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HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

EXERCISES

Section 18.1 The First Law of Thermodynamics

15. INTERPRET We identify the system as the water in the insulated container. The problem involves calculating the

work done to raise the temperature of a system, so the first law of thermodynamics comes into play.

DEVELOP Because the container is a perfect thermal insulator, no heat enters or leaves the water inside of it.

Thus, Q = 0 and the first law of thermodynamics in Equation 18.1 gives ,U Q W WΔ = + = where W is the work

done by shaking the container. The change in the internal energy of the water is determined from its temperature

rise, U mc TΔ = Δ (see comments in Section 16.1 on internal energy). Combining these two expressions for the

internal energy change allows us to find the work done.

EVALUATE The work done on the water is

( ) ( ) ( )1.0 kg 4.184 kJ/ kg K 7.0 K 29.3 kJ

W U mc T= Δ = Δ

⎡ ⎤= ⋅ =⎣ ⎦

ASSESS According to the convention adopted for the first law of thermodynamics, positive work signifies that

work is done on the water.

16. INTERPRET This problem involves finding the mechanical work that is required to raise the given mass of water

by 3.0°C, which we can calculate by using the first law of thermodynamics.

DEVELOP The change in the internal energy of the water is ,U mc TΔ = Δ where ΔT = 3.0°C = 3.0 K and the

work done on it is W = 9.0 kJ. Inserting this into the first law of thermodynamics (Equation 18.1) gives Q = ΔU −

W = mcΔT − W. For part (b), note that if the water had been in perfect thermal isolation, no heat would have been

transferred, so Q′ = 0 and we can solve for the work W′ as for part (a).

EVALUATE (a) Inserting the given quantities into the expression for the heat transferred gives

( ) ( ) ( )0.50 kg 4.184 kJ/ kg K 3.0 K 9.0 kJ 2.7 kJQ mc T W ⎡ ⎤= Δ − = ⋅ − = −⎣ ⎦

(b) If Q = 0, then the work needed is

( ) ( ) ( )0.50 kg 4.184 kJ/ kg K 3.0 K 6.3 kJW mc T ⎡ ⎤= Δ = ⋅ =⎣ ⎦

ASSESS The fact that Q < 0 for part (a) indicates that heat is transferred from the water to the environment, as expected.

17. INTERPRET The system of interest is the gas that undergoes expansion. The problem involves calculating the

change in internal energy of a system, so the first law of thermodynamics comes into play.

DEVELOP The heat added to the gas is (40 W)(25 s) 1000 J.Q Pt= = = In addition, the system does 750 J of

work on its surroundings, so the work done by the surroundings on the system is 750 J.W = − The change in

internal energy can be found by using the first law of thermodynamics given in Equation 18.1.

EVALUATE Using Equation 18.1 we find

1000 J 750 J 250 JU Q WΔ = + = − =

ASSESS Since 0,UΔ > we conclude that the internal energy has increased.

18

18-2 Chapter 18

© Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may


18. INTERPRET This problem involves calculating the rate of heat flow into the system given the rate at which the

internal energy is increasing and the rate at which the system is doing work on its surroundings.

DEVELOP The dynamic form of the first law of thermodynamics (Equation 18.2) is

dQ dU dWdt dt dt

= −

The problem states that dU/dt = 45 W and dW/dt = −165 W (where the negative sign is appropriate because the

system is doing work).

EVALUATE Inserting the given quantities gives

45 W 165 W 210 WdQdt

= + =

ASSESS The rate of heat flow into the system is positive because the system is doing work on its surroundings and increasing its internal energy.

19. INTERPRET This problem is about heat and mechanical energy, which are related by the first law of

thermodynamics. The system of interest is the automobile engine.

DEVELOP Since we are dealing with rates, we make use of the dynamic form of the first law of thermodynamics,

Equation 18.2:

dU dQ dWdt dt dt

= +

If we assume that the engine system operates in a cycle, then / 0.dU dt = The engine’s mechanical power output

dW/dt can then be calculated once the heat output is known.

EVALUATE The above conditions yield ( )out/ 68 kWdQ dt = and ( ) ( )in/ 0.17 / .dW dt dQ dt= Equation 18.2 then

gives

out in out

10.17

dW dQ dQ dQ dQ dWdt dt dt dt dt dt

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

or

( )( )

out11

/ 68 kW 14 kW1 (0.17) 1 0.17

dQ dtdWdt −−= = = −

− −

ASSESS We find the mechanical power output /dW dt to be proportional to the heat output, out( / ) .dQ dt In

addition, /dW dt also increases with the percentage of the total energy released in burning gasoline that ends up as

mechanical work. The mechanical output is negative because the system is doing work on its surroundings.

Section 18.2 Thermodynamic Processes

20. INTERPRET This problem involves an ideal gas that changes from an initial pressure-volume state to a final

pressure-volume state by traversing the given pressure-volume curve. From this, we are to find the work done by

the gas during this process.

DEVELOP The work done by the gas is given by the negative of Equation 18.3, which is

( )2

1

by gas

V

V

W p V dV= ∫

This is the area under the line segment AB in Figure 18.19. We can find this area from simple geometrical

arguments, without having to invoke calculus.

EVALUATE The work done by the gas is the area of the trapezoid under line AB, which is

( )( ) ( )( )by gas 1 2 2 1 1 1 1 1 1 11 1 32 22 2 2

W W P P V V P P V V PV= = + − = + − =

Heat, Work, and The First Law of Thermodynamics 18-3



ASSESS The work can also be obtained from Equation 18.3. On the path AB,

( ) ( )( )( )

1 2 11

2 1

V V P Pp V P

V V− −

= +−

so

( ) ( ) ( )2

1

22 1by gas 1 2 1 2 1 1 1

2 1

1 3 .2 2

V

V

P PW p V dV P V V V V PVV V⎛ ⎞−= = − + − =⎜ ⎟−⎝ ⎠

∫

21. INTERPRET The expansion of the ideal gas involves two stages: an isochoric (constant-volume) process and an

isobaric (constant-pressure) process. We are asked to find the total work done by the gas.

DEVELOP For an isochoric process, ΔV = 0 so W1 = 0 (see Equation 18.3 and Table 18.1). On the other hand, for

an isobaric process, the work done is W2 = pΔV which is the negative of Equation 18.7 because we are interested in

the work done by the gas (Equation 18.7 gives the work done on the gas).

EVALUATE Summing the two contributions to find the total work gives

( ) ( )tot 1 2 2 2 1 1 1 1 1 12 2 2W W W p V V p V V pV= + = − = − =

ASSESS In the pV diagram of Fig. 18.19, the area under AC is zero, and that under CB, a rectangle, is 1 12 .p V The

work done by the gas is the area under the pV curve.

22. INTERPRET This problem involves the isothermal (i.e., the temperature is constant) expansion of a balloon as it

rises. We are asked to find the work done by the helium as the balloon rises.

DEVELOP During an isothermal expansion, the work done by a given amount of ideal gas is

( )2 1 lnW nRT V V=

which is the negative of Equation 18.3 because that form expresses the work done on the gas.

EVALUATE Inserting the given quantities yields

( ) ( ) ( )0.30 mol 8.314 J/ mol K 300 K ln(5.0) 1.2 kJW ⎡ ⎤= ⋅ =⎣ ⎦

ASSESS The result is positive because positive work is done by the gas to expand its volume.

23. INTERPRET The constant temperature of 300 K indicates that the process is isothermal. We are to find the

increase in volume of the balloon and the work done by the gas, given the pressure difference in experiences.

DEVELOP We assume the gas to be ideal and apply the ideal-gas law given in Equation 17.2: pV = nRT. For an

isothermal process, T = constant, so we obtain 1 1 2 2pV p V= , from which we can find the fractional volume increase.

The total work done by the gas can be calculated using the negative of Equation 18.4:

2

1

ln VW nRTV

⎛ ⎞= ⎜ ⎟

⎝ ⎠

because we are interested in the work done by the gas, not on the gas.

EVALUATE (a) For the isothermal expansion process, the volume increases by a factor of

2 1

1 2

100 kPa 475 kPa 3

V pV p

= = =

(b) Using Equation 18.4, the work done by the gas is

( ) ( ) ( )2

1

4ln 0.30 mol 8.314 J/ mol K 300 K ln 220 J3

VW nRTV

⎛ ⎞ ⎛ ⎞⎡ ⎤= = ⋅ =⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎝ ⎠⎝ ⎠

to two significant figures.

ASSESS Because 2 1V V> , we find the work to be positive, W > 0. This makes sense because the gas inside the

balloon must do positive work to expand outward.

24. INTERPRET This problem involves an ideal gas that we compress in an isothermal process to half its volume. We

are to find the work needed to do this.

18-4 Chapter 18



DEVELOP In an isothermal compression of a fixed quantity of ideal gas, work is done on the gas so we use

Equation 18.4, W = −nRTln(V1/V2), where V1 = 2V2 and T = 300 K for this problem.

EVALUATE Inserting the given quantities into Equation 18.4 gives

( ) ( ) ( ) ( ) ( )2 1ln / 2.5 mol 8.314 J/ mol K 300 K ln 1/2 4.3 kJ.W nRT V V ⎡ ⎤= − = − ⋅ = +⎣ ⎦

ASSESS The work is positive because positive work must be expended by an external agent other than the gas to compress the gas.

25. INTERPRET The thermodynamic process here is adiabatic, so no heat flows between the system (the gas) and its

environment. We are to find the volume change needed to double the temperature.

DEVELOP In an adiabatic process, Q = 0, so the first law of thermodynamics (Equation 18.1) becomes .U WΔ =

The temperature and volume are related by Equation 18.11b: 1 1

1 1 2 2constant =TV T Vγ γ− −=

The temperature doubles, so T2 = 2T1 and γ = 1.4, so we can solve for the fractional change in volume.

EVALUATE From the equation above, we have

1 1

1/ ( 1)

2 11 1 2 2

1 2

V TTV T VV T

γ γγ

− −

−⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

Thus, for the temperature to double, the volume change must be 1/ ( 1) 1/(1.4 1)

2 1

1 2

1 0.1772

V TV T

γ − −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

ASSESS We see that increasing the temperature along the adiabat is accompanied by a volume decrease. In

addition, since constant,pV γ = the final pressure is also increased: 1.4

12 1 1 1

2

1 11.30.177

Vp p p pV

γ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠


environment. We are to find the temperature increase as the gas volume is reduced.


