chapter 15 thermodynamics

CHAPTER 15 THERMODYNAMICS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ____________________________________________________________________________________________ 1. (c) This sign convention for Q and W is discussed in Section 15.3 (see, in particular,

Equation 15.1). 2. ΔU = −9.3 × 105 J 3. (c) According to the discussion in Section 15.4, the area under a pressure−volume graph is

the work W for any kind of process. Since the graph shows the gas being compressed, work is done on the gas.

4. (b) The work done by a gas is the area under the pressure-volume graph. The areas under

curves A and B are the same (4 ‘squares’ each), and each is greater than that under curve C (3 ‘squares’).

5. W = +1.2 × 105 J 6. (a) The first law of thermodynamics states that the heat Q is related to the change ΔU in the

internal energy and the work W by Q = ΔU + W. Since all three paths start at A and end at B, the change in the internal energy of each gas is the same. Therefore, the path that involves the greatest amount of work is the one that has greatest amount of heat added to the gas. The drawing shows that the work (which is the area under the pressure-volume graph) is greatest for path 1 and smallest for path 3.

7. (e) The first law of thermodynamics, Q = (UB − UA) + W, can be used to find the heat Q,

since UB and UA are given and W is the area under the pressure-volume graph. 8. (e) The internal energy of a monatomic ideal gas depends directly on its temperature (see

Equation 14.7). In an isothermal process the temperature does not change. Therefore, the internal energy of the gas does not change.

9. W = −1.81 × 104 J 10. (c) The temperature of the gas increases when its internal energy increases. According to

the first law of thermodynamics, ΔU = Q − W. Both heat and work can change the internal energy and, hence, the temperature of the gas. If Q = 0 J, ΔU can still increase if W is a negative number, which means that work is done on the gas.

11. (d) The work done on a monatomic gas during an adiabatic compression is given by (See

Equation 15.4) ( )3i f2W nR T T= − .

730 THERMODYNAMICS

12. (a) The change in temperature is greatest when the change ΔU in the internal energy of the gas is greatest. According to the first law of thermodynamics, ΔU = Q − W, the change in the internal energy is greatest when heat Q is added and no work (W = 0 J) is done by the gas. The gas does no work when its volume remains constant, so its change in temperature is greater than if the volume had changed.

13. (c) A more-efficient engine produces more work from the same amount of input heat, as

expressed by W = eQH (Equation 15.11). Part of the heat QH from the hot reservoir is used to perform work W, and the remainder QC is rejected to the cold reservoir. The conservation of energy states that QH = W + QC (Equation 15.12). If QH is constant and W increases, the heat QC must decrease.

14. (a) The efficiency e of a heat engine depends on the ratio QC/QH through the relation

e = 1 − QC/QH (Equation 15.13). Doubling QC and QH does not change this ratio. 15. (d) The efficiency of a Carnot engine is given by e = 1 − (TC/ TH), Equation 15.15, where

TC and TH are the Kelvin temperatures of the cold and hot reservoirs. The efficiencies of engines C, B, and A, are, respectively, 0.50, 0.20, and 0.11.

16. W2/ W1 = 1.5 17. (c) The refrigerator uses the work W done by the electrical energy to remove heat QC from

its interior and deposit heat QH into the room. In accordance with the conservation of energy, QH = QC + W. Therefore, the heat delivered into the room is greater than the electrical energy used to produce the work.

18. (e) According the conservation of energy, the heat QH delivered to the room equals the heat

QC removed from the interior of the refrigerator plus the work W; QH = QC + W. The coefficient of performance is the heat QC removed from the refrigerator divided by the work W done by the refrigerator; COP = QC /W (Equation 15.16).

19. (b) The sun loses heat, so its entropy decreases. The earth, on the other hand, gains heat so

its entropy increases. The transfer of heat from the sun to the earth, like the flow of heat from a hot reservoir to a cold reservoir, is irreversible. Therefore, the entropy of the sun-earth system increases.

20. (e) The change ΔS in entropy of the gas is given by ΔS = (Q/T)R (Equation 15.18), where Q

is the heat added and T is its temperature (which remains constant since the expansion is isothermal).

Chapter 15 Problems 731

CHAPTER 15 THERMODYNAMICS PROBLEMS 1. REASONING Since the student does work, W is positive, according to our convention.

Since his internal energy decreases, the change ΔU in the internal energy is negative. The first law of thermodynamics will allow us to determine the heat Q.

SOLUTION a. The work is W = +1.6 × 104 J . b. The change in internal energy is ΔU = –4.2 × 104 J . c. Applying the first law of thermodynamics from Equation 15.1, we find that

Q = ΔU +W = −4.2 ×104 J( ) + 1.6 ×104 J( )= −2.6 ×104 J

2. REASONING The system undergoes two processes. For the first or outgoing process, we know both the heat and the work. For the second or return process, we know only the heat and wish to determine the work. We will apply the first law of thermodynamics to both processes. To determine the work in the return process, we will use the fact that the return process brings the system back to its initial state.

SOLUTION a. As applied to the return process, the first law of thermodynamics (Equation 15.1)

indicates that ( ) ( )return return return returnreturn return or U Q W W Q UΔ = − = − Δ (1)

We know that return 114 JQ = − . However, to use Equation (1) to calculate the Wreturn, we

also need a value for ( )returnUΔ . To find this value, we begin by applying the first law of thermodynamics to the outgoing process:

( ) ( ) ( )outgoing outgoingoutgoing 165 J 312 J 147 JU Q WΔ = − = − = −

Now we use the fact that the return process brings the system back to its initial state and express the overall change in the internal energy due to the outgoing and the return process as follows:

( ) ( ) ( ) ( ) ( )overall outgoing return return outgoing0 or U U U U UΔ = Δ + Δ = Δ = − Δ (2)

732 THERMODYNAMICS

In obtaining Equation (2), we have recognized that internal energy is a function only of the state of the system, so that ( )overall 0UΔ = , because the system begins and ends in the same state. We can now substitute Equation (2) into Equation (1) and obtain that

( ) ( ) ( ) ( )return return returnreturn outgoing 114 J 147 J 261 JW Q U Q U= − Δ = + Δ = − + − = −

b. Since Wreturn is negative, work is done on the system . 3. SSM REASONING Energy in the form of work leaves the system, while energy in the

form of heat enters. More energy leaves than enters, so we expect the internal energy of the system to decrease, that is, we expect the change ΔU in the internal energy to be negative. The first law of thermodynamics will confirm our expectation. As far as the environment is concerned, we note that when the system loses energy, the environment gains it, and when the system gains energy the environment loses it. Therefore, the change in the internal energy of the environment must be opposite to that of the system.

SOLUTION a. The system gains heat so Q is positive, according to our convention. The system does

work, so W is also positive, according to our convention. Applying the first law of thermodynamics from Equation 15.1, we find for the system that

ΔU =Q −W = 77 J( )− 164 J( )= −87 J

As expected, this value is negative, indicating a decrease. b. The change in the internal energy of the environment is opposite to that of the system, so

that ΔU environment J= +87 .

4. REASONING The first law of thermodynamics, which is a statement of the conservation

of energy, states that, due to heat Q and work W, the internal energy of the system changes by an amount ΔU according to ΔU = Q − W (Equation 15.1). This law can be used directly to find ΔU.

SOLUTION Q is positive (+7.6 × 104 J) since heat flows into the system; W is also positive (+7.6 × 104 J) since work is done by the system. Using the first law of thermodynamics from Equation 15.1, we obtain

( ) ( )4 4 47.6 10 J 4.8 10 J = 2.8 10 JU Q WΔ = − = + × − + × + ×

The plus sign indicates that the internal energy of the system increases.


5. REASONING In both cases, the internal energy ΔU that a player loses before becoming exhausted and leaving the game is given by the first law of thermodynamics: U Q WΔ = − (Equation 15.1), where Q is the heat lost and W is the work done while playing the game. The algebraic signs of the internal energy change ΔU and the heat loss Q are negative, while that of the work W is positive. SOLUTION a. From the first law of thermodynamics, the work W that the first player does before leaving the game is equal to the heat lost minus the internal energy loss:

( )5 5 56.8 10 J 8.0 10 J 1.2 10 JW Q U= −Δ = − × − − × = + ×

b. Again employing the first law of thermodynamics, we find that the heat loss Q experienced by the more-warmly-dressed player is the sum of the work done and the internal energy change:

( )5 5 52.1 10 J 8.0 10 J 5.9 10 JQ W U= +Δ = × + − × = − ×

As expected, the algebraic sign of the heat loss is negative. The magnitude of the heat loss

is, therefore, 55.9 10 J× . 6. REASONING According to the discussion in Section 14.3, the internal energy U of a

monatomic ideal gas is given by 32U nRT= (Equation 14.7), where n is the number of

moles, R is the universal gas constant, and T is the Kelvin temperature. When the temperature changes to a final value of Tf from an initial value of Ti, the internal energy changes by an amount

Uf −U i

ΔU

= 32 nR Tf −Ti( )

Solving this equation for the final temperature yields f i2 .3

T U TnR

⎛ ⎞= Δ +⎜ ⎟⎝ ⎠ We are given n

and Ti , but must determine ΔU. The change ΔU in the internal energy of the gas is related to the heat Q and the work W by the first law of thermodynamics, ΔU = Q − W (Equation 15.1). Using these two relations will allow us to find the final temperature of the gas.

SOLUTION Substituting ΔU = Q − W into the expression for the final temperature gives

( )

( ) ( ) ( )

f i2

3

2 2438 J 962 J 345 K = 436 K3 3.00 mol 8.31 J/ mol K

T Q W TnR

⎛ ⎞= − +⎜ ⎟⎝ ⎠

⎧ ⎫⎪ ⎪⎡ ⎤= + − − +⎨ ⎬⎣ ⎦⎡ ⎤⋅⎪ ⎪⎣ ⎦⎩ ⎭

734 THERMODYNAMICS

Note that the heat is positive (Q = +2438 J) since the system (the gas) gains heat, and the

work is negative (W = −962 J), since it is done on the system.

7. SSM REASONING The change ΔU in the weight lifter’s internal energy is given by the first law of thermodynamics as U Q WΔ = − (Equation 15.1), where Q is the heat and W is the work. The amount of heat is that required to evaporate the water (perspiration) and is mLv (see Equation 12.5), where m is the mass of the water and Lv is the latent heat of vaporization of perspiration.

SOLUTION a. The heat required to evaporate the water is energy that leaves the weight lifter’s body

along with the evaporated water, so this heat Q in the first law of thermodynamics is negative. Therefore, we substitute vQ mL= − into the first law and obtain

( )( ) ( )v

6 5 50.150 kg 2.42 10 J/kg 1.40 10 J 5.03 10 J

U Q W mL WΔ = − = − −

= − × − × = − ×

b. Since 1 nutritional Calorie = 4186 J, the number of nutritional calories is

55.03 10 J×( ) 1Calorie4186 J21.20 10 nutritional Calories

⎛ ⎞= ×⎜ ⎟

⎝ ⎠

8. REASONING We will determine the heat by applying the first law of thermodynamics to

the overall process. This law is ΔU = Q – W (Equation 15.1). We will add the changes in the internal energy for the two steps in the process to obtain the overall change ΔU and similarly add the work values to get the overall work W.

SOLUTION Using the first law of thermodynamics from Equation 15.1, we have

or U Q W Q U WΔ = − = Δ + (1) In both steps the internal energy increases, so overall we have ΔU = 228 J + 115 J = +343 J.

In both steps the work is negative according to our convention, since it is done on the system. Overall, then, we have W = –166 J – 177 J = –343 J. With these values for ΔU and W, Equation (1) reveals that

( ) ( )343 J 343 J 0 JQ U W= Δ + = + + − =

Since the heat is zero, the overall process is adiabatic .


9. SSM REASONING According to Equation 15.2, W = PΔV, the average pressure P of the expanding gas is equal to P W V= / Δ , where the work W done by the gas on the bullet can be found from the work-energy theorem (Equation 6.3). Assuming that the barrel of the gun is cylindrical with radius r, the volume of the barrel is equal to its length L multiplied by the area (πr2) of its cross section. Thus, the change in volume of the expanding gas is ΔV L r= π 2 .

SOLUTION The work done by the gas on the bullet is given by Equation 6.3 as

2 2 3 21 1final initial2 2( ) (2.6 10 kg)[(370 m/s) 0] 180 JW m v v −= − = × − =

The average pressure of the expanding gas is, therefore,

73 2

180 J 1.2 10 Pa(0.61 m) (2.8 10 m)

WPV π −= = = ×

Δ ×

10. REASONING Equation 15.2 indicates that work W done at a constant pressure P is given

by W = PΔV. In this expression ΔV is the change in volume; ΔV = Vf – Vi, where Vf is the final volume and Vi is the initial volume. Thus, the change in volume is

WVP

Δ = (1) The pressure is known, and the work can be obtained from the first law of thermodynamics

as W = Q – ΔU (see Equation 15.1). SOLUTION Substituting W = Q – ΔU into Equation (1) gives

( ) ( ) 3 35

2780 J +3990 J9.60 10 m

1.26 10 PaW Q UVP P

−+ −− ΔΔ = = = = − ××

Note that Q is positive (+2780 J) since the system gains heat; ΔU is also positive (+3990 J)

since the internal energy of the system increases. The change ΔV in volume is negative, reflecting the fact that the final volume is less than the initial volume.

11. REASONING The work done in an isobaric process is given by Equation 15.2, W P V= Δ ;

therefore, the pressure is equal to /P W V= Δ . In order to use this expression, we must first determine a numerical value for the work done; this can be calculated using the first law of thermodynamics (Equation 15.1), U Q WΔ = − .

SOLUTION Solving Equation 15.1 for the work W, we find

31500 J–(+4500 J) = 3.0 10 JW Q U= −Δ = − ×

736 THERMODYNAMICS

Therefore, the pressure is

35

33.0 10 J 3.0 10 Pa0.010 m

WPV

− ×= = = ×Δ −

The change in volume ΔV, which is the final volume minus the initial volume, is negative

because the final volume is 0.010 m3 less than the initial volume. 12. REASONING The work done in the process is equal to the “area” under the curved line

between A and B in the drawing. From the graph, we find that there are about 78 “squares” under the curve. Each square has an “area” of

( )( )4 3 3 12.0 10 Pa 2.0 10 m 4.0 10 J−× × = × SOLUTION a. The work done in the process has a magnitude of

( )( )178 4.0 10 J 3100 JW = × = b. The final volume is smaller than the initial volume, so the gas is compressed. Therefore,

work is done on the gas so the work is negative . 13. SSM REASONING AND SOLUTION a. Starting at point A, the work done during the first (vertical) straight-line segment is

W1 = P1ΔV1 = P1(0 m3) = 0 J For the second (horizontal) straight-line segment, the work is

W2 = P2ΔV2 = 10(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1200 J For the third (vertical) straight-line segment the work is

W3 = P3ΔV3 = P3(0 m3) = 0 J For the fourth (horizontal) straight-line segment the work is

W4 = P4ΔV4 = 15(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1800 J The total work done is

W = W1 + W2 + W3 + W4 = 33.0 10 J+ × b. Since the total work is positive, work is done by the system .

14. REASONING We will determine the heat by applying the first law of thermodynamics to the overall process. This law is ΔU = Q – W (Equation 15.1). To determine the heat Q,


values for the change ΔU in the internal energy and the work W will be needed. We will use 32

U nRT= (Equation 14.7) for the internal energy of an ideal gas and W P V= Δ

(Equation 15.2) for the work done at constant pressure. In Equation 14.7, we do not know the Kelvin temperature T. However, we can use the ideal gas law PV nRT= (Equation 14.1) to deal with this lack of information.

SOLUTION a. Using the first law of thermodynamics from Equation 15.1, we have

or U Q W Q U WΔ = − = Δ + (1)

Using Equation 14.7 for the internal energy and Equation 14.1 for the ideal gas law, we have 3 3 32 2 2

or U nRT PV U P V= = Δ = Δ (2)

where we have taken advantage of the fact that the pressure P is constant. Using Equation (2) and W P V= Δ (Equation 15.2) to substitute into Equation (1), we obtain

( )( )

3 52 2

5 5 3 3 32

2.6 10 Pa 6.2 10 m 4.0 10 J

Q U W P V P V P V

−

= Δ + = Δ + Δ = Δ

= × + × = + ×

b. The heat Q is positive. Therefore, according to our convention, heat flows into the gas .

