· 484 CHAPTER 14 THERMODYNAMICS 14 Thermodynamics Answers to Discussion Questions •14.1 Air is...

484 CHAPTER 14 THERMODYNAMICS

14 Thermodynamics

Answers to Discussion Questions

•14.1Air is drawn from A -B as the piston moves out. From B -C, it is compressed adiabaticallyand rises in temperature. Fuel is sprayed in at C and immediately explodes because of the hightemperature. C -D is the isobaric combustion leg when heat enters. At D, the fuel is burntand the piston continues to move outward. D -E is the rapid and therefore adiabatic expansionwhere the hot gas does work. E is the end of the cycle and it’s there that the exhaust valveopens. E -B is an isovolumic drop to the outside pressure, accompanied by the loss of heat viathe exhausting of hot gas. With the exhaust valve still open, the remaining gas is ejected asthe piston moves inward from B to A.

•14.2To blow up a balloon in the air you must push against the atmospheric pressure, i.e., youmust perform a positive amount of work. This is not necessary in space, however, since thereis no atmosphere to push against up there. Similarly, as an ideal gas expands freely into anevacuated chamber it is not pushing against anything so W = 0; and since the expansionprocess is rapid Q = 0 as well, meaning that the gas’s internal energy (and therefore itstemperature) stays the same as a result of the free expansion: ∆T ∝ ∆U = Q + W = 0. Ifthe gas expands into the air, however, it must do work on the environment in pushing againstthe atmosphere (W < 0), whereby decreasing its internal energy (and hence temperature):∆T ∝ ∆U = Q + W = W < 0.

CHAPTER 14 THERMODYNAMICS 485

•14.3Steam (blue arrows) enters behind the piston and drives it to the left as the slide valve movesto the right. Steam in the left half of the cylinder (red arrows) escapes through the exhaustport to the low-pressure condenser. The valve on the right then closes, the steam expands, andthe piston moves left. Now the valve on the right opens to the exhaust, steam enters via thevalve on the left and the piston moves right.

•14.4Eq. (14.14) suggests that no heat would be exhausted to a reservoir at zero kelvin by a Carnotengine. Thus, converting all the heat-in to work, it would thereby violate the Second Law. Notetoo that at low temperatures both the isothermals and the adiabats flatten out and run nearlyhorizontal, parallel with the volume axis in P -V diagrams. The implication being that a Carnotengine would have to have longer and longer piston strokes to sustain the area enclosed as thetemperature dropped toward 0 K.

•14.5Warm, moist low-density air rises (via thermals) doing work on the atmosphere as it expands.Accordingly, since the process occurs quickly, and gases are very poor conductors, it will beessentially adiabatic. The internal energy of the uprush of air will decrease and its temperaturewill drop. Water will then condense out, forming a cloud.

•14.6The scheme violates the Second Law by converting all the heat extracted into work — there isno low-temperature sink. It might be worth considering as an investment, but it will certainlynot sustain the temperature indefinitely. It is impossible to maintain the system at constantentropy by performing entropy increasing processes.

•14.7The Sun provides the energy stored chemically in wood, coal, and oil. Order increases locallyas the plants and animals storing solar energy grow, ultimately to form fossil fuels, but increasein the disorder of the Sun overbalances that.

•14.8He will die because the quality of the energy available to him has declined to the point where itis of little use. Energy is conserved in the room but entropy is always increasing and because theroom is isolated the entropy cannot be dumped out into the environment. Life must terminateinevitably under such conditions.

•14.9The Sun evaporates water that comes down to the rivers and reservoirs as rain, thereafter topour down a waterfall and drive a turbine. In effect, the Sun does work against gravity.


•14.10The coal radiates, cools, and decreases in entropy. The chamber warms and increases in entropy.Since the system is isolated its net entropy will either increase, or at best stay the same.

•14.11The process is adiabatic and so isentropic. Work is done by the gas on the piston. Therefore,the internal energy decreases, and the temperature (T ) drops. The volume obviously doublesand the pressure decreases. The entropy must stay constant since Q = 0.

•14.12The net energy (overlooking any radiation out to space) is constant; redistributed but constant.The entropy, however, has increased. Ordered fuel has burned and in its place disorderedexhaust was blown into the atmosphere.

•14.13The engine actually exhausts gas at a much higher temperature than 300 K. Fuel is incompletelyburned, heat is conducted and radiated from the engine, and friction is present as well.

•14.14Technology generally accelerates the rate of increase of entropy.

•14.15Yes it is possible, but with so many molecules the likelihood of all of them being in one corner isminute. Of course, if there were only 1 molecule flying around, the chance of “all the molecules”being in one corner would be fairly high. It’s less likely with 10 molecules and much less stillwith 1028.

•14.16The disordered mayhem of high temperature gas in the chamber becomes partially ordered asit passes through the nozzle and streams out in the backward direction. Thermal energy isconverted to organized kinetic energy and the temperature of the gas drops.


Answers to Multiple Choice Questions

1. b 2. b 3. d 4. e 5. d 6. d 7. b8. c 9. e 10. b 11. b 12. e 13. c 14. d

15. b 16. c 17. b 18. a 19. c 20. d 21. a22. c 23. d 24. a 25. b 26. a

Solutions to Problems

14.1Use Eq. (14.1), ∆U = Q + W , with Q = +500 J and W = −250 J:

∆U = Q + W = (+500 J) + (−250 J) = +250 J .