The temperature and volume are related by Equation 18.11b: 1 1

1 1 2 2TV T Vγ γ− −=

The final volume is 1/4 of the initial, so V2 = V1/4 and γ = 1.4, so we can solve for the change in temperature.

EVALUATE From the equation above, we have

1 1

1

1.4 111 1 2 2 2 1

2

(291 K)(4) 507 KVTV T V T TV

γ γγ

− −

−

−⎛ ⎞= ⇒ = = =⎜ ⎟

⎝ ⎠

Thus, the temperature change is

2 1 507 K 291 K 216 KT T TΔ = − = − =

ASSESS We see that decreasing the volume along the adiabat is accompanied by a temperature increase. In

addition, since constant,pV γ = the final pressure is also increased:

( )1.412 1 1 1

2

4 6.96Vp p p pV

γ⎛ ⎞

= = =⎜ ⎟⎝ ⎠

27. INTERPRET This problem involves an ideal gas that we compress in an isothermal process to half its volume. We

are to find the work needed to do this.

DEVELOP In an isothermal compression of a fixed quantity of ideal gas, work is done on the gas so we use

Equation 18.4, W = −nRTln(V1/V2), where V2/V1 = 0.05 for this problem.




EVALUATE Inserting the given quantities into Equation 18.4 gives

( ) ( ) ( ) ( )5 3 62 1 1 1 2 1ln / ln / (1.013 10 Pa) 6.8 m ln 0.050 2.1 10 J.W nRT V V pV V V= − = − = − × = ×

ASSESS The work is positive because positive work must be expended by an external agent other than the gas to compress the gas.

Section 18.3 Specific Heats of an Ideal Gas

28. INTERPRET This problem involves finding the molar specific heat at constant volume and constant pressure for

the given gas mixture, which contains the given quantities of diatomic O2 and monatomic Ar.

DEVELOP This problem in similar to Example 18.5, so we will use the same approach as outlined there.

EVALUATE Because O2 has 5 degrees of freedom, so its average energy per molecule is 5kT/2. For Ar, the

average energy per molecule is 3kT/2 because it is monatomic. The total internal energy is therefore

( ) ( ) ( )2O Ar

5 3 5 32.5 mol 3 .0 mol 10.75 mol2 2 2 2

U n RT n RT RT RT⎡ ⎤= + = + =⎢ ⎥⎣ ⎦

From Equation 18.6, the molar specific heat at constant volume is

( )( )V

10.75 mol1 2.02.5 3.0 mol

RdUC Rn dT

= = =+

to two significant figures. From Equation 18.9, the molar specific heat at constant pressure is

P V 3.0C C R R= + =

ASSESS We find that the molar specific heat at constant volume is between that of a diatomic molecule (2.5 R) and a monatomic molecule (1.5 R), as expected since we have a mixture of these two types of molecules.

29. INTERPRET The problem is about the specific heat of a mixture of gases. We want to know what fraction of the

molecules is monatomic, given the its specific-heat ratio.

DEVELOP The internal energy of a mixture of two ideal gases is

1 1 2 2U f NE f NE= +

where if is the fraction of the total number of molecules, N, of type i, and iE is the average energy of a molecule

of type i. Classically, ( )12 ,E g kT= where g is the number of degrees of freedom (the equipartition theorem).

From Equation 18.6, the molar specific heat at constant volume is

( )V 1 1 2 2 1 1 2 21 1 1 1

2 2 2dU dC nR f g T f g T R f g f g

n dT dT⎛ ⎞= = + = +⎜ ⎟⎝ ⎠

Suppose that the temperature range is such that g1 = 3 for the monatomic gas, and g2 = 5 for the diatomic gas, as

discussed in Section 18.3. Then

( ) ( )V 1 2 11 3 5 2.52

C R f f R f= + = −

where 2 11f f= − since the sum of the fractions of the mixture is unity. Now, CV can also be specified by the ratio

P V

V V V

V

= 1

1

C C R RC C C

RC

γ

γ

+= = +

=−

Equating the two expressions allows us to solve for 1.f

EVALUATE Solving, we find

1 11 12.5 57.7%

1 0.52f f

γ− = = ⇒ =

−

ASSESS From the equation above, we see that the specific-heat ratio can be written as

1

112.5 f

γ = +−

18-6 Chapter 18



In the limit where all the gas molecules are monatomic, 1 1,f = and 1.67.γ = On the other hand, if all the

molecules are diatomic, then 1 0f = and the specific-heat ratio is 1.4.γ = The equation yields the expected results

in both limits.

30. INTERPRET For this problem, we are to find the specific-heat ratio of a gas consisting of the given ratio of

different gas molecules.

DEVELOP By generalizing the result of the previous problem, the molar specific heat of a mixture of three gases

is

1 2 31 2 3 .V V V VC f C f C f C= + +

For each gas and the mixture, use ( )1VC R γ= − to obtain

( ) ( ) ( ) ( )1 1 1 11 1 2 2 3 31 1 1 1f f fγ γ γ γ− − − −− = − + − + −

which we can evaluate to find γ.

EVALUATE When the data specified in the problem is inserted, one gets

1 0.50 0.30 0.20 2.771 0.29 0.40 0.67

1.36γ

γ

= + + =−

=

ASSESS Because we have some triatomic molecules, the specific-heat ratio is less than that for a pure diatomic gas (see previous problem).

31. INTERPRET The thermodynamic process is adiabatic, and we want to know the temperature change when work is

done on a monatomic gas and a diatomic gas.

DEVELOP In an adiabatic process, Q = 0, so from the first law of thermodynamics (Equation 18.1) ,U WΔ =

where W is the work done on the gas. From Equation 18.6, ,VU nC TΔ = Δ the change in temperature is

V V

U WTnC nCΔΔ = =

If the work done per mole on the gas is 2.5 kJ/mol,W n = then ( )2.5 kJ/mol VT CΔ =

EVALUATE (a) For an ideal monatomic gas, ( )3 32 2 8.314 J/ mol K ,VC R ⎡ ⎤= = ⋅⎣ ⎦ so 200 K.TΔ =

(b) For an ideal diatomic gas (with five degrees of freedom), 52VC R= so 120 K.TΔ =

ASSESS Since the diatomic gas has a greater specific heat CV, its temperature change is less than that of the

monatomic gas.

PROBLEMS

32. INTERPRET The constant temperature of 440 K indicates that the process is isothermal. We are given the amount

of work done by the gas, and are asked to find the heat it absorbs and how many moles of gas there are.

DEVELOP Apply the ideal-gas law given in Equation 17.2: .pV nRT= For an isothermal process, T = constant,

so we obtain 1 1 2 2.pV p V= Since 0UΔ = for an isothermal process, the heat absorbed is the negative of the total

work done on the gas (Equation 18.4).

2

1

ln VQ W nRTV

⎛ ⎞= − = ⎜ ⎟

⎝ ⎠

For this problem, the gas does 3.3 kJ of work on its surroundings, so the surroundings do −3.3 kJ of work on the

gas, so W = −3.3 kJ.

EVALUATE (a) Using the equation above, the heat absorbed is 3.3 kJ.Q W= − =

(b) Solving the expression above for the number n of moles gives

( ) ( ) ( ) ( )2 1

3.3 kJ 0.39 molln / 8.314 J/ mol K 440 K ln 10

WnRT V V

−= = =⎡ ⎤⋅⎣ ⎦

ASSESS The heat absorbed by the gas is equal to the work done by the gas on its surrounding as it expands, and

there is no change in temperature.




33. INTERPRET We’re asked to find the rate that heat is produced in the body when cycling. This involves the first law

of thermodynamics.

DEVELOP We’re dealing with the rates of work and heat production, so we’ll use Equation 18.2: .dQdU dWdt dt dt= +

If the body is releasing stored food energy, that corresponds to a decrease in the body’s internal energy: / 500 W.dU dt = − Likewise, the mechanical power quoted is for work done by the body, so / 120 W.dW dt = − We

are looking for the rate at which the body produces heat, which is technically heat that it is losing (–dQ). EVALUATE The rate of heat production is the negative heat absorbed, so

( ) ( )500 W 120 W 380 WdQ dU dWdt dt dt

− = − + = − − + − =

ASSESS All the signs can be confusing, but essentially the body burns stored energy and some of it is used to do

work and the rest is released as heat.

34. INTERPRET This problem deals with isothermal compression, so the temperature is constant in this process. We

are to find the work done to compress the given quantity of ideal gas from 3.5 L to 3.0 L.

DEVELOP Apply Equation 18.4,

( )2 1lnQ W nRT V V= − =

where W is the work done by the gas, V1 = 3.5 L, and V2 = 3.0 L. Since we do 61 J of work on the gas to compress

it, W = 61 J, so we can solve this expression for the temperature T.

EVALUATE The temperature of the ideal gas is

( ) ( ) ( ) ( )2 1

61 J 190 Kln 0.25 mol 8.314 J/ mol K ln 3.0 /3.5

WTnR V V

− −= = =⎡ ⎤⋅⎣ ⎦

ASSESS This gas is about 100 K below room temperature.

35. INTERPRET Assume the air inside the spherical bubble behaves like an ideal gas at constant temperature, so the

process is isothermal. We need to find the diameter of the bubble at maximum pressure and the work done on the

gas in compressing it.

DEVELOP Apply the ideal-gas law given in Equation 17.2: .pV nRT= For an isothermal process, T = constant,

which leads to 1 1 2 2.p V p V= Since the volume of a spherical bubble of diameter d is 3 34

3 2 6d dV π π⎛ ⎞= =⎜ ⎟

⎝ ⎠

the relationship between the diameter and the pressure is 3 31 2

1 2

1/3

2 1

1 2

6 6d dp p

d pd p

π π⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

EVALUATE (a) Using the equation above, we find the diameter at the maximum pressure to be

( )( ) ( )

1/31/3

12 1

2

80 760 mm of Hg1.52 mm 1.49 mm

125 760 mm of Hgpd dp

⎡ ⎤+⎛ ⎞= = =⎢ ⎥⎜ ⎟ +⎢ ⎥⎝ ⎠ ⎣ ⎦

(b) The work done on the air is given by Equation 18.4, or

( ) ( )

2 1 2on air 1 1 1 1

1 2 1

3

ln ln ln

101.3 kPa 885 mm of Hg840 mm of Hg 1.52 mm ln760 mm of Hg 6 840 mm of Hg

10.7 μJ

V p pW nRT pV pVV p p

π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞

= ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

ASSESS Positive work is done by the blood in compressing the air bubble.