15. REASONING AND SOLUTION The work done by the expanding gas is

W = Q − ΔU = 2050 J − 1730 J = 320 J The work, according to Equation 6.1, is also the magnitude F of the force exerted on the

piston times the magnitude s of its displacement. But the force is equal to the weight mg of the block and piston, so that the work is W = Fs = mgs. Thus, we have

s = Wmg

= 320 J135 kg( ) 9.80 m/s2( ) = 0.24 m

16. REASONING AND SOLUTION The rod’s volume increases by an amount ΔV = βV0ΔT,

according to Equation 12.3. The work done by the expanding aluminum is, from Equation 15.2,

W = PΔV = PβV0ΔT = (1.01 × 105 Pa)(69 × 10–6/C°)(1.4 × 10–3 m3)(3.0 × 102 C°) = 2.9 J

17. SSM REASONING The pressure P of the gas remains constant while its volume

increases by an amount ΔV. Therefore, the work W done by the expanding gas is given by W P V= Δ (Equation 15.2). ΔV is known, so if we can obtain a value for W, we can use this

738 THERMODYNAMICS

expression to calculate the pressure. To determine W, we turn to the first law of thermodynamics U Q WΔ = − (Equation 15.1), where Q is the heat and ΔU is the change in the internal energy. ΔU is given, so to use the first law to determine W we need information about Q. According to Equation 12.4, the heat needed to raise the temperature of a mass m of material by an amount ΔT is Q cm T= Δ where c is the material’s specific heat capacity.

SOLUTION According to Equation 15.2, the pressure P of the expanding gas can be

determined from the work W and the change ΔV in volume of the gas according to

WPV

=Δ

Using the first law of thermodynamics, we can write the work as W Q U= −Δ

(Equation 15.1). With this substitution, the expression for the pressure becomes

W Q UPV V

−Δ= =Δ Δ

(1)

Using Equation 12.4, we can write the heat as Q cm T= Δ , which can then be substituted

into Equation (1). Thus,

( ) ( )( )35

3 3

1080 J/ kg C 24.0 10 kg 53.0 C 939 J3.1 10 Pa

1.40 10 m

Q U cm T UPV V

−

−

− Δ Δ −Δ= =Δ Δ

⎡ ⎤⋅ ° × ° −⎣ ⎦= = ××

18. REASONING AND SOLUTION According to the first law of thermodynamics, the change

in internal energy is ΔU = Q – W. The work can be obtained from the area under the graph. There are sixty squares of area under the graph, so the positive work of expansion is

W = 60 1.0 × 104 Pa( ) 2.0 × 10–3 m3( ) = 1200 J

Since Q = 2700 J, the change in internal energy is ΔU Q W= = =– 2700 J –1200 J 1500 J

19. REASONING AND SOLUTION Since the pan is open, the process takes place at constant

(atmospheric) pressure P0. The work involved in an isobaric process is given by Equation 15.2: W = P0ΔV. The change in volume of the liquid as it is heated is given according to Equation 12.3 as ΔV = βV0ΔT, where β is the coefficient of volume expansion. Table 12.1


gives β = 207 × 10–6 C°( )–1

for water. The heat absorbed by the water is given by

Equation 12.4 as Q = cmΔT, where ( )4186 J/ kg C°c = ⋅ is the specific heat capacity of liquid water according to Table 12.2. Therefore,

0 0 0 0 0

0( / )P V P V T P PW

Q cm T cm T c m V cβ β β

ρΔ Δ

= = = =Δ Δ

where ρ = ×1 00 103. kg / m3 is the density of the water (see Table 11.1). Thus, we find

( )( )( ) ( )5 –6 –1

603 3

1.01 10 Pa 207 10 C4 99 10

4186 J/ kg C 1.00 10 kg/m

PWQ c

βρ

× × °= = = ×

⎡ ⎤° ×⎣ ⎦⋅–.

20. REASONING The work W done when an ideal gas expands isothermally at a Kelvin

temperature T is f

ilnV

W nRTV

⎛ ⎞= ⎜ ⎟

⎝ ⎠ (Equation 15.3), where Vf and Vi are the final and initial

volumes of the gas, respectively, ( )8.31 J/ mol KR = ⋅ is the universal gas constant, and n is

the number of moles of the gas. Solving this expression for f

i

VV

gives the desired ratio.

SOLUTION Rearranging Equation 15.3, we have

f f

i iln or lnV V WW nRTV V nRT

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (1)

As Section 14.1 discusses, the number of moles n is given by the mass m divided by the

mass per mole: 6.0 g 1.5 mol

Mass per mole 4.0 g/molmn = = =

Using Equation (1), we obtain

( ) ( ) ( )f

i

9600 Jln 2.081.5mol 8.31 J/ mol K 370 K

V WV nRT

⎛ ⎞= = =⎜ ⎟ ⋅⎡ ⎤⎝ ⎠ ⎣ ⎦

Therefore, Vf /Vi = e2.08 = 8.0 .

21. SSM REASONING Since the gas is expanding adiabatically, the work done is given by

Equation 15.4 as ( )3i f2W nR T T= − . Once the work is known, we can use the first law of

thermodynamics to find the change in the internal energy of the gas.

740 THERMODYNAMICS

SOLUTION a. The work done by the expanding gas is

( ) ( ) ( ) ( ) 33 3i f2 2 5.0 mol 8.31J/ mol K 370 K 290 K 5.0 10 JW nR T T= − = ⋅ − = + ×⎡ ⎤⎣ ⎦

b. Since the process is adiabatic, Q = 0 J, and the change in the internal energy is

3 30 5.0 10 J = 5.0 10 JU Q WΔ = − = − × − × 22. REASONING When n moles of an ideal gas change quasistatically to a final volume Vf

from an initial volume Vi at a constant temperature T, the work W done is (see Equation 15.3)

f

i f

i

ln or ln

V WW nRT TV V

nRV

⎛ ⎞= =⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎜ ⎟

⎝ ⎠

(1)

where R is the universal gas constant. To determine T from Equation (1), we need a value for the work, which we do not have. However, we do have a value for the heat Q. To take advantage of this value, we note that Section 14.3 discusses the fact that the internal energy of an ideal gas is directly proportional to its Kelvin temperature. Since the temperature is constant (the neon expands isothermally), the internal energy remains constant. According to the first law of thermodynamics (Equation 15.1), the change ΔU in the internal energy is given by ΔU = Q – W. Since the internal energy U is constant, ΔU = 0, so that W = Q.

SOLUTION Substituting W = Q into the expression for T in Equation (1), we find that the

temperature of the gas during the isothermal expansion is

( ) ( )

3

3f

3i

4.75 10 J 208 K0.250 m3.00 mol 8.31 J/ mol K lnln0.100 m

QTV

nRV

×= = =⎛ ⎞ ⎛ ⎞

⋅⎡ ⎤⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠⎝ ⎠

23. REASONING We can use the first law of thermodynamics, ΔU = Q − W (Equation 15.1) to

find the work W. The heat is Q = −4700 J, where the minus sign denotes that the system (the

gas) loses heat. The internal energy U of a monatomic ideal gas is given by 32U nRT=

(Equation 14.7), where n is the number of moles, R is the universal gas constant, and T is the Kelvin temperature. If the temperature remains constant during the process, the internal energy does not change, so ΔU = 0 J.

SOLUTION The work done during the isothermal process is

= = 4700 J + 0 J = 4700 JW Q U−Δ − − The negative sign indicates that work is done on the system.


24. REASONING An adiabatic process is one for which no heat enters or leaves the system, so

Q = 0 J. The work is given as W = +610 J, where the plus sign denotes that the gas does work, according to our convention. Knowing the heat and the work, we can use the first law of thermodynamics to find the change ΔU in internal energy as ΔU = Q – W (Equation 15.1). Knowing the change in the internal energy, we can find the change in the temperature by recalling that the internal energy of a monatomic ideal gas is U = 32 nRT,

according to Equation 14.7. As a result, it follows that ΔU = 32 nRΔT. SOLUTION Using the first law from Equation 15.5 and the change in internal energy from

Equation 14.7, we have Δ ΔU Q W nR T Q W= − = − or 3

2 Therefore, we find

ΔT =

2 Q −W( )3nR

=2 0 J( )− 610 J( )⎡⎣ ⎤⎦

3 0.50 mol( ) 8.31 J/ mol ⋅K( )⎡⎣ ⎤⎦= −98 K

The change in temperature is a decrease. 25. SSM REASONING When the expansion is isothermal, the work done can be calculated

from Equation (15.3): ( )f i ln /W nRT V V= . When the expansion is adiabatic, the work done

can be calculated from Equation 15.4: 3i f2 ( )W nR T T= − .

Since the gas does the same amount of work whether it expands adiabatically or

isothermally, we can equate the right hand sides of these two equations. We also note that since the initial temperature is the same for both cases, the temperature T in the isothermal expansion is the same as the initial temperature Ti for the adiabatic expansion. We then have

3fi i f2

i ln ( )

VnRT nR T T

V⎛ ⎞

= −⎜ ⎟⎜ ⎟⎝ ⎠ or

3i ff 2

i i

( )ln

T TVV T

⎛ ⎞ −=⎜ ⎟⎜ ⎟⎝ ⎠

SOLUTION Solving for the ratio of the volumes gives

3 32 2i f if

i

( ) / (405 K 245 K)/(405 K) 1.81T T TVV

e e− −= = =

26. REASONING According to the first law of thermodynamics U Q WΔ = − (Equation 15.1),

where ΔU is the change in the internal energy, Q is the heat, and W is the work. This expression may be solved for the heat. ΔU can be evaluated by remembering that the internal energy of a monatomic ideal gas is 3

2U nRT= (Equation 14.7), where n is the

742 THERMODYNAMICS

number of moles, R = 8.31 J/(mol⋅K) is the universal gas constant, and T is the Kelvin temperature. Since heat is being added isothermally, the temperature remains constant and so does the internal energy of the gas. Therefore, ΔU = 0 J. To evaluate W we use

( )f iln /W nRT V V= (Equation 15.3), where Vf and Vi are the final and initial volumes of the gas, respectively.

SOLUTION According to the first law of thermodynamics, as given in Equation 15.1, the

heat added to the gas is Q U W= Δ +

Using the fact that ΔU = 0 J for an ideal gas undergoing an isothermal process and the fact

that f

ilnV

W nRTV

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

(Equation 15.3), we can rewrite the expression for the heat as

follows:

f

ilnV

Q U W W nRTV

⎛ ⎞= Δ + = = ⎜ ⎟⎜ ⎟⎝ ⎠

Since the volume of the gas doubles, we know that Vf = 2 Vi. Thus, it follows that

( ) ( ) ( ) if

i

2ln 2.5 mol 8.31 J/ mol K 430 K ln

VVQ nRT

V⎛ ⎞

⎡ ⎤= = ⋅⎜ ⎟ ⎣ ⎦⎜ ⎟⎝ ⎠ iV6200 J

⎛ ⎞⎜ ⎟ =⎜ ⎟⎝ ⎠

27. REASONING During an adiabatic process, no heat flows into or out of the gas (Q = 0 J).

For an ideas gas, the final pressure and volume (Pf and Vf) are related to the initial pressure

and volume (Pi and Vi) by ( )i i f f Equation 15.5 ,PV PVγ γ= where γ is the ratio of the specific

heat capacities at constant pressure and constant volume (γ = 75 in this problem). The initial and final pressures are not given. However, the initial and final temperatures are known, so we can use the ideal gas law, PV = nRT (Equation 14.1) to relate the temperatures to the pressures. We will then be able to find Vi/Vf in terms of the initial and final temperatures.

SOLUTION Substituting the ideal gas law, PV = nRT, into i i f fPV PVγ γ= gives

11i fi f i f f

i f or i

nRT nRTV V T V T V

V Vγ γ γγ −−⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Solving this expression for the ratio of the initial volume to the final volume yields

11i f

f i

V TV T

γ −⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠


The initial and final Kelvin temperatures are Ti = (21 °C + 273) = 294 K and Tf = (688 °C + 273) = 961 K. The ratio of the volumes is

( )75

11i f

f i

11961 K= = 19.3

294 K

V TV T

γ − − ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

744 THERMODYNAMICS

28. REASONING a. The work done by the gas is equal to the area under the pressure-versus-volume curve. We will measure this area by using the graph given with the problem. b. Since the gas is an ideal gas, it obeys the ideal gas law, PV = nRT (Equation 14.1). This implies that A A A B B B/ = /P V T P V T . All the variables except for TB in this relation are known. Therefore, we can use this expression to find the temperature at point B.

c. The heat Q that has been added to or removed from the gas can be obtained from the first law of thermodynamics, Q = ΔU + W (Equation 15.1), where ΔU is the change in the internal energy of the gas and W is the work done by the gas. The work W is known from part (a) of the problem. The change ΔU in the internal energy of the gas can be obtained from Equation 14.7, ( )3

B A B A2 ,U U U nR T TΔ = − = − where n is the number of moles, R is the universal gas constant, and TB and TA are the final and initial Kelvin temperatures. We do not know n, but we can use the ideal gas law (PV = nRT ) to replace nRTB by PBVB and to replace nRTA by PAVA.

SOLUTION a. From the drawing we see that the area under the curve is 5.00 “squares,” where each square has an area of ( )( )5 3 52.00 10 Pa 2.00 m 4.00 10 J.× = × Therefore, the work W done

by the gas is

( )( )5 65.00 squares 4.00 10 J/square 2.00 10 JW = × = ×

b. In the Reasoning section, we have seen that A A A B B B/ = /P V T P V T . Solving this relation for the temperature TB at point B, using the fact that PA = PB (see the graph), and taking the values for VB and VA from the graph, we have that

( )3

B B BB A A 3

A A A

10.0 m 185 K 925 K2.00 m

P V VT T T

P V V⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

c. From the Reasoning section we know that the heat Q that has been added to or removed

from the gas is given by Q = ΔU + W. The change ΔU in the internal energy of the gas is ( )3

B A B A2U U U nR T TΔ = − = − . Thus, the heat can be expressed as

Volume, m3

2.00 4.00 6.00 8.00 0 10.0 12.0

2.00

4.00

6.00

0 Pres

sure

(×10

5 Pa)

A B


( )3

B A2 = + + Q U W nR T T WΔ = − We now use the ideal gas law (PV = nRT ) to replace nRTB by PBVB and nRTA by PAVA.