The internal energy of the system increases, as ∆U > 0.

14.2Use Eq. (14.1), ∆U = Q + W . Here Q = +1000 J and ∆U = +250 J, so

W = ∆U − Q = 250 J − 1000 J = −750 J .

14.3Use Eq. (14.1), ∆U = Q + W . Here W = +150 J (where the positive sign is because the workis done on, rather than by, the system), and ∆U = −300 J; so

Q = ∆U − W = −300 J − 150 J = −450 J .

14.4Apply Eq. (14.1), with Q = −600 J and W = −400 J:

∆U = Q + W = −600 J + (−400 J) = −1.00 × 103 J = −1.00 kJ .

Here Q < 0, indicating that heat flows out of the system.


14.5The tank is sealed so its volume is a constant: ∆V = 0. The work done on the gas is thenW = −P∆V = 0. It follows from Eq. (14.1) that

Q = ∆U − W = ∆U = +500 J .

14.6

(a) Tf = 24.9◦C, the same as the temperature of the heat reservoir.

(b) Heat flows from the high-temperature region (the reservoir) to the low-temperature region(the water), causing the water temperature to rise until thermal equilibrium is reached, whenthe temperature of the water equals that of the reservoir.

(c) Use Q = cWmW(Tf − Ti), where mW = 75.1 kg, Ti = 14.9◦C, and Tf = 24.9◦C:

Q = cWmW(Tf − Ti) = (4.186 kJ/kg·K)(75.1 kg)(24.9◦C − 14.9◦C) = 3.14 MJ .

(d) The heat flowing into the water ends up as thermal energy, and as a result the watermolecules on average speed up, increasing the mean value of the molecular kinetic energy asthe temperature of the water rises.

(e) Since no work is done the First Law for the water system becomes ∆U = Q + W = Q =3.14 MJ, i.e., the internal energy of the water increases by 3.14 MJ.

14.7

(a) The volume of the gas increases as it is heated up.

(b) The cross-sectional area of the piston is A = 0.20 m2. As the piston travels a distanceof s = 8.00 cm = 0.080 0 m, the change in volume for the gas in the cylinder is ∆V = As =(0.20 m2)(0.080 0 m) = 1.6 × 10−2 m3.

(c) Since the cylinder expands work is done by the gas, not on the gas. From Eq. (14.3)

W = −P∆V = −(16.0 kPa)(1.6 × 10−2 m3) = −256 J ,

where by our convention the negative sign means that work is done by the gas.

(d) Apply the First Law to the gas: ∆U = Q + W . Here Q = +400 J (as heat is added to thegas) and W = −256 J; so

∆U = Q + W = 400 J + (−256 J) = 144 J .

14.8

(a) Since the tank is sealed it volume V is constant, and so no work is done on or by the gas:W = −P∆V = 0.


(b) If heat flows out of the gas while no work is done on the gas, then its temperature (and,therefore, internal energy) would decrease. In fact, with Q = −400 J and W = 0, the FirstLaw reads

∆U = Q + W = −400 J + 0 = −400 J < 0 .

14.9By definition in an isovolumic process the volume V of the gas is constant, and so no work isdone on or by the gas: W = −P∆V = 0. Moreover, the net amount of heat flowing into thegas is Q = 5.0 kJ − 1.0 kJ = 4.0 kJ, which is positive, so both P and T increase. According tothe First Law

∆U = Q + W = 4.0 kJ + 0 = 4.0 kJ .

14.10Since the temperature of the ideal gas remains fixed so does its internal energy: ∆U = 0. Thework W done by the gas is then obtained from Eq. (14.1) to be

W = Q − ∆U = Q = −500 J .

The gas does negative work on the environment, which in turn does an equal amount of positivework (+500 J) on the gas.

14.11The change in internal energy, ∆U , is related to the heat Q that enters the system and thework W performed by the system via Eq. (14.1): ∆U = Q + W . In this case Q = 100 cal =(100 cal)(4.186 J/cal) = 418.6 J and W = 100.4 N·m = 100.4 J. (Note that W > 0 since thework is done on, rather than by, the system). Thus

∆U = Q + W = 418.6 J + 100.4 J = +519 J .

14.12The volume of m = 1.00 kg of ice (I) is VI = mρI , where ρI = 920 kg/m3. When it melts intowater (W), its new volume is VW = m/ρW , where ρW = 1000 kg/m3. The work done on themelting ice by the atmosphere at pressure P is then

W = −P∆V = P (VI − VW) = P

(m

ρI

− m

ρW

)

= (1.013 × 105 Pa)(

1.00 kg

920 kg/m3− 1.00 kg

1000 kg/m3

)

= 8.81 J .


The heat absorbed by the same amount of ice while melting is

Q = mLf = (1.00 kg)(333.7 kJ/kg) = 334 kJ >> W .