18-8 Chapter 18



36. INTERPRET This problem involves compressing a gas in an isothermal process, so the temperature is constant.

We are to find how much work it takes to reduce the volume by a factor of 22 given the work it takes to decrease

the volume by a factor of 2.

DEVELOP Apply Equation 18.4 to both situations. To compress a gas by a factor of two, we have

( )2 ln 2W nRT− =

and to compress it by a factor of 22 gives

( )22 ln 22W nRT− =

Given that T is constant, we can take the ratio of these expressions to find the work W22 needed to compress the gas

by a factor of 22.

EVALUATE The work needed to compress the gas by a factor of 22 is

( )( ) ( ) ( )

( )22 2

ln 22 ln 221.5 k J 6.7 kJ

ln 2 ln 2W W= = =

ASSESS Note that the work required is logarithmic in volume.

37. INTERPRET The thermodynamic process here is adiabatic, with no heat flowing between the system (the gas) and

its environment. We are to find the specific-heat ratio γ given the fraction increase in pressure of the gas.

DEVELOP In an adiabatic process, Q = 0 and the first law of thermodynamics becomes .U WΔ = The pressure

and volume are related by Equation 18.11a: constant.PV γ = This implies

2 11 1 2 2

1 2

p VpV p Vp V

γγ γ ⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

so we can solve for γ.

EVALUATE Taking the natural logarithm on both sides of the above to solve for ,γ we obtain

( )( )

( )( )

2 12 1

1 2 1 2

ln ln 2.55ln ln 1.35

ln ln 2p pp V

p V V Vγ γ

⎛ ⎞ ⎛ ⎞= ⇒ = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

ASSESS The value of γ indicates that gas consists of polyatomic molecules (see Problems 29 and 30).

38. INTERPRET The problem involves compressing a gas adiabatically.

DEVELOP For an adiabatic process, the initial pressure and volume are related to the final pressure and volume

through Equation 18.11a: 1 1 2 2 .p V p Vγ γ= We can use this to rewrite Equation 18.12 for the work done on the gas in

going from the initial to the final volume: 1

2 2 1 1 1 1 1

2

11 1

p V pV pV VWV

γ

γ γ

−⎡ ⎤⎛ ⎞− ⎢ ⎥= = −⎜ ⎟− − ⎢ ⎥⎝ ⎠⎣ ⎦

EVALUATE (a) The final pressure in this case is

( )1.40

12 1

2

6.25 L98.5 kPa 173 kPa4.18 L

Vp pV

γ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

(b) The work done on the gas in compressing it from V1 = 6.25 L to V2 = 4.18 L is

( )( ) 1.40 198.5 kPa 6.25 L 6.25 L 1 269 J1.40 1 4.18 L

W−⎡ ⎤⎛ ⎞= − =⎢ ⎥⎜ ⎟− ⎝ ⎠⎢ ⎥⎣ ⎦

ASSESS The volume decreases by a factor of 1.5, but the pressure increases by a factor of 1.76. The pressure

increases due to both the increase in the density (N/V) and an increase in the internal energy of the gas

( ).VU nC TΔ = Δ

39. INTERPRET This problem involves a cyclic process. The three processes that make up the cycle are: isothermal

(AB), isochoric (BC), and isobaric (CA). We are given the pressure at point A and are to find the pressure at point B

and the net work done on the gas.




DEVELOP Along the isotherm AB T = constant, so the ideal-gas law (Equation 17.2, pV = nRT) gives

.A A B Bp V p V= For an isothermal process, the work W done on the gas is (Equation 18.4):

2

1

ln VW nRTV

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

EVALUATE (a) If AB is an isotherm, with 5 LAV = and 1 L,BV = then the ideal-gas law gives

( )5 60 kPa 300 kPa1

AB A

B

VP PV

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

(b) The work W done on the gas in the isothermal process AB is

( ) 1.0 Lln ln 300 J ln 483 J5.0 L

B BAB A A A

A A

V VW nRT p VV V

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

The process BC is isochoric (constant volume) so 0.BCW = The process CA is isobaric (constant pressure) so the

work done by the gas is (see Table 18.1)

( ) ( )( )60 kPa 5.0 L 1.0 L 240 JCA A A CW p V V′ = − = − =

or the work done on the gas is WCA = −240 J. The total work done by the gas is the sum of these three

contributions, or

483 J 0 240 J 243 J 240 JABCA AB BC CAW W W W= + + = + − = ≈


ASSESS Since the process is cyclic, the system returns to its original state, there’s no net change in internal

energy, so 0.UΔ = This implies that 240 J.ABCAQ W= − = − That is, 240 J of heat must come out of the system.

40. INTERPRET This problem is similar to the previous one, except that the isotherm curve AB is now to be

considered as an adiabat. We are given the specific-heat ratio and are to find the pressure change in going from

point A to point B, and the net work done by the cycle.

DEVELOP For an adiabat, Equation 18.11a is valid. Applying this to points A and B, and taking the ratio, gives

A B

B A

p Vp V

γ⎛ ⎞

= ⎜ ⎟⎝ ⎠

which allows us to find the pressure at point B. The calculation for part (b) then proceeds as for Problem 18.39,

with the help of Equation 18.12, which gives the work done on the gas in an adiabatic process.

EVALUATE (a) The pressure at point B is

( )1.4

5.0 L60 kPa 570 kPa1.0 L

AB A

B

Vp pV

γ⎛ ⎞ ⎛ ⎞

= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠


(b) The work done on the gas between points A and B is

( )( ) ( )( )571 kPa 1.0 L 60 kPa 5.0 L678 J

1 0.4B B A A

ABP V P VW

γ−−= = =

−

The process BC is isochoric (constant volume), so WBC = 0. The process CA is the same as for Problem 18.37, so

WCA = −240 J. Summing all three contributions gives a net work done on the gas of

678 J 0 240 J 4 40 J.ABCA AB BC CAW W W W= + + = + − = −


ASSESS Because the net work done on the gas is negative, the gas does work on its environment.

41. INTERPRET We identify the thermodynamic process here as adiabatic compression.

DEVELOP In an adiabatic process, Q = 0, and the first law of thermodynamics becomes .U WΔ = The

temperature and volume are related by Equation 18.11b:

1 constantTV γ − =

18-10 Chapter 18



From the equation above, we obtain

1 1

1

11 1 2 2 2 1

2

VTV T V T TV

γ γγ

− −

−⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

where 1 2/V V is the compression ratio (for T and V at maximum compression).

EVALUATE Inserting the values given gives

( )( )1

0.412 1

2

303 K 8.5 713 K 440 CVT TV

γ −⎛ ⎞

= = = = °⎜ ⎟⎝ ⎠

ASSESS Note that the temperature T appearing in the gas laws is the absolute temperature. The higher the

compression ratio 1 2/ ,V V the greater the temperature at the maximum compression, and hence a higher thermal

efficiency.

42. INTERPRET This problem involves an adiabatic process that results in a change of pressure and volume. We are

to find the fractional change in volume given that the pressure increases by a factor of 2.

DEVELOP Apply Equation 18.11a, pVγ = constant. Using subscripts 1 and 2 to indicate before and after the

process, respectively, this gives

1 1 2 21

2 1

1 2

constantpV p V

V pV p

γ γ

γ

= =

⎛ ⎞= ⎜ ⎟⎝ ⎠

which we can solve for the fractional change in volume, using γ = 1.4 and p2 = 2p1.

EVALUATE Inserting the given quantities into the expression above gives

( )1

1 1.42 1

1 2

0.5 0.61V pV p

γ⎛ ⎞

= = =⎜ ⎟⎝ ⎠

ASSESS Thus, the new volume is almost 50% of the original volume, which seems reasonable given that the pressure has doubled.

43. INTERPRET We identify the thermodynamic process here as adiabatic compression. We are to find the

temperature and pressure in the cylinder when it is at maximum compression.

DEVELOP In an adiabatic process, Q = 0, and the first law of thermodynamics becomes .U WΔ = Because we

are dealing with an adiabatic process, the temperature and volume are related by Equation 18.11b:

1 constantTV γ − =

Applying this expression before (subscript 1) and after (subscript 2) compression gives

1 1

1

11 1 2 2 2 1

2

VTV T V T TV

γ γγ

− −

−⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

where 1 2 10.2V V = is the compression ratio (for T and V at maximum compression) and γ = 1.4. In addition, since

constantpV γ = (Equation 18.11a), the final pressure is ( )2 1 1 2p p V V γ= .

EVALUATE (a) Substituting the values given in the problem statement, we find the air temperature at the

maximum compression to be 1

0.412 1

2

(320 K)(10.2) 810 KVT TV

γ −⎛ ⎞

= = =⎜ ⎟⎝ ⎠

(b) The corresponding pressure is

( ) ( )( )1.42 1 1 2 101.3 kPa 10.2 2.62 MPa 25.8 atmp p V V γ= = = =

ASSESS The higher the compression ratio 1 2/ ,V V the greater the temperature and pressure at the maximum

compression, and hence, a higher thermal (fuel) efficiency.





environment. We are to find the change in temperature and volume as the pressure is reduced.


The temperature, pressure and volume are related by Equation 18.11: 1 1

1 1 2 2 1 1 2 2,TV T V pV p Vγ γ γ γ− −= =

The initial pressure is p1 = 1.0 atm, and the final pressure is p2 = 0.340 atm. With γ = 5/3, we can solve for the new

volume and temperature.