The result is ( )3

B B A A2 Q P V P V W= − + Taking the values for PB, VB, PA, and VA from the graph and using the result from part a

that W = 2.00 × 106 J, we find that the heat is ( )

( )( ) ( )( )

3B B A A2

5 3 5 3 6 632 2.00 10 Pa 10.0 m 2.00 10 Pa 2.00 m 2.00 10 J 4.40 10 J

Q P V P V W= − +

⎡ ⎤= × − × + × = ×⎣ ⎦

29. SSM REASONING AND SOLUTION Step A → B The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is

( ) ( ) ( )3 32 2 1.00 mol 8.31 J/ mol K 800.0 K – 400.0 K 4990 JU nR T ⎡ ⎤Δ = Δ = =⎣ ⎦⋅

The work for this constant pressure step is W = PΔV. But the ideal gas law applies, so

( ) ( ) ( )1.00 mol 8.31 J/ mol K 800.0 K – 400.0 K 3320 JW P V nR T ⎡ ⎤= Δ = Δ = =⎣ ⎦⋅ The first law of thermodynamics indicates that the heat is

( ) ( ) ( )

32

52 1.00 mol 8.31 J/ mol K 800.0 K – 400.0 K 8310 J

Q U W nR T nR T= Δ + = Δ + Δ

⎡ ⎤= =⎣ ⎦⋅

Step B → C The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is

( ) ( ) ( )3 32 2 1.00 mol 8.31 J/ mol K 400.0 K –800.0 K 4990 JU nR T ⎡ ⎤Δ = Δ = =⎣ ⎦⋅ –

The volume is constant in this step, so the work done by the gas is W 0 J= . The first law of thermodynamics indicates that the heat is

4990 JQ U W U= Δ + = Δ = –

746 THERMODYNAMICS

Step C → D The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is

( ) ( ) ( )3 32 2 1.00 mol 8.31 J/ mol K 200.0 K – 400.0 K 2490 JU nR T ⎡ ⎤Δ = Δ = =⎣ ⎦⋅ –

The work for this constant pressure step is W = PΔV. But the ideal gas law applies, so

( ) ( ) ( )1.00 mol 8.31 J/ mol K 200.0 K – 400.0 K 1660 JW P V nR T ⎡ ⎤= Δ = Δ = =⎣ ⎦⋅ –

The first law of thermodynamics indicates that the heat is

( ) ( ) ( )

32

52 1.00 mol 8.31 J/ mol K 200.0 K – 400.0 K 4150 J

Q U W nR T nR T= Δ + = Δ + Δ

⎡ ⎤= =⎣ ⎦⋅ –

Step D → A The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is

( ) ( ) ( )3 32 2 1.00 mol 8.31 J/ mol K 400.0 K – 200.0 K 2490 JU nR T ⎡ ⎤Δ = Δ = =⎣ ⎦⋅

The volume is constant in this step, so the work done by the gas is 0 JW = The first law of thermodynamics indicates that the heat is

2490 JQ U W U= Δ + = Δ = 30. REASONING The cylinder containing the gas is perfectly insulated, so no heat can enter or

leave. Therefore, the compression of the gas is adiabatic, and the initial and final pressures (Pi, Pf) and volumes (Vi, Vf) of the gas are related by i i f fPV PVγ γ= (Equation 15.5), where

53γ = because the gas is monatomic and ideal. The final Kelvin temperature Tf of the gas is

twice the initial temperature Ti, so we have that Tf = 2Ti. We will use the ideal gas law PV nRT= (Equation 14.1) to determine the final pressure Pf of the gas from its volume and Kelvin temperature. In Equation 14.1, n is the number of moles of the gas and R = 8.31 J/(mol�K) is the universal gas constant. SOLUTION In order to make use of the ideal gas law PV nRT= (Equation 14.1), we rewrite i i f fPV PVγ γ= (Equation 15.5) so that the product PV is raised to the power γ :

( ) ( )1 1 1 1

i i i f f f i i i f f f or P P V P P V P PV P PVγ γγ γ γ γ γ γ γ γ− − − −= = (1)


In Equation (1), we have used the fact that P(a + b) = PaPb. Substituting i i iPV nRT= and

f f fPV nRT= (Equation 14.1) into Equation (1) yields

1iP nRγ− ( ) 1

i fT P nRγ γ−= ( ) 1 1

f i i f f or T P T P Tγ γ γγ γ− −= (2)

Solving Equation (2) for 1

fPγ− and then raising both sides to the power ( )1 1 γ− , we obtain

( )11

1 1i i i if i f i

f ff or

P T T TP P P P

T TT

γ γ γγ γγ γ

γ

−−− − ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (3)

Equation (3) makes use of the fact that ( )1 1aa a aP P P P= = = . Substituting Tf = 2Ti and 53γ = into Equation (3), we find that

( ) i5f 1.50 10 Pa

TP = ×

i2T

( ) ( )( )

( )5 3 1 5 3 5 25 511.50 10 Pa 8.49 10 Pa

2

⎡ − ⎤⎣ ⎦ −⎛ ⎞ ⎛ ⎞⎜ ⎟ = × = ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

31. REASONING AND SOLUTION a. Since the curved line between A and C is an isotherm, the initial and final temperatures

are the same. Since the internal energy of an ideal monatomic gas is U = (3/2)nRT, the initial and final energies are also the same, and the change in the internal energy is ΔU = 0. The first law of thermodynamics, then, indicates that for the process A→B→C, we have

0 or U Q W Q WΔ = = − =

The heat is equal to the work. Determining the work from the area beneath the straight line

segments AB and BC, we find that

( )( )5 3 3 4– 4.00 10 Pa 0.400 m – 0.200 m –8.00 10 JQ W= = × = ×

b. The minus sign is included because the gas is compressed, so that work is done on the gas. Since the answer for Q is negative, we conclude that heat flows out of the gas .

32. REASONING AND SOLUTION a. The final temperature of the adiabatic process is given by solving Equation 15.4 for Tf.

[ ]f i 3 32 2

825 J393 K – 327 K(1.00 mol) 8.31 J/(mol K)

WT TnR

= − = =⋅

748 THERMODYNAMICS

b. According to Equation 15.5 for the adiabatic expansion of an ideal gas, i i f fPV PVγ γ= . Therefore,

if i

f

PV V

Pγ γ ⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

From the ideal gas law, PV nRT= ; therefore, the ratio of the pressures is given by

i i f

f f i

P T VP T V

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Combining the previous two equations gives

i ff i

f i

T VV V

T Vγ γ ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Solving for Vf we obtain

( ) ( )1 1f i i if i

f f i f or

V T V TV V

V T V T

γ γγ γ− −⎛ ⎞⎛ ⎞ ⎛ ⎞

= =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

( )( ) ( )1/ 1 1/ 1

1i if i i

f f

T TV V V

T T

γ γγ

− −−⎡ ⎤⎛ ⎞ ⎛ ⎞

= =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Therefore,

1/( 1) 1/(2/3) 3/ 23 3 3i

f if

393 K 393 K (0.100 m ) (0.100 m ) 0.132 m327 K 327 K

TV V

T

γ −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

33. SSM REASONING AND SOLUTION Let the left be side 1 and the right be side 2. Since

the partition moves to the right, side 1 does work on side 2, so that the work values involved satisfy the relation 1 2W W= − . Using Equation 15.4 for each work value, we find that

( ) ( )3 3

2 21i 1f 2i 2f

21f 2f 1i 2i

– – or

525 K 275 K 8.00 10 K

nR T T nR T T

T T T T

= −

+ = + = + = ×

We now seek a second equation for the two unknowns 1f 2fandT T . Equation 15.5 for an

adiabatic process indicates that 1i 1i 1f 1fP V P Vγ γ= and 2i 2i 2f 2fP V P Vγ γ= . Dividing these two equations and using the facts that 1i 2i 1f 2fandV V P P= = , gives

1i 1i 1f 1f 1i 1f

2i 2f2i 2i 2f 2f

orP V P V P V

P VP V P V

γγ γ

γ γ

⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠


Using the ideal gas law, we find that

γ γ1i 1f 1i 1i 1f 1f

2i 2i 2i 2f 2f2f

/ /= becomes =

/ /P V nRT V nRT PP nRT V nRT PV

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Since 1i 2i 1f 2fandV V P P= = , the result above reduces to

1/ 1/1 1f 1f 1i

2 2f 2f 2i

525 Kor 1.474275 K

i

i

T T T TT T T T

γ γ γ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

Using this expression for the ratio of the final temperatures in 2

1f 2f 8.00 10 KT T+ = × , we find that

1f 2fa. 477 K and b. 323KT T= =

34. REASONING The heat Q that must be added to n moles of a monatomic ideal gas to raise its temperature by ΔT Kelvin degrees under conditions of constant pressure is PQ C n T= Δ

(Equation 15.6), where 52PC R= (Equation 15.7) is the molar specific heat capacity of a

monatomic ideal gas and ( )8.31 J/ mol KR = ⋅ is the universal gas constant. SOLUTION Substituting Equation 15.7 into Equation 15.6 shows that the heat is

( )52PQ C n T R n T= Δ = Δ (1)

As Section 14.1 discusses, the number of moles n is given by the mass m divided by the

mass per mole: 8.0 g 0.20 mol

Mass per mole 39.9 g/molmn = = =

With this value for n, Equation (1) gives

( ) ( ) ( )( )5 52 2 8.31 J/ mol K 0.20 mol 75 K 310 JpQ C n T R n T= Δ = Δ = ⋅ =⎡ ⎤⎣ ⎦

____________________________________________________________________________________________ 35. SSM REASONING AND SOLUTION According to the first law of thermodynamics

(Equation 15.1), f iU U U Q WΔ = − = − . Since the internal energy of this gas is doubled by the addition of heat, the initial and final internal energies are U and 2U, respectively. Therefore,

f i 2U U U U U UΔ = − = − = Equation 15.1 for this situation then becomes U Q W= − . Solving for Q gives

750 THERMODYNAMICS

Q U W= + (1) The initial internal energy of the gas can be calculated from Equation 14.7:

( ) ( ) ( )3 3 42 2

2.5 mol 8.31 J/ mol K 350 K 1.1 10 JU nRT= = ⋅ = ×⎡ ⎤⎣ ⎦ a. If the process is carried out isochorically (i.e., at constant volume), then W = 0, and the

heat required to double the internal energy is

40 1.1 10 JQ U W U= + = + = ×

b. If the process is carried out isobarically (i.e., at constant pressure), then W P V= Δ , and Equation (1) above becomes

Q U W U P V= + = + Δ (2) From the ideal gas law, PV nRT= , we have that P V nR TΔ = Δ , and Equation (2) becomes Q U nR T= + Δ (3) The internal energy of an ideal gas is directly proportional to its Kelvin temperature. Since

the internal energy of the gas is doubled, the final Kelvin temperature will be twice the initial Kelvin temperature, or ΔT = 350 K. Substituting values into Equation (3) gives

4 41.1 10 J (2.5 mol)[8.31 J/(mol K)](350 K) 1.8 10 JQ = × + ⋅ = ×

36. REASONING Under constant-pressure conditions, the heat QP required to raise the

temperature of an ideal gas is given by P PQ C n T= Δ (Equation 15.6), where CP is the molar specific heat capacity at constant pressure, n is the number of moles of the gas, and ΔT is the change in the temperature. The molar specific heat capacity CP at constant pressure is greater than the molar specific heat capacity at constant volume CV, as we see from

P VC C R= + (Equation 15.10), where R = 8.31 J/(mol�K). Equation 15.6 also applies to the constant-volume process, so we have that V VQ C n T= Δ , where QV = 3500 J is the heat required for the constant-volume process. We will use these relations, and the fact that the change in temperature and the number of moles of the gas are the same for both processes, to find the heat QP required for the constant-pressure process. SOLUTION We begin by substituting P VC C R= + (Equation 15.10) into P PQ C n T= Δ (Equation 15.6) to obtain ( )P P VQ C n T C R n T= Δ = + Δ (1)


Solving V VQ C n T= Δ (Equation 15.6) for CV yields VV

QC

n T=

Δ. Substituting this result

into Equation (1), we find that

( ) VP V V

QQ C R n T R n T Q Rn T

n T⎛ ⎞

= + Δ = + Δ = + Δ⎜ ⎟Δ⎝ ⎠ (2)

Therefore, the heat required for the constant-pressure process is

( ) ( )[ ]( )P 3500 J 1.6 mol 8.31 J mol K 75 K 4500 JQ = + ⋅ = 37. SSM REASONING When the temperature of a gas changes as a result of heat Q being

added, the change ΔT in temperature is related to the amount of heat according to ( ) Equation 15.6 ,Q Cn T= Δ where C is the molar specific heat capacity, and n is the

number of moles. The heat QV added under conditions of constant volume is

V V V,Q C n T= Δ where CV is the specific heat capacity at constant volume and is given by

( )3V 2 Equation 15.8C R= and R is the universal gas constant. The heat QP added under

conditions of constant pressure is P P P ,Q C n T= Δ where CP is the specific heat capacity at

constant pressure and is given by ( )5P 2 Equation 15.7 .C R= It is given that QV = QP, and

this fact will allow us to find the change in temperature of the gas whose pressure remains constant.

SOLUTION Setting QV = QP, gives

CVnΔTV

QV

= CPnΔTP

QP

Algebraically eliminating n and solving for ΔTP, we obtain

( )3

V 2P V 5

P 275 K 45 K

RCT T

C R⎛ ⎞⎛ ⎞

Δ = Δ = =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

38. REASONING The gas is in a rigid container, so we conclude that the heating takes place at

constant volume. For a constant-volume process involving an ideal monatomic gas, the amount Q of heat transferred is given by VQ C n T= Δ (Equation 15.6), where 3

V 2C R= (Equation 15.8) is the molar specific heat capacity at constant volume, R is the universal gas constant, n is the number of moles of gas, and ΔT is the change in temperature. SOLUTION Solving VQ C n T= Δ (Equation 15.6) for the number n of moles of the gas, we obtain

752 THERMODYNAMICS

V

QnC T

=Δ

(1)

Substituting 3

V 2C R= (Equation 15.8) and ΔT = Tf − Ti into Equation (1) yields

( )( ) ( )( )

( ) ( )

3V f if i2

23

2 8500 J11 mol

3 8.31 J/ mol K 279 K 217 K

Q Q QnC T R T TR T T

= = =Δ −−

= =⋅ −⎡ ⎤⎣ ⎦

39. REASONING AND SOLUTION a. The amount of heat needed to raise the temperature of the gas at constant volume is given

by Equations 15.6 and 15.8, Q = n CV ΔT. Solving for ΔT yields

( )( )3

23

v 2

5.24 10 J 1.40 10 K3.00 mol

QTnC R

×Δ = = = ×

b. The change in the internal energy of the gas is given by the first law of thermodynamics

with W = 0, since the gas is heated at constant volume:

3 35.24 10 J 0 = 5.24 10 JU Q WΔ = − = × − ×

c. The change in pressure can be obtained from the ideal gas law,

( ) ( )23

3

3.00 mol 1.40 10 K2.33 10 Pa

1.50 m

RnR TPV

×ΔΔ = = = ×

40. REASONING According to Equations 15.6 and 15.7, the heat supplied to a monatomic

ideal gas at constant pressure is PQ C n T= Δ , with 5P 2C R= . Thus, 5

2Q nR T= Δ . The percentage of this heat used to increase the internal energy by an amount ΔU is

52

Percentage 100 % 100 %U UQ nR T

⎛ ⎞⎛ ⎞Δ Δ= × = ×⎜ ⎟⎜ ⎟ ⎜ ⎟Δ⎝ ⎠ ⎝ ⎠ (1)

But according to the first law of thermodynamics, U Q WΔ = − . The work W is W P V= Δ ,

and for an ideal gas P V nR TΔ = Δ . Therefore, the work W becomes W P V nR T= Δ = Δ and the change in the internal energy is 5 3

2 2U Q W nR T nR T nR TΔ = − = Δ − Δ = Δ . Combining this expression for ΔU with Equation (1) yields a numerical value for the percentage of heat that is used to increase its internal energy.


SOLUTION a. The percentage is

32

5 52 2

3Percentage 100 % 100 % 100 % 60.0 %5

nR TUnR T nR T

⎛ ⎞ ⎛ ⎞ΔΔ ⎛ ⎞= × = × = × =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟Δ Δ ⎝ ⎠⎝ ⎠ ⎝ ⎠

b. The remainder of the heat, or 40.0% , is used for the work of expansion. 41. REASONING AND SOLUTION The change in volume is ΔV = – sA, where s is the

distance through which the piston drops and A is the piston area. The minus sign is included because the volume decreases. Thus,

s VA

= –Δ The ideal gas law states that ΔV = nRΔT/P. But 5

2pQ C n T Rn T= Δ = Δ . Thus,

ΔT = Q / 5

2 Rn( ) . Using these expressions for ΔV and ΔT, we find that

s = –nRΔT / PA

=–nR Q / 5

2 Rn( )⎡⎣

⎤⎦

PA= –Q

52 PA

=– –2093 J( )

52 1.01× 105 Pa( ) 3.14 × 10–2 m2( ) = 0.264 m

42. REASONING The heat added is given by Equation 15.6 as Q = CV n ΔT, where CV is the

molar specific heat capacity at constant volume, n is the number of moles, and ΔT is the change in temperature. But the heat is supplied by the heater at a rate of ten watts, or ten joules per second, so Q = (10.0 W)t, where t is the on-time for the heater. In addition, we know that the ideal gas law applies: PV = nRT (Equation 14.1). Since the volume is constant while the temperature changes by an amount ΔT, the amount by which the pressure changes is ΔP. This change in pressure is given by the ideal gas law in the form (ΔP)V = nR(ΔT).