14.13The net work W done by an ideal gas in a cyclic process is equal to the area enclosed by thepath representing the process in the corresponding P -V diagram. In this case the area enclosedassumes the shape of a triangle so its area is 1

2base×height, where base = CB = 2Vi −Vi = Vi

and height = AC = Pi − 12Pi = 1

2Pi . Hence

W = −12

CB × AC = −12Vi

(12Pi

)= −1

4PiVi .

14.14According to Eq. (14.3) the work W done on the gas at a constant pressure P is −P∆V as itsvolume expands by ∆V . But we also know from Eq. (14.1) that W = ∆U −Q. Equate the twoexpressions for W to obtain −P∆V = ∆U − Q. Plug in Q = (+1.00 kcal)(4.186 kcal/kJ) =4.186 kJ, ∆U = +286 J, and ∆V = (15 liters)(10−3 m3/liter) = 0.15 m3, and solve for P :

P =Q − ∆U

∆V=

+4186 J − (+286 J)0.15 m3

= 2.6 × 104 Pa = 0.26 MPa .

14.15As the gas expands from the small container into the plastic bag it is pushing against theatmospheric pressure outside the bag, while its volume increases from nearly zero back to20.0 m3. Thus the work it does on the atmosphere in the process is

W = −P∆V = −(1.013 × 105 Pa)(20.0 m3) = 2.03 × 106 J = −2.03 MJ .

Since the gas does work while taking in virtually no amount of heat (as the process is rapid),∆U = Q + W = W < 0; so its temperature will decrease as a result of its expansion, whetherit is released into the bag or directly into the air.

14.16Since the gas remains in thermal contact with a thermal reservoir while being compressed veryslowly, its temperature remains unchanged, and so does its internal energy: ∆U ∝ ∆T = 0.The work done by the gas in the process is W = 40 J, and the heat it absorbs is Q = ∆U−W =−W = −40 J. Since Q < 0 the gas actually releases 40 J of heat.


14.17The work W done on the gas is equal in magnitude to the area under the curve representingthe process in the corresponding P -V diagram. In this case the area in question can be foundsimply by counting the “squares” formed by the grid lines under the curve. Each such “square”represents (∆P )(∆V ) = (1.0 MPa)(0.50 cm3) = (1.0 × 106 Pa)(0.50 × 10−6 m3) = 0.50 Jof work; and since there are 24 such squares under the curve, the total work done is W =−(∆P )(∆V ) = −24(0.50 J) = −12 J. Here W < 0, indicating that the gas performs apositive amount of work on the environment.

14.18The volume of the brass sphere expands by ∆V as its temperature changes by ∆T : ∆V =βV0∆T . Here β = 56 × 10−6 K−1, V0 = 4

3πR3 with R = 15 cm, and ∆T = 280◦C − 0◦C =280 C◦ = 280 K. Thus the work W done by the sphere on the atmosphere at pressure P is

W = −P∆V = −43πPβR3∆T

= −43π(1.013 × 105 Pa)(56 × 10−6 K−1)(0.15 m)3(280 K)

= −2.2 J .

14.19Similar to Problem (14.17), we find the work W on the gas by evaluating the area enclosedby the curve representing the process in the corresponding P -V diagram. Here the area is arectangle with side lengths AB = 2Vi − Vi = Vi and AD = Pi − 1

2Pi = 12Pi , so

W = −(∆P )(∆V ) = AB × AD = −12PiVi .

14.20

(a) Work is done by the gas as it expands, and on the gas if it is compressed. From the figurewe see that in following path-1 the volume of the gas increases, so work is done by the gasrather than on the gas.

(b) The concept in question is the Conservation of Energy, which requires that the work doneon the gas, combined with the heat flowing into it, to be equal the change in its internal energy:∆U = Q + W . This is the First Law of Thermodynamics.

(c) Solve for Q from the First Law, with W = −458 J (negative since the work is done by thegas) and ∆U = −178 J (negative since the internal energy of the gas decreases):

Q = ∆U − W = −178 J − (−458 J) = 280 J .


Since Q > 0 heat actually flows into the gas.

(d) The internal energy U is a function of the state of the gas. As long as the gas proceeds fromstate-A to state-B, ∆U = UB − UA is always equal to −178 J, regardless of the path taken.

(e) Now Q = +50.0 J while ∆U = −178 J, so from the First Law

W = ∆U − Q = −178 J − 50.0 J = −228 J ,

which is negative, indicating that the work is done by the gas. This is expected from thePV -diagram as the gas expands isobarically along the second part of path-2.

14.21

(a) The process is isovolumic, i.e., ∆V = 0; so W ∝ ∆V = 0.

(b) The gas can be placed in a sealed chamber (so as to fix its volume) while being cooleddown, drawing heat out of it and therefore lowering its pressure (and temperature).

(c) Apply the First Law to the process D-A: ∆U = UA −UD = Q + W , where UD = 25.0 kJ,Q = 4.5 kJ (positive as heat is added into the gas), and W = 0. Thus

UA = UD + Q + W = 25.0 kJ + 4.5 kJ + 0 = 29.5 kJ .

(d) Point-B has the same pressure as point-A but a greater volume. So from Charles’s Law,T/V = constant, we have

TB

TA

=VB

VA

> 1 ,

or TB > TA .

(e) Since the internal energy of a given amount of ideal gas is directly proportional to itsabsolute temperature,

UB

UA

=TB

TA

> 1 ,

or UB > UA .