EVALUATE (a) From the equation above, we have

( ) ( )3/5

1/ 3 3 3 32 1 1 2

1.0 atm1.75 10 m 3.43 10 m0.34 atm

V V p p γ ⎛ ⎞= = × = ×⎜ ⎟⎝ ⎠

(b) The final temperature is 1 (5/3) 13 3

12 1 3 3

2

3.43 10 m(285 K) 185 K 88 C1.75 10 m

VT TV

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

ASSESS We see that decreasing the pressure along the adiabat is accompanied by a temperature decrease. In

addition, since constant,pV γ = the final volume increases.

45. INTERPRET This problem involves an adiabatic compression of a gas, for which we know the initial and final

temperatures. We are to find the ratio of the final pressure to the initial pressure of the gas.

DEVELOP For an adiabatic process, the temperature and pressure are related by Equation 18.11a:

constantpV γ =

Rearranging this equation and using the ideal-gas law (Equation 17.2, pV = nRT) gives

( ) ( )1 1 /(1 ) / (1 )1 1 2 2constantpV p pV p nRT p T p Tγ γγ γ γ γ γ γ γ− − − −= = = ⇒ =

EVALUATE Inserting the given quantities into the expression above for temperature gives 5/3

1 (5/3) 12 2

1 1

293 K 35428 K

p Tp T

γγ − −⎛ ⎞ ⎛ ⎞

= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

ASSESS The pressure increase is enormous in order to accompany the increase in temperature along the adiabat. The corresponding change in volume change is

1 3/522 1

1 2

1 2.96 10354

V pV p

γ−⎛ ⎞ ⎛ ⎞= = = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

The final volume is only about 3% of the original.

46. INTERPRET This problem explores how different ways of compressing volume (isothermal, adiabatic, or

isobaric) affects the internal energy of the system. The internal energy of n moles of diatomic gas is 52 ,U nRT= by

equipartition theorem. DEVELOP In an isothermal process, the temperature T is kept constant, so 0UΔ = . In an isobaric process, p = constant, and 2 1 2 1/ / .T T V V= Finally, in an adiabatic process, Q = 0 and

1

2 1

1 2

T VT V

γ −⎛ ⎞

= ⎜ ⎟⎝ ⎠

EVALUATE (a) For isothermal compression, ΔT = 0, so 0UΔ = . There is no change in internal energy. (b) For isobaric compression,

2 2 2 2

1 1 1 1

5 / 2 1 .5 / 2 2

U nRT T VU nRT T V

= = = =

(c) For adiabatic compression, γ = 7/5 for diatomic gas,

( )1

1.4 12 2 2 1

1 1 1 2

5 / 2 2 1.325 / 2

U nRT T VU nRT T V

γ −−⎛ ⎞

= = = = =⎜ ⎟⎝ ⎠

18-12 Chapter 18



ASSESS As the gas volume is halved, the internal energy of the gas decreases if compression is isobaric, but increases if the compression is adiabatic. There is no change if the compression is isothermal.

47. INTERPRET In the problem we compare the work done in adiabatic compression to the work done in isothermal

compression.


through Equation 18.11a: 1 1 2 2 .pV p Vγ γ= We can use this to rewrite Equation 18.12 for the work done on the gas in


2 2 1 1 1 1 1adiabatic

2

11 1

p V pV pV VWV

γ

γ γ

−⎡ ⎤⎛ ⎞− ⎢ ⎥= = −⎜ ⎟− − ⎢ ⎥⎝ ⎠⎣ ⎦

On the other hand, for an isothermal process, the work W done on the gas is (Equation 18.4):

2 2 1isothermal 1 1 1 1

1 1 2

ln ln lnV V VW nRT pV pVV V V

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

EVALUATE (a) For 2 1/ 1 / 2,V V = and γ = 5/3, from the above expressions, we find the ratio of the work done

for the two processes to be

( )1

2/3adiabatic 1

isothermal 1 2 2

1 11 2 1 1.27( 1)ln( / ) (2 / 3)ln(2)

W VW V V V

γ

γ

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

(b) The extra work goes into internal energy, raising the temperature of the gas.

ASSESS For diatomic gas where γ = 5/3, the ratio would be

( )1

0.4adiabatic 1

isothermal 1 2 2

1 11 2 1 1.15( 1)ln( / ) (0.4)ln(2)

W VW V V V

γ

γ

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

48. INTERPRET This problem involves first an isothermal expansion, then an isobaric compression, then finally an

isochoric process to return to the original state. We are to draw a pV diagram of this process, then calculate the

work done over the entire cycle, then finally, find the heat transfer between the given stages.

DEVELOP The isothermal expansion is along the curve AB in the figure below, which is given by PV = constant.

Because the gas expands, it starts at the lower volume VA at point A, and expands to occupy the greater volume VB

at point B. For the isobaric compression, the gas follows the horizontal line BC, with the pressure remaining

constant at PB = PC. Finally, for the isochoric process, the gas follows the vertical line CA, with the volume

remaining constant at VC = VA. To find the total work done on or by the gas during this cycle, sum the work done

over each interval,

tot AB BC CAW W W W= + +

where we consider work done on the system to be positive. For the isotherm (process AB), ΔU = 0 and the heat

transferred to the system is QAB = 35 J. From the first law of thermodynamics (Equation 18.1), this gives WAB =

−QAB = −35 J. For process BC, the work done on the gas is WBC = 22 J, and WCA = 0 because the volume of the gas

does not change (see Equation 18.7). To find the heat that enters or leaves the system during the stages BCA, recall

that the change in internal energy is zero for the whole cycle, so

tot tot tot 0U Q WΔ = + =

and that the total heat is composed of two parts:

tot AB BCAQ Q Q= +

so we can solve for QBCA.




EVALUATE (a) See figure above.

(b) Summing the various contributions to the work gives

tot AB BC CA 35 J 22 J 0 J 13 JW W W W= + + = − + + = −

so the gas does 13 J of work on its environment.

(c) The heat transferred to the system is

tot AB BCA tot

BCA tot AB 13 J 35 J 22 JQ Q Q WQ W Q

= + = −= − − = − = −

so 22 J of heat is transferred out of the gas during the process BCA.

ASSESS The total heat transferred to the system over the cycle is Qtot = 35 J − 22 J = 13 J, which provides the energy for the 13 J of work done by the system over this same cycle.

49. INTERPRET This problem explores how different ways of adding heat (isothermal, isochoric, or isobaric) affects

the final temperature of the system. Starting with the given quantity of gas at the given temperature, we are to find

the work done by the gas upon adding heat to the gas via these different processes.

DEVELOP In an isothermal process, the temperature T is kept constant, so 0UΔ = . The first law of

thermodynamics (Equation 18.1) therefore gives Q = −W. In an isochoric process, ΔV = 0 and W = 0 (see Equation

18.7), so the first law of thermodynamics gives .VQ U nC T= Δ = Δ Finally, in an isobaric process, Δp = 0 and

( )P VQ nC T n C R T= Δ = + Δ

Use these expressions to solve for ΔT and W for each case.

EVALUATE (a) With ΔU = 0, W = −Q = −1.75 kJ. For an isothermal process, the temperature is constant so T2 = 255 K.

(b) Solving the expression above for ΔT, we find for an isochoric process

( )( )1.75 kJ 24 K

3.5 mol 5 2V

QTnC R

Δ = = =

Therefore, 2 255 K 279 KT T= + Δ = . Because the volume does not change, W = 0.

(c) In an isobaric process,

( )( )1.75 kJ 17 K

( ) 3.5 mol 5 2p V

Q QTnC n C R R R

Δ = = = =+ +

and T2 = 255 K + 17 K = 272 K. The work done by the gas is

( )( )2 1.75 kJ

500 J7 2 7p

R RW p V nR T Q QC R

= Δ = Δ = = = =

ASSESS Comparing all three cases, we find

: isothermal < isobaric < isochoric: isochoric < isobaric < isothermal

TW

Δ

The results agree with that illustrated in Table 18.1.

50. INTERPRET This problem is an exercise to show the functional relationship in terms of pressure and volume

between an adiabatic process and an isothermal process.

DEVELOP The equations of an adiabat and an isotherm, passing through the point 0 0( , )V p in the pV diagram are

( ) ( )0 0 0 0andp V p V V p V p V Vγ γ= =

respectively. Differentiate these expressions and evaluate the results at (V0, p0) to find the desired relationship.

18-14 Chapter 18



EVALUATE Differentiating the above expression for an adiabat and evaluating the result at V = V0 and p = p0 gives

( )00

1 00 0

0adiabatp pV V

dp pp V VdV V

γ γ γγ − −==

−⎛ ⎞ = − =⎜ ⎟⎝ ⎠

Differentiating the expression above for an isotherm and evaluating the result at V = V0 gives

0

0 0 02

0isotherm V V

dp p V pdV V V=

⎛ ⎞ = − = −⎜ ⎟⎝ ⎠

Comparing the two results gives

adiabat isotherm

dp dpdV dV

γ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ASSESS We have found the expression given in the problem statement. Because γ > 0, we see that the slope of the pV curve for an adiabat is greater than that for an isotherm, so the pressure of an adiabatic process changes more rapidly with volume compared with an isothermal process, as shown in Fig. 18.11.

51. INTERPRET The problem involves a cyclic process with three separate stages: adiabatic, isochoric, and

isothermal. We are given the initial volume and pressure of the gas, and its adiabatic exponent γ, and are to find the

pressure at points B and C and the net work done on the gas.

DEVELOP For the adiabatic process AB, Q = 0 and the first law of thermodynamics becomes .U WΔ = − The

pressure and volume are related by Equation 18.11a: constantpV γ = , which gives

AA A B B B A

B

Vp V p V p pV

γγ γ ⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

Point C lies on an isotherm (constant temperature) with A, so the ideal-gas law (Equation 17.2) yields

A AC

C

p VpV

=

To find the net work done on the gas, sum the contributions from each stage of the cycle. The contribution WAB

may be found from Equation 18.12 for an adiabatic process:

B B A AAB 1

p V p VWγ

−=−

For an isochoric (constant volume) process WBC = 0 (see Equation 18.7), and for the isothermal process

ACA A

C

ln VW nRTV

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

EVALUATE (a) From the equation above, the pressure at point B is

( )1.67

AB A

B

1250 kPa 40 kPa3

Vp pV

γ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

(b) The pressure at point C is

AC A

C

250 kPa 83 kPa.3

Vp pV

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

(c) The net work done by the gas is

ABCA AB BC CA.W W W W= + +

The work ABW for the adiabatic segment is

( )( ) ( )( )3 3B B A A

AB

39.9 kPa 3.0 m 250 kPa 1.00 m194 kJ

1 0.67p V p VW

γ−−= = = −

−

BCW is for an isochoric process and equals zero. Finally, CAW is for an isothermal process and is

( )ACA A

C

1ln 250 kJ ln 275 kJ3

VW nRTV

⎛ ⎞ ⎛ ⎞= − = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠




Summing these contributions gives

ABCA AB BC CA 194 kJ 0 275 kJ 80 kJW W W W= + + = − + + =

so the work done on the gas 80 kJ.

