF), and compare this to the human body’s

efficiency for converting chemical energy into work (approximately 20%). Does this efficiency comparison contradict the second law of thermodynamics?

(b) From the result of Part (a), and a general knowledge of the conditions under which most warm-blooded organisms exist, give a reason why no warm-blooded organisms have evolved heat engines to increase their internal energies.

Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures.

(a) Express the maximum efficiency of an engine operating between body temperature and 70°F: h c C 1 T T

−

=

ε