walker4 ism ch18

8/10/2019 Walker4 Ism Ch18

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18 1

Chapter 18: The Laws of Thermodynamics

Answers to Even-Numbered Conceptual Questions

2. (a)Yes. Heat can flow into the system if at the same time the system expands, as in an isothermal expansionof a gas. (b)Yes. Heat can flow out of the system if at the same time the system is compressed, as in anisothermal compression of a gas.

4. No. The heat might be added to a gas undergoing an isothermal expansion. In this case, there is no changein the temperature.

6. Yes. In an isothermal expansion, all the heat added to the system to keep its temperature constant appears aswork done by the system.

8. The final temperature of an ideal gas in this situation is T; that is, there is no change in temperature. Thereason is that as the gas expands into the vacuum, it does no work because it has nothing to push against.The gas is also insulated, so no heat can flow into or out of the system. It follows that the internal energy ofthe gas is unchanged, which means that its temperature is unchanged as well.

10. This would be a violation of the second law of thermodynamics, which states that heat always flows from ahigh-temperature object to a low-temperature object. If heat were to flow spontaneously between objects ofequal temperature, the result would be objects at different temperatures. These objects could then be used torun a heat engine until they were again at the same temperature, after which the process could be repeatedindefinitely.

12. Yes. In fact, the heat delivered to a room is typically 3 to 4 times the work done by the heat pump.

14. The law of thermodynamics most pertinent to this situation is the second law, which states that physicalprocesses move in the direction of increasing disorder. To decrease the disorder in one region of spacerequires work to be done, and a larger increase in disorder in another region of space.

Solutions to Problems and Conceptual Exercises

1. Picture the Problem: Thermodynamic systems change their internal energy when heat flows and when work is done.

Strategy:Use equation 18-3 to find the change in internal energy for each system.

Solution:1. (a) Calculate U: 50 J 50 J 0 JU Q W = = =

2. (b) Calculate U: ( )50 J 50 J 0 JU Q W = = =

3. (c) Calculate U: ( )50 J 50 J 100 JU Q W = = =

Insight: In part (c) 50 J of heat energy flows out of the system, and the system does work on the external world,

removing an additional 50 J of energy from the system. The net effect is 100 J of energy is removed from the system.

2. Picture the Problem: A gas expands, doing 100 J of work while its internal energy increases by 200 J.

Strategy:Use equation 18-3 to find the heat flow into the system.

Solution:Solve equation 18-3 for Q: 100 J 200 J 300 JQ W U= + = + =

Insight: The work is positive because the system is doing work on the external world. The heat flow into the systemprovides the energy to do this work plus an additional 200 J for the system to store as internal energy.


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Chapter 18: The Laws of Thermodynamics James S. Walker, Physics, 4thEdition


18 2

3. Picture the Problem: A swimmer does work and gives off heat during a workout.

Strategy: The heat and work are given in the problem. Use equation 18-3 to find the change in internal energy.

Solution:1. Write down the heat: 54.1 10 JQ=

2. Write down the work: 56.7 10 JW=

3. Find the change in internal energy: 5 5 54.1 10 J 6.7 10 J 10.8 10 JU Q W = = =

Insight: The heat is negative because the system (the swimmer) loses heat. Qis positive when the system gains heat.

4. Picture the Problem: The temperature of one mole of an ideal gas increases as heat is added.

Strategy: Use the first law of thermodynamics to determine the work, recognizing that the change in internal energy of

a monatomic gas is given by 32

.U n R T =

Solution: Solve equation 18-3 for W:

( ) ( ) ( )

3

2

3

21210 J 1mol 8.31 J/ mol K 276 K 272 K 1160 J

W Q U Q nR T = =

= =

Insight: Of the 1210 J of heat flow into the gas, only 50 J was converted to internal energy. The rest was energy forthe gas to do work on the external world.

5. Picture the Problem: Three different processes act on a system, resulting in different final states.

Strategy:Use the first law of thermodynamics (equation 18-3) to solve for the unknown quantity in each process.

Solution:1. (a)Apply equation 18-3 directly: 77 J ( 42 J) 119 JU Q W = = =

2.(b) Apply equation 18-3 directly: 77 J 42 J 35 JU Q W = = =

3. (c)Solve equation 18-3 for Q: 120 J 120 J 0Q U W= + = + =

Insight:Conservation of energy requires that the net energy flow (Qand W) into the system must equal the change ininternal energy, irregardless of the process by which the energy exchange occurs.

6. Picture the Problem: A cycle of four processes is shown on the pressurevolume diagram at right.

Strategy:Set the sum of the changes in internal energy equal to zero. Then

solve for CDU .

Solution:1. Sum thechanges in internal energy:

AB BC CD DA0U U U U + + + =

2.Solve for CDU : CD AB BC DAU U U U =

3. Insert the numeric values: ( )CD 82 J 15 J 56 J 41 JU = =

Insight:In a complete cycle the system returns to its original state, which means that the internal energy must return to

its initial value. Therefore the net change in internal energy must be zero.


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18 3

7. Picture the Problem: In a basketball game, the player does work and gives off heat in the form of perspiration.

Strategy: Use the first law of thermodynamics (equation 18-3) to solve for the change in internal energy. Calculate theheat loss from the latent heat of vaporization of the perspiration.

Solution:1. (a)Write equation 18-3using the latent heat of vaporization:

U Q W mL W = =

2.Enter the given values: ( )( )6 50.110 kg 2.26 10 J/kg 2.43 10 J 492 kJU = =

3. (b)Convert the change in energyfrom Joules to Calories:

( )( )492 kJ 0.239 kcal/kJ 117 kcal 117 CalU = = =

Insight: There are about 150 Cal in a 1.0 oz bag of potato chips.

8. Picture the Problem: A monatomic ideal gas undergoes a process in which heat is absorbed and work is done by the

gas. The process results in a change in temperature for the gas.

Strategy: Use the change in energy for a monatomic gas ( )3 f i2U n R T T = with the first law of thermodynamics to

solve for the final temperature.

Solution:1. (a)Replace Uin the first law withthe change in internal energy for a monatomic gas: ( )3 f i2

U Q W

nR T T Q W

=

=

2.Solve for the final temperature: ( )f i2

3

Q WT T

nR

= +

3. Insert the numeric values:( )

( ) ( )f

2 3280 J 722 J263 K 468 K

3 1mol 8.31 J/ mol K T

= + =

4. (b)When there are more molecules to share the energy, the average energy gained per molecule is smaller, resulting

in a decrease in the final temperature found in part (a).

Insight:Because the heat added was greater than the work done, the net change in internal energy was positive. A

positive change in internal energy results in a temperature increase. If the internal energy had decreased, thetemperature would have dropped.

9. Picture the Problem: As a car operates, its engine converts the internal energy of the gasoline into both work and heat.

Strategy:Use the first law of thermodynamics to calculate the heat given off, rel .Q

Solution:1. (a)Solve the firstlaw for the heat released.

( )rel8

5

rel

1.19 10 J/gal5.20 10 J/mi 4.24 MJ mi

25.0 mi/gal

U Q W

Q U W

=

= = =

2.(b) Increasing miles per gallon improves efficiency, resulting in a decrease of heat released to the atmosphere.

Insight:The gas mileage of a car is a measure of how much work (miles transported) the car can do for a given amount

of available energy (in gallons of gas). The maximum amount of work would occur if all of the energy was converted

to work and no heat was ejected. However, the Second Law of Thermodynamics says this is not possible.

10. Picture the Problem: A monatomic gas undergoes a process in which work is done on the gas, resulting in an increase

in temperature. During the process heat may enter or leave the gas.Strategy:Combine the change in energy for a monatomic gas ( )3 f i2U n R T T = with the first law of

thermodynamics (equation 18-3) to solve for the heat flow.

Solution:Solve equation 18-3 for Q: ( )

( ) ( ) ( )

3f i2

3

2560 J 4 mol 8.31 J/ mol K 130 C 5.9 kJ

Q W U W n R T T = + = +

= + =

Insight:It was not necessary to convert the temperature difference from Celsius degrees to kelvins in this problem

because temperature differences are the same in both scales.


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18 4

11. Picture the Problem: The three-process cycle for an ideal gas is shown by

the pressurevolume plot at right.

Strategy: For each process in which two of the parameters Q, W, or Uare known, use the first law of thermodynamics to solve for the third

parameter. Process A B is at constant volume, so use AB 0W = .

Because the three processes form a complete cycle, we can find the change

in internal energy for Process C A by setting the net change in internal

energy equal to zero.