ASSESS Since the process is cyclic, the system returns to its original state, there’s no net change in internal

energy, so ΔU = 0. This implies that ABCA 80 kJ.Q W= − = − That is, 80 kJ of heat must come out of the system.

52. INTERPRET This problem involves a cyclic process with an adiabatic compression, an isobaric compression, and

an isothermal expansion. We are to find the net work done on the gas and its minimum volume.

DEVELOP The pV diagram for this problem is different than that of Figure 18.14; only point A is the same (see

figure below). Note that pAVA = 400 J and VA = 2VB. To find the net work done on the gas, sum up the

contributions from each stage. For the adiabatic stage AB, the work done on the gas is given by Equation 18.12:

B B A AAB 1

p V p VWγ

−=−

For the isobaric process BC, the work done on the gas is given by Equation 18.7:

( )BC B B C BW p V p V V= − Δ = − −

Via the adiabatic process, we can find the product pBVB (using Equation 18.11), which gives

AB A

B

1

A

BB B A A

Vp pV

Vp V p VV

γ

γ −

⎛ ⎞= ⎜ ⎟

⎝ ⎠

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Via the isothermal process CA, we know that C is at the same temperature as A. We also know that pC = pB

because BC is isobaric. Thus, from the ideal-gas law (Equation 17.2), we have

C C B C A Ap V p V p V= =

Therefore, the work WBC is

1

ABC A A

B

1VW p VV

γ −⎡ ⎤⎛ ⎞⎢ ⎥= −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

The work done on the gas for the isothermal process CA is given by Equation 18.4, which gives (using the ideal-

gas law pV = nRT)

ACA A A

C

ln VW p VV

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

The minimum volume of the gas occurs at point C, and is given by 1

A A BC B

B A

p V VV Vp V

γ −⎛ ⎞

= = ⎜ ⎟⎝ ⎠

18-16 Chapter 18



EVALUATE (a) Summing the work done on the gas for each process gives

( ) ( )( )1 0.4A BB B A A

AB A A

400 J 2 11319.5 J

1 1 0.4V Vp V p VW p V

γ

γ γ

− −−−= = = =− −

where we have used the expression above relating pBVB to pAVA from the adiabatic process. Continuing, we have

for process BC

( )( )1

0.4ABC A A

B

1 400 J 2 1 127.8 JVW p VV

γ −⎡ ⎤⎛ ⎞⎢ ⎥= − = − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

and for process CA we have

( )( ) ( )A ACA A A A A

C B

ln ln 400 J 1.4 ln 2 388.2 JV VW p V p VV V

γ⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − = − = − = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎣ ⎦

Summing these, the total work done on the gas is tot 319.5 J 127.8 J 388.2 J 59.1 JW = + − = , or about 59 J.

(b) The minimum volume is

( )1 0.412.0 L 1.5 L

2B

C BA

VV VV

γ −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

ASSESS The total work is positive, so work is done on the gas in this cycle.

53. INTERPRET We are to find the work done in the given heat cycle, which consists of an isochoric doubling in

pressure, and adiabatic compression, an isochoric cooling to 300 K, and an isothermal expansion.

DEVELOP The net (or total) work done on the gas is the sum of the work done for each process in the cycle (see

figure below), which we give here:

(AB) It is heated at constant volume until the pressure is doubled. Because ΔV = 0, WAB = 0 (see Equation 18.7).

(BC) It is compressed adiabatically until its volume is 14 the initial value. This is analogous to the process AB of

Problem 18.46, so the result is

( ) 1B C

BC B B

11

V VW p V

γ

γ

− −=

−

where VB/VC = 1/4 and pBVB = 800 J (see figure below).

(CD) It is cooled at constant volume to a temperature of 300 K. No work is done because ΔV = 0, so WCD = 0.

(DA) It is expanded isothermally until it returns to the original state. This is analogous to the process CA of

Problem 18.46, so we use that result:

ADA A A

D

ln VW p VV⎛ ⎞

= − ⎜ ⎟⎝ ⎠

where VA/VD = 4 and pAVA = 400 J (see figure).




EVALUATE Summing the nonzero contributions to the work gives

( )

( ) ( ) ( )

1B C A

tot BC DA B B A AD

0.4

1ln

1

4 1800 J 400 J ln 4 930 J0.4

V V VW W W p V p VV

γ

γ

− − ⎛ ⎞= + = − ⎜ ⎟− ⎝ ⎠

−= − =

ASSESS Work is done on the gas. From the figure above, we see that the area under the entire curve is negative,

because the gas goes around the cycle counterclockwise. As per Figure 18.6, the work done by the gas is the

negative of the area under the pV curve.

54. INTERPRET This problem is similar to the preceding two problems. It involves again a thermodynamic cycle, this

time involving an isothermal compression, followed by an isochoric increase in pressure, then an adiabatic

expansion. We are to find the total work done on or by the gas over this cycle.

DEVELOP We begin by drawing the pV curve for this cycle (see figure below). Sum the contributions to the work

from each process in the cycle. For the isothermal process AB, apply Equation 18.4 and the ideal-gas law

(Equation 17.2, pV = nRT). The work done on the gas is

BAB A A

A

ln VW p VV⎛ ⎞

= − ⎜ ⎟⎝ ⎠

where pAVA = 1.25 kJ and VB/VA = 1/3 (see figure below). For the isochoric process BC, no work is done because

ΔV = 0 (see Equation 18.7). For the adiabatic process CA, the work done is

( ) 1A C

CA A A

11

V VW p V

γ

γ

−−=

−

where pAVA = 1.25 kJ and VA/VC = 1/3.

EVALUATE Summing the nonzero contributions to the work gives

( )

( ) ( ) ( )

1A CB

tot AB CA A A A AA

0.67

1ln

1

1.00 3.001250 J ln 3 1250 J 656 kJ0.670

V VVW W W p V p VV

γ

γ

− −⎛ ⎞= + = − +⎜ ⎟ −⎝ ⎠

−= + = −

ASSESS The work is negative, which means that the gas does work on the environment, which is consistent with the clockwise direction of the cycle about the pV curve.

55. INTERPRET We’re asked to derive the relation between pressure and temperature in an adiabatic process.

DEVELOP We just need to combine Equations 18.11a and 18.11b: constant,pV γ = and 1 constant.TV γ − =

EVALUATE Isolating the volume in both equations and matching the exponents gives

( )( )

( )

( )1 1

11

1

constantconstant

constant

p T p TV

p

VT

γ γ γ γ

γ γγ

γ γγ

− − −

−−

−

= =⎫⎪= ⎪⎬⎪

= ⎪⎭

18-18 Chapter 18



The “constant” term is just a place-holder. You can raise the constant to a power and it’s still just a constant. Or

you can multiply the constant by another constant, and the result is still just a constant.

ASSESS For 1,γ > the result says the pressure and temperature will increase or decrease together during an

adiabatic process.

56. INTERPRET The problem involves a cyclic process with three separate stages, which are (in order) adiabatic,

isochoric, and isothermal. We are to find the total work done in the process. As seen in Problems 18.46–18.48,

there will be net work on the gas by the environment because the system proceeds in a counterclockwise manner

around the cycle.

DEVELOP We begin by sketching the pV curve for this cycle (see figure below). The total, or net, work done is

the sum of the work done in the individual processes of the cycle. For the adiabatic process AB, Q =0, and the first

law of thermodynamics (Equation 18.1) becomes .U WΔ = − The pressure and volume are related by Equation

18.11a: constant,pV γ = and the work done by the gas is given by Equation 18.12:

B B A AAB 1

p V p VWγ

−=−

Since BC 0VΔ = for the isochoric process, BC 0.W = Similarly, for an isothermal process, the work done is

(Equation 18.4)

A ACA A A A

C C

ln lnV VW nRT p VV V

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

where we have used the ideal-gas law (Equation 17.2) pV = nRT. The minimum volume attained is VB = VC, which

is related to VA by the adiabatic compression, because pVγ = constant for an adiabatic process (Equation 18.11a),

we have 1/

AC B A

B

pV V Vp

γ⎛ ⎞

= = ⎜ ⎟⎝ ⎠

where pA/pB = 1/3 (see figure below) and VA = 25 L.

EVALUATE The solutions below are presented in the reverse order.

(a) The work done on the gas is the sum of the work done on the gas during each process:

( ) ( )

( )( ) ( ) ( )

0B B A A A

on gas ABCA AB BC CA A AC

1 1B A

A A

1 1.67 1

ln1

1 ln1

1 3 1 ln 350 kPa 25 L

1.67 1 1.67

211 J

A B

p V p V VW W W W W p VV

p p p pp V

γ

γ

γ γ

=

−

−

⎛ ⎞−= = + + = − ⎜ ⎟− ⎝ ⎠⎡ ⎤−⎢ ⎥= −

−⎢ ⎥⎣ ⎦⎡ ⎤−⎢ ⎥= −

−⎢ ⎥⎣ ⎦=

to two significant figures, and where we have used the relationship between VA, VB, and VC derived for part (b) to

find the minimum volume.




(b) Inserting the volume and pressure in to the expression for the minimum volume gives

( )1/ 1/1.67125 L 13 L

3A

C B AB

pV V Vp

γ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

(c) See the figure above for the pV cycle.

ASSESS Since the process is cyclic, the system returns to its original state and there is no net change in internal

energy, so 0.UΔ = This implies that ABCA 211 J.Q W= − = − That is, 211 J of heat must come out of the system.