SOLUTION Using Equation 15.6 and the expression Q = (10.0 W)t for the heat delivered

by the heater, we have

( ) or 10.0 W or 10.0 WV

V VC n T

Q C n T t C n T tΔ

= Δ = Δ = Using the ideal gas law in the form (ΔP)V = nR(ΔT), we can express the change in

temperature as ΔT = (ΔP)V/nR. With this substitution for ΔT, the expression for the time becomes

754 THERMODYNAMICS

t =

CV n ΔP( )V10.0 W( )nR

According to Equation 15.8, CV = 32 R for a monatomic ideal gas, so we find

( )( )

( )( )

( )( )( )

4 3 33 5.0 10 Pa 1.00 10 m33 7.5 s2 10.0 W 2 10.0 W 2 10.0 W

P VRn P VtnR

−× ×ΔΔ= = = =

43. REASONING AND SOLUTION Let P, V, and T represent the initial values of pressure,

volume, and temperature. The first process is isochoric, so

Q1 = CVn ΔT1 = (3R/2)n ΔT1 The ideal gas law for this process gives ΔT1 = 2PV/(nR), so Q1 = 3PV. The second process is isobaric, so

Q2 = CPn ΔT2 = (5R/2)n ΔT2 The ideal gas law for this process gives ΔT2 = 3PV/nR, so Q2 = (15/2) PV. The total heat is

Q = Q1 + Q2 = (21/2) PV. But at conditions of standard temperature and pressure (see Section 14.2), P = 1.01 × 105 Pa

and V = 22.4 liters = 22.4 × 10–3 m3, so

( )( )5 3 3 421 212 2 1.01 10 Pa 22.4 10 m 2.38 10 JQ PV −= = × × = ×

44. REASONING The efficiency of the engine is H/e W Q= (Equation 15.11), where W is

the magnitude of the work and HQ is the magnitude of the input heat. In addition, energy

conservation requires that H CQ W Q= + (Equation 15.12), where CQ is the rejected heat.

SOLUTION a. Solving Equation 15.11 for HQ , we can find the magnitude of the input heat:

HH

5500 J or 8600 J0.64

W We Q

Q e= = = =


b. Solving Equation 15.12 for CQ , we can find the magnitude of the rejected heat:

H C C H or 8600 J 5500 J 3100 JQ W Q Q Q W= + = − = − = ______________________________________________________________________________ 45. SSM REASONING

a. The efficiency e of a heat engine is given by H

We

Q= (Equation 15.11), where |QH| is the

magnitude of the input heat needed to do an amount of work of magnitude |W|. The “input energy” used by the athlete is equal to the magnitude |ΔU| of the athlete’s internal energy change, so the efficiency of a “human heat engine” can be expressed as

or W W

e UU e

= Δ =Δ

(1)

b. We will use the first law of thermodynamics U Q WΔ = − (Equation 15.1) to find the magnitude |Q| of the heat that the athlete gives off. We note that the internal energy change ΔU is negative, since the athlete spends this energy in order to do work. SOLUTION a. Equation (1) gives the magnitude of the internal energy change:

455.1 10 J 4.6 10 J

0.11W

Ue

×Δ = = = ×

b. From part (a), the athlete’s internal energy change is 54.6 10 JUΔ = − × . The first law of dynamics U Q WΔ = − (Equation 15.1), therefore, yields the heat Q given off by the athlete:

5 4 54.6 10 J 5.1 10 J 4.1 10 JQ U W= Δ + = − × + × = − ×

The magnitude of the heat given off is, thus, 54.1 10 J× .

46. REASONING According to the definition of efficiency given in Equation 15.11, an engine

with an efficiency e does work of magnitude HW e Q= , where HQ is the magnitude of the input heat. We will apply this expression to each engine and utilize the fact that in each case the same work is done. We expect to find that engine 2 requires less input heat to do the same amount of work, because it has the greater efficiency.

SOLUTION Applying Equation 15.11 to each engine gives

W = e1 QH1

Engine 1

and W = e2 QH2

Engine 2

Since the work is the same for each engine, we have

756 THERMODYNAMICS

11 H1 2 H2 H2 H1

2 or

ee Q e Q Q Q

e⎛ ⎞

= = ⎜ ⎟⎜ ⎟⎝ ⎠

It follows, then, that

( )1H2 H1

2

0.18 5500 J 3800 J0.26

eQ Q

e⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

which is less than the input heat for engine 1, as expected.

47. REASONING According to Equation 15.11, the efficiency of a heat engine is H/e W Q=, where W is the magnitude of the work and HQ is the magnitude of the input heat. Thus,

the magnitude of the work is HW e Q= . We can apply this result before and after the tune-up to compute the extra work produced.

SOLUTION Using Equation 15.11, we find the work before and after the tune-up as

follows:

Before Before H After After H and W e Q W e Q= = Subtracting the “before” equation from the “after” equation gives

( ) ( )After Before After H Before H After Before H 0.050 1300 J 65 JW W e Q e Q e e Q− = − = − = =

48. REASONING The efficiency e of a heat engine is given by H/e W Q= (Equation 15.11),

where W is the magnitude of the work and HQ is the magnitude of the input heat, which

is the 4.1 × 106 J of energy generated by the climber’s metabolic processes. The work Wnc done in climbing upward is related to the vertical height of the climb via the

work-energy theorem (see Equation 6.8), which is

Wnc = KEf + PEf

Final total mechanical energy

− KE0 + PE0( )

Initial total mechanical energy

Here, Wnc is the net work done by nonconservative forces, in this case the work done by the

climber in going upward. Since the climber starts at rest and finishes at rest, the final kinetic energy KEf and the initial kinetic energy KE0 are zero. As a result, we have Wnc = W = PEf – PE0, where PEf and PE0 are the final and initial gravitational potential energies, respectively. Equation 6.5 gives the gravitational potential energy as PE = mgh, where m is the mass of the climber, g is the magnitude of the acceleration due to gravity,


and h is the vertical height of the climb. Taking the height at her starting point to be zero, we then have Wnc = W = mgh.

SOLUTION Using H/e W Q= (Equation 15.11) for the efficiency of a heat engine and

relating the magnitude W of the work to the height h of the climb via the work-energy theorem as W = mgh, we find that

( )( )( )2

6H H

52 kg 9.80 m/s 730 m0.091

4.1 10 JW mgheQ Q

= = = =×

49. SSM REASONING The efficiency e of an engine can be expressed as (see Equation

15.13) C H1 /e Q Q= − , where CQ is the magnitude of the heat delivered to the cold

reservoir and HQ is the magnitude of the heat supplied to the engine from the hot reservoir.

Solving this equation for CQ gives ( )C H1Q e Q= − . We will use this expression twice, once for the improved engine and once for the original engine. Taking the ratio of these expressions will give us the answer that we seek.

SOLUTION Taking the ratio of the heat rejected to the cold reservoir by the improved

engine to that for the original engine gives ( )( )

C, improved improved H, improved

C, original original H, original

1

1

Q e Q

Q e Q

−=

−

But the input heat to both engines is the same, so H, improved H, originalQ Q= . Thus, the ratio

becomes

C, improved improved

originalC, original

1 1 0.42 0.751 1 0.23

Q e

eQ

− −= = =− −

50. REASONING The efficiency of either engine is given by Equation 15.13,

( )C H1 /e Q Q= − . Since engine A receives three times more input heat, produces five times more work, and rejects two times more heat than engine B, it follows that

HA HB3Q Q= , A B5W W= , and CA CB2Q Q= . As required by the principle of energy conservation for engine A (Equation 15.12),

758 THERMODYNAMICS

QHA

3QHB

= QCA

2 QCB

+ WA

5WB

Thus, HB CB B3 2 5Q Q W= + (1)

Since engine B also obeys the principle of energy conservation (Equation 15.12),

HB CB BQ Q W= + (2) Substituting HBQ from Equation (2) into Equation (1) yields

CB B CB B3( ) 2 5Q W Q W+ = +

Solving for BW gives

1B CB2W Q=

Therefore, Equation (2) predicts for engine B that

3HB CB B CB2Q Q W Q= + =

SOLUTION a. Substituting CA CB2Q Q= and HA HB3Q Q= into Equation 15.13 for engine A, we

have

( )CA CB CB

A 3HA HB CB2

2 2 4 51 1 1 19 93 3

Q Q Qe

Q Q Q= − = − = − = − =

b. Substituting 3

HB CB2Q Q= into Equation 15.13 for engine B, we have

CB CBB 3

HB CB2

2 11 1 13 3

Q Qe

Q Q= − = − = − =

51. REASONING We will use the subscript “27” to denote the engine whose efficiency is

27.0% (e27 = 0.270) and the subscript “32” to denote the engine whose efficiency is 32.0% (e32 = 0.320). In general, the efficiency eCarnot of a Carnot engine depends on the Kelvin temperatures, TC and TH, of its cold and hot reservoirs through the relation (see Equation 15.15) eCarnot = 1 − (TC/TH). Solving this equation for the temperature TC, 32 of the engine whose efficiency is e32 gives TC, 32 = (1 − e32)TH, 32. We are given e32, but do not know the temperature TH, 32. However, we are told that this temperature is the same as that of the hot reservoir of the engine whose efficiency is e27, so TH, 32 = TH, 27. The temperature


TH, 27 can be determined since we know the efficiency and cold reservoir temperature of this engine.

SOLUTION The temperature of the cold reservoir for engine whose efficiency is e32 is TC, 32 = (1 − e32)TH, 32. Since TH, 32 = TH, 27, we have that

TC, 32 = (1 − e32)TH, 27 (1)

The efficiency e27 is given by Equation 15.15 as e27 = 1 − (TC, 27/TH, 27). Solving this equation for the temperature TH, 27 of the hot reservoir and substituting the result into Equation 1 yields

( )32C, 32 C, 27

27

1 1 0.320 275 K 256 K1 1 0.270e

T Te

⎛ ⎞− −⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠

52. REASONING The maximum efficiency of the engine is the efficiency that a Carnot engine would have operating with the same hot and cold reservoirs. Thus, the maximum efficiency

is CCarnot

H1T

eT

= − (Equation 15.15), where TC and TH are the Kelvin temperatures of the

cold and the hot reservoir, respectively. SOLUTION Using Equation 15.15 for eCarnot and recognizing that the engine has an

efficiency e that is three-fifths the maximum or Carnot efficiency, we obtain 3 3 3C

Carnot5 5 5H

620 K1 1 0.21950 K

Te e

T⎛ ⎞ ⎛ ⎞= = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

______________________________________________________________________________ 53. SSM REASONING The efficiency e of a Carnot engine is given by Equation 15.15,

C H1 ( / )e T T= − , where, according to Equation 15.14, C H C H/ /Q Q T T= . Since the

efficiency is given along with CT and CQ , Equation 15.15 can be used to calculate HT .

Once HT is known, the ratio C H/T T is thus known, and Equation 15.14 can be used to

calculate HQ . SOLUTION a. Solving Equation 15.15 for HT gives

CH

378 K 1260 K1– 1–0.700T

Te

= = =

760 THERMODYNAMICS

b. Solving Equation 15.14 for HQ gives

4HH C

C

1260 K (5230 J) 1.74 10 J378 K

TQ Q

T⎛ ⎞ ⎛ ⎞= = = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

54. REASONING We seek the ratio C, f C, iT T of the final Kelvin temperature of the cold

reservoir to the initial Kelvin temperature of the cold reservoir. The Kelvin temperature TH of the hot reservoir is constant, and it is related to the temperature of the cold reservoir by

the Carnot efficiency: CCarnot

H1T

eT

= − (Equation 15.15). We will solve Equation 15.15 for

the temperature TC of the cold reservoir, and use the resulting expression to determine the ratio C, f C, iT T . Because the temperature of the cold reservoir increases, we expect the ratio to have a value greater than 1.

SOLUTION Solving CCarnot

H1T

eT

= − (Equation 15.15) for the cold-reservoir-temperature

TC yields

( )CCarnot C H Carnot

H1 or 1

Te T T e

T= − = − (1)

Applying Equation (1) to both the initial temperature and the final temperature, we find that ( ) ( )C,i H Carnot,i C,f H Carnot,f1 and 1T T e T T e= − = − (2)

Taking the ratio of Equations (2), we obtain

HC,f

C,i

TTT

=( )Carnot,f

H

1 e

T

−

( )Carnot,f

Carnot,iCarnot,i

1 1 0.70 1.21 1 0.751

eee

− −= = =− −−

55. REASONING The smallest possible temperature of the hot reservoir would occur when the

engine is a Carnot engine, since it has the greatest efficiency of any engine operating between the same hot and cold reservoirs. The efficiency eCarnot of a Carnot engine is (see Equation 15.15) eCarnot = 1 − (TC/TH), where TC and TH are the Kelvin temperatures of its cold and hot reservoirs. Solving this equation for TH gives TH = TC/(1 − eCarnot). We are given TC, but do not know eCarnot. However, the efficiency is defined as the magnitude W

of the work done by the engine divided by the magnitude HQ of the input heat from the hot

reservoir, so Carnot H/e W Q= (Equation 15.11). Furthermore, the conservation of energy


requires that the magnitude HQ of the input heat equals the sum of the magnitude W of

the work done by the engine and the magnitude CQ of the heat it rejects to the cold

reservoir, H CQ W Q= + . By combining these relations, we will be able to find the temperature of the hot reservoir of the Carnot engine.

SOLUTION From the Reasoning section, the temperature of the hot reservoir is TH = TC/(1 − eCarnot). Writing the efficiency of the engine as Carnot H/e W Q= , the expression for the temperature becomes

C CH

Carnot

H

= 1 1

T TT

WeQ

=− −

From the conservation of energy, we have that H CQ W Q= + . Substituting this expression

for HQ into the one above for TH gives

C CH

C

285 K 1090 K18 500 J11 1 18 500 J 6550 J

T TT

W WW W Q

= = = =−− − ++

56. REASONING The efficiency eCarnot of a Carnot engine is CCarnot

H1T

eT

= −

(Equation 15.15), where TC and TH are, respectively, the Kelvin temperatures of the cold and hot reservoirs. After the changes are made to the temperatures, this same equation still applies, except that the variables must be labeled to denote the new values. We will use a “prime” for this purpose. From the original efficiency and the information given about the changes made to the temperatures, we will be able to obtain the new temperature ratio and, hence, the new efficiency.

SOLUTION After the reservoir temperatures are changed, the engine has an efficiency that,

according to Equation 15.15, is C

CarnotH

1T

eT′

′ = −′

where the “prime” denotes the new engine. Using unprimed symbols to denote the original

engine, we know that C C2T T′ = and H H4T T′ = . With these substitutions, the efficiency of the new engine becomes

1C C CCarnot 2

H H H

21 1 1

4T T T

eT T T′ ⎛ ⎞

′ = − = − = − ⎜ ⎟⎜ ⎟′ ⎝ ⎠ (1)

762 THERMODYNAMICS

To obtain the original ratio C H/T T , we use Equation 15.15:

C C

Carnot CarnotH H

1 or 1T T

e eT T

= − = −

Substituting this original temperature ratio into Equation (1) gives

( ) ( ) ( )1 1 1 1CCarnot Carnot Carnot2 2 2 2

H1 1 1 1 1 0.40 0.70

Te e e

T⎛ ⎞

′ = − = − − = + = + =⎜ ⎟⎜ ⎟⎝ ⎠

57. REASONING AND SOLUTION The efficiency of the engine is e = 1 − (TC/TH) so (i) Increase TH by 40 K; e = 1 − [(350 K)/(690 K)] = 0.493 (ii) Decrease TC by 40 K; e = 1 − [(310 K)/(650 K)] = 0.523 The greater improvement is made by lowering the temperature of the cold reservoir. 58. REASONING AND SOLUTION The amount of work delivered by the engines can be

determined from Equation 15.12, H CQ W Q= + . Solving for W for each engine gives:

1 H1 C1 2 H2 C2 and W Q Q W Q Q= − = − The total work delivered by the two engines is

( ) ( )1 2 H1 C1 H2 C2W W W Q Q Q Q= + = − + −

But we know that H2 C1Q Q= , so that

( ) ( )H1 C1 C1 C2 H1 C2W Q Q Q Q Q Q= − + − = − (1)

Since these are Carnot engines,

( )C1 3C1 C1C1 H1

H1 H1H1

670 K 4800 J 3.61 10 J890 K

Q T TQ Q

T TQ⎛ ⎞= ⇒ = = = ×⎜ ⎟⎝ ⎠

Similarly, noting that H2 C1Q Q= and that TH2 = TC1, we have

( )3 3C2 C2C2 H2 C1

H2 C1

420 K3.61 10 J 2.26 10 J670 K

T TQ Q Q

T T⎛ ⎞= = = × = ×⎜ ⎟⎝ ⎠

Substituting into Equation (1) gives


3 3 4800 J 2.26 10 J 2.5 10 JW = − × = × 59. SSM REASONING AND SOLUTION The temperature of the gasoline engine input is

T1 = 904 K, the exhaust temperature is T2 = 412 K, and the air temperature is T3 = 300 K. The efficiency of the engine/exhaust is

e1 = 1 − (T2/T1) = 0.544

The efficiency of the second engine is

e2 = 1 − (T3/T2) = 0.272 The magnitude of the work done by each segment is 1 1 H1W e Q= and

2 2 H2 2 C1W e Q e Q= = since

H2 C1Q Q= Now examine ( )1 2 1/W W W+ to find the ratio of the total work produced by both engines

to that produced by the first engine alone.