(f) The path from A-B is isobaric, with P = 0.80 MPa. From Eq. (14.3) the work done on thegas is

W = −P∆V = −P (VB − VA) = −(0.80 MPa)(0.200 m3 − 0.100 m3) = −80 kJ ,

which is negative, indicating that work is actually done by the gas.

(g) UA = 29.5 kJ [see part (c) above].

P (atm)

V (liters)5 10 15 200

10

15

20

25

AB

C

T=TB

T=TA=TC

5

30


14.22

A P -V diagram depicting the process is shown above. The gas starts from state A, withPA = (3 atm)(1.013× 105 Pa/atm) = 3.039 Pa and VA = (2 liters)(10−3 m3/liter) = 0.020 m3;proceeds to B, with PB = 3 atm and VB = 2.0 liters; then reaches point C, with VC = VB =2.0 liters and TC = TA . Since the volume of the gas does not change from B to C the gas doeswork only from A to B:

WAC = WAB = −PA∆VAB = −PA(VB − VA)

= −(3.039 × 105 Pa)(0.020 m3 − 0.002 0 m3)

= 5.5 × 103 J = 5.5 kJ .

The temperature of the gas does not change from A to C so ∆UAC = UC − UA = 0, and fromEq. (14.1)

QAC = ∆UAC − WAC = −WAB = −5.5 kJ .

Here WAC > 0 and QAC < 0, indicating that work is done on the system, which loses heat inthe process.


14.23The work W done in an isobaric process is given by Eq. (14.3): W = −P∆V . Here P =0.30 MPa and ∆V = A∆h, with A = 0.50 m2 and ∆h = 6.5 cm; so

W = −P∆V = −PA∆h

= −(0.30 × 106 Pa)(0.50 m2)(0.065 m) = −9.8 × 103 Pa·m3

= −9.8 kJ ,

and from Eq. (14.1)

∆U = Q + W = 10.0 kJ + (−9.8 kJ) = +0.2 kJ .

14.24Use Eq.(14.1), ∆U = Q+W , where Q = +10 000 J and W = −P∆V , with P = 1.013×105 Paand ∆V = +1.00 m3. Thus

∆U = Q + W = +10 000 J − (1.013 × 105 Pa)(1.00 m3) = −9.1 × 104 J ,

or −91 kJ. Here ∆U < 0, meaning that the gas loses part of its internal energy in the process.

14.25Again, apply Eq. (14.1), ∆U = Q + W . In this case Q = −30 kJ and W = −P∆V , withP = 1.013 × 105 Pa and ∆V = −(0.25 m)(1.00 m2) = −0.25 m3. Thus

∆U = Q + W = −30 kJ − (1.013 × 105 Pa)(−0.25 m3) = −5 kJ .

Since ∆U < 0 the gas in the cylinder loses energy in the process. Here we have assumed thatthe gas pressure does not change.

14.26Since the water (of mass m) is in a vacuum chamber (where P = 0) it does not perform anywork on the environment even as it expands: W = −P∆V = 0. Then from Eq. (14.1)

∆U = Q + W = Q = cWm∆T = (4186 J/kg·C◦)(5.0 kg)(10 C◦) = +2.1 × 105 J .

14.27The cylinder is thermally insulated so no heat flows in or out of it: Q = 0. Thus according toEq. (14.1) the change in internal energy of the water-steam mixture inside the cylinder, ∆U , isrelated to the work W done by the steam as ∆U = Q+W = W = −P∆V , where P (= 99 kPa)is the pressure the piston is pushing against and ∆V = +4950 cm3 = +4.950 × 10−3 m3.


Now, the temperature inside the cylinder remains at 100◦C so there is no change in the averagemolecular KE of the system, and so ∆U is solely due to change in molecular PE as a resultof the phase change as the steam condenses into water: ∆U = −mSLv , where mS is themass of the steam which condenses into water in the process and Lv = 2259 kJ/kg is the heatof vaporization for water. Here we noted that ∆U < 0 as steam loses thermal energy whilecondensing into water.

Equate the two expressions obtained above for ∆U to obtain ∆U = −P∆V = −mSLv , whichgives

mS =P∆V

Lv

=(99 × 103 Pa)(4.950 × 10−3 m3)

2259 × 103 J/kg= 2.2 × 10−4 kg = 0.22 g .

14.28For isobaric processes Eq. (14.3) holds true: W = −P∆V . Here ∆V = Vf −Vi = − 1

2Vi , whereaccording to the Ideal Gas Law PVi = nRTi ; so the work done on the gas is

W = −P∆V =12PVi =

12nRTi

=12(2.0 mol)(8.314 4 J/mol·K)(273 K)

= 2.3 × 103 J = 2.3 kJ .

14.29Consider a certain amount of water (W) of mass m, which is turned into steam (S). Its volumechange in the vaporization process is ∆V = VS−VW = m/ρS−m/ρW , where ρS = 0.598 kg/m3

and ρW = 1.000 × 103 kg/m3. The work done by the water as it expands into steam isW = −P∆V = −mP (1/ρS − 1/ρW), where P = 1.01 × 105 Pa. The fraction of the heat ofvaporization (Q = mLv) that is expended in doing this much work is then

|W |Q

=mP (1/ρS − 1/ρW)

mLv

=P

Lv

(1ρS

− 1ρW

)

=(

1.01 × 105 Pa

2259 kJ/ kg

)(1

0.598 kg/m3− 1

1.000 × 103 kg/m3

)

= 0.074 7 = 7.47% .