Solution:1. (a) Find the constant volume work: AB AB (0) 0W P V P= = =

2.(b) Use the first law to solve for

the change in internal energy: AB AB AB53 J 0 53 JU Q W = = =

3. (c)Use the first law to solve forthe change in internal energy:

( )BC BC BC 280 J 130 J 150 JU Q W = = =

4. (d) Set the net change in

internal energy equal to zero: AB BC CA0U U U + + =

5. Solve for CAU : ( ) ( )CA AB BC 53 J 150 J 200 JU U U = = =

6.(e) Solve the first law for the heat transfer: CA CA CA 200 J 150 J 350 JQ U W= + = + =

Insight:This three-step process converts 17 J of heat ( )AB BC CAQ Q Q+ + into 17 J of work ( )AB BC CAW W W+ + . Any

time a cycle forms a clockwise loop on a pressurevolume diagram, heat will be transformed into work. If the cycle is

counterclockwise, the net result will be work transformed into heat.

12. Picture the Problem: The PVplots at right show three different multi-stepprocesses, labeled A, B, and C.

Strategy:The work done by a gas is equal to the area under the PVplot as

long as the gas is expanding (the volume increases). In each of the

depicted processes the gas expands so the work is positive in each case.

Find the area under each plot to determine the ranking of the work done.

Solution:1.Find WA:

( ) ( )A 1 1 2 2 3 3

30 4 kPa 1 m 0 4 kJ

W P V P V P V = + + = + + =

2.(b)Find WB: ( )( ) ( )( )3 31B 1 1 2 2 3 3 21 kPa 1 m 1 3 kPa 2 m 0 5 kJW P V P V P V = + + = + + + =

3. (c)Find WC: ( )( )3C 1 1 2 2 3 3 0 1 kPa 3 m 0 3 kJW P V P V P V = + + = + + =

4. (d) By comparing the values of the work done in each process we arrive at the ranking WC< WA< WB.

Insight:Even if process B had begun at 0 kPa of pressure and increased linearly to the point (5 m3, 3 kPa) to form a

triangle shape, the work done would be ( )( )312 3 kPa 3 m 4.5 kJ= and the ranking would remain the same.

13. Picture the Problem: An ideal gas absorbs the same amount of heat during two different constant pressure processes.

Strategy: Use the first law of thermodynamics (equation 18-3) and the work at constant pressure (equation 18-4) to

solve for the change in volume.

Solution:1. Solve equation 18-3 for the work: U Q W W Q U = =

2. Replace the work with equation 18-4

and solve for the change in volume:

Q UP V Q U V

P

= =

3. (a)Insert the numeric values with 920 J:U = 3

920 J 920 J0

110 10 PaV

= =


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18 5

4.(b) Insert the numeric values 360 J:U = 3 33

920 J 360 J5.1 10 m

110 10 PaV

= =

Insight:When the heat absorbed and change in internal energy were equal, no work was done. Therefore the volume

remained constant. If the change in internal energy were greater than the heat absorbed, then the volume would have

decreased, as work was done on the gas.

14. Picture the Problem: An ideal gas is compressed at constant pressure to one-half of its initial volume.

Strategy: Because this is a constant pressure process, solve equation 18-4 for the initial volume.

Solution:1. Write equation 18-4 in

terms of initial and final volumes:( ) ( )1f i i i i2

2

WP V V W P V V W V

P

= = =

2. Insert the numeric values:( ) 3

i 3

2 790 J0.013 m

120 10 PaV

= =

Insight:If the initial volume were larger than 0.013 m3, more work would be needed to compress the gas to half its

volume at the same pressure. For instance, it would require 1200 J to compress a gas with an initial volume of 0.020 m3to 0.010 m3at a constant pressure of 120 kPa.

15. Picture the Problem: The ideal gas expands at constant pressure.

Strategy: Because this is a constant pressure process, solve equation 18-4 for the pressure.

Solution:1. Solve equation 18-4 for the pressure: ( )f if i

W

P V V W P V V = =

2.Insert the given values:3 3

93 J60 Pa

2.3 m 0.74 mP= =

Insight:The pressure is proportional to the amount of work done. For example, compressing a gas at ten times thepressure (or 600 Pa) between the same two volumes requires ten times the work (or 930 J).

16. Picture the Problem: A monatomic ideal gas expands to twice its original volume, while the temperature is heldconstant.

Strategy:An ideal gas undergoing an isothermal process obeys the relation constant.PV= Use this relation to calculatethe ratio of final to initial pressure.

Solution:1. Write the isothermal relation in terms

of initial and final conditions and solve for f iP P :f i

i i f f

i f

P VPV PV P V= =

2. Set f i2V V= and solve:f i

i i

1

2 2

P V

P V= =

Insight:Because the pressure and volume are inversely proportional to each other, increasing one by a given factor willreduce the other by the same factor. For example, to decrease the volume to one-third its original volume, the pressure

would have to be increased by a factor of three.

17. Picture the Problem: A system that is thermally isolated from its surroundings undergoes a process in which its

internal energy increases.

Strategy:Use the first law of thermodynamics to calculate the work done during an adiabatic process ( 0Q= ).

Solution:1. Solve the first law for W, setting Q= 0: U Q W

W U =

=

2. (a) Since the internal energy increased ( 0U > ) the work must be negative, which means it is done on the system.

3.(b) Solve numerically: ( )670 J 670 JW U= = =

Insight: When no heat can enter or leave a system, any change in internal energy is equal to the work done on thesystem.


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18 6

18. Picture the Problem: The pressurevolume diagram shows five

processes through which a system passes in going from state A tostate C. The area under the curve is divided into four geometric

shapes to simplify the area measurement.

Strategy:The area under the line on the pressurevolume diagramis the work. Break the area into rectangles and triangles and add up

their areas to find the work. Use the ideal gas law to find the final

temperature and use the first law of thermodynamics to find theheat.

Solution:1. (a)Break the area into four

regions and sum the areas to get the work: AC rect tri,1 tri,2 squareW P V A A A A= = + + +

2. Find each area: ( )( )3rect 10 m 2 m 200 kPa 1600 kJA lw 3= = =

( )( )3 31 1tri,1 2 2 6 m 4 m 600 kPa 200 kPa 400 kJA bh= = =

( )( )3 31 1tri,2 2 2 8 m 6 m 600 kPa 400 kPa 200 kJA bh= = =

( )( )3 3square 8 m 6 m 400 kPa 200 kPa 400 kJA wh= = =

3. Sum all four areas:AC

1600 kJ 400 kJ 200 kJ 400 kJ 2.6 MJW = + + + =

4. (b) Use the ideal gas law to find a ratio

between volumes and temperatures:i f

i i f f

i f

,V VnR

PV nRT PV nRT T P T

= = = =

5. Solve the ratio for the final temperature: ( )3

ff i 3

i

10 m220 K 1100 K

2 m

VT T

V= = =

6. (c)Solve the first law for the heat: 32

Q U W nR T W = + = +

7. Use the ideal gas law to eliminate nR : i i

i

3

2

PVQ T W

T

= +

8. Solve numerically: ( ) ( ) ( )3

6200 kPa 2 m3 880 K 2.60 10 J 5.0 MJ2 220 K

Q = + =

Insight:The work done in an isobaric process between points A and C would be the area of the rectangle only.

Increasing the pressure as the gas expands increases the work done by the gas.

19. Picture the Problem: The pressurevolume diagram shows the two processesthrough which a system passes in going from state A to state B. The area under

the curve is divided into a rectangle and triangle to simplify the area

measurement.

Strategy:Find the area under the curve in the pressurevolume diagram,because this is equal to the work.

Solution:1. (a)Set the work equalto the area under the curve:

AB rect triW P V A A= = +

2. Find the area of the rectangle:

( )( )rect

3 3

rect

6 m 2 m 200 kPa

800 kJ

A lw

A

=

=

=


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3. Find the area of the triangle: ( ) ( )3 31 1tri 2 2 6 m 4 m 600 kPa 200 kPa 400 kJA bh= = =

4.Add the two areas:AB 800 kJ 400 kJ 1200 kJW = + =

5.(b)No, the work only depends on the area under the PVplot.

Insight:The area under a PVplot is always equal to the work done by the system. It is independent of the type of

system.

20. Picture the Problem: A monatomic ideal gas expands at constant temperature.

Strategy: Use equation 18-5 to find the work done in the isothermal process. Then

solve the first law of thermodynamics (equation 18-3) for the heat.

Solution:1. (a)Apply equation 18-5 directly: f

i

ln V

W nRT V

=

2. Insert the given data:

( ) ( )

4.33L8.00 mol 8.31 J/ mol K 245 K ln 22.0 kJ

1.12LW

= =

3.(b) Solve equation 18-3for Q: 0 22.0 kJ 22.0 kJQ U W= + = + = The heat flow is positive

so this is heat flow into the gas.

4. (c)Both answers increase by a factor of 2 because the work done is proportional to number of moles of gas.