1 H1 2 C1 C11 2 2

1 1 H1 1 H11

e Q e Q QW W eW e Q e Q

⎛ ⎞++ ⎛ ⎞= = + ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

But, ( )1 C1 H11 /e Q Q= − , so that C1 H1 1/ 1Q Q e= − . Therefore,

( )1 2 2 21 2

1 111 1 1 1 0.500 0.272 1.23

W W e ee e

e eW+

= + − = + − = + − =

60. REASONING The maximum efficiency e at which the power plant can operate is given

by Equation 15.15, e T TH= −1 ( / )C . The power output is given; it can be used to find the magnitude W of the work output for a 24 hour period. With the efficiency and W

known, Equation 15.11, H/e W Q= , can be used to find HQ , the magnitude of the input

heat. The magnitude CQ of the exhaust heat can then be found from Equation 15.12,

H CQ W Q= + . SOLUTION a. The maximum efficiency is

764 THERMODYNAMICS

C

H

323 K1 1 0.360505 K

Te

T= − = − =

b. Since the power output of the power plant is P = 84 000 kW, the required heat input

HQ for a 24 hour period is

7

13H

(8.4 10 J/s)(24 h) 3600 s 2.02 10 J0.360 1 h

W PtQe e

× ⎛ ⎞= = = = ×⎜ ⎟⎝ ⎠

Therefore, solving Equation 15.12 for CQ , we have

13 12 13

C H 2.02 10 J 7.3 10 J = 1.3 10 JQ Q W= − = × − × × 61. SSM REASONING The expansion from point

a to point b and the compression from point c to point d occur isothermally, and we will apply the first law of thermodynamics to these parts of the cycle in order to obtain expressions for the input and rejected heats, magnitudes HQ and CQ , respectively. In order to simplify the resulting expression for C H/Q Q , we will then use the fact that the expansion from point b to point c and the compression from point d to point a are adiabatic.

SOLUTION According to the first law of thermodynamics, the change in internal energy

ΔU is given by U Q WΔ = − (Equation 15.1), where Q is the heat and W is the work. Since the internal energy of an ideal gas is proportional to the Kelvin temperature and the temperature is constant for an isothermal process, it follows that 0 JUΔ = for such a case.

The work of isothermal expansion or compression for an ideal gas is ( )f iln /W nRT V V=(Equation 15.3), where n is the number of moles, R is the universal gas constant, T is the Kelvin temperature, Vf is the final volume of the gas, and Vi is the initial volume. We have, then, that

f f

i i or 0 ln or ln

V VU Q W Q nRT Q nRT

V V⎛ ⎞ ⎛ ⎞

Δ = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Applying this result for Q to the isothermal expansion (temperature = TH) from point a to

point b and the isothermal compression (temperature = TC) from point c to point d, we have

b dH H C C

a cln and lnV V

Q nRT Q nRTV V

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

a

d c

b TH

TC

Volume →

Pres

sure

→


where f bV V= and i aV V= for the isotherm at TH and f dV V= and i cV V= for the isotherm at TC. In this problem, we are interested in the magnitude of the heats. For QH, this poses

no difficulty, since b aV V> , ( )b aln /V V is positive, and we have

bH H

alnV

Q nRTV

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

(1)

However, for QC, we need to be careful, because c dV V> and ( )d cln /V V is negative. Thus, we write for the magnitude of QC that

d cC C C

c dln lnV V

Q nRT nRTV V

⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2)

According to Equations (1) and (2), the ratio of the magnitudes of the rejected and input

heats is

c cC C

C d d

H b bH H

a a

ln ln

ln ln

V VnRT T

Q V VQ V V

nRT TV V

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(3)

We now consider the adiabatic parts of the Carnot cycle. For the adiabatic expansion or compression of an ideal gas the initial pressure and volume (Pi and Vi) are related to the final pressure and volume (Pf and Vf) according to

i i f fPV P Vγ γ= (15.5) where γ is the ratio of the specific heats at constant pressure and constant volume. It is also

true that /P nRT V= (Equation 14.1), according to the ideal gas law. Substituting this expression for the pressure into Equation 15.5 gives

1 1i fi f i i f f

i f or

nRT nRTV V TV T V

V Vγ γ γ γ− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Applying this result to the adiabatic expansion from point b to point c and to the adiabatic compression from point d to point a, we obtain

1 11 1H b C c C d H a and T V T V T V T Vγ γγ γ− −− −= =

Dividing the first of these equations by the second shows that

1 11 1H b C c b c b c

1 1 1 1H a a a dC d d

or or T V T V V V V VT V V V VT V V

γ γγ γ

γ γ γ γ

− −− −

− − − −= = =

With this result, Equation (3) becomes

766 THERMODYNAMICS

cC

dC

H

lnV

TVQ

Q

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠=b

Ha

lnV

TV

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

C

H

TT

=

62. REASONING AND SOLUTION The efficiency e of the power plant is three-fourths its

Carnot efficiency so, according to Equation 15.15,

C

H

40 K 273 K0.75 1 0.75 1 0.33285 K 273 K

Te

T⎛ ⎞ +⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠

The power output of the plant is 91.2 10× watts. According to Equation 15.11,

( )H H/ Power /e W Q t Q= = ⋅ . Therefore, at 33% efficiency, the magnitude of the heat input per unit time is

9

H 9Power 1.2 10 W 3.6 10 J/s0.33

Qt e

×= = = × From the principle of conservation of energy, the heat output per unit time must be

C H 9 Power 2.4 10 J/sQ Qt t

= − = × The rejected heat is carried away by the flowing water and, according to Equation 12.4,

CQ cm T= Δ . Therefore,

C C or Q Qcm T mc Tt t t t

Δ ⎛ ⎞= = Δ⎜ ⎟⎝ ⎠

Solving the last equation for Δt, we have

( ) ( )9/C C

52.4 10 J/s 5.7 C

( / ) ( / ) 4186 J/ kg C 1.0 10 kg/s

Q Q tT

tc m t c m t×Δ = = = = °

⎡ ⎤⋅ ° ×⎣ ⎦

63. SSM REASONING The coefficient of performance of an air conditioner is C /Q W ,

according to Equation 15.16, where CQ is the magnitude of the heat removed from the

house and W is the magnitude of the work required for the removal. In addition, we know

that the first law of thermodynamics (energy conservation) applies, so that H CW Q Q= − ,

according to Equation 15.12. In this equation HQ is the magnitude of the heat discarded

outside. While we have no direct information about CQ and HQ , we do know that the air


conditioner is a Carnot device. This means that Equation 15.14 applies: C H C H/ /Q Q T T= . Thus, the given temperatures will allow us to calculate the coefficient of performance.

SOLUTION Using Equation 15.16 for the definition of the coefficient of performance and

Equation 15.12 for the fact that H CW Q Q= − , we have

C C C H

H C C H

/Coefficient of performance

1 /Q Q Q QW Q Q Q Q

= = =− −

Equation 15.14 applies, so that C H C H/ /Q Q T T= . With this substitution, we find

( ) ( )

C H C H

C HC H

C

H C

/ /Coefficient of performance

1 /1 /

297 K 21311 K 297 K

Q Q T TT TQ Q

TT T

= =−−

= = =− −

64. REASONING The magnitude |QC| of the heat removed from the inside compartment (cold

reservoir) of the refrigerator is found from the first law of thermodynamics:

C HQ Q W= − (Equation 15.12), where |QH| is the magnitude of the heat deposited in the kitchen (hot reservoir) and |W| is the magnitude of the work done. For a Carnot refrigerator,

the magnitudes of the heat removed and deposited are related by C C

HH

Q TTQ

= (Equation

15.14), where TC and TH are, respectively, the Kelvin temperatures of the inside compartment and the kitchen.

SOLUTION Solving C C

HH

Q TTQ

= (Equation 15.14) for |QH| yields

HH C

C

TQ Q

T⎛ ⎞= ⎜ ⎟⎝ ⎠

(1)

Substituting Equation (1) into C HQ Q W= − (Equation 15.12), we obtain

H

C C CHC

C

or 1

T WQ Q W Q

TTT

⎛ ⎞= − =⎜ ⎟ ⎛ ⎞⎝ ⎠ −⎜ ⎟⎝ ⎠

Therefore, the magnitude of the heat that can be removed from the inside compartment is

768 THERMODYNAMICS

4

CH

C

2500 J 3.1 10 J299 K 11 277 K

WQ

TT

= = = ×⎛ ⎞ −−⎜ ⎟⎝ ⎠

65. REASONING The coefficient of performance COP is defined as CCOP /Q W=

(Equation 15.16), where CQ is the magnitude of the heat removed from the cold reservoir

and W is the magnitude of the work done on the refrigerator. The work is related to the

magnitude HQ of the heat deposited into the hot reservoir and CQ by the conservation of

energy, H CW Q Q= − . Thus, the coefficient of performance can be written as (after some algebraic manipulations)

H

C

1COP = 1

QQ

−

The maximum coefficient of performance occurs when the refrigerator is a Carnot refrigerator. For a Carnot refrigerator, the ratio H C/Q Q is equal to the ratio H C/T T of the

Kelvin temperatures of the hot and cold reservoirs, H C H C/ /Q Q T T= (Equation 15.14).

SOLUTION Substituting H C H C/ /Q Q T T= into the expression above for the COP gives

H

C

1 1COP = 13296 K 11275 K

TT

= =−−

66. REASONING According to the first law of thermodynamics (see Equation 15.12), the

magnitude W of the work required is given by H CW Q Q= − , where HQ is the

magnitude of the heat deposited in the hot reservoir and CQ is the magnitude of the heat removed from the cold reservoir. Since the air conditioners are Carnot devices, we know that the ratio C H/Q Q is equal to the ratio C H/T T of the Kelvin temperatures of the cold

and hot reservoirs: C H C H/ /Q Q T T= (Equation 15.14). SOLUTION Combining Equations 15.12 and 15.14, we have

H HH C C C

CC1 1

Q TW Q Q Q Q

TQ

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠


Applying this result to each air conditioner gives

( )

( )

309.0 K 1 4330 J 220 J294.0 K

309.0 K 1 4330 J 120 J301.0 K

W

W

⎛ ⎞= − =⎜ ⎟⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎝ ⎠

Unit A

Unit B

We can find the heat deposited outside directly from Equation 15.14 by solving it for HQ .

C C HH C

H CH or

Q T TQ Q

T TQ⎛ ⎞

= = ⎜ ⎟⎜ ⎟⎝ ⎠

Applying this result to each air conditioner gives

( )

( )

H

H

309.0 K 4330 J 4550 J294.0 K

309.0 K 4330 J 4450 J301.0 K

Q

Q

⎛ ⎞= =⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

Unit A

Unit B

67. REASONING AND SOLUTION We know that

C H 14 200 J 800 J 13 400 JQ Q W= − = − = Therefore,

( )CC H

H

13 400 J301 K 284 K14 200 J

QT T

Q

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

68. REASONING The efficiency of the engine is H/e W Q= (Equation 15.11), where W is

the magnitude of the work and HQ is the magnitude of the input heat. The coefficient of

performance COP of the heat pump is HCOP /Q W= , according to Equation 15.17. SOLUTION Using H/e W Q= (Equation 15.11) to substitute into Equation 15.17 for the

coefficient of performance, we obtain

H

H

1 1 1COP 1.8/ 0.55

QW W Q e

= = = = =

770 THERMODYNAMICS

69. REASONING Since the refrigerator is a Carnot device, we know that C C

HH

Q TTQ

=

(Equation 15.14). We have values for TH (the temperature of the kitchen) and CQ (the magnitude of the heat removed from the food). Thus, we can use this expression to determine TC (the temperature inside the refrigerator), provided that a value can be obtained

for HQ (the magnitude of the heat that the refrigerator deposits into the kitchen). Energy

conservation dictates that H CQ W Q= + (Equation 15.12), where W is the magnitude of the work that the appliance uses and is known.

SOLUTION From Equation 15.14, it follows that

C CCC H

HH H or

Q QTT T

TQ Q

⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠

Substituting H CQ W Q= + (Equation 15.12) into this result for TC gives

( )C CC H H

H C

2561 J301 K 275 K241 J 2561 J

Q QT T T

Q W Q

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ++ ⎝ ⎠⎝ ⎠ ⎝ ⎠

70. REASONING The mass m of ice that the ice maker can produce in one day depends upon

the magnitude |QC| of the heat it can extract from water as the water cools from Ti = 15.0 °C to Tf = 0 °C and then freezes at 0 °C. To cool a mass m of water from Ti to Tf requires extracting an amount of heat 1Q cm T= Δ (Equation 12.4) from it, where c is the specific heat capacity of water (see Table 12.2 in the text), and ΔT = 15 C° is the difference between the higher initial temperature Ti and the lower final temperature Tf. For the freezing process, the heat loss is given by 2 fQ mL= (Equation 12.5), where Lf is the latent heat of fusion of water (see Table 12.3 in the text). Therefore, the magnitude |QC| of the total heat to be extracted from the water each day is

QC = cm ΔTHeat removed

to cool the water

+ mLfHeat removed

to freeze the water

(1)

The ice maker operates in the same way as a refrigerator or an air conditioner, so the ice

maker’s coefficient of performance (COP) is given by CCOPQW

= (Equation 15.16), where

|W| is the magnitude of the work done by the ice maker in extracting the heat of magnitude |QC| from the water. Solving Equation 15.16 for |QC|, we find that


( )C COPQ W= (2)

The wattage P of the ice maker is the rate at which it does work. This rate is given by W

Pt

= (Equation 6.10a). Therefore, the magnitude |W| of the total amount of work the ice

maker performs in one day is W Pt= (3) SOLUTION Solving Equation (1) for the desired mass m of ice, we find that

( ) Cf C

f or

Qm c T L Q m

c T LΔ + = =

Δ + (4)

Substituting Equation (2) into Equation (4), then, yields

( )C

f f

COPQ Wm

c T L c T L= =

Δ + Δ + (5)

Lastly, Substituting Equation (3) into Equation (5), we obtain

( )f f

COP COPW Ptm

c T L c T L= =

Δ + Δ +

The elapsed time of t = 1 day is equivalent to t = 8.64×104 s (see the inside of the front cover of the text). Therefore, the maximum amount of ice that the ice maker can produce in one day is

( ) ( )( )( )( ) ( )

4

4f

225 W 8.64 10 s 3.60COP176 kg

4186 J/ kg C 15.0 C 33.5 10 J/kg

Ptm

c T L

×= = =

Δ + ⎡ ⎤⋅ + ×⎣ ⎦o o

71. SSM REASONING The net heat added to the room is H CQ Q− , where HQ is the

magnitude of the heat put into the room by the unit and CQ is the magnitude of the heat removed by the unit. To determine this net heat, we will use the fact that energy conservation applies, so that H CQ W Q= + (Equation 15.12). We will also use the fact

that the coefficient of performance COP of the air conditioner is CCOP /Q W=

(Equation 15.16). In Equations 15.12 and 15.16, W is the amount of work needed to operate the unit.