57. INTERPRET The pump handle is pressed rapidly, so you can assume that there’s no time for heat to flow into or out of

the gas in the pump. This means that process is adiabatic.

DEVELOP You want to check that the temperature rise in the pump is less than 75°C. Equation 18.11b tells you

how to relate the final temperate and volume to the initial temperature and volume: 1 10 0 .TV T Vγ γ− −= Therefore, the

temperature rise is 1

00 0 1VT T T T

V

γ −⎡ ⎤⎛ ⎞Δ = − = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

The gas in the cylinder is air ( )1.4γ = initially at o0 18 C 291 K.T = =

EVALUATE Since the cross-sectional area of the cylinder remains unchanged during the compression, the volume

ratio is just equal to the ratio in the cylinder’s height:

( )1 1.4 1

00

32 cm1 291 K 1 93 K16 cm

VT TV

γ − −⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞Δ = − = − =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

This does not meet your 75°C criteria.

ASSESS There’s no obvious way to avoid this temperature rise. The cylinder can be made of a material with a

high thermal conductivity, such as aluminum, so that heat can flow out as fast as possible.

58. INTERPRET We’re given experimental data and are asked to estimate the work involved in inflating a human lung.

DEVELOP We want calculate the work done by the gas in expanding the lungs: .W p VΔ = Δ In the given graph, we

can take the volume change between successive points as ,VΔ and the pressure at the midpoint between successive

points as the average pressure .p To find the total work done, we sum over these small individual expansions.

EVALUATE There are seven main volume expansions that correspond to pressure increases of 5 kPa. We add the work

involved in each expansion to get the total work. (Note: to save space, we leave off the units until the end of the sum):

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

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

11 12

1 1 12 2 2

1 1 12 2 2

12

0.5 5 125 0 5 10 260 125 10 15 520 260

15 20 800 520 20 25 1100 800 25 30 1350 1100

30 35 1550 1350 kPa mL

30 J

i i i iW p V p p V V− −= Δ = + −

⎡= + − + + − + + −⎣+ + − + + − + + −

⎤+ + − ⋅⎦=

∑ ∑

ASSESS It may seem counterintuitive that the pressure increases as the volume increases, but in this case air is

rushing into the lungs from the outside. When a person breathes, their diaphragm contracts to open up the lung

cavity. This lowers the pressure on the lungs, so they inflate like two balloons. The air rushing into the lungs does

the work to expand the lung volume; the diaphragm is only indirectly responsible. Afterwards, the diaphragm

expands to squeeze on the lungs, which increases the pressure and thus forces air out.

59. INTERPRET For this problem, we are given the heat transferred from a monatomic gas to the surroundings and

the change in internal energy of the gas, and we are to find the work done in the process.

DEVELOP Apply the first law of thermodynamics, Equation 18.1, ΔU = Q + W, where Q = −15 kJ. The internal

energy change can be calculated using the constant-volume specific heat for a monatomic gas, which is CV = 3R/2

(see Equation 18.13). Thus,

VU nC TΔ = Δ

where n = 21 mol and ΔT = 160 K.

18-20 Chapter 18



EVALUATE Combining the above expressions and inserting the given quantities gives

( ) ( ) ( )321 mol 8.314 J/ mol K 160 K 15 kJ 57 kJ2VW U Q nC T Q ⎛ ⎞ ⎡ ⎤= Δ − = Δ − = ⋅ + =⎜ ⎟ ⎣ ⎦⎝ ⎠

ASSESS The work is positive so the work is done on the gas.

60. INTERPRET The thermodynamic process here involves two stages: isothermal compression followed by an

adiabatic compression. We are to find the final temperature of the gas.

DEVELOP During the first stage, the gas is compressed isothermally so ΔT = 0 and there is no change in the

temperature of the system (i.e., T0 = T1), but the volume is reduced from V0 to V1 = V0/3. During the next stage of

adiabatic compression, the temperature and volume are related by Equation 18.11b: 1 constantTV γ − =

which we apply to the points 1 and 2 (i.e., before and after the adiabatic compression, respectively) to find

1 1

1

11 1 2 2 2 1

2

VTV T V T TV

γ γγ

− −

−⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

where the final temperature is T2.

EVALUATE Substituting the values given, we find T2 to be

( )7/5 11 0.4

1 02 1 0

2 0

/3 5273 K 335 K/5 3

V VT T TV V

γ −−⎛ ⎞⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

ASSESS Since 1 constantTV γ − = for an adiabatic process, the compression results in an increase of the final

temperature.

61. INTERPRET This problem involves an adiabatic compression of a gas, for which we know the initial temperature

and pressure and the final pressure. We are to find the final temperature of the gas.

DEVELOP For an adiabatic process, the temperature and pressure are related by Equation 18.11a:

constantpV γ =

Rearranging this equation and using the ideal-gas law (Equation 17.2, pV = nRT) gives

( ) ( )( )

1 1

1

constant

another constant

pV p pV p nRT

pT

γ γγ γ γ

γ γ

− −

−

= = =

=

Applying this to the gas both before and after the compression gives (1/ ) 10 0( / )T T P P γ −=

EVALUATE Inserting the given quantities into the expression above for temperature gives

( )( )1 1.3 1

(1/ ) 10 0

100 kPa( / ) 273 K 334 K240 kPa

T T P P γ−

− ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

or 23.3 10 K× to two significant figures.

ASSESS Because the gas is compressed adiabatically, its temperature rises.

62. INTERPRET The problem involves a cyclic process with three separate stages of the cycle: isochoric (AB),

isobaric (BC), and isothermal (CA). We are to calculate the net work done on the gas over the entire cycle, and the

heat transfer over segment AB.

DEVELOP The work done on the gas in each segment of the cycle is summarized in Table 18.1. For the isochoric

process (path AB), ΔV = 0 so W = 0. For the isobaric process (path BC), the work done is

( )BC B C BW p V V= − −

Finally, for the isothermal process (CA), the work done on the gas is given by Equation 18.4:

ln ACA A

C

VW nRTV⎛ ⎞

= − ⎜ ⎟⎝ ⎠




Segment AB is isochoric, so QAB = nCVΔT (see Table 18.1). To eliminate CV, note that

= 11

P VV

V V V

C C R R RCC C C

γγ

+= = + ⇒ =−

which follows from the definition of the adiabatic ratio γ (see discussion for Equation 18.11a). Inserting this into

the expression for the heat transfer gives

( )1 1

B A B B A AAB

nR T T p V p VQγ γ

− −= =− −

where the second equality follows from the ideal-gas law (Equation 17.2), pV = nRT.

EVALUATE (a) Using the equations above, we obtain

( )( )

( )( ) ( )

0 (isochoric)( ) 250 kPa 1.00 L 5.00 L 1.00 kJ (isobaric)

ln ln 50.0 kPa 5.00 L ln 5.00 402 J (isothermal)

AB

BC B C B

A ACA A A A

C C

WW p V V

V VW nRT p VV V

== − − = − − =

⎛ ⎞ ⎛ ⎞= − = − = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Adding up all the contributions, the net work done on the gas is

0 1000 J 402 J 598 JABCA AB BC CAW W W W= + + = + − =

(b) The heat transferred is

( )( )250 kPa 50 kPa 5.0 L2.5 kJ

1 1.4 1.0B B A A

ABp V p VQ

γ−−= = =

− −

Since 0,ABQ > heat is transferred into the gas.

ASSESS At constant volume, the gas must be heated in order to raise its pressure.

= 11

P VV

V V V

C C R R RCC C C

γγ

+= = + → =−

63. INTERPRET This problem is similar to the preceding one, except that the order of segments in the cycle are

changed. The cycle proceeds along the 350-K isotherm from 50 kPa to 250 kPa (AC), then decreases its pressure to

50 kPa via an isochoric process (CD), then returns to the starting point via a 50-kPa isobar (DA). We are to find the

total work done on the gas for the complete cycle and the heat transferred in segment CD.

DEVELOP The formulas are the same as for the previous problem, but the initial and final point for each segment

are different. The work done on the gas over segment AC will be the opposite of the result we found for the

segment CA in Problem 18.56, so WAC = 402 J. The work done over segment CD is zero because the volume is

constant (see Equation 18.7). The work done over segment DA is

( )BC D A DW p V V= − −

The heat transfer can be calculated as per Problem 18.56, which gives

1D D C C

CDp V p VQ

γ−=−

EVALUATE (a) The total work is

( )( )0

( ) 402 J 50.0 kPa 5.00 L 1.00 L 202 JACD AC CD DA AC D A DW W W W W p V V=

= + + = − − = − − =

(b) The heat transfer over segment CD is

( )( )50 kPa 250 kPa 1 L500 J

1 0.4D D C C

CDp V p VQ

γ−−= = = −

−

Because QCD < 0, we interpret the result as 500 J of heat transferred out of the gas.

ASSESS At constant volume, the gas must be cooled to lower its pressure.

18-22 Chapter 18



64. INTERPRET This problem deals with an adiabatic expansion of a gas mixture. The gas contains monatomic and

diatomic molecules, and we are given the volume doubles and the pressure decreases by one third because of the

adiabatic expansion, and we are to find the fraction of each molecule type.

DEVELOP In an adiabatic process (AB), Q = 0 and the pressure and volume are related by Equation 18.11a:

constant.pV γ = Applying this to the gas before and after the expansion gives

2 11 1 2 2

1 2

p VpV p Vp V

γγ γ ⎛ ⎞

= ⇒ = ⎜ ⎟⎝ ⎠

Taking the natural logarithm of both sides of the above and solving for γ, we obtain

2 1 2 1

1 2 1 2

ln( / ) ln(1/3) ln(3)ln ln 1.58ln( / ) ln(1/2) ln(2)

p V p pp V V V

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

To find the fraction of the molecules that are argon, we use the result from Exercise 29:

112.5

1f

γ= −

−

EVALUATE Substituting 1.58γ = into the equation above gives

Ar1 12.5 2.5 0.790 79.0%

1 1.58 1f

γ= − = − = =

− −

ASSESS In the limit where all the gas molecules are monatomic, 1 1,f = and monatomic 1.67.γ = On the other hand,

if all the molecules are diatomic, then 1 0f = and the specific-heat ratio is diatomic 1.4.γ = Our ratio of γ = 1.58 is

closer to 1.67. This implies that the gas mixture is predominantly monatomic.