14.30The rate at which heat flows into the system is Q/t = (35%)(0.83 kW/m2)(1.0 m2) = 0.29 kW,while the rate at which work is done on the system is W/t = −150 J/s = −0.150 kW. Thusfrom Eq. (14.1) the rate of change in internal energy for the system is

∆U

t=

Q

t+

W

t= 0.29 kW − 0.150 kW = 0.14 kW .


14.31

(a) The temperature remains unchanged, since the process is isothermal.

(b) The internal energy of an ideal gas is proportional to its absolute temperature. Since Tdoes not change, neither does U , i.e., ∆U = 0.

(c) and (d) Apply the First Law: ∆U = Q + W , with ∆U = 0 and W = 10.5 kJ, to obtainQ = ∆U − W = 0 − 10.5 kJ = −10.5 kJ. Since Q < 0 heat was transferred out of the gas.

14.32In an adiabatic process Q = 0. Since the system delivers 10.0 kJ of work into the environmentW = −10.0 kJ. The First Law becomes

∆U = Q + W = 0 − 10.0 kJ = −10.0 kJ ,

meaning that the internal energy of the system decreases by 10.0 kJ.

14.33Use Eq. (14.5) for the work W done on an ideal gas undergoing isothermal expansion:

W = −nRT lnVf

Vi

= −PiVi lnVf

Vi

= −[(10.0 atm)(1.013 Pa/ atm)](6000 × 10−6 m3) ln(

12 000 cm3

6000 cm3

)

= −4.21 × 103 J = −4.21 kJ .

Here the Ideal Gas Law PV = nRT has been used.

14.34Apply Eq.(14.5), W = −nRT ln(Vf/Vi) = −PiVi ln(Vf/Vi), to solve for Pi . Here Vi = 4.00 m3,Vf = Vi + 2.00 m3 = 6.00 m3, and W = −2.00 kJ; so

Pi = − W

Vi ln(Vf/Vi)= − −2.00 kJ

(4.00 m3) ln(6.00 m3/4.00 m3)= 1.23 kPa .

14.35The initial volume of the gas is Vi = (10.0 liters)(10−3 m3/ liter) = 0.010 0 m3. Thus fromEq. (14.5) the work done on the gas as its volume changes from Vi to 1

2Vi is

W = −PiVi lnVf

Vi

= −(0.101 3 × 106 Pa)(0.010 0 m3) ln

12Vi

Vi

= 702 J .


14.36The work W done on an ideal gas undergoing an isothermal expansion is W = −PiVi ln(Vf/Vi),where i and f denote the state of the gas before and after the expansion, respectively. This givesPi , the initial pressure of the gas prior to the expansion:

Pi = − W

Vi ln(Vf/Vi)= − −1.00 kJ

(5.00 m3) ln(3Vi/Vi)= 182 Pa .

14.37First find the final volume Vf of the gas from PiVi = PfVf :

Vf = Vi

(Pi

Pf

)= (3.00 m3)

(0.300 MPa

0.900 MPa

)= 1.00 m3 .

Thus the work done on the gas is

W = −PiVi lnVf

Vi

= −(0.300 MPa)(3.00 m3) ln(

1.00 m3

3.00 m3

)= 0.989 MJ .

14.38The work done on an ideal gas in an isothermal process is given by Eq. (14.5). In our casen = 5.0 mol, T = (27 + 273)K = 300 K, and Vf/Vi = 2; so

W = −nRT lnVf

Vi

= −(5.0 mol)(8.314 4 J/mol·K)(300 K) ln 2 = −8.6 kJ .

14.39

(a) According to the problem statement Vf = 12Vi , i.e., Vf/Vi = 1

2 .

(b) As the gas is slowly compressed it maintains thermal contact with the environment, keepingits temperature constant (the same as that of the environment). (This would not be possibleif it were to be compressed quickly, since heat exchange through the pump takes longer time.)For the isothermal process PiVi = PfVf , and so

Pi

Pf

=Vf

Vi

=12

.

(c) From the result above Pf = 2Pi = 2(1.00 atm) = 2.00 atm.

(d) Since the gas is compressed work is done on it.


(e) According to Eq. (14.5) the work done on the gas is

W = −nRT lnVf

Vi

= −PiVi lnVf

Vi

,

where in the last step we applied the Ideal Gas Law, PV = nRT , to the initial state of the gas.

(f) Plug Pi = 1.00 atm = 1.013 × 105 Pa, Vi = (0.60 liters)(10−3 m3/liter) = 6.0 × 10−4 m3,and Vf = 1

2Vi into the formula for W above to obtain the work done on the gas:

W = −PiVi lnVf

Vi

= −(1.013 × 105 Pa)(6.0 × 10−4 m3) ln12

= 42 J .