SOLUTION Rearranging Equation 15.12 for energy conservation, we have

H C H C or Net heat =Q W Q Q Q W= + − = (1)

772 THERMODYNAMICS

Thus, the net heat is just the work W needed to operate the unit. We can obtain this work directly from the coefficient of performance as specified by Equation 15.16:

C CCOP or COP

Q QW

W= = (2)

Substituting W from Equation (2) into Equation (1), we find that

4C 4

H C7.6 10 JNet heat = 3.8 10 J

COP 2.0Q

Q Q W ×− = = = = × 72. REASONING Power is energy per unit time. Therefore, the heat Qspace heater that the space

heater would put into the kitchen is the heater’s power output P times the time t during which the heater is on: space heaterQ Pt= (Equation 6.10b). The magnitude of the heat that

the refrigerator puts into the kitchen is HQ . To determine this heat, we will use the fact

that energy conservation applies, so that H CQ W Q= + (Equation 15.12). We will also

use the fact that the coefficient of performance COP of the refrigerator is CCOP /Q W=

(Equation 15.16). In Equations 15.12 and 15.16, W is the amount of work needed to

operate the refrigerator and CQ is the magnitude of the heat that must be removed from the water to make the ice. Heat must be removed from the water for two reasons. First, the water must be cooled by 20.0 Cº (from 20.0 ºC to 0.0 ºC). The amount of heat removed to accomplish the cooling is coolingQ cm T= Δ (Equation 12.4), where 4186 J/(kg C )c = ⋅ ° is the specific heat capacity of water as given in Table 12.2, m is the mass of the water, and ΔT is the amount of the temperature drop. After the water reaches 0.0 ºC, additional heat must be removed to make the water freeze. The amount of heat removed to accomplish the freezing is freezing fQ mL= (Equation 12.5), where 4

f 33.5 10 J/kgL = × is the latent heat of fusion of water as given in Table 12.3.

SOLUTION According to Equation 6.10b, the run-time t of the heater is

space heater /t Q P= . We know that space heater HQ Q= , so that the run time is

space heater HQ Qt

P P= = (1)

Using energy conservation in the form of Equation 15.12, we know that H CQ W Q= + , so that Equation (1) becomes

CH W QQt

P P+

= = (2)


Since the coefficient of performance COP of the refrigerator is CCOP /Q W=

(Equation 15.16), we can substitute ( )C / COPW Q= into Equation (2) and obtain

( )C CC C/ COP 1 1COP

Q QW Q Qt

P P P

⎡ ⎤ ++ ⎛ ⎞⎣ ⎦= = = +⎜ ⎟⎝ ⎠ (3)

Using coolingQ cm T= Δ (Equation 12.4) and freezing fQ mL= (Equation 12.5), we can write

the magnitude CQ of the heat that must be removed from the water to make the ice in the

following way: C fQ cm T mL= Δ + . Substituting this result into Equation (3), we find that

[ ]( )( ) ( )( )

C f

4

3

1 11 1COP COP

4186 J/(kg C ) 1.50 kg 20.0 C 1.50 kg 33.5 10 J/kg1 1 279 s3.00 3.00 10 W

Q cm T mLt

P PΔ +⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎧ ⎫⋅ ° ° + ×⎪ ⎪⎛ ⎞= + =⎨ ⎬⎜ ⎟ ×⎝ ⎠ ⎪ ⎪⎩ ⎭

______________________________________________________________________________ 73. SSM REASONING Let the coefficient of performance be represented by the symbol

COP. Then according to Equation 15.16, CCOP = /Q W . From the statement of energy

conservation for a Carnot refrigerator (Equation 15.12), H CW Q Q= − . Combining Equations 15.16 and 15.12 leads to

( ) ( )C C C

H C H C C H C

/ 1COP / / 1

Q Q QQ Q Q Q Q Q Q

= = =− − −

Replacing the ratio of the heats with the ratio of the Kelvin temperatures, according to

Equation 15.14, leads to

H C

1COP/ 1T T

=−

(1) The heat CQ that must be removed from the refrigerator when the water is cooled can be

calculated using Equation 12.4, CQ cm T= Δ ; therefore,

C

COP COPQ cm TW Δ= = (2)

SOLUTION a. Substituting values into Equation (1) gives

774 THERMODYNAMICS

( )( )

1

H

C

1 1COP 2.0 1020.0 273.15 K1 16.0 273.15 K

TT

= = = ×+− −+

b. Substituting values into Equation (2) gives

( ) ( )( ) 4

1

4186 J/ kg C 5.00 kg 20.0 C 6.0 C= 1.5 10 J

COP 2.0 10cm TW

⎡ ⎤⋅ ° ° − °Δ ⎣ ⎦= = ××

74. REASONING The efficiency of the Carnot engine is, according to Equation 15.15,

C

H

842 K 11 11684 K 2

Te

T= − = − =

The magnitude of the work delivered by the engine is, according to Equation 15.11,

1

H H2W e Q Q= =

The heat pump removes an amount of heat HQ from the cold reservoir. Thus, the amount

of heat Q′ delivered to the hot reservoir of the heat pump is

1 3H H H H2 2

Q Q W Q Q Q′ = + = + =

Therefore, H/ 3/ 2Q Q′ = . According to Equation 15.14, H C/ /Q Q T T′ ′= , so C/ 3 / 2T T′ = . SOLUTION Solving for T ′ gives

3 3 3

C2 2(842 K) = 1.26 10 KT T′ = = ×

75. REASONING The total entropy change universeSΔ of the universe is the sum of the entropy

changes of the hot and cold reservoirs. For each reservoir, the entropy change is given by

R

QST

⎛ ⎞Δ = ⎜ ⎟⎝ ⎠ (Equation 15.18). As indicated by the label “R,” this equation applies only to

reversible processes. For the irreversible engines, therefore, we apply this equation to an imaginary process that removes the given heat from the hot reservoir reversibly and rejects the given heat to the cold reservoir reversibly. According to the second law of thermodynamics stated in terms of entropy (see Section 15.11), the reversible engine is the one for which universe 0 J/KSΔ = , and the irreversible engine that could exist is the one for which universe 0 J/KSΔ > . The irreversible engine that could not exist is the one for which

universe 0 J/KSΔ < .


SOLUTION Using Equation 15.18, we write the total entropy change of the universe as the sum of the entropy changes of the hot (H) and cold (C) reservoirs.

CHuniverse

H C

QQS

T T−

Δ = +

In this expression, we have used HQ− for the heat from the hot reservoir because that

reservoir loses heat. We have used CQ+ for the heat rejected to the cold reservoir because that reservoir gains heat. Applying this expression to the three engines gives the following results:

CHuniverse

H C

CHuniverse

H C

CHuniverse

H C

1650 J 1120 J 0.4 J/K550 K 330 K

1650 J 990 J 0 J/K550 K 330 K

1650 J 6 550 K

QQS

T T

QQS

T T

QQS

T T

− −Δ = + = + = +

− −Δ = + = + =

− −Δ = + = +

Engine I

Engine II

Engine III 60 J 1.0 J/K330 K

= −

Since universe 0 J/KSΔ = for Engine II, it is reversible. Since universe 0 J/KSΔ > for

Engine I, it is irreversible and could exist. Since universe 0 J/KSΔ < for Engine III, it is irreversible and could not exist.

76. REASONING Equation 15.19 gives the unavailable work as unavailable 0 universeW T S= Δ , where T0 is the Kelvin temperature of the coldest reservoir into which heat can be rejected and ΔSuniverse is the total entropy change of the universe. We can use this expression to find Q, because ΔSuniverse involves Q. In this case, the “universe” means the reservoir where the heat originates and the reservoir into which the heat spontaneously flows. The reservoir out of which the heat flows loses entropy, while the reservoir into which the heat flows gains entropy. ΔSuniverse is the sum of the two changes.

SOLUTION To determine each of the two contributions to ΔSuniverse, we use R

QST

⎛ ⎞Δ = ⎜ ⎟⎝ ⎠

(Equation 15.18). In this expression the subscript “R” reminds us that we must imagine a reversible process by which the heat exits the reservoir at 394 K and arrives in the reservoir at 298 K. Thus, we have

776 THERMODYNAMICS

ΔSuniverse =−Q

394 KNegative entropy changedue to loss of heat from

original reservoir

+ Q

298 KPositive entropy changedue to gain of heat by

target reservoir

With this expression for ΔSuniverse, Equation 15.19 for the unavailable work becomes

unavailable 0 universe 0 394 K 298 KQ QW T S T −⎛ ⎞= Δ = +⎜ ⎟⎝ ⎠

Solving this equation for Q reveals that

( )4unavailable

0

2800 J 1.4 10 J1 1 1 1248 K

394 K 298 K 394 K 298 K

WQ

T= = = ×

− −⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

77. SSM REASONING AND SOLUTION The change in entropy ΔS of a system for a

process in which heat Q enters or leaves the system reversibly at a constant temperature T is given by Equation 15.18, R( / )S Q TΔ = . For a phase change, Q mL= , where L is the latent heat (see Section 12.8).

a. If we imagine a reversible process in which 3.00 kg of ice melts into water at 273 K, the

change in entropy of the water molecules is

( )5

3f

R R

(3.00 kg) 3.35 10 J/kg3.68 10 J/K

273 KmLQS

T T

×⎛ ⎞⎛ ⎞Δ = = = = ×⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

b. Similarly, if we imagine a reversible process in which 3.00 kg of water changes into

steam at 373 K, the change in entropy of the water molecules is

( )6

4v

R R

(3.00 kg) 2.26 10 J/kg1.82 10 J/K

373 KmLQS

T T

×⎛ ⎞⎛ ⎞Δ = = = = ×⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

c. Since the change in entropy is greater for the vaporization process than for the fusion

process, the vaporization process creates more disorder in the collection of water molecules.

78. REASONING According to the discussion on Section 15.11, the change ΔSuniverse in the

entropy of the universe is the sum of the change in entropy ΔSC of the cold reservoir and the change in entropy ΔSH of the hot reservoir, or ΔSuniverse = ΔSC + ΔSH. The change in entropy of each reservoir is given by Equation 15.18 as ΔS = (Q/T)R, where Q is the heat removed


from or delivered to the reservoir and T is the Kelvin temperature of the reservoir. In applying this equation we imagine a process in which the heat is lost by the house and gained by the outside in a reversible fashion.

SOLUTION Since heat is lost from the hot reservoir (inside the house), the change in entropy is negative: ΔSH = −QH/TH. Since heat is gained by the cold reservoir (the outdoors), the change in entropy is positive: ΔSC = +QC/TC. Here we are using the symbols QH and QC to denote the magnitudes of the heats. The change in the entropy of the universe is

C Huniverse C H

C H

24 500 J 24 500 J + 11.6 J/K258 K 294 K

Q QS S S

T TΔ = Δ Δ = − = − =

In this calculation we have used the fact that TC = 273 − 15 °C = 258 K and TH = 273 + 21 °C = 294 K.

79. REASONING The change ΔSuniverse in entropy of the universe for this process is the sum

of the entropy changes for (1) the warm water (ΔSwater) as it cools down from its initial temperature of 85.0 °C to its final temperature Tf , (2) the ice (ΔSice) as it melts at 0 °C, and (3) the ice water (ΔSice water ) as it warms up from 0 °C to the final temperature Tf:

ΔSuniverse = ΔSwater + ΔSice + ΔSice water

To find the final temperature Tf , we will follow the procedure outlined in Sections 12.7 and 12.8, where we set the heat lost by the warm water as it cools down equal to the heat gained by the melting ice and the resulting ice water as it warms up. The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount ΔT is Q = cmΔT (Equation 12.4), where c is the specific heat capacity. The heat that must be supplied to melt a mass m of a substance is Q = mLf (Equation 12.5), where Lf is the latent heat of fusion.

SOLUTION a. We begin by finding the final temperature Tf of the water. Setting the heat lost equal to

the heat gained gives

cmwater 85.0 °C−Tf( )Heat lost by water

= miceLf

Heat gained by melting ice

+ micec Tf − 0.0 °C( )

Heat gained by ice water

Solving this relation for the final temperature Tf yields

778 THERMODYNAMICS

( )( )( ) ( )( ) ( )( )

( ) ( )

water ice ff

ice water

4

85.0 C

4186 J/ kg C 6.00 kg 85.0 C 3.00 kg 33.5 10 J/kg30.0 C

4186 J/ kg C 3.00 kg 6.00 kg

cm m LT

c m m° −

=+

⎡ ⎤⋅ ° ° − ×⎣ ⎦= = °⎡ ⎤⋅ ° +⎣ ⎦

We have taken the specific heat capacity of ( )4186 J/ kg C⋅ ° for water from Table 12.2 and

the latent heat of 433.5 10 J/kg× from Table 12.3. This temperature is equivalent to Tf = (273 + 30.0 °C) = 303 K.

The change ΔSuniverse in the entropy of the universe is the sum of three contributions:

[Contribution 1]

( )fwater water

i

303 K = ln = (6.00 kg)[4186 J/ kg C ] ln = 4190 J/K358 K

TS m c

T⎛ ⎞ ⎛ ⎞Δ ⋅ ° −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

where Ti = 273 + 85.0 °C = 358 K. [Contribution 2]

( )( )4f

ice

3.00 kg 33.5 10 J/kg +3680 J/K

273 KmLQS

T T

×Δ = = = =

[Contribution 3]

( )fice water ice

i

303 K = ln = (3.00 kg)[4186 J/ kg C ] ln = 1310 J/K273 K

TS m c

T⎛ ⎞ ⎛ ⎞Δ ⋅ ° +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The change in the entropy of the universe is

ΔSuniverse = ΔSwater + ΔSice + ΔSice water =

28.0 10 J/K+ × b. The entropy of the universe increases , because the mixing process is irreversible. 80. REASONING We think of the entire universe as divided into two parts: the sun and the rest

of the universe. As the sun spontaneously radiates heat into space, the change in its entropy each second is ΔSsun , and the change in the entropy of the rest of the universe is ΔSrest each second. The total entropy change of the universe in one second is the sum of the two entropy changes universe sun restS S SΔ = Δ +Δ (1)


The spontaneous thermal radiation of the sun is an irreversible process, but if we imagine a reversible process by which the sun loses an amount Q of heat, and a reversible process by

which the rest of the universe absorbs this heat, then we can make use of R

QST

⎛ ⎞Δ = ⎜ ⎟⎝ ⎠

(Equation 15.18) to calculate the entropy changes of the sun and of the rest of the universe:

sun restsun rest

and Q Q

S ST T− +

Δ = Δ = (2)

In Equations (2), Tsun is the sun’s surface temperature, and Trest is the average temperature of the rest of the universe. The explicit algebraic signs (− and +) in Equations (2) indicate, respectively, that the sun loses heat and the rest of the universe gains heat. The heat Q radiated by the sun to the rest of the universe in a time t = 1.0 s is given by Equation 13.2: 4

sunQ e T Atσ= (13.2)

In Equation 13.2, e is the emissivity of the sun (assumed to be 1, because the sun is taken to be a perfect blackbody), ( )8 2 45.67 10 J/ s m Kσ −= × ⋅ ⋅ is the Stefan-Boltzmann constant, and A is the sun’s surface area. Because the sun is a sphere with a radius r, its surface area is

( )22 8 18 24 4 6.96 10 m 6.09 10 mA rπ π= = × = × (3) SOLUTION Substituting Equations (2) into Equation (1), we obtain

universe sun restsun rest rest sun

1 1Q QS S S Q

T T T T− ⎛ ⎞Δ = Δ +Δ = + = −⎜ ⎟

⎝ ⎠ (4)

Substituting Equation 13.2 into Equation (4) yields

4

universe sunrest sun rest sun

1 1 1 1S Q e T AtT T T T

σ⎛ ⎞ ⎛ ⎞Δ = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Therefore, the net entropy change of the entire universe due to one second of thermal radiation from the sun is

( ) ( ) ( ) ( )( )

universe

48 2 4 18 2

26

1 11 5.67 10 J/ s m K 5800 K 6.09 10 m 1.0 s2.73 K 5800 K

1.4 10 J/K

S

−

Δ =

⎛ ⎞⎡ ⎤= × ⋅ ⋅ × −⎜ ⎟⎣ ⎦ ⎝ ⎠

= ×

81. REASONING

780 THERMODYNAMICS

a. According to the discussion in Section 15.11, the change ΔSuniverse in the entropy of the universe is the sum of the change in entropy ΔSH of the hot reservoir and the change in entropy ΔSC of the cold reservoir, so ΔSuniverse = ΔSH + ΔSC. The change in entropy of each reservoir is given by Equation 15.18 as ΔS = (Q/T)R . The engine is irreversible, so we must imagine a process in which the heat Q is added to or removed from the reservoirs reversibly. T is the Kelvin temperature of a reservoir. Since heat is lost from the hot reservoir, the change in entropy is negative: ΔSH = − HQ /TH. Since heat is gained by the cold reservoir,

the change in entropy is positive: ΔSC = + CQ /TC. The change in entropy of the universe is

CHuniverse H C

H C +

QQS S S

T TΔ = Δ Δ = − +

b. The magnitude W of the work done by any engine depends on its efficiency e and input

heat HQ via HW e Q= (Equation 15.11). For a reversible engine, the efficiency is related

to the Kelvin temperatures of its hot and cold reservoirs by e = 1 − (TC/TH), Equation 15.15. Combining these two relations will allow us to determine W .

c. The difference in the work produced by the two engines is labeled Wunavailable in Section15.11, where Wunavailable = Wreversible − Wirreversible. The difference in the work is related to the change in the entropy of the universe by Wunavailable = T0 ΔSuniverse (Equation 15.19), where T0 is the Kelvin temperature of the coldest heat reservoir. In this case T0 = TC.