Insight:The internal energy of a monatomic ideal gas depends only on the temperature of the gas. In an isothermalprocess the temperature remains constant, and therefore so does the internal energy.

21. Picture the Problem: A monatomic ideal gas expands at constant

temperature.

Strategy: Use the Ideal Gas Law to solve for the constant temperature.

Then use equation 18-5 to solve for the work.

Solution:1. (a)Solve the IdealGas Law for the temperature:

PVPV nRT T

nR= =

2. Insert the numeric values:( )

( )

3 3

i f

100 10 Pa 4.00 m

145 mol 8.31 J/ mol K

332 K

T T

= =

=

3.(b) Write equation 18-5,

using the Ideal Gas Law toreplace nRT PV= :

( )3

3f fi i 3

i i

4.00 mln ln 400 kPa 1.00 m ln 555 kJ

1.00 m

V VW nRT PV

V V

= = = =

Insight:If the gas had expanded at constant pressure to the same final volume, and then cooled at constant volume to

the final pressure, the total work done would have been 1200 kJ. Isothermal expansion does less work than isobaric

expansion.


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18 8


temperature.

Strategy: Use equation 18-5 and the first law of thermodynamics tocalculate the heat between any two volumes of the isothermal expansion.

Solution 1. Solve the first law for heat: Q U W= +

2. Set 0U = and use equation 18-5 for W:f

i0 ln

V

Q nRT V

= +

3 (a) Because the gas expands ( f iV V> ) heat must be positive, so heat

enters the system.

4.(b) Heat input equals the work done; work done is equal to the area under the PVplot. The area from 31.00 m to3

2.00 m is greater than the area from3

3.00 m to3

4.00 m .

5. (c) Insert the numeric values for

the expansion from3

1.00 m to3

2.00 m :( )

33

3

2.00 m400 kPa 1.00 m ln 277 kJ

1.00 mQ

= =

6.(d)Insert the numeric values for

the expansion from3

3.00 m to3

4.00 m :( )

33

3

4.00 m100 kPa 4.00 m ln 115 kJ

3.00 mQ

= =

Insight:The heat absorbed in the expansion from31.00 m to 32.00 m is equal to the heat absorbed in the expansion

from3

2.00 m to3

4.00 m , because the heat absorbed is a function of the relative volume expansion. In both of these

cases the volume doubles.

23. Picture the Problem: A monatomic ideal gas expands under constant pressure.

Strategy: Use the first law of thermodynamics to solve for the heat transfer. The

change in internal energy can be obtained from equation 17-16, and the work from

the area under the PV curve.

Solution:1. Solve the first law for Q: 32

Q U W nR T P V = + = +

2. Use the ideal gas law to eliminate nR T : ( ) ( )3 52 2

Q P V P V P V = + =

3. (a) From the heat equation we see that heat must be added to the system for the volume to expand.

4.(b) Insert given values for the heat: ( )( )3 35

130 kPa 0.93 m 0.76 m 55 kJ2

Q= =

Insight:When a monatomic ideal gas expands, the density of particles within the gas decreases. In order for thepressure to remain constant, the temperature of the gas must then increase. This increase in internal energy of the gas

necessitates a heat flow into the gas.

24. Picture the Problem: A monatomic ideal gas is thermally isolated from its surroundings. As the gas expands its

temperature decreases.

Strategy: Use the first law of thermodynamics (equation 18-3) to solve for the work done by the gas. Use

equation 17-16 to find the change in internal energy.

Solution:1. Solve the first law for W: W Q U=

2.(a) Because the process is adiabatic, Q= 0.

3. (b) Substitute 32

U n R T = (equation 17-16): 320W nR T =


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18 9

4. Insert the given values: ( )( )( )32 3.92 mol 8.31 J mol K 205 C 485 C

13.7 kJ

W=

=

5. (c)Solve the first law for thechange in internal energy:

13.7 kJU W = =

Insight:When a monatomic ideal gas expands adiabatically it does work on the external world. The energy to do this

work cannot come from the absorption of heat, so the internal energy, and therefore the temperature, must decrease.

25. Picture the Problem: An ideal gas completes a three-process cycle.

Strategy: Find the net work done by the system by measuring the areaenclosed by the PVplot. Then use the first law of thermodynamics to

calculate the heat absorbed.

Solution:1. (a)Find the area of

the triangle on the PV-plot:( )( )312 4 1 m 150 50 kPa

= 150 kJ

W=

2.(b) 0U = over a complete cycle.

3. (c)Solve the first law for the heat: 150 kJQ W U= + =

Insight:Because the change in energy for a complete cycle is zero, the work done is equal to the net heat absorbed.

26. Picture the Problem: A monatomic ideal gas undergoes a process in which its

volume increases while the pressure remains constant.

Strategy: Use the area under thePVplot to find the work done by the gas.

Calculate the initial and final temperatures from the ideal gas law. Use the

change in temperature to calculate the change in internal energy. Finally,calculate the heat added to the system using the first law of thermodynamics.

Solution:1. (a) Multiply Pby :V ( )3 3 5210 kPa 1.9 m 0.75 m 2.4 10 JW P V= = =

2.(b) Solve the ideal gas law for T:PV

TnR

=

3. Insert initial conditions:( )

( )

3 3

2

i

210 10 Pa 0.75 m3.9 10 K

49 mol 8.31 J/ mol K T

= =

4. Insert the final conditions:( )( )

3 3

2

f

210 10 Pa 1.9 m9.8 10 K

49 mol 8.31 J/ mol K T

= =

5. (c) Use equation 17-15 to solve for :U ( ) ( ) ( )3 32 25

49 mol 8.31 J/ mol K 980 K 390 K

3.6 10 J

U n R T = =

=

6.(d) Solve the first law for heat: 5360 kJ 240 kJ 6.0 10 JQ U W= + = + =

Insight:The work done by the gas is proportional to the pressure. The temperature, and thus the change in internal

energy, is also proportional to the pressure. Therefore the heat absorbed must also be proportional to the pressure. Ifthe initial pressure was 420 kPa (double that given in the problem) then the work done would be 480 kJ, the change in

internal energy 720 kJ, and the heat 1200 J.


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18 10

27. Picture the Problem: An ideal gas completes a three-process cycle.

Strategy:Use the ideal gas law to calculate the temperature at each state.

Solve the first law of thermodynamics for the heat absorbed in eachprocess.

Solution:1. (a)Solve the ideal

gas law for temperature:

PVT

nR=

2. Solve for AT : ( )( )( )( )

3 3

A150 10 Pa 1.00 m

67.5 mol 8.31 J mol K

267 K

T =

=

3.Solve for BT :( ) ( )

( )( )

3 3

B

50 10 Pa 4.00 m357 K

67.5 mol 8.31 J mol KT

= =

4. Solve for CT :( )( )

( )( )

3 3

C

50 10 Pa 1.00 m89.1 K

67.5 mol 8.31 J mol KT

= =

5.(b) A B : The temperature rises and the gas does work, so heat enters the system.

B C : The temperature drops and work is done on the gas, so heat leaves the system.

C A : The temperature rises and no work is done on or by the gas, so heat enters the system.

6. (c) Solve the first law for Q: 32

Q U W nR T W = + = +

7. Insert values for A B: ( )( )( )

( )( )

3

2

31

2

67.5 mol 8.31 J mol K 357 K 267 K

150 kPa 50 kPa 3.00 m 376 kJ

Q=

+ + =

8. Insert values for B C : ( )( ) ( )

( )

3

2

3

67.5 mol 8.31 J mol K 89.1 K 357 K

50 kPa 3.00 m 375 kJ

Q=

+ =

9. Insert values for C :A ( )( ) ( )32 67.5 mol 8.31 J mol K 267 K 89.1 K 0 150 kJQ= + =

Insight:The net work done during this cycle is equal to the net heat absorbed, or W

=

150 kJ. Although the sum of theheats in part (d) appears to be 151 kJ, that is an artifact of significant digits and rounding issues, its exactly 150 kJ.

28. Picture the Problem: A gas with initial volume iV either expands at

constant pressure to twice its initial volume, or contracts to 1/3 its original

volume.

Strategy: Use the area under the PVplot to find the work done by the gas.

Solution:1. (a) Set the work equalto the pressure times volume:

( )

( )

f i

i i

i

2

W P V V

P V V

W PV

=

=

=

2. Insert the numeric values: ( )3140 kPa 0.66 m 92 kJW= =

3.(b) Set the work equal to the

pressure times volume:( ) if i ii

2

3 3

VW P V V P V PV

= = =

4. Insert the numeric values: ( )( )32

140 kPa 0.66 m 62 kJ3

W= =

Insight:The work is positive when the volume increases and negative when the volume decreases.