65. INTERPRET For this problem, we are to find the ratio of triatomic molecules to monatomic molecules that will

result in a gas that has the specific heat at constant volume of a diatomic gas.

DEVELOP The specific heat of a mixture of two gases is 1 21 2 ,V V VC f C f C= + where the fi are the number fractions

of the gases. Gas 1 is monatomic 32( ),VC R= gas 2 is triatomic (with 3 ,VC R= as given in the problem statement),

and we wish to have mixture with 52VC R= , as for a diatomic gas. In this case, then

5 31 22 2 3 ,R R f R f= + or 1 25 3 6 .f f= +

Given that f1 + f2 = 1 and f1 = 10 mol, we can solve for f2.

EVALUATE From the equations above, we find 1 1 3f = and 2 2 3f = . With 10 mol of monatomic gas, one needs

20 mol of triatomic gas.

ASSESS There is twice the number of triatomic molecules as there are monatomic molecules.

66. INTERPRET This problem is about melting, and it involves heat of fusion. The source of energy is the

gravitational potential energy of the rock that is dropped into the water.

DEVELOP The gravitational potential energy of the rock melts the ice and no heat energy is transferred (Q = 0) to

the environment. The first law of thermodynamics (Equation 18.1) then takes the form ΔU = W. The work is W =

mrockgh and the internal energy of the ice is raised miceLf (see Equation 17.5), where Lf = 334 kJ/kg (from Table

17.1). Therefore, we have

rock ice fm gh m L=

EVALUATE Thus, we find the height from which the rock must be dropped is

( )( )( )( )

ice2

rock

0.0063 kg 334 kJ/kg25 m

8.5 kg 9.8 m/sfm L

hm g

= = =

ASSESS In this problem, the rock did positive work on the ice-water system. Therefore, the work done on the

system is positive, so W > 0.

67. INTERPRET This problem deals with an isothermal expansion of a gas from an unknown pressure to the ambient

pressure of 1 atm. The expansion extracts heat from the surrounding ice-water bath so that 10 g of ice are created

in the process. We are to find the original pressure of the gas.




DEVELOP For an isothermal expansion of an ideal gas, Equation 18.4 and the ideal-gas law (PV = nRT = constant

for an isothermal expansion) give

2 2

1 1

ln lnV pQ nRT nRTV p

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

From this, /

1 2Q nRTp p e=

where p1 is the original gas pressure, p2 = 1 atm, and n = 0.30 mol. The heat Q is provided by the freezing of 10 g

of ice that is already at 0°C, which is (see Equation 17.5)

( )( )ice f 0.01 kg 334 kJ/kg 3.34 kJQ m L= = =

EVALUATE Inserting the known quantities into the expression for p1 gives

( ) ( ) ( )4.91

13.34 kJ(1.0 atm)exp (1.0 atm) 140 atm

0.30 mol 8.314 J/ mol K 273 Kp e

⎧ ⎫⎪ ⎪= = =⎨ ⎬⎡ ⎤⋅⎪ ⎪⎣ ⎦⎩ ⎭


ASSESS Notice that the original quantity of ice is not involved, provided we know the ice-water bath is at 0°C and the quantity of the ice that is produced by the energy requirements of the isothermal expansion.

68. INTERPRET The problem concerns the work done in expanding a frog’s lung.

DEVELOP Equation 18.3, 2

1,

V

VW pdV= ∫ gives the work done by a gas as its volume goes from 1V to 2.V We’ll

plug the formula for the pressure in the frog’s lung, 3 210 67 220 ,p v v v= − + where p is in Pa and / mL.v V=

EVALUATE If the frog’s lung inflates from 0 to 4.5 mL, the work done by the gas is

( )( ) ( ) ( )

4.5 3 2

0

4 3 21 1 1 64 3 2

10 67 220 Pa mL

10 4.5 67 4.5 220 4.5 10 J 1.2 mJ

W v v v dv

−

⎡ ⎤= − + ⋅⎢ ⎥⎣ ⎦⎡ ⎤= − + × =⎣ ⎦

∫

ASSESS This is much smaller than our result in Problem 18.52, but there we were considering a human lung

inflating to over 1500 mL.

69. INTERPRET For this problem, we are to derive the work done by an adiabatic process by applying Equation 18.3,

which describes the work done by changes in a volume of gas.

DEVELOP For an adiabatic process, pressure and volume are related by Equation 18.11a, pVγ = constant.

Applying this to a gas before and after an arbitrary adiabatic process

1 1

1 1

pV pV

p VpV

γ γ

γ

γ

=

=

Insert this expression into Equation 18.3 to find the work done by an adiabatic process.

EVALUATE The work in going from a pressure p1 to a pressure p2 via an adiabatic process is

( )2 2

1 1

1 1 1 12 1 2 2 2 1 1 1

12 1 1 1 1

2 2 1 1

1 1

1

V V

V V

dV V V p V V pV VW pdV pV pVV

PV PV

γ γ γ γ γ γγ γ

γ γ γ

γ

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

∫ ∫

which is Equation 18.12 (note that we have used 1 1 2 2pV p Vγ γ= to obtain the last equality of the first line).

ASSESS From the general expression for work done by a change in volume of a gas, we have derived the expression for the work done by an adiabatic change in volume of a gas.

70. INTERPRET The system of interest is the horizontal piston-cylinder system that contains an ideal gas. We want to

demonstrate that the piston undergoes simple harmonic motion when slightly displaced.

18-24 Chapter 18



DEVELOP Since the piston-cylinder system is horizontal (see figure below), we do not need to consider the force

of gravity on the piston. At equilibrium, the pressure forces from inside and outside the piston are equal, so the gas

pressure at the equilibrium position of the piston is p0. We also assume that the gas temperature at equilibrium is

T0, so 0 0 0 ,p V nRT= where 0 0V Ax= is the piston volume at equilibrium. When the piston is displaced from its

equilibrium position by an amount Δx (positive to the right), the horizontal force on it is

( )0 0 ,F pA p A p p A= − = − and Newton’s second law gives an acceleration of

( ) ( )20

2

d x p p Adt M

Δ −=

For isothermal expansions and compressions of the gas,

( )0 0 0 0 0pV p V pA x x p Ax= ⇒ + Δ =

For small displacements 0 ,x xΔ << the pressure is

( ) ( )00 0 0 0

0 0

1 11

xp p p p x xx x x x

⎡ ⎤= = ≈ − Δ⎣ ⎦+ Δ + Δ

(see the binomial approximation in Appendix A). Substituting the expression for p into Newton’s second law

equation allows us to show that the piston executes simple harmonic motion.

EVALUATE Combining the two equations above yields

( ) ( )20 20

20

d x p p A p A x xdt M M x

ωΔ − Δ= = − = − Δ

where 20 0.p A Mxω = This is the equation for simple harmonic motion of the piston, about its equilibrium position

0,x with angular frequency 0 0 .p A Mxω = Since 0 0 0 0 0 ,p V p Ax nRT= = we may eliminate x0 to obtain

0

0

p AMnRT

ω =

ASSESS In order for the gas temperature to remain constant, as assumed above, heat must flow into and out of the

gas. This requires time, so the motion of the piston must be very slow. If the motion is rapid (or if the cylinder is

thermally insulated), there is no time for heat transfer in the gas and the expansions and compressions are adiabatic. In

this case,

0 0 0 0 0( )pV p V pA x x p A xγ γ γ γ γ γ= ⇒ + Δ =

or ( )0 01p p x x γ−= + Δ For small displacements 0,x xΔ << we have ( )0 01p p x xγ≈ − Δ and

( ) ( )20 20

20

( )d x p p A p A x xdt M M x

γ ω⎛ ⎞Δ − Δ= = − = − Δ⎜ ⎟⎝ ⎠

where 20 0p A Mxω γ= This represents simple harmonic motion with

00

0 0

p A p AMx MnRT

γ γω = =

71. INTERPRET We are to derive the relationship between temperature and volume for adiabatic processes.

DEVELOP For an adiabatic, the relationship between pressure and volume is constant.pV γ = Use this and the

ideal-gas law (Equation 17.2) ,pV nRT= to derive equation 18.11b, 1 constant.TV γ − =

EVALUATE From the ideal-gas law,

nRTpV nRT pV

= ⇒ =




Substitute this into constantpV γ = to obtain

( ) 1

1

constant

constant a different constant

nRT V nR TVV

TVnR

γ γ

γ

−

−

= =

= =

ASSESS We have shown what was required. Note that the constants involved in Equations 18.11a and 18.11b are

different constants.

72. INTERPRET In this problem we are asked to analyze the data of the pressures of an ideal gas as a function of

volumes in a thermodynamic process, and deduce whether the process is isothermal or adiabatic.

DEVELOP For an adiabatic process, the relationship between pressure and volume is constant.pV γ = Taking the

logarithm on both sides leads to ln ln lnp V cγ+ = , where c is a constant. Therefore, if the process is adiabatic,

plotting lnp versus lnV gives a straight line with slope equal to –γ. On the other hand, if the process is isothermal,

then constant.pV = Taking the logarithm on both sides leads to ln ln lnp V c+ = , where c is a constant. Plotting

lnp versus lnV would give a straight line with slope equal to –1 for the isothermal process.

EVALUATE The log-log plot is shown below.

The slope is –1.4. This implies that the process is adiabatic.

ASSESS The value 1.4γ = implies that the gas is diatomic.

73. INTERPRET We are given the relationship between pressure and volume for compressing a given volume of air,

and we are to find the work done during this reversible process.

DEVELOP The most general equation for work done by a gas is Equation 18.3: 2

1.V

VW pdV= − ∫ We are given the

initial pressure p0 = 1.0 atm, the initial volume V0 = 17 m3, the final pressure pf = 1.4 atm, and the formula 2

0 0

00

p Vp V

Vp pV

⎛ ⎞=⎜ ⎟

⎝ ⎠

=

Insert this into Equation 18.3 and integrate from V0 to Vf, where Vf is given by 2

f0

0f

pV Vp

⎛ ⎞= ⎜ ⎟

⎝ ⎠

EVALUATE Performing the integration gives

( )( )

ff3 3 3 3 3

2 2 2 2 2

00

3

0 0 0 f0 f 0 0 0

0 00 0 0

3 3 30 0 f

0

2 2 23 3 3

2 101 10 Pa 17 m2 1.4 atm1 1 2.0 MJ3 3 1.0 atm

V V

VV

V p p P pW p dV V V V V VV pV V V

p V pp

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

∫

ASSESS Because the work is negative, the gas does 2.0 MJ of work on its environment in this process.