SOLUTION a. From part a of the REASONING, the change in entropy of the universe is

CHuniverse

H C

QQS

T TΔ = − +

The magnitude CQ of the heat rejected to the cold reservoir is related to the magnitude

HQ of the heat supplied to the engine from the hot reservoir and the magnitude W of the

work done by the engine via C HQ Q W= − (Equation 15.12). Thus, ΔSuniverse becomes

H Huniverse

H C

1285 J 1285 J 264 J 1.74 J/K852 K 314 K

Q Q WS

T T− −Δ = − + = − + = +

b. The magnitude W of the work done by an engine depends on its efficiency e and input

heat HQ via HW e Q= (Equation 15.11). For a reversible engine, the efficiency is related


to the temperatures of its hot and cold reservoirs by e = 1 − (TC/TH), Equation 15.15. The work done by the reversible engine is

( )Creversible H H

H

314 K1 1 1285 J 811 J852 K

TW e Q Q

T⎛ ⎞ ⎛ ⎞= = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

c. According to the discussion in part c of the REASONING, the difference between the work done by the reversible and irreversible engines is

Wreversible − Wirreversible

Wunavailable

= TCΔSuniverse = 314 K( ) 1.74 J/K( ) = 546 J

82. REASONING During an adiabatic process, no heat flows into or out of the gas (Q = 0 J).

For an ideas gas, the final pressure and volume (Pf and Vf) are related to the initial pressure

and volume (Pi and Vi) by ( )i i f f Equation 15.5 ,PV PVγ γ= where γ is the ratio of the specific

heat capacities at constant pressure and constant volume (γ = 53 in this problem). We will use this relation to find Vf /Vi.

SOLUTION Solving i i f fPV PVγ γ= for Vf /Vi and noting that the pressure doubles

(Pf /Pi = 2.0) during the compression, we have

( )1 1

5/3f i

i f

1 0.662.0

V PV P

γ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

83. SSM REASONING According to the first law of thermodynamics (Equation 15.1),

ΔU Q W= − . For a monatomic ideal gas (Equation 14.7), U nRT= 32 . Therefore, for the

process in question, the change in the internal energy is Δ ΔU nR T= 32 . Combining the last

expression for ΔU with Equation 15.1 yields

32 nR T Q WΔ = −

This expression can be solved for ΔT . SOLUTION a. The heat is Q = +1200 J , since it is absorbed by the system. The work is W = +2500 J ,

since it is done by the system. Solving the above expression for ΔT and substituting the values for the data given in the problem statement, we have

ΔTQ WnR

=−

= −⋅

= ×32

32

101200 J 2500 J(0.50 mol)[8.31 J / (mol K)]

–2.1 2 K

782 THERMODYNAMICS

b. Since ΔT T T= −final initial is negative, Tinitial must be greater than Tfinal; this change

represents a decrease in temperature. Alternatively, one could deduce that the temperature decreases from the following physical

argument. Since the system loses more energy in doing work than it gains in the form of heat, the internal energy of the system decreases. Since the internal energy of an ideal gas depends only on the temperature, a decrease in the internal energy must correspond to a decrease in the temperature.

84. REASONING According to Equation 15.11, the efficiency e of a heat engine is

H/e W Q= , where W is the magnitude of the work done by the engine and HQ is the

magnitude of the input heat that the engine uses. HQ is given, but W is unknown.

However, energy conservation requires that H CQ W Q= + (Equation 15.12), where CQ is the magnitude of the heat rejected by the engine and is given. From this equation, therefore, a value for W can be obtained.

SOLUTION According to Equation 15.11, the efficiency is

H

We

Q=

Since H CQ W Q= + (Equation 15.12), we can solve for W to show that H CW Q Q= − .

Substituting this result into the efficiency expression gives 4 4

H C4

H H

5.6 10 J 1.8 10 J 0.685.6 10 J

Q QWe

Q Q− × − ×= = = =

×

85. REASONING When a gas expands under isobaric conditions, its pressure remains

constant. The work W done by the expanding gas is W = P (Vf − Vi), Equation 15.2, where P is the pressure and Vf and Vi are the final and initial volumes. Since all the variables in this relation are known, we can solve for the final volume.

SOLUTION Solving W = P (Vf − Vi) for the final volume gives

3 3 3 3f i 5

480 J 1.5 10 m 4.5 10 m1.6 10 Pa

WV VP

− −= + = + × = ××


86. REASONING The magnitude |W| of the work done by a heat engine is related to its

efficiency e by H

We

Q= (Equation 15.11), where |QH| is the magnitude of heat extracted

from the hot reservoir. Solving Equation 15.11 for |W|, we obtain

HW e Q= (1)

The conservation of energy requires that the magnitude |QH| of the extracted heat must be equal to the magnitude |W| of the work done plus the magnitude |QC| of the heat rejected by the engine: H CQ W Q= + (15.12)

Because we know the efficiency e and |QC|, we can use Equations (1) and (15.12) to determine the magnitude |W| of the work the engine does in one second. SOLUTION Substituting Equation (15.12) into Equation (1) yields ( )H CW e Q e W Q= = + (2)

Solving Equation (2) for the magnitude |W| of the work done by the engine, we obtain

( ) ( )C

C C or 1 or 1e Q

W eW e Q W e e Q We

= + − = =−

(3)

The engine rejects |QC| = 9900 J of heat every second, at an efficiency of e = 0.22, so the magnitude of the work it does in one second is, by Equation (3),

( )( )C 0.22 9900 J

2800 J1 1 0.22e Q

We

= = =− −

87. SSM REASONING AND SOLUTION a. Since the energy that becomes unavailable for doing work is zero for the process, we

have from Equation 15.19, unavailable 0 universe 0W T S= Δ = . Therefore, universe 0SΔ = and

according to the discussion in Section 15.11, the process is reversible . b. Since the process is reversible, we have (see Section 15.11)

universe system surroundings 0S S SΔ = Δ +Δ = Therefore,

surroundings system –125 J/KS SΔ = −Δ =

784 THERMODYNAMICS

88. REASONING AND SOLUTION Suppose this device were a Carnot engine instead of a

heat pump. We know that its efficiency e would be

e = 1 − (TC/TH) = 1 − [(265 K)/(298 K)] = 0.111 The efficiency, however, is also given by

H

We

Q=

Since the heat pump’s coefficient of performance COP is HCOP /Q W= , we have that

H 1 1COP 9.030.111

QW e

= = = =

89. REASONING AND SOLUTION Equation 15.14 holds for a Carnot air conditioner as well

as a Carnot engine. Therefore, solving Equation 15.14 for CQ , we have

( )5 5CC H

H

299 K 6.12 10 J 5.86 10 J312 K

TQ Q

T⎛ ⎞ ⎛ ⎞= = × = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

90. REASONING According to Equation 15.11, the efficiency e of a heat engine is given by

H/e W Q= , where W is the magnitude of the work and HQ is the magnitude of the input

heat. Thus, the magnitude of the work is HW e Q= . The efficiency eCarnot of a Carnot engine is given by Equation 15.15 as eCarnot = 1 – TC/TH,

where TC and TH are, respectively, the Kelvin temperatures of the cold and hot reservoirs. This expression can be used to determine TC.

SOLUTION Using Equation 15.11, we find the work delivered by each engine as follows:

( )( )

( )( )

H

H

0.60 1200 J 720 J

0.80 1200 J 960 J

W e Q

W e Q

= = =

= = =

Engine A

Engine B

Equation 15.15, which gives the efficiency eCarnot of a Carnot engine, can be solved for the

temperature TC of the cold reservoir:

( )CCarnot C Carnot H

H1 or 1T

e T e TT

= − = −


Applying this result to each engine gives

( )( )

( )( )

C

C

1 0.60 650 K 260 K

1 0.80 650 K 130 K

T

T

= − =

= − =

Engine A

Engine B

91. SSM REASONING For segment AB, there is no work, since the volume is constant. For

segment BC the process is isobaric and Equation 15.2 applies. For segment CA, the work can be obtained as the area under the line CA in the graph.

SOLUTION a. For segment AB, the process is isochoric, that is, the volume is constant. For a process in

which the volume is constant, no work is done, so W = 0 J . b. For segment BC, the process is isobaric, that is, the pressure is constant. Here, the

volume is increasing, so the gas is expanding against the outside environment. As a result, the gas does work, which is positive according to our convention. Using Equation 15.2 and the data in the drawing, we obtain

W = P Vf −Vi( )

= 7.0×105 Pa( ) 5.0×10−3 m3( )− 2.0×10−3 m3( )⎡⎣

⎤⎦ = +2.1×103 J

c. For segment CA, the volume of the gas is decreasing, so the gas is being compressed and

work is being done on it. Therefore, the work is negative, according to our convention. The magnitude of the work is the area under the segment CA. We estimate that this area is 15 of the squares in the graphical grid. The area of each square is

(1.0 × 105 Pa)(1.0 × 10–3 m3) = 1.0 × 102 J

The work, then, is W = – 15 (1.0 × 102 J) = − ×1 5 103. J

______________________________________________________________________________

92. REASONING AND SOLUTION a. The work is the area under the path ACB. There are 48 "squares" under the path, so that

( )( )4 –3 348 2.0 10 Pa 2.0 10 m 1900 JW = − × × = − The minus sign is included because the gas is compressed, so that work is done on it. Since

there is no temperature change between A and B (the line AB is an isotherm) and the gas is ideal, ΔU = 0, so

786 THERMODYNAMICS

Q = ΔU + W = W = 1900 J− b. The negative answer for W means that heat flows out of the gas. 93. SSM REASONING The change in the internal energy of the gas can be found using the

first law of thermodynamics, since the heat added to the gas is known and the work can be calculated by using Equation 15.2, W = P ΔV. The molar specific heat capacity at constant pressure can be evaluated by using Equation 15.6 and the ideal gas law.

SOLUTION a. The change in the internal energy is

( )( )4 4 3 4 331.4 J 1.40 10 Pa 8.00 10 m 3.00 10 m 24.4 J

U Q W Q P V

− −

Δ = − = − Δ

= − × × − × =

b. According to Equation 15.6, the molar specific heat capacity at constant pressure is

CP = Q/(n ΔT). The term n ΔT can be expressed in terms of the pressure and change in volume by using the ideal gas law:

P ΔV = n R ΔT or n ΔT = P ΔV/R

Substituting this relation for n ΔT into CP = Q/(n ΔT), we obtain

( )( )4 4 331.4 J 37.3 J/(mol K)

1.40 10 Pa 5.00 10 m

8.31 J/(mol K)

PQC P VR

−= = = ⋅Δ × ×

⋅

94. REASONING According to the conservation of energy, the work W done by the electrical

energy is H CW Q Q= − , where HQ is the magnitude of the heat delivered to the outside

(the hot reservoir) and CQ is the magnitude of the heat removed from the house (the cold reservoir). Dividing both sides of this relation by the time t, we have

CH = QQW

t t t−

The term /W t is the magnitude of the work per second that must be done by the electrical

energy, and the terms H /Q t and C /Q t are, respectively, the magnitude of the heat per second delivered to the outside and removed from the house. Since the air conditioner is a Carnot air conditioner, we know that H C/Q Q is equal to the ratio H C/T T of the Kelvin


temperatures of the hot and cold reservoirs, H C H C/ /Q Q T T= (Equation 15.14). This

expression, along with the one above for /W t , will allow us to determine the magnitude of the work per second done by the electrical energy.

SOLUTION Solving the expression H C H C/ /Q Q T T= for HQ , substituting the result into

the relation CH QQWt t t

= − , and recognizing that C /Q t = 10 500 J/s, give

( )

C H

C CH C

C 2H

C

306.15 K 1 10 500 J/s 1 5.0 10 J/s292.15 K

Q TQ QQ TW

t t t t t

Q Tt T

= − = −

⎛ ⎞⎛ ⎞ ⎛ ⎞= − = − = ×⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

In this result we have used the fact that TH = 273.15 + 33.0 °C = 306.15 K and the fact that TC = 273.15 + 19.0 °C = 292.15 K.

95. REASONING AND SOLUTION The total heat generated by the students is

Q = (200)(130 W)(3000 s) = 7.8 × 107 J For the isochoric process.

Q = CvnΔT = (5R/2)n ΔT The number of moles of air in the room is found from the ideal gas law to be

( )( )( ) ( )

5 34

1.01 10 Pa 1200 m5.0 10 mol

8.31 J/ mol K 294 KPVnRT

×= = = ×

⎡ ⎤⋅⎣ ⎦

Now

( ) ( )7

5 452 2

7.8 10 J 75 K8.31J/ mol K 5.0 10 mol

QTRn

×Δ = = =⎡ ⎤⋅ ×⎣ ⎦

96. REASONING When an amount Q of heat flows spontaneously from the hot reservoir at

temperature TH = 373 K to the cold reservoir at temperature TC = 273 K, the amount of energy rendered unavailable for work is

unavailable 0 universeW T S= Δ (15.19)

In Equation 15.19, T0 is the Kelvin temperature of the lowest-temperature reservoir (T0 = 173 K) that is available, and ΔSuniverse is the change in entropy of the universe as a

788 THERMODYNAMICS

result of the heat transfer. This entropy change is the sum of the entropy change ΔSH of the hot reservoir and the entropy change ΔSC of the cold reservoir:

universe H CS S SΔ = Δ +Δ (1)

To find the entropy changes of the hot and cold reservoirs, we employ R

QST

⎛ ⎞Δ = ⎜ ⎟⎝ ⎠

(Equation 15.18), imagining reversible processes where an amount of heat Q is transferred from the hot reservoir to the cold reservoir. The entropy changes given by Equation 15.18 will be the same as for the irreversible process that occurs as heat Q flows from the hot reservoir, through the copper rod, and into the cold reservoir, because entropy is a function of state. Equation 15.18, then, yields

H CH C

and Q Q

S ST T− +

Δ = Δ = (2)

The hot reservoir loses heat, and the cold reservoir gains heat, as heat flows from the hot reservoir to the cold reservoir. The explicit algebraic signs (− and +) in Equation (2) reflect these facts. Because the heat flow from the hot reservoir to the cold reservoir occurs as a process of thermal conduction, the heat Q that flows through the rod in a time t is found from

( ) kA T tQ

LΔ

= (13.1)

Here, ( )390 J s m Ck = ⋅ ⋅ o is the thermal conductivity of copper (see Table 13.1 in the

text), A and L are the cross-sectional area and length of the rod, respectively, and ΔT is the difference in temperature between the ends of the rod: ΔT = TH − TC. SOLUTION Substituting Equation (1) into Equation 15.19 yields ( )unavailable 0 universe 0 H CW T S T S S= Δ = Δ +Δ (3)

Substituting Equations (2) into Equation (3) and simplifying, we find that

( ) 0 0unavailable 0 H C 0

H C H C

T TQ QW T S S T Q

T T T T⎛ ⎞ ⎛ ⎞−

= Δ + Δ = + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(4)

Lastly, substituting Equation 13.1 into Equation (4), we obtain a final expression for the amount of energy that becomes unavailable for doing work in this process:

( )0 0 0 0unavailable

H C H C

T T T TkA T tW Q

T T L T T⎛ ⎞ ⎛ ⎞Δ

= − + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(5)


Expressed in seconds, the elapsed time is 2.0 mint = ( ) 60 s1 min

120 s⎛ ⎞ =⎜ ⎟⎝ ⎠

. In addition, the

temperature difference ΔT = 373 K − 273 K = 1.00×102 K is equivalent to 1.00×102 C°, because one kelvin is equivalent to one Celsius degree. Equation (5), then, yields the desired result:

( ) ( )( )( )4 2 2

unavailable

390 J s m C 9.4 10 m 1.00 10 C 120 s 173 K 173 K0.35 m 373 K 273 K

2100 J

W−⎡ ⎤⋅ ⋅ × × ⎛ ⎞⎣ ⎦= − +⎜ ⎟⎝ ⎠

=

o o

97. REASONING The power rating P of the heater is equal to the heat Q supplied to the gas

divided by the time t the heater is on, /P Q t= (Equation 6.10b). Therefore, /t Q P= . The heat required to change the temperature of a gas under conditions of constant pressure is given by ( )P Equation 15.6 ,Q C n T= Δ where CP is the molar specific heat capacity at constant pressure, n is the number of moles, and ΔT = Tf − Ti is the change in temperature. For a monatomic ideal gas, the specific heat capacity at constant pressure is 5

P 2 ,C R= Equation (15.7), where R is the universal gas constant. We do not know n, Tf and Ti, but we can use the ideal gas law, PV = nRT , (Equation 14.1) to replace nRTf by PfVf and to replace nRTi by PiVi, quantities that we do know.