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18 11

29. Picture the Problem: A system expands at constant pressure.

Strategy:Use the first law of thermodynamics to find the heat absorbed

Solution:1. Solve the first law for heat: Q U W U P V = + = +

2. (a) Find Qfor 65 J:U = ( )( )3 365 J 125 10 Pa 0.75 m

94 kJ; into the gas

Q= +

=

3.(b) Find Qfor 1850 J:U = ( )3 31850 J 125 10 Pa 0.75 m 92 kJ; into the gasQ= + =

Insight:For heat to flow out of the gas, the internal energy would need to decrease by at least 93.8 kJ, which is equal to

the amount of work done by the gas.

30. Picture the Problem: A monatomic ideal gas is adiabatically compressed.

Strategy:Use the ideal gas law to relate the initial and final conditions of the gas. Then use equation 18-9 to eliminate

the unknown volumes from the equation.

Solution:1 (a)Doing work on the system must increase the internal energy and therefore the temperature when no heat

flows into or out of the system.

2. (b)Write the ideal gas law in

terms of the initial and final states:

i i f f

i f

PV PV

PV nRT nR

T T

= = =

3. Solve for the final temperature: f ff i

i i

P VT T

P V

=

4.Now solve equation 18-9

for the ratio of volumes:

1 1

f i f

i f i

i i f f V P P

PV P V V P P

= = =

, where 53

=

5.Insert this ratio into the equation for fT : ( )( )1 1 351 1

f f f

f i i

i i i

140 kPa280 K 310 K

110 kPa

P P PT T T

P P P

= = = =

Insight:As predicted, when work was done on the gas in an adiabatic process the temperature increased.

31. Picture the Problem: A monatomic ideal gas expands where the increasein pressure is proportional to the increase in volume.

Strategy: Calculate the work done from the area under the PVplot. Use

equations 17-16 and the ideal gas law to find the change in internal energy.Lastly, use the first law to calculate the heat flow into the gas.

Solution:1. (a)Determine the work

from the area under the graph:( )( )1 i i i i2

i i

3 2

3

W V V P P

P V

= +

=

2.(b) Write the change in internalenergy in terms of equation 17-16:

f i

3 3f i2 2

U U U

nRT nRT

=

=

3. Use the ideal gas law to write in

terms of pressure and volume: ( )3 3 3f f i i f f i i2 2 2U P V P V P V P V = =

4. Write in terms of initial conditions: ( )( )3 15i i i i i i2 22 3U P V P V P V = =

5. (c)Solve the first law for heat: 15 21i i i i i i2 23Q U W P V P V P V = + = + =

Insight:Suppose the pressure were to triple as the volume triples. In this case, the work would increase to i i4P V , the

change in internal energy would increase to i i12P V , and the heat would increase to i i16P V .


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18 12

32. Picture the Problem: A gas expands to double its initial volume in three different

processes: constant pressure, isothermal, and adiabatic.

Strategy: Use the PVdiagram to rate the work because the work is the area under

the curve. Determine the ranking of final temperatures from the ideal gas law.

Solution:1. (a)The area under the constant-pressure curve is the greatest, hence this

process does the most work.

2.(b) The area under the adiabatic curve is the smallest, hence this process does the

least work.

3. Solve the ideal gas law for temperature and note that

the final temperature is proportional to the final pressure:0

f f

2VPVT P

nR nR

= =

4.(c)The constant-pressure expansion has the highest final temperature because it has the highest final pressure.

5.(d) The adiabatic expansion has the lowest final temperature, because at final volume, it has the lowest pressure.

Insight:In the isothermal process the temperature remained constant. From the graph we see that the final temperature

in the constant pressure is higher than the temperature in the isothermal process, therefore the temperature increased

during the constant pressure process. In the adiabatic process the temperature decreased.

33. Picture the Problem: You plan to add a certain amount of heat to a gas in order to raise its temperature.

Strategy:Use the principles concerned with constant volume and constant pressure processes to answer the question.

Solution:1. (a)At constant pressure, the gas expands and does work as the heat is added. Hence, only part of the heat

goes into increasing the internal energy. When heat is added at constant volume no work is done. In this case, all the

heat goes into increasing the internal energy. We conclude that if you add the heat at constant volume, the increase in

temperature is greater than the increase in temperature if you add the heat at constant pressure.

2.(b) The best explanation is II. All the heat goes into raising the temperature when added at constant volume; none

goes into mechanical work. Statement I ignores the work done during gas expansion at constant pressure, and

statement III is false.

Insight:More heat is required to increase the temperature at constant pressure because part of the input heat is

converted to the work of the expanding gas.

34. Picture the Problem: Heat is added to monatomic ideal gas to increase its temperature. The heat is added either at

constant volume or constant pressure.

Strategy:Use the molar specific heat at constant volume VC (equation 18-6) and at constant pressure PC (equation

18-7) to solve for heat.

Solution:1. (a)Write Qin terms of PC : ( ) ( ) ( )5 5

p 2 23.5 mol 8.31 J/ mol K 23 K 1.7 kJQ nR T = = =

2.(b) Write Qin terms of VC : ( ) ( ) ( )3 3

v 2 23.5 mol 8.31 J/ mol K 23 K 1.0 kJQ nR T = = =

Insight:More heat is required to increase the temperature at constant pressure because part of the input heat is

converted to the work of the expanding gas.


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18 13

35. Picture the Problem: Heat is added to a monatomic gas resulting in a temperature increase. This process is performed

at constant volume and then again at constant pressure.

Strategy:Use the molar specific heat at constant volume (equation 18-6) and at constant pressure (equation 18-7) tosolve for the temperature change for a given heat input.

Solution:1. (a)Write Qin terms

of VC and solve for :T ( )

( ) ( )

3v 2

v2 535 J2

0.95 K3 3 45 mol 8.31 J/ mol K

Q nR T

QT

nR

=

= = =

2.(b) Qin terms of PC and solve for :T

( )

( ) ( )

5p 2

p2 2 535 J0.57 K

5 5 45 mol 8.31 J/ mol K

Q nR T

QT

nR

=

= = =

Insight: A greater temperature change occurs at constant volume, because all of the heat goes into increasing the

temperature. In the constant pressure case, some of the heat is converted to work as the gas expands.

36. Picture the Problem: Heat is added to a monatomic ideal gas until the internal energy has doubled. This process is

first done at constant pressure and then at constant volume.

Strategy: Use equation 17-15 to determine the relationship between initial and final temperatures when the internal

energy doubles. Then use the molar heat at constant volume (equation 18-6) and at constant pressure (equation 18-7) tocalculate the heat added.

Solution:1. (a)Divide equation 17-15 bytemperature and equate initial and final terms:

i f

i f

3

2

U UnR

T T= =

2. Set f i2U U= and solve for Tf:f i

f i i i

i i

22

U UT T T T

U U

= = =

3. Write the heat in terms of molar

heat capacity at constant pressure:( )5P f i2Q nR T T =

4. Write final temperature in

terms of initial temperature:( )5 5P i i i2 22Q nR T T nRT = =

5. Solve numerically: ( ) ( ) ( )5P 2 2.5 mol 8.31 J/ mol K 325 K 17 kJQ = =

6.(b) Write the heat in terms of molar

heat capacity at constant volume:3 3

V i2 2Q nR T nRT = =

7. Solve numerically: ( ) ( ) ( )3V 2 2.5 mol 8.31 J/ mol K 325 K 10 kJQ = =

Insight:At 325 K the internal energy of the gas was 10 kJ. Adding 10kJ at constant volume is sufficient to double the

internal energy, but it is not sufficient at constant pressure.

37. Picture the Problem: Heat is added to monatomic ideal gas to increase its temperature. The heat is added at constant

pressure and then at constant volume.

Strategy:Use the molar specific heat at constant volume (equation 18-6) and at constant pressure (equation 18-7) tosolve for heat.

Solution:1. (a)Solve 5P 2Q nR T = for :T ( )

( )( )P

2 170 J22.9 K

5 5 2.8 mol 8.31 J mol K

QT

n R = = =

2.(b) Solve 3V 2Q nR T = for :T ( )

( )( )V

2 170 J24.9 K

3 3 2.8 mol 8.31 J mol K

QT

n R = = =

Insight: A greater temperature change occurs at constant volume, because all of the heat goes into increasing thetemperature. In the constant pressure case some of the heat is converted to work as the gas expands.


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18 14

38. Picture the Problem: A monatomic ideal gas is expanded at constant

pressure by a fixed change in volume. This process is done twice withdiffering initial volumes.

Strategy: Use the area under the PVplot to calculate the work done in

each process.

Solution:1. (a)Write the work

as area under the PVplot:

W P V=

2. Solve numerically: ( )( )6 3

3 3 3

3

1 10 m160 10 Pa 8600 cm 5400 cm 0.51 kJ

cmW

= =

3.(b) Work is directly proportional to the change in volume. Therefore, the work done by the gas in the second

expansion is equal to that done in the first expansion.