14.40

(a) This is an isobaric process, for which W = −P∆V = −P (Vf − Vi). From Fig. P40,P = 2.00 atm = 2.00(1.013× 105 Pa) = 2.026× 105 Pa, Vi = 1.00 liter = 1.00× 10−3 m3, andVf = 4.00 liter = 4.00 × 10−3 m3. Thus

W = −P (Vf − Vi) = −(2.026 × 105 Pa)(4.00 × 10−3 m3 − 1.00 × 10−3 m3) = −608 J .

(b) Since W < 0 work is done by the gas.

(c) In an isobaric expansion the temperature of the gas rises. Since the internal energy U ofan ideal gas is proportional to its absolute temperature, U must also increase, i.e., UB > UA .

(d) Write the First Law, ∆U = Q + W , or Q = ∆U − W . Since ∆U > 0 and W < 0 (seeabove), Q > 0, meaning that heat must flow into the gas.

(e) W = 0, as the volume of the gas is unchanged.

(f) The temperature of the gas decreases from point-B to point-C, since its pressure decreaseswhile its volume stays the same. Therefore its internal energy U , being proportional to theabsolute temperature, must also decrease: ∆U = UC − UB < 0.

(g) Apply the First Law to the process B-C: ∆U = Q+ W , or Q = ∆U −W . Since ∆U < 0and W = 0 (see above), Q = ∆U < 0, meaning that heat must flow out of the gas.

(h) This is an isothermal process (as you can check for yourself that PV = constant alongthe curve from point-C to point-A.) So we may apply Eq. (14.5), with the initial state beingpoint-C and the final state being point-A:

W = −nRT lnVf

Vi

= −PfVf lnVf

Vi

= −(2.026 × 105 Pa)(1.00 × 10−3 m3) ln(

1.00 liters

4.00 liter

)

= 281 J ,

which is indeed substantially less than WB→A = 608 J. [Note that what we calculated in part(a) was WA→B which was −608 J.]


(i) Again, the process C-A is an isothermal process, which leaves the internal energy U of anideal gas unchanged as U ∝ T = constant.

(j) and (k) Apply the First Law to the process C-A: ∆U = Q + W , or Q = ∆U − W . Since∆U = 0 and W > 0 (see above), Q = −W = −281 J. Since Q < 0 heat must flow out of thegas.

14.41Since the gas maintains good thermal contact with a reservoir its temperature is constant,i.e., it undergoes an isothermal process, for which the work W done on it satisfies Eq. (14.5),W = −nRT ln(Vf/Vi). We are given W = −28.0 J (negative since work is done by the gas),T = 283.1 K, Vi = 2.0 liters, and n = 0.119 mol, and we wish to find Vf . First find ln(Vf/Vi):

lnVf

Vi

= − W

nRT= − (−28.0 J)

(0.119 mol)(8.314 4 J/mol·K)(283.1 K)= 0.099 96 .

Now raise e to the power of each side of the equation above and use the identity eln x = x(where x = Vf/Vi) to undo the logarithm:

eln

VfVi =

Vf

Vi

= e0.099 96 = 1.105 ,

and so Vf = 1.015Vi = 1.015(2.0 liters) = 2.2 liters.

14.42For an isothermal process from an initial state (i) to a final state (f), PiVi = PfVf . Also, forthe initial state PiVi = nRT , so Pi = nRT/Vi . Thus or

Vf

Vi

=Pi

Pf

=nRT/Vi

Pf

.

Substitute this into Eq. (14.5) to obtain

W = −nRT lnVf

Vi

= −nRT ln(

nRT

PfVi

).

Plug in W = −13.9 J, n = 0.119 mol, T = 10.0◦C = (10.0 + 273.15) K = 283.15 K, andVi = 1.29 liters = 1.29 × 10−3 m3 to obtain

ln(

nRT

PfVi

)= − W

nRT= − −13.9 J

(0.119 mol)(8.314 4 J/mol·K)(283.15 K)= 0.049 62 .

Now raise e to the power of each side:

exp[ln

(nRT

PfVi

)]=

nRT

PfVi

= e0.049 62 = 1.050 9 ,


which yields

Pf =nRT

1.0509Vi

=(0.119 mol)(8.314 4 J/mol·K)(283.15 K)

1.0509(1.29 × 10−3 m3)= 2.06 × 105 Pa ,

which is (2.06 × 105 Pa)/(1.013 × 105 Pa/ atm) = 2.04 atm.

14.43Start with Eq.(14.5), which gives the work done by an ideal gas in an isothermal process: W =−nRT ln(Vf/Vi). But since the process is isothermal PiVi = PfVf = nRT , so Vf/Vi = Pi/Pf .Substitute this expression into Eq. (14.5) to obtain

W = −nRT lnVf

Vi

= −nRT lnPi

Pf

.

14.44Use the result of the previous problem, with n = 2.00 mol, T = 400 K, Pi = 12 atm, andPf = 1.5 atm:

W = −nRT lnPi

Pf

= −(2.00 mol)(8.314 4 J/mol·K)(400 K) ln(

12 atm

1.5 atm

)= −14 kJ .