18-26 Chapter 18



74. INTERPRET We are to derive an expression for the work done by a non-ideal gas in an isothermal process that follows the Van der Waals model. We are given the equation relating pressure and volume for a Van der Waals gas. DEVELOP Solving the Van der Waals expression for pressure gives

2nRT np aV nb V

⎛ ⎞= − ⎜ ⎟− ⎝ ⎠

Insert this expression for pressure into Equation 18.3 and integrate from V1 to V2 to find the work done by the gas..

EVALUATE Performing the integration gives

( )22 2

1 1 1

2 2

21

2 1 2

ln

1 1ln

VV V

V V V

nRT n anW pdV a dV nRT V nbV nb V V

V nbnRT anV nb V V

⎡ ⎤ ⎡ ⎤⎛ ⎞= − = − − = − − +⎢ ⎥ ⎢ ⎥⎜ ⎟− ⎝ ⎠⎢ ⎥ ⎣ ⎦⎣ ⎦⎛ ⎞ ⎛ ⎞−= + −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

∫ ∫

ASSESS For an ideal gas, a = b = 0, and this simplifies to ( )2 1lnW nRT V V= − (Equation 18.4), as expected.

75. INTERPRET An ideal gas expands along a given path, and we are to find the work done by the gas. We can do

this by using the general expression for work done by a gas that changes pressure and volume, Equation 18.3. DEVELOP The gas goes from (p1, V1) to (p2, V2) where p2 = 2p1 and V2 = 2V1. The path it takes is along

2

11 2

1

1 V Vp pV

⎡ ⎤⎛ ⎞−⎢ ⎥= + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Insert this expression into the integrand of Equation 18.3 and integrate from V1 to V2 to find the work done on the gas. The work done by the gas will be the negative of this result. EVALUATE The work done on the gas is

( )

( ) ( ) ( )

2 1 1

1 1 1

1

1

22 21 2 21

1 1 12 21 1

23 2 2 3 3 2 2 21 1

1 1 1 1 1 1 1 1 1 12 21 1

1 1 1 1

1 2 2

1 12 8 4 2 23 3

7 43 23 3

V V V

V V V

V

V

V V pW pdV p dV V VV V dVV V

p pV VV V V V V V V V V V VV V

pV p V

⎡ ⎤−⎡ ⎤⎢ ⎥= − = − + = − − +⎣ ⎦⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤= − − + = − − − − + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎛ ⎞= − − + = −⎜ ⎟⎝ ⎠

∫ ∫ ∫

so the work done by the gas is 4p1V1/3.

ASSESS The units are pressure times volume, which is joules, as expected.

76. INTERPRET We calculate the rate /dT dy at which air cools as it rises, approximating the process as adiabatic.

DEVELOP To express dT in terms of dp, let’s first differentiate the ideal-gas law: .Vdp pdV nRdT+ = For

adiabatic processes, ,U WΔ = − which we can write in differential form as ,VnC dT pdV= − and insert into the

differentiated ideal-gas law to obtain

( )V

1V VdT dp dpn C R nR

γγ

⎛ ⎞−= = ⎜ ⎟+ ⎝ ⎠

where we have used ( )/ 1 .VC R γ= −

The hydrostatic equation shows how the pressure increases with depth: ,dp gdhρ= but since we are considering

altitude (the opposite of depth), we include a negative sign: ,dp gdyρ= − so that the pressure decreases with

increasing altitude.

EVALUATE Combining the temperature/pressure relation to hydrostatic equation gives:

1 1dT Vg Mgdy nR R

ρ γ γγ γ

⎛ ⎞ ⎛ ⎞− − −= = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠




where M is the molar mass (the mass of one mole of the gas). We’re told to assume that 1.4,γ = and the average

molecular weight is 29u, which is equivalent to saying that the molar mass is 29 g/mol.M = So the adiabatic

lapse rate is:

( )( )20.029 kg/mol 9.8 m/s 1.4 1 9.8 K/km8.314 J/K mol 1.4

dTdy

−⎛ ⎞= − = −⎜ ⎟⋅ ⎝ ⎠

ASSESS The result seems reasonable: the temperature of an air mass drops by 10°C for every kilometer it rises.

Note that the static temperature gradient of the atmosphere is actually smaller than this (about a 6°C drop for every

kilometer increase in altitude). This is because other effects, like condensation of water molecules, contribute.

77. INTERPRET We are to calculate the efficiency and waste heat of a power plant, given the heat input and the

power produced.

DEVELOP The efficiency is

power outpower in

e =

The waste heat is the difference between the power in and the power out. We need to know that the rate of waste

heat produced is less than 400 MW, and the power plant has efficiency is 50%.e ≥

EVALUATE The power-plant efficiency is

360 MW 54%.670 MW

e = =

Heat is lost at a rate of lost 670 MW 360 MW 310 MW 400 MWP = − = < .

ASSESS This power plant meets the requirements.

78. INTERPRET We are to find the pressure and volume of air within a diving bell, given that the temperature varies

in such a way that the pressure and volume are related by the given equation.

DEVELOP The equation relating pressure and temperature is

00

Vp pV

=

where V0 = 17 m3 and p0 = 1 atm. The maximum allowable pressure is max 1.5 atm.p = We can plug these values

into the equation, and see what is the resulting volume V. If it’s greater than 3min 8.7 m ,V = the design is ok.

EVALUATE Solving the given equation for V gives

( ) ( )( )

2 323 30 0 0

0 22

1.0 atm 17 m7.6 m 8.0 m

1.5 atmV p Vp p VV p

= ⇒ = = = <

ASSESS The design needs work. Note also that since the pressure appears only in a ratio, we do not need to

convert to SI units.

79. INTERPRET The problem involves the power needed to extract at 320 K and 1 atm and then compress it adiabatically

to 350 atm.


through Equation 18.11a: 1 1 2 2 .pV p Vγ γ= We can use this to rewrite Equation 18.12 for the work done on the gas in


2 2 1 1 1 1 1

2

11 1

p V p V p V VWV

γ

γ γ

−⎡ ⎤⎛ ⎞− ⎢ ⎥= = −⎜ ⎟− − ⎢ ⎥⎝ ⎠⎣ ⎦

Using γ = 1.3, we obtain

( )1/

1/1.31 2

2 1

350 90.6V pV p

γ⎛ ⎞

= = =⎜ ⎟⎝ ⎠

18-28 Chapter 18



which then leads to

( )1.3 11 11 1 190.6 1 9.55 9.55( )

1.3 1p VW pV nRT−⎡ ⎤= − = =⎣ ⎦−

On the other hand, the rate of CO2 emission is 61100 tonnes/h 1.1 10 kg/h 305.6 kg/sR = = × =

which is equivalent to 3/ 6.944 10 mol/sdn dt = × noting that the molar mass of CO2 is 44 g. The power required for this process is

19.55dW dnRTdt dt

⎛ ⎞= ⎜ ⎟⎝ ⎠

EVALUATE Substituting the values given, we obtain

3 819.55 9.55(8.314 J/ mol K)(320 K)(6.944 10 mol/s) 1.76 10 W 0.176 GWdW dnRT

dt dt⎛ ⎞= = ⋅ × = × =⎜ ⎟⎝ ⎠

Since the plant produces electrical energy at a rate of 1.0 GW, the power required is about 18% of the power output. ASSESS This is a substantial fraction of the energy the power plant produces. In addition, it doesn’t include the energy cost of separating the CO2 from other stack gases or of transporting it to the compression site.

80. INTERPRET We consider the physics behind warm winds called Chinooks.

DEVELOP We’re told that the wind sweeping down from the mountain has no time to exchange heat with the

surroundings.

EVALUATE No heat exchange means the process is adiabatic.

The answer is (d). ASSESS Recall that air has one of the lowest thermal conductivities in Table 16.2. So it’s perhaps not surprising

that a large mass of air might need a lot of time to exchange an appreciable amount of heat with its surroundings.


DEVELOP For an adiabatic process, U WΔ = from Equation 18.10. Here, W is the work done on the air mass.

EVALUATE We’re told that the pressure increases by 50% as the air descends from the mountain. If the pressure

increases, there is positive work being done on the air (the opposite of Figure 8.11), so the change in the internal

energy, ,UΔ is positive.

The answer is (a). ASSESS One can verify this result by combining Equations 18.11a and 18.12 to obtain: 1 1/ 1 1/

2 1 .W p pγ γ− −∝ − Since

2 1p p> and γ is always greater than one, the work done on the air mass is positive.


DEVELOP From 18.11a, the pressure and volume in the mountains is related to that in the plains by

1 1 2 2 .pV p Vγ γ= We’ll assume 1.4.γ =

EVALUATE Given the initial and final pressure, we can solve for the relative change in the volume: 1/ 1/1.4

2 1

1 1 2

60 kPa1 1 1 25%90 kPa

V V pV V p

γ⎛ ⎞Δ ⎛ ⎞= − = − = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

This implies the volume decreases by less than 50%.

The answer is (d). ASSESS It makes sense that the volume of air decreases when exposed to the higher pressures at the base of the

mountain.


DEVELOP One could use the result from the previous problem and apply it to Equation 18.11b: 1 constant.TV γ − = But we will use the expression derived in Problem 18.49, relating the pressure and temperature

for an adiabatic process, 1 constant.p Tγ γ− =




EVALUATE We know the initial temperature up in the mountains is 260 K, so the temperature down in the plains

must be:

( )1 1/ 1 1/1.4

22 1

1

90 kPa260 K 290 K60 kPa

pT Tp

γ− −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

The answer is (c). ASSESS As the name implies, the Chinook is a warm wind. In this case, there is a 30°C increase in going from the

mountains (–13°C) down to the plains (17°C).

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