4. (c)Solve numerically: ( )( )6 3

3 3 3

3

1 10 m160 10 Pa 5400 cm 2200 cm 0.51 kJ

cmW

= =

Insight:Because the work is proportional to the change in volume, increasing the volume by 3200 cm3from any initial

volume will produce the same amount of work.

39. Picture the Problem: A monatomic ideal gas that is thermally isolated from its surroundings expands to double its

initial volume.

Strategy:Use equation 18-9 to calculate the ratio of the initial and final pressures. Use that result and the ideal gas lawto determine the ratio of the initial and final temperatures.

Solution:1. (a)Solve equation 18-9 forthe ratio of final to initial pressure:

f if f i i

i f

P V

P V P V P V

= =

2. Set final volume as twice the initial:

5 3 5 3

f i

i i

10.315

2 2

P V

P V

= = =

3.(b) Calculate the ratio of initial to finaltemperatures using the ideal gas law: ( )( )

f f f f f

i i i i i

0.315 2 0.630T P V nR P V T P V nR P V

= = = =

4 (c)Calculate fP from equation 18-9:

5/ 33

i

f i 3

f

1.2 m330 kPa 100 kPa

2.4 m

VP P

V

= = =

5. Calculate the ratio of pressures: f

i

104 kPa0.315

330 kPa

P

P= = , which matches step 2.

6.Calculate iTfrom the ideal gas law:( )( )

3 3

i i

i

330 10 Pa 1.2 m353 K

135 mol 8.31 J/ mol K

PVT

nR

= = =

7. Calculate fT from the ideal gas law: ( )( )

3 3f f

f

104 10 Pa 2.4 m222 K

135 mol 8.31 J/ mol K

PVT

nR

= = =

8. Calculate the ratio of temperatures: f

i

222 K0.630

353 K

T

T= = , which matches step 3.

Insight:When a monatomic ideal gas expands to double its volume, its pressure decreases rapidly to less than a third ofits initial pressure while its temperature decreases to a little under two-thirds of the initial Kelvin temperature.


15/36



18 15

40. Picture the Problem: A thermally isolated monatomic ideal gas is compressed causing its pressure and temperature to

increase.

Strategy: Use equation 18-9 to calculate the final volume in terms of the initial and final pressures. Then combineequation 18-9 and the ideal gas law to derive an equation for the final volume in terms of the temperatures.

Solution:1. (a)Solve equa-

tion 18-9 for the final volume: ( )1

3/ 5

3 3ii i f f f i

f

105 kPa 0.0750 m 0.0618 m

145 kPa

PP V P V V V

P

= = = =

2.(b) Combine the Ideal

Gas Law with equation 18-9

and solve for final volume:

( )

( )53

i i f f

i f

i f

i f

11/ 1

13 3f

f i

i

295 K0.0750 m 0.0835 m

317 K

P V P V

nRT nRT V V

V V

TV V

T

=

=

= = =

Insight:When a monatomic ideal gas is compressed adiabatically, its volume decreases while its temperature and

pressure increase. Because the temperature in part (b) is less than the initial temperature, the gas had to expand as itcooled. This is seen in the final volume being greater that the initial volume.

41. Picture the Problem: A monatomic ideal gas is heated at constant

volume and then expanded at constant pressure.

Strategy: Calculate the three temperatures ( i 1 f, andT T T ) using the

Ideal Gas Law. Use these temperatures together with VC and PC to

calculate the heat transferred during each process. Calculate the work

during the constant pressure process from the area under the PVplot.

Finally, use the first law of thermodynamics to find the change ininternal energy.

Solution:1. (a)Solve the Ideal Gas Law for iT:( )

( )

3 3

i ii

106 10 Pa 1.00 m212.6 K

60.0 mol 8.31 J/ mol K

P VT

nR

= = =

2. Solve the Ideal Gas Law for 1T:( )

( )

3 3

1 1

1

212 10 Pa 1.00 m425.2 K

60.0 mol 8.31 J/ mol K

P VT

nR

= = =

3.Solve the Ideal Gas Law for fT :( )

( )

3 3

f ff

212 10 Pa 3.00 m1275.6 K

60.0 mol 8.31 J/ mol K

P VT

nR

= = =

4. Find the heat for process 1: ( )

( ) ( ) ( )

3v 1 i2

3

260.0 mol 8.31 J/ mol K 212.6 K 159 kJ

Q nR T T =

= =


( ) ( ) ( )

5P f 12

5

260.0 mol 8.31 J/ mol K 850.4 K 1060 kJ

Q nR T T =

= =

6. (b) Find the work from the area underthe PVplot: ( )( )

3 3 3212 10 Pa 3.00 m 1.00 m 424 kJW P V= = =

7. (c) Use the first law to determine the totalchange in internal energy:

159 kJ 1060 kJ 424 kJ 795 kJU Q W = = + =

Insight: The change in internal energy is independent of the processes between the initial and final states. If the initial

process had been at constant pressure and the final at constant volume, the net work and heat absorbed would be lessthan in this problem, but the change in internal energy would still be the same.


16/36



18 16


pressure and then is heated at constant volume.

Strategy:Calculate the three temperatures ( i 2 f, andT T T ) using the

Ideal Gas Law. Use these temperatures together with VC and PC to

calculate the heat transferred during each process. Calculate the workduring the constant pressure process from the area under the PVplot.

Finally, use the first law of thermodynamics to find the change in

internal energy.

Solution:1. (a)Solve the Ideal Gas Law for iT:( )

( )

3 3

i i

i

106 10 Pa 1.00 m212.6 K

60.0 mol 8.31 J/ mol K

P VT

nR

= = =

2. Solve the Ideal Gas Law for 2T :( )

( )

3 3

2 22

106 10 Pa 3.00 m637.8 K

60.0 mol 8.31 J/ mol K

P VT

nR

= = =

3.Solve the Ideal Gas Law for fT :( )

( )

3 3

f f

f

212 10 Pa 3.00 m1275.6 K

60.0 mol 8.31 J/ mol K

P VT

nR

= = =


( ) ( ) ( )

5p 2 i2

5

260.0 mol 8.31 J/ mol K 425.2 K 530 kJ

Q nR T T =

= =


( ) ( ) ( )

3v f 22

3

260.0 mol 8.31 J/ mol K 637.8 K 477 kJ

Q nR T T =

= =

6. (b) Find the work from thearea under the PVplot:

( )( )3 3 3106 10 Pa 3.00 m 1.00 m 212 kJW P V= = =

7. (c) Use the first law to determine

the total change in internal energy:530 kJ 477 kJ 2 12 kJ 795 kJU Q W = = + =

Insight:The change in internal energy is independent of the processes between the initial and final states. If the initial

process had been at constant volume and the final at constant pressure, the net work and heat absorbed would be greaterthan in this problem, but the change in internal energy would still be the same.

43. Picture the Problem: A Carnot engine operates between a hot reservoir at the Kelvin temperature Thand a coldreservoir at the Kelvin temperature Tc.

Strategy:Use the expression for the efficiency of a Carnot heat engine to answer the conceptual questions.

Solution:1. (a)The efficiency of a Carnot heat engine (emax= 1 Tc/Th) (equation 18-13) depends on the ratio of Tc toTh. Doubling both temperatures leaves this ratio unchanged. We conclude that if both temperatures are doubled, the

efficiency of the engine will stay the same.

2.(b) Adding the same temperature to both Tcand Thmeans that the ratio Tc/Thwill have a value that is closer to 1.

Therefore, the efficiency of the engine will decrease.

Insight: The best way to increase the efficiency of a Carnot engine is to increase the temperature difference between

the two reservoirs. In real life the cold reservoir is usually the environment (near 300 K) and cannot be changed, so thebest option is to increase the temperature of the hot reservoir.


17/36



18 17

44. Picture the Problem: A Carnot engine can be operated with one of four sets of reservoir temperatures

Strategy:Use the expression for the efficiency of a Carnot heat engine (equation 18-13) to determine the ranking of the

efficiencies when the engine is operated between the stipulated temperature reservoirs.

Solution:1.Find eA:c, A

A

h, A

400 K1 1 0.50

800 K

Te

T= = = 2.Find eB:

c, B

B

h, B

400 K1 1 0.33

600 K

Te

T= = =

3.Find eC:c, C

Ch, C

800 K1 1 0.33

1200 K

Te

T= = = 4.Find eD:

c, D

Dh, D

800 K1 1 0.20

1000 K

Te

T= = =

5. By comparing the efficiencies we arrive at the ranking eD< eB= eC< eA.

Insight:The larger the temperature difference between the two reservoirs, the greater the efficiency of a Carnot heat

engine. A Carnot engine operating between 300 K and 1200 K would have an efficiency of 75%.