14.45For process A-B-C-A the work done is WABCA = WAB + WBC + WCA , where WAB = 0 sincethere is no volume change for the gas from A to B, WBC = −P2(V2 − V1) (isobaric process),and WCA = −nRTA ln(V1/V2) = −P2V2 ln(V1/V2). Thus

WABCA = WAB + WBC + WCA = −P2V2 + P2V1 − P2V2 lnV1

V2

.

Similarly for process C-D-E-C the work done is

WCDEC = WCD + WDE + WEC = −P3V3 + P3V2 − P3V3 lnV2

V3

.

Now, since points B and D are on the same isothermal curve P2V1 = P3V2 ; and since pointsC and E are on the same isothermal curve P2V2 = P3V3 . Combine the two equations above toobtain P3/P2 = V1/V2 = V2/V3 . Substitute these results into the two expressions for WABCA

and WCDEC above to obtain WABCA = WCDEC .

P

V

P

P1

P3

P2

V1 V2

A

B

C

adiabatic

isothermal


14.46The initial and final states in an adiabatic process of an ideal gas are related by Eq. (14.6):PiV

γi

= PfVγf

. Here Pi = 1.0 atm, Vf/Vi = 1/2, and γ = 1.40 for diatomic gases; so

Pf = Pi

(Vi

Vf

)γ

= (1.00 tam) 21.40 = 2.64 atm .

Plug this result into the Ideal Gas Law,PfVf/Tf = PiVi/Ti , which we solve for Tf , with Ti =273 K:

Tf = Ti

(Pf

Pi

)(Vf

Vi

)= (273 K)

(2.64 atm

1.00 atm

)(Vi/2Vi

)= 360 K .

14.47

The P -V diagram depicting the process for the ideal gas in question is shown above. Thegas starts from state A, where the pressure is P1 = 1.00 atm and V1 = 22.4 liters; proceedingadiabatically to state B where its volume gets doubled: V2 = 2V1 , before reaching state Cisothermally with V3 = V1 = 22.4 liters.

(a) During the adiabatic expansion from A to B the pressure of the system decreases from P1

to P2 . As the gas gets compressed isothermally from B to C its pressure increases from P2 toP3 . So clearly the lowest pressure the system attains is P2 . Since process A → B is adiabaticP1V

γ1

= P2Vγ2

, so

P2 = P1

(V1

V2

)γ

= P1

(V1

2V1

)1.40

= (1.00 atm)(

12

)1.40

= 0.379 atm ,


which is equivalent to (0.379 atm)(0.101 3 MPa/atm) = 0.038 4 MPa.

(b) As the gas expands adiabatically from A to B its temperature drops from T1 (= 273 K) toT2 , its value in state B. The temperature of the system remains at T2 throughout the rest ofthe process since B → C is isothermal. So the lowest temperature T2 is first attained in stateB. Apply the Ideal Gas Law to states A and B: P1V1/T1 = P2V2/T2 , which we solve for T2 :

T2 = T1

(P2

P1

)(V2

V1

)= (273 K)

(0.379 atm

1.00 atm

)(2Vi

Vi

)= 207 K .

Here we noted that 1.00 mol of any ideal gas occupies 22.4 liters at STP, so the initial temper-ature of the gas must be T1 = 273 K (0◦C).

14.48Refer to the previous problem for the P -V diagram depicting the process from A to B to C.For the isothermal process from B to C the work done on the gas is given by Eq. (14.5):

WBC = −nRT2 lnV1

V2

= −(1.00 mol)(8.314 4 J/mol·K)(207 K) ln(

V1

2V1

)= 1.19 kJ ,

i.e., 1.19 kJ of work must be done by the environment on the gas to return it to its originalvolume.

The final temperature T3 of the system is the same as T2 , its value at B, since B → C isisothermal. Thus T3 = 207 K.

Now consider states A and C, which satisfy P1/T1 = P3/T3 . Solve for P3 , the final pressure instate C:

P3 =P1T3

T1

=(1.00 atm)(207 K)

273 K= 0.758 atm .

14.49Since the process is rapid it must be adiabatic. The initial values of P and V are then relatedto the final values by PiV

γi

= PfVγf

. Also, from Ideal Gas Law PiVi/Ti = PfVf/Tf . Combinethe two equations above to yield

Pf

Pi

=Tf

Ti

(Vi

Vf

)=

Tf

Ti

(Pf

Pi

)1/γ

,

or

Tf = Ti

(Pf

Pi

)1−1/γ

= [(273 + 27)K](

1.0 atm

4.5 atm

)1−1/1.40

= 195 K ,

which is equivalent to −78◦C. Notice that we didn’t have to know the actual initial volume.


14.50Denote the state of the gas before and after the adiabatic expansion with subscripts i andf, respectively. Then PiVi/Ti = PfVf/Tf (Ideal Gas Law) and PiV

γi

= PfVγf

(for adiabaticprocesses). Solve for Pf from the last equation:

Pf = Pi

(Vi

Vf

)γ

= (150 kPa)(

40.0 cm3

1.00 cm3

)1.67

= 7.104 5 × 104 kPa .

Now solve for Tf from the Ideal Gas Law:

Tf = Ti

(PfVf

PiVi

)= [(273.15 + 10.0)K]

(7.1045 × 106 kPa)(1.00 cm3)(150 kPa)(40.0 cm3)

= 3.35 × 103 K .