SOLUTION Substituting ( )P P f iQ C n T C n T T= Δ = − into /t Q P= and using the fact that

5P 2C R= give

( ) ( )5P f i f i2C n T T Rn T TQt

P P P− −

= = = Replacing RnTf by PfVf and RnTi by PiVi and remembering that Pi = Pf , we find

( )5i f i2 P V V

tP−

= Since the volume of the gas increases by 25.0%, Vf = 1.250Vi. The time that the heater is on

is ( ) ( ) ( )

( )( )( )

5 5 5i f i i i i i i2 2 2

5 3 352

1.250 0.250

7.60 10 Pa 0.250 1.40 10 m44.3 s

15.0 W

P V V P V V P Vt

P P P

−

− −= = =

× ×= =

790 THERMODYNAMICS

98. REASONING The first law of thermodynamics states that due to heat Q and work W, the internal energy of the system changes by an amount ΔU according to ΔU = Q − W (Equation 15.1). This law will enable us to find the various quantities for the three processes. Process A→B There is no work done for the process A→B. The reason is that the volume is constant (see the drawing), which means that the change ΔV in the volume is zero. Thus, the area under the plot of pressure versus volume is zero, and WA→B = 0 J. Since Q is known, the first law can then be used to find ΔU.

Process B→C Since the change ΔUB→C in the internal energy of the gas and the work WB→C are known for the process B→C, the heat QB→C can be determined directly by using the first law of thermodynamics: QB→C = ΔUB→C + WB→C (Equation 15.1).

Process C→A For the process C→A it is possible to find the change in the internal energy of the gas once the changes in the internal energy for the processes A→B and B→C are known. The total change ΔUtotal in the internal energy for the three processes is ΔUtotal = ΔUA→B + ΔUB→C + UC→A, and we can use this equation to find ΔUC→A, with the aid of values for ΔUA→B and ΔUB→C. This is possible because ΔUtotal is the change in the internal energy for the total process A→B→C→A. This process begins and ends at the same place on the pressure-versus-volume plot. Therefore, the value of the internal energy U is the same at the start and the end, with the result that ΔUtotal = 0 J.

SOLUTION

a. Since the change in volume is zero (ΔV = 0 m3), the area under the plot of pressure versus volume is zero, with the result that the work is WA→B = 0 J .

b. The change in the internal energy of the gas for the process A→B is found from the first law of thermodynamics:

ΔUA→B = QA→B − WA→B = +561 J − 0 J = 561 J+

c. According to the first law of thermodynamics, the change in the internal energy of the gas for the process B→C is ΔUB→C = QB→C − WB→C. Thus, the heat for this process is

QB→C = ΔUB→C + WB→C = +4303 J + 3740 J = 8043 J+

Volume

Pres

sure

A

B C


Since this heat is positive, it is added to the gas.

d. The change ΔUtotal in the total internal energy for the three processes is ΔUtotal = ΔUA→B + ΔUB→C + ΔUC→A. Solving this equation for ΔUC→A gives ΔUC→A = ΔUtotal − ΔUA→B − ΔUB→C. As discussed in the REASONING (Process C→A), ΔUtotal = 0 J, since the third process ends at point A, which is the start of the first process. Therefore,

ΔUC→A = 0 J − 561 J − 4303 J = 4864 J−

e. According to the first law of thermodynamics, the change in the internal energy of the gas for the process C→A is ΔUC→A = QC→A − WC→A. The heat for this process is

QC→A = ΔUC→A + WC→A = −4864 J + (−2867 J) = 7731 J− Since this heat is negative, it is removed from the gas.

99. SSM REASONING AND SOLUTION We wish to find an expression for the overall

efficiency e in terms of the efficiencies 1e and 2e . From the problem statement, the overall efficiency of the two-engine device is

1 2

H

W We

Q+

= (1)

where HQ is the input heat to engine 1. The efficiency of a heat engine is defined by

Equation 15.11, H/e W Q= , so we can write 1 1 HW e Q= (2) and

2 2 H2W e Q= Since the heat rejected by engine 1 is used as input heat for the second engine, H2 C1Q Q= ,

and the expression above for 2W can be written as 2 2 C1W e Q= (3) According to Equation 15.12, we have C1 H 1–Q Q W= , so that Equation (3) becomes

792 THERMODYNAMICS

( )2 2 H 1 W e Q W= − (4)

Substituting Equations (2) and (4) into Equation (1) gives

( ) ( )1 H 2 H 1 1 H 2 H 1 H

H H

e Q e Q W e Q e Q e Qe

Q Q

+ − + −= =

Algebraically canceling the HQ 's in the right hand side of the last expression gives the

desired result: 1 2 1 2e e e e e= + −

100. REASONING We seek the difference Wdiff in work between the work Wad done by the gas

during the adiabatic process and the work Walt done by the gas during the alternative process: diff ad altW W W= − (1) Considering the alternative process first, we bear in mind that, according to ( )f iW P V V= − (Equation 15.2), a gas only does work if its volume increases. As the pressure of the gas decreases to Pf from Pi during the first step, its volume remains constant, and, therefore, the gas does no work. Consequently, the total work Walt done by the gas during the entire alternative process occurs during the second step, when the gas expands to a volume Vf from a volume Vi at a constant pressure Pf: ( )alt f f iW P V V= − (2) Only the initial volume Vi is given, so the final volume Vf in Equation (2) must be determined. Because the gas is ideal and monatomic, and can undergo an adiabatic process that takes it from the initial volume Vi to the final volume Vf , we will use i i f fPV PVγ γ=

(Equation 15.5), with 53γ = , to determine the final volume Vf . Solving Equation 15.5 for

the final volume yields i if

f

PVV

P

γγ = , or

( )351 1 5

3 3 2 3i i if i 4

f f

2.20 10 Pa6.34 10 m 1.15 10 m8.15 10 Pa

PV PV V

P P

γ γγ− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞×= = = × = ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠⎝ ⎠


In the adiabatic process, the gas does an amount of work Wad given by ( )3

ad i f2W nR T T= − (15.4) In Equation 15.4, n is the number of moles of the gas, R is the universal gas constant, and Ti and Tf are the initial and final Kelvin temperatures of the gas. In order to determine the initial and final temperatures, we will employ the ideal gas law: PV nRT= (Equation 14.1). Solving Equation 14.1 for the initial and final Kelvin temperatures yields

i i f fi f and PV PV

T TnR nR

= = (3)

SOLUTION Substituting Equations (3) into Equation 15.4 gives

( )3 3ad i f2 2W nR T T nR= − = i iPV

nRf fPVnR

− ( )3i i f f2 PV PV⎛ ⎞ = −⎜ ⎟

⎝ ⎠ (4)

Substituting Equations (4) and (2) into Equation (1), we obtain

( ) ( )

( )

3diff ad alt i i f f f f i2

3 3 3 5i i f f f f f i i i f f f i2 2 2 2

3 5i i f i f2 2

W W W PV PV P V V

PV PV PV PV PV PV PV

PV P V V

= − = − − −

= − − + = − +

= + −

Substituting in the given values of Pi , Pf , Vi , and the calculated value of Vf , yields

( )

( )( )( ) ( )

3 5diff i i f i f2 2

5 3 332

4 3 3 2 352

2.20 10 Pa 6.34 10 m

8.15 10 Pa 6.34 10 m 1.15 10 m 270 J

W PV P V V

−

− −

= + −

= × ×

⎡ ⎤+ × × − × =⎣ ⎦

101. SSM REASONING The conservation of energy dictates that the heat delivered into the

house (the hot reservoir) equals the energy from the work done by the heat pump plus the energy in the form of heat taken from the cold outdoors (the cold reservoir). The heat delivered into the house is given, so that we can use energy conservation to determine the work, provided that we can obtain a value for the heat taken from the outdoors. Since we are dealing with an ideal heat pump, we can obtain this value by using Equation 15.14, which

794 THERMODYNAMICS

relates the ratio of the magnitudes of the heats for the cold and hot reservoirs to the ratio of the reservoir temperatures (in kelvins).

The energy-conservation principle requires that

QH = W + QC (Equation 15.12), where

QH , W , and

QC are, respectively, the magnitudes of the heat delivered into the house

(the hot reservoir), the work done by the heat pump, and the heat taken from the cold outdoors (the cold reservoir). Solving for W gives Equation (1) below.

W = QH − QC (1)

In this result, we have a value for

QH but we need to find an expression for

QC . To

obtain a value for QC , we will use the information given about the temperatures TH for the

hot reservoir and TC for the cold reservoir. According to Equation 15.14, QC / QH = TC /TH ,

which can be solved for QC to show that

QC = QHTCTH

⎛⎝⎜

⎞⎠⎟

We can now substitute this result into Equation (1) above.

W = QH − QC = QH − QH

TCTH

⎛

⎝⎜

⎞

⎠⎟ = QH 1−

TCTH

⎛

⎝⎜

⎞

⎠⎟

It follows that the magnitude of the work for the two given outdoor temperatures is

Outdoor temperature of 273 K W = QH 1−

TCTH

⎛

⎝⎜

⎞

⎠⎟ = (3350 J) 1− 273 K

294 K⎛⎝⎜

⎞⎠⎟= 239 J

Outdoor temperature of 252 K W = QH 1−

TCTH

⎛

⎝⎜

⎞

⎠⎟ = (3350 J) 1− 252 K

294 K⎛⎝⎜

⎞⎠⎟= 479 J

More work must be done when the outdoor temperature is lower, because the heat is pumped up a greater temperature “hill.”

_____________________________________________________________________________ 102. CONCEPTS (i) Sublimation is the process whereby a solid phase changes directly into a

gas phase in response to the input of heat. The heat per kilogram needed to cause the phase change is called the latent heat of sublimation Ls. The heat Q needed to bring about the


sublimation of a mass m of solid material is given by Q = mLs. (ii) For a given mass of material, gases generally have greater volumes than solids do, so the volume of the material

increases. The increase in volume isΔV =Vgas −Vsolid . Since the volume of the solidVsolid is

negligibly small in comparison to the volume of the gasVgas , we haveΔV =Vgas . Using the

ideal gas law, it follows thatVgas = nRT P , so thatΔV = nRT P . In this result, n is the number of moles of material, R is the universal gas constant, and T is the Kelvin temperature. (iii) To make room for itself, the expanding material must push against the environment and, in doing so, does work on the environment. Since the pressure remains constant, the work done by the material is given byW = PΔV . SinceΔV = nRT P , the work

becomesW = P nRT P( ) = nRT . (iv) According to the conservation-of-energy principle, energy can neither be created nor destroyed, but can only be converted to one form or another. Therefore, part of the heat Q is used forΔU and part for W, with the result that Q = ΔU +W . (v) The first law of thermodynamics is ΔU =Q −W . Rearranging this equation gives Q = ΔU +W , which is identical to the result obtained from the conservation-of-energy principle. CALCULATIONS Using the facts that Q = ΔU +W , Q = mLs , and W = nRT we have that Solving forΔU gives In this result, n is the number of moles of the ideal gas. The number of moles of gaseous zinc is the mass m of the sample divided by the mass per mole of zinc or n = m 0.0654 kg mol( ) . Therefore, we find

_____________________________________________________________________________

Q = ΔU +W or mLs = ΔU + nRT

ΔU = mLS − nRT

ΔU = mLS − nRT

= 1.50 kg( ) 1.99 ×106 Jkg

⎛⎝⎜

⎞⎠⎟− 1.50 kg

0.0654 kg mol⎛⎝⎜

⎞⎠⎟

8.31 Jmol ⋅K

⎛⎝⎜

⎞⎠⎟ 6.00 ×102 K( )

= 2.87 ×106 J

796 THERMODYNAMICS

103. SSM CONCEPTS (i) The change is greater with engine 2. Kinetic energy isKE= 12 mv

2 , where m is the mass of the crate and v is its speed. The change in the kinetic energy is the final minus the initial value, orKEf −KE0 . Since the crate starts from rest, it has zero initial kinetic energy. Thus the change is equal to the final kinetic energy. Since engine 2 gives the crate the greater final speed, it causes the greater change in kinetic energy. (ii) The work-energy theorem states that the net work done on an object equals the change in the object’s kinetic energy, or W = KEf −KE0 . The net work is the work done by the net force. The surface on which the crate moves is horizontal, and the crate does not leave it. Therefore, the upward normal force that the surface applies to the crate must balance the downward weight of the crate. Furthermore, the surface is frictionless, so there is no friction force. The net force acting on the crate, then, consists of the single force due to the tension in the rope, which arises from the action of the engine. Thus, the work done by the engine is, in fact, the net work done on the crate. But we know that engine 2 causes the crate’s kinetic energy to change by the greater amount, so that the engine must do more work. (iii) The temperature of the hot reservoir for engine 2 is greater. We know that engine 2 does more work, but each engine receives the same 1450 J of input heat. Therefore, engine 2 derives more work from the input heat. In other words, it is more efficient. But the efficiency of a Carnot engine depends only on the Kelvin temperatures of its hot and cold reservoirs. Since both engines use the same cold reservoir whose temperature is 275 K, only the temperatures of the hot reservoirs are different. Higher temperatures for the hot reservoir are associated with greater efficiencies, so the temperature of the hot reservoir for engine 2 is greater. CALCULATIONS The efficiency e of a heat engine is the magnitude of the work Wdivided by the magnitude of the input heat QH , or e = W QH . The efficiency of a Carnot engine is eCarnot = 1−TC TH , where TC and TH are, respectively, the temperatures of the cold and hot reservoirs. Combining these two equations, we have But W is the magnitude of the net work done on the crate, and it equals the change in the crate’s kinetic energy, or W = KEf −KE0 = 1

2 mv2 .

With this substitution, the efficiency expression becomes Solving for the temperature TH , we find

1− TCTH

=WQH

1− TCTH

=12 mv2

QH

TH = TC

1− mv2

2 QH


Using this expression, we can calculate the temperature of the hot reservoir for each engine: Engine 1 Engine 2 As expected, the value of TH for engine 2 is greater.

_____________________________________________________________________________

TH = 275 K

1−125 kg( ) 2.00 m s( )2

2 1450 J( )

= 332 K

TH = 275 K

1−125 kg( ) 3.00 m s( )2

2 1450 J( )

= 449 K

chapter 15 thermodynamics

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Transcript of chapter 15 thermodynamics