45. Picture the Problem: A heat engine does work as it extracts heat from the hot reservoir and

rejects heat to the cold reservoir.

Strategy: Use equation 18-10 to calculate hQ and then equation 18-11 to calculate the

efficiency:

Solution:1. Solve equation 18-10 for h :Q h c

h c

W Q Q

Q W Q

=

= +

2. Insert the expression from

step 1 into equation 18-11:h c

340 J0.28

340 J 870 J

W We

Q W Q= = = =

+ +

Insight:Because the work Wis always less than cW Q+ the efficiency must always be less than one.

46. Picture the Problem: A heat engine does work as it extracts heat from the hot reservoir and

rejects heat to the cold reservoir.

Strategy: Use equation 18-10 to calculate Wand then equation 18-11 to calculate the

efficiency.

Solution:1. (a)Calculate workfrom equation 18-10: h c 690 J 430 J 260 JW Q Q= = =

2.(b) Calculate efrom equation 18-11:h

260 J0.38

690 J

We

Q= = =

Insight:Decreasing cQ while keeping hQ constant will always increase the work done and increase the efficiency.

47. Picture the Problem: A Carnot heat engine does 2500 J of work as it extracts heat from thehot reservoir and rejects heat to the cold reservoir.

Strategy: Use equation 18-14 to solve for the heat input h ,Q and then calculate the heat

rejected cQ from equation 18-10.

Solution:1. (a)Solve

equation 18-14 for h :Q h

c h

2500 J8.5 kJ

1 1 290 K 410 K

WQ

T T= = =

2.(b) Solve equation 18-10 for c :Q 3

c h 8.5 10 J 2500 J 6.0 kJQ Q W= = =

Insight:Equation 18-14 shows that for two fixed reservoirs the work done by a Carnot

engine is proportional to the heat absorbed. Equation 18-10 shows that the heat rejected is

also proportional to the work.


18/36



18 18

48. Picture the Problem: A heat engine does work at a rate of 250 MW as it extracts heat from

the hot reservoir at a rate of 820 MW and rejects heat to the cold reservoir.

Strategy: Solve equation 18-10 for cQ and divide by time to determine the rate that heat is

discarded from the heat engine. Then find the efficiency from equation 18-11.

Solution:1. (a)Solve eq. 18-10

for cQ and divide by time:c h

c h Q Q W

Q Q Wt t t

= =

2. Insert the numeric values: c 838 MW 253 MW 585 MWQt

= =

3.(b) Write equation 18-11 in terms of rates:h h

253 MW0.302

838 MW

W W te

Q Q t= = = =

Insight:Typically in power plants it is the rate of heat flow and power output that is important. As shown in this

problem, the same equations used for heat and work can be used to determine heat flow rates and power output.

49. Picture the Problem: A heat engine produces power as it extracts heat from a hot temperature reservoir and rejects

some heat to a cold temperature reservoir.

Strategy: Combine equations 18-10 and 18-11 to solve for the rate of heat rejected to the cold reservoir in terms of the

efficiency and power output. Then use equation 18-11 to solve for the rate that heat must be supplied.

Solution:1. (a)Combine equations

18-10 and 18-11 to find cQ :h c

c c

1 1

W We

Q W Q

WW Q Q W

e e

= =+

+ = =

2. Divide by the time to find the exhaust rate: ( )c

1 11 548 MW 1 1.16 GW

0.320

Q W

t t e

= = =

3.(b) Write equation 18-11 as a

rate equation and solve for hQ t:h 548 MW 1.71 GW

0.32

Q W t

t e= = =

Insight:Typically in power plants it is the rate of heat flow and power output that is important. As shown in this

problem, the same equations used for heat and work can be used to determine heat flow rates and power output.

50. Picture the Problem: A heat engine produces 2700 J of work as it extracts heat from a hot

temperature reservoir and rejects heat to a cold temperature reservoir.

Strategy:Use equation 18-11 to calculate hQ and then equation 18-10 to calculate cQ :

Solution:1. (a)Solve equation 18-11 for hQ : hW

Qe

=

2. Insert the given values:h

2700 J15 kJ

0.18Q = =

3. (b): Solve equation 18-19 for cQ :4

c h 1.50 10 J 2700 J 12 kJQ Q W= = =

4. (c)Higher efficiency means less heat input is needed to produce the same work. Consequently, less heat is lost to the

surroundings. The answers in parts (a) and (b) will decrease.

Insight:As an example of higher efficiency, if 0.24e= and W= 2700 J, then h 10.0 kJQ = and c 7.6 kJ.Q =

51. Picture the Problem: The efficiency of a Carnot engine is increased by lowering the cold temperature reservoir.

Strategy:Use equation 18-13 to solve for the temperature of the cold temperature reservoir using the initial efficiency,and then the new efficiency.

Solution:1. (a)Solve equation 18-13 for cT : ( ) ( )c h max1 545 K 1 0.300 382 K T T e= = =


19/36



18 19

2.(b) The efficiency of a heat engine increases as the difference in temperature of the hot and cold reservoirs increases.

Therefore, the temperature of the low temperature reservoir must be decreased.

3. (c) Insert the new efficiency: ( )c 545 K 1 0.400 327 K T = =

Insight:The efficiency of a Carnot engine is increased when the temperature difference between the two reservoirs

increases. When the hot temperature reservoir is fixed, the efficiency can be increased by lowering the temperature of

the cold reservoir.

52. Picture the Problem: A Carnot engine produces 2200 J of work as it extracts 2500 J of heatfrom a hot temperature reservoir and rejects heat to a cold temperature reservoir.

Strategy: Use equation 18-11 to determine the efficiency. Solve equation 18-10 for the heatexhausted at the low temperature. Finally, solve equation 18-13 for the ratio of reservoir

temperatures.

Solution:1. (a)Calculate the

efficiency from equation 18-11:h

2200 J0.88

2500 J

We

Q= = =

2.(b) Solve equation 18-10 for cQ : c h 2500 J 2200 J 300 JQ Q W= = =

3. (c)Solve equation 18-13

for temperature ratio:c h

max

h c max

1 11 8.3

1 1 0.88

T Te

T T e= = = =

Insight:A heat engine with 88% efficiency would need to have the hot temperature reservoir at least 8.3 times hotterthan the cold temperature reservoir. For instance, if the cold reservoir is at room temperature (300 K) the hot

temperature reservoir would have to be at least 2500 K, plenty hot enough to melt steel!

53. Picture the Problem: A Carnot engine operates between two temperature reservoirs where the hot reservoir is 55 K

warmer than the cold reservoir.

Strategy: Insert the efficiency and reservoir temperatures into equation 18-13 and solve for the cold temperature.

Solution:1. Write the hot temperature

as h c 55T T= + in equation 18-13.c

max

c

155

Te

T=

+

2.Solve for c :T 2

c

max

55 K 55 K 55 K 55 K 445 K 4.5 10 K

0.11T

e= = = =

3.Now find hT from the given relation:2

h c 55 K = 445 K + 55 K 500 K 5.0 10 K T T= + = =

Insight:The efficiency is determined by the ratio of the temperatures, not by their differences. For example, a second

set of temperatures c = 200 KT and h = 255 KT are separated by 55 K, but a Carnot engine operating between these

two temperatures has an efficiency of 22%.

54. Picture the Problem: A Carnot engine produces work in a cycle where it exhausts 2/3 of the

input heat to the cold reservoir ( 2c h3Q Q= ).

Strategy:Calculate the efficiency from equation 18-12 and the fact that 2c h3 .Q Q= Then

use equation 18-13 to determine the reservoir temperature ratio.

Solution:1. (a) Write equation 18-12:2

hc 3

h h

11 13

QQeQ Q

= = =

2. (b) Solve equation 18-13

for the temperature ratio:c

max

h

1 21 1

3 3

Te

T= = =

Insight:In a Carnot engine, the ratio of heat exhausted to heat input will always be equal to the ratio of the temperature

of cold reservoir to the temperature of the hot reservoir, as in this case c h c h 2 3Q Q T T = = .


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18 20

55. Picture the Problem: A dozen ice cubes are frozen in a freezer that is located in a kitchen with a variable temperature.

Strategy:Determine the temperature dependence of the coefficient of performance of an ideal refrigerator in order to

answer the conceptual question.

Solution:1. (a) The coefficient of performance of an ideal refrigerator is given by cQ W(equation 18-15). Use

h cW Q Q= (equation 18-10) together with the Carnot relationshiph h

c c

Q T

Q T= to determine the temperature dependence

of the COP of an ideal refrigerator.

2.Substitute for Wand h cQ Q

in equation 18-15:

c c c

h c h c h c h c

1 1COP

1 1

Q Q T

W Q Q Q Q T T T T = = = = =

3. From this expression we can see that the smaller the temperature difference Th Tc, the greater the coefficient ofperformance. Cooling the kitchen reduces the temperature difference between the freezer compartment and the kitchen.

We conclude that if the temperature in the kitchen is decreased, the cost (work needed) to freeze a dozen ice cubes is

less than it was before the kitchen was cooled

4. (b) The best explanation is I.The difference in temperature between the inside and the outside of the refrigerator is

decreased, and hence less work is required to freeze the ice. Statements II and III are each false.

Insight:The derived result suggests that air conditioners become less efficient the larger the temperature differencebetween indoors and outdoors. This is one reason why electrical demand is so high on very hot days. Many air

conditioners are turned on during such a heat wave, and each of them operates less efficiently than they would on acooler day.

56. Picture the Problem: A refrigerator extracts heat from the cold reservoir and ejects the heat

to the hot reservoir. This process requires work to be input into the system.

Strategy:Solve equation 18-10 for the heat exhausted to the hot reservoir. Then useequation 18-15 to solve for the coefficient of performance of the refrigerator.

Solution:1. (a)Solve equation 18-10 for h :Q h cQ Q W= +

2. Insert the given values:h

110 J 480 J 0.59 kJQ = + =

3.(b) Insert the values into equation 18-15: c110 J

COP 0.23480 J

Q

W= = =

Insight:This rather inefficient refrigerator exhausts to the room over five times the amount of heat extracted from

inside the refrigerator. The additional heat comes from the work done by the refrigerator.

57. Picture the Problem: A refrigerator extracts heat from the cold reservoir and ejects the heat

to the hot reservoir. This process requires work to be input into the system.

Strategy: Solve equation 18-15 for the work needed to run the refrigerator. Then use

equation 18-10 to determine the heat exhausted to the hot reservoir.

Solution:1. (a)Solve equation 18-15 for W:

4

c 3.45 10 J 19.7 kJCOP 1.75

QW

= = =

2. (b)Solve equation 18-10 for h :Q : h c4 4

3.45 10 J 1.97 10 J 54.2 kJ

Q Q W= +

= + =

Insight:Increasing the coefficient of performance decreases work required to extract the

same heat from the inside of the refrigerator. For example, if the COP were increased to

2.75, only 12.5 kJ of work would be required to extract the 34.5 kJ.


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18 21

58. Picture the Problem: A heat pump injects 3240 J of heat into a warm room by pulling heat

from the cold outside. This process requires 345 J of work.

Strategy:Use equation 18-10 to calculate the heat removed from the outside air. Then use

the Carnot relation, h c h c= ,Q Q T T to determine the outside temperature.

Solution:1. (a)Solve

equation 18-10 for cQ :c h

3240 J 345 J 2.90 kJ

Q Q W=

= =

2.Solve the Carnot relation

for cold temperature:[ ]cc h

h

2895 J273.15 21.0 K 262.8 K

3240 J

QT T

Q

= = + =

3. Convert to Celsius:c 262.8 K 273.15 K 10 CT = =

Insight: Decreasing the outside temperature decreases the coefficient of performance, thereby increasing the work

necessary to produce the same heat output. When the external temperature drops to 20C, the same 3240 J of heatoutput requires 455 J of work.

59. Picture the Problem: An air conditioner extracts heat from a room at a rate of 11 kW.

Strategy:Modify equation 18-10 and the Carnot relation to be rate equations by dividingeach energy term by time. Then use the resulting equations to determine the mechanical

power.

Solution:1. Solve equation 18-10 for hQ t: h cQ t W t Q t = +

2. Write the Carnot relation as a rate equation: h h

c c

=Q t T

Q t T

3. Substitute the expression for hQ tfrom step 1:c h

c c

=W t Q t T

Q t T

+

4 Solve for mechanical power: ( )h

c

c1

T

P W t Q t T

= =

5. Insert numerical values: 273.15 32C11 kW 1 0.41 kW

273.15 21CP

+= =

+

Insight: The coefficient of performance for the air conditioner can be found by dividing the rate of heat flow by the

mechanical power giving 26.8.COP=

60. Picture the Problem: A reversible refrigerator can operate as a refrigerator or as a heat engine.

Strategy: Combine equations 18-10, 18-11, and 18-15 to solve for efficiency in terms of the coefficient of

performance.

Solution:1. Combine equations 18-10 and

18-11, then divide top and bottom by W:h c c

1

1

W We Q W Q Q W = = =+ +

2.Replace cQ Wusing equation 18-15:1 1

0.09091 COP 1 10.0

e= = =+ +

Insight:Writing the efficiency as1

1 COPe=

+ shows that the larger the coefficient of performance, the smaller the

efficiency. It also shows that while the COP can range from zero to infinity, the efficiency ranges from zero to one.


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18 22

61. Picture the Problem: A freezer (refrigerator) extracts heat from water to freeze liquid water into ice cubes. Electrical

energy must be supplied to the freezer to extract the heat.

Strategy:Calculate the heat that must be extracted from the water to cool the water to freezing, freeze the water into

ice, and cool the ice to 5C from the specific heats and latent heat of fusion. Then use equation 18-15 to calculate therequired electrical work.

Solution:1. Calculate the heat

necessary to freeze the water:( ) ( ) ( ) ( )c w f ice w f ice1 2 1 2Q mc T mL mc T m c T L c T = + + = + +

2.Now substitute the expressionfrom step 1 into equation 18-15:

( ) ( )

( ) ( )

( ) ( )

c

w f ice1 2

4

5

COP COP

4186 J/ kg K 15 C 33.5 10 J/kg1.5 kg1.5 10 J

4.0 2090 J/ kg K 5.0 C

Q mW c T L c T = = + +

+ = =

+

Insight:The heat extracted from the water is equal to the electrical power times the COP or 600 kJ.

62. Picture the Problem: A reversible engine can operate in reverse as a heat pump.

Strategy: Combine equations 18-11 and 18-17 to solve for efficiency in terms of the coefficient of performance.

Solution: Combine equations 18-11 and 18-17:1 1

COP 4.30.23

hQ

W e= = = =

Insight:The coefficient of performance of the heat pump is the inverse of the reversible engines efficiency. Becausethe efficiency ranges from zero to one, the coefficient of performance can range from one to infinity.

63. Picture the Problem: You rub your hands together and friction converts the mechanical energy into thermal energy.

Strategy:Use the definition of entropy change to answer the conceptual question.

Solution:1. (a) The definition of entropy change S Q T = (equation 18-18) indicates that whenever heat is transferred

there is a corresponding entropy change. In this case the mechanical energy of the hands is converted into thermal

energy Qthat is deposited into the hands, increasing their entropy. No heat is subtracted from anywhere, so we

conclude that the entropy of the universe will increase if you rub your hands together.

2.(b)The best explanation is III. The heat produced by rubbing raises the temperature of your hands and the air, which

increases the entropy. Statements I and II are each false.

Insight:All such frictional processes produce an increase in the entropy of the universe, because low-entropy

mechanical energy is converted into high-entropy thermal energy.

64. Picture the Problem: An ideal gas is expanded slowly while the temperature is held constant.


Solution:1. (a) The definition of entropy change S QT = (equation 18-18) indicates that whenever heat is transferred

there is a corresponding entropy change. In this case heat is added to the gas so that it might maintain a constantinternal energy while it expands and does work on the external world. We conclude that the entropy of the gas will

increase as it is expanded slowly and isothermally.

2.(b)The best explanation is I. Heat must be added to the gas to maintain a constant temperature, and this increases the

entropy of the gas. Statement II mistakenly supposes entropy is a function of temperature only, and statement III is

partially true (the temperature will decrease) but ignores the increase in disorder and entropy that results from the

expansion.

Insight: In this case the entropy change of the universeis zero because the process is reversible. Any entropy gained by

the gas is lost by the reservoir when the heat is extracted.

65. Picture the Problem: A gas is expanded reversibly and adiabatically.


Solution:1. (a) The definition of entropy change S QT = (equation 18-18) indicates that whenever heat is transferred

there is a corresponding entropy change. In this case no heat is exchanged because the expansion process is adiabatic.

We conclude that the entropy of the gas will stay the same as it is expanded reversibly and adiabatically.


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18 23

2.(b)The best explanation is I. The process is reversible, and no heat is added to the gas. Therefore, the entropy of the

gas remains the same. Statement II is partly true (when randomly moving molecules occupy a larger volume they

become more disordered) but the work required for an adiabatic expansion comes at the expense of internal energy, and

the temperature of the gas will decrease, making the system more ordered. Statement III is partially true (the

temperature will decrease) but entropy is not a function of temperature only.

Insight:One type of irreversibleadiabatic expansion is the free expansion of a gas into a vacuum. In this case no work

is done on the external world (because the vacuum offers no resistance to the expansion) so the internal energy and thetemperature of the gas remain the same. The gas now occupies a larger volume, however, and has become more

disordered, so the entropy increases during this process. Entropy always increases during irreversible processes.

66. Picture the Problem: Heat is added to water at its boiling point, converting the water entirely to steam.

Strategy:Use the latent heat of vaporization of water to calculate the heat added to the water. Then calculate thechange in entropy from equation 18-18.

Solution:Combine equa-tions 17-20 and 18-18:

( )54v

1.85 kg 22.6 10 J/kg1.12 10 J/K 11.2 kJ/K

100 273.15 K

m LQS

T T

= = = = =

+

Insight:As heat is added to the water, the entropy of the water increases.

67. Picture the Problem: Heat is removed from water at its freezing point, converting the water to ice.

Strategy: Use the latent heat of fusion of water to calculate the heat extracted from the water. Then calculate thechange in entropy from equation 18-18.

Solution: Combine equations 17-20 and 18-18:( )4

f3.1 kg 33.5 10 J/kg

3.8 kJ K273.15 0 K

mLQS

T T

= = = =

+

Insight:As heat is taken from the water, the entropy of the water decreases.

68. Picture the Problem: You heat a pan of water on the stove.

Strategy: Use Q mc T = (equation 16-13) to determine the amount of heat added to the water for each temperature

change, and then use avS Q T = (equation 18-18) to determine the entropy change, where Tavis the average

temperature of the water during the heating process.

Solution: Combining equations 16-13 and 18-18 we see that the entropy change, av ,S mc T T = increases linearlywith the temperature change Tand is inversely proportional to the average temperature, Tav. Noting that processes Cand D involve only a 5C temperature change, we arrive at the following ranking: D < C < B < A.

Insight:A 10C temperature increase near 30C temperature (process A) produces a larger entropy increase than a

10C increase near 40C (process B) because the average temperature Tavis smaller for process A. Using Tavis only an

approximation; the exact solution requires calculus and reveals that ( )f iln .S mc T T = A comparison of the two

approaches reveals that they differ by less than 0.01% over the temperature intervals given in this question.

69. Picture the Problem: As heat flows out of a house, the entropy of the house decreases and the entropy of the outside

increases.

Strategy:Use equation 18-18 to sum the entropy changes for inside and outside the house. The resulting equation can

be divided by time in order to determine the rate of entropy change.

Solution: Sum the inside and outside changes

in entropy, and divide each term by the time:

( )

inside outside

20.0 kW 20.0 kW9.6 W K

273.15 22 K 273.15 14.5 K

S Q t Q t

t T T

= +

= + =

+ +

Insight:Although the change in entropy of the inside was negative, the net change in entropy is positive. This will betrue for all physical situations.


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18 24

70. Picture the Problem: While descending at constant speed, the parachutists gravitational potential energy is converted

to heat, resulting in an increase in entropy.

Strategy: Use equation 18-18 to calculate the change in entropy with the heat given by the change in gravitationalpotential energy.

Solution:1. Set the heat in equation 18-18 equal

to the change in gravitational potential energy:

Q U mghS

T T T

= = =

2.Solve numerically for the change in entropy: ( )( )288 kg 9.81 m/s 380 m

1100 J/K 0.11 kJ/K21 273.15 K

S = = =+

Insight: In the absence of air resistance, the parachutists gravitational potential energy would have been converted tokinetic energy, not heat, with no increase in entropy.

71. Picture the Problem: Imperfect insulation in a house allows heat Qcto leak into the house from the outside. An airconditioner extracts that same heat from the house and expels slightly more heat outside, where the temperature is

warmer. The entropy change of the universe takes is the sum of the changes in entropy due to the heat entering the

house and the air conditioner expelling the heat outside.

Strategy: Use equation 18-18 to sum the entropy changes for inside and outside the house. The resulting equation can

be divided by time to determine the rate of entropy change. Use equation 18-10 to determine the rate that the air

conditioner expels heat to the external world.

Solution:1. (a)The entropy of the universe will increase. The air conditioner has the efficiency of a Carnot engine, so

it does not increase the entropy of the universe, but the heat leaking into the house does produce a net entropy increase.

2. (b) Sum the inside and outside changesin entropy due to the air conditioner and the

leaking insulation: in out in outAC Leak

c h c c

c h c h

Q Q Q QS

T T T T

= + + +

3. Eliminate the indoor changes in entropy

because they are equal and opposite:c h ch

h h h

Q Q QQS

T T T

= + =

4. Use equation 18-10 to eliminate h ,Q

and divide by time to obtain the rate

of entropy increase:

h c

h h h

0.41 kW 1.3 W/K

32 273.15 K

Q Q W S W t S

T T t T

= = = = =

+

Insight:The heat exchange through the air conditioner does not change the entropy of the universe, but it does increase

due to the heat flow of 11 kW into the house through the windows and doors (see problem 69).

72. Picture the Problem: A heat engine extracts 6400 J from a hot reservoir and performs

2200 J of work. The remainder of the heat is exhausted to the cold reservoir.

Strategy:Calculate the net change in entropy by summing the changes in entropy at the hot

and cold reservoirs. Use equation 18-18 to write the entropy change in terms of the heat

transfers and temperature. Finally, use equation 18-10 to find the heat exchange at the cold

reservoir.

Solution:1. Sum the entropy changes:h c

Q QST T

= +

2. Write in terms from the hot and cold reservoirs: ch

h c

QQS

T T

= +

3. Use equation 18-10 to eliminate c :Q h h

h c

Q Q WS

T T

= +


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18 25

4. Substitute numerical values: 6400 J 6400 J 2200 J 2.6 J K610 K 320 K

S

= + =

Insight:This engine operates at less than maximum efficiency. The Carnot efficiency of this engine is 0.475 and the

actual efficiency is 0.344. If the engine were running at maximum efficiency the net change in entropy would be zero.

73. Picture the Problem: An ideal gas is held in an insulated container at the temperature T. All the gas is initially in one-

half of the container, with a partition separating the gas from the other half of the container, which is a vacuum. Thepartition then ruptures, and the gas expands to fill the entire container.

Strategy:Use the first law of thermodynamics to determine the change in the internal energy of the gas. For an idealgas the internal energy depends only upon the temperature, so any change in the internal energy will produce a

corresponding change in temperature.

Solution: The change in the internal energy of the ideal gas is given by U Q W = (equation 18-3). The container is

insulated so no heat is exchanged with the external world and Q= 0. Furthermore, no work is done on the externalworld (because the vacuum offers no resistance to the expansion) so W= 0. We conclude that the internal energy does

not change and neither does the temperature. The final temperature is therefore equal to T.

Insight:This process can be considered an irreversible adiabatic expansion of the gas into a vacuum. Although the

internal energy and the temperature of the gas remain the same, the gas now occupies a larger volume and has become

more disordered, so the entropy increases during this process. Entropy always increases during irreversible processes.

74. Picture the Problem: A system is taken through the three-process cycle shown in

the PVplot shown at right.

Strategy:Whenever a system expands (volume is increasing) at nonzero pressure itdoes positive work on the external world. Likewise, if the system is compressed,

work is done on the system and the work has a negative value. Use these principles

to determine the correct sign of the work for each of the indicated processes.

Solution:1. (a)In the process A B the volume decreases, work is done on thesystem, and the work is negative.

2. (b)In the process B C the volume increases, work is done by the system, and the work is positive.

3. (c)In the process C A the volume decreases, work is done on the system, and the work is negative.

Insight:The two negative works occur at higher pressure than the positive work, so the net work done by the entire

cycle is negative. Another way to remember this is to note that for complete cycles depicted on a PVplot,counterclockwise cycles correspond to negative work and clockwise cycles correspond to positive work.

75. Picture the Problem: An ideal gas has the pressure and volume indicated by point I

in the PVplot at right. At this point its temperature is 1T.The temperature of the gas

can be increased to 2T by using the constant-volume process, I II ,or the

constant-pressure process, I III.

Strategy:The change in entropy for each process equals av ,S Q T = where Qis

the heat transferred and Tavis the average temperature. Tavis the same for the

constant-volume process as it is for the constant pressure process, so the process

with the largest heat transfer Qresults in the largest increase in entropy.

Solution:In the process I II the volume remains constant so no work is done.In the process I III the system does positive work on the external world, so theheat flow must both increase the internal energy (because of the increase in

temperature) and provide the energy to do the work. We conclude that the heat flow

and the entropy change for the process I II is less than the entropy change forthe process I III.

Insight:More heat flow is always required for such a constant pressure process than is required for a constant volume

process because of the work done on the external world during the expansion.


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Chapter 18: The Laws of Thermodynamics James

walker4 ism ch18

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