14.51For an ideal gas undergoing an adiabatic process from an initial state i to a final state f, bothEqs. (14.6) and the Ideal Gas Law hold true: PiV

γi

= PfVγf

, PiVi/Ti = PfVf/Tf . Rewritethe adiabatic equation as Pi/Pf = (Vf/Vi)

γ , and the Ideal Gas Law equation as Ti/Tf =(Pi/Pf)(Vi/Vf). Combine the last two equations to obtain

Ti

Tf

=(

Pi

Pf

)(Vi

Vf

)=

(Vf

Vf

)γ(Vi

Vf

)=

(Vf

Vi

)γ(Vf

Vf

)−1

=(

Vf

Vi

)γ−1

.

14.52Use the result of the previous problem to find the final temperature Tf , with Ti = 25.0◦C =(25.0 + 273.15)K = 298.15 K and Vf/Vi = 1/20:

Tf = Ti

(Vf

Vi

)1−γ

= (298.15 K)(

Vi/20Vi

)1−1.35

= 8.5 × 102 K .

14.53Use the result of Problem(14.51): Ti/Tf = (Vf/Vi)

γ−1. In this case Vi/Vf = 7.5 and γ = 1.35,so

Ti

Tf

=(

Vf

Vi

)γ−1

=(

Vi

Vf

)1−γ

= (7.5)1−1.35 = 0.49 .

Note that here Tf > Ti . This is expected, since the gas goes through an adiabatic compressionin which no heat enters or leaves the system while work is being done on it, causing an increasein its internal energy and hence temperature.

14.54Apply the First Law of Thermodynamics and the Ideal Gas Law to each process.


From A to B: Since T = constant, so is U , i.e., ∆U = 0. Since the gas expands W < 0, i.e.,work is done by the gas. Now, the First Law requires that ∆U = Q + W , so Q = ∆U −W =−W > 0, so heat flows into the gas.

From B to C: Since ∆V = 0, W = 0. Also, as the pressure of the gas decreases while itsvolume is unchanged, T must decrease, as T = PV/nR ∝ P . Thus U , which is proportionalto T , must also decrease: ∆U < 0. Now, Q = ∆U − W = ∆U < 0, meaning that heat flowsout of the gas.

From C to D: By definition Q = 0, since the process is adiabatic. As the gas expands workis done by it, so W < 0. Now ∆U = Q + W = W < 0, so the internal energy of the gasdecreases.

From D to E: Since the gas is being compressed work is done on it, i.e., W > 0. Also,as its volume decreases while its pressure is unchanged, its temperature must decrease, asT = PV/nR ∝ V . Therefore U must also decrease, i.e., ∆U < 0. Now Q = ∆U −W < 0 (asboth ∆U and −W are negative), so heat flows out of the gas.

From E to A: Again by definition Q = 0, since the process is adiabatic. As the gas is beingcompressed work is done on it, so W > 0. Now ∆U = Q+W = W > 0, so the internal energyof the gas increases.

14.55From Fig. P55 we count the region enclosed by the P -V curve to be about 22 “boxes”, with atotal area of 22(1.0 × 105 Pa)(1.0 × 10−5 m3) = 22 J. Note that the magnitude of the workdone by the system in one cycle equals to the area enclosed. So W = −22 J, where the negativesign indicates that positive work is done by the gas on the environment.

14.56Use Eq. (14.10), with QH = 100 kJ and QL = 75 kJ:

e = 1 − QL

QH

= 1 − 75 kJ

100 kJ= 0.25 = 25% .

14.57Imagine a heat engine which operates between a high-temperature reservoir at TH and a low-temperature reservoir at TL . The maximum possible efficiency of such an engine is that ofa Carnot engine, given by Eq. (14.11): ec = 1 − TL/TH . In this case TH = 98.6 ◦F =[(5/9)(98.6 − 32) + 273] K = 310 K and TC = 20◦C = (20 + 273)K = 293 K, so the maxi-mum efficiency is

ec = 1 − TL

TH

= 1 − 293 K

310 K= 0.055 = 5.5% .


14.58Use Eq.(14.11) for the maximum possible efficiency, with TL = 100◦C = (100+273)K = 373 Kand TL = 400◦C = (400 + 273)K = 673 K:

ec = 1 − TL

TH

= 1 − 373 K

673 K= 0.446 = 44.6% .

14.59Again, use Eq. (14.11) for the maximum possible efficiency:

ec =TH − TL

TH

=(180 + 273)K − (80.0 + 273)K

(180 + 273)K= 0.221 = 22.1% .

14.60According to Eq. (14.11) the theoretical efficiency is given by

ec =TH − TL

TH

.

Increasing TH results in an increase in both the numerator of the denominator; while decreasingTL decreases the numerator only and therefore has the more profound effect. (As a numericalexample, consider TH = 400 K and TL = 300 K, which yields ec = 25%. Raising TH by 10 Kwould increase ec to 26.8%, while lowering TL by 10 K would increase ec more, to 27.5%.)

14.61

(a) The internal energy is a function of the state of the system. For a cyclic process ∆U = 0at the end of each cycle, since the system returns to the original state after going through thecycle.

(b) During each cycle 2.01 kJ of heat flows into the system while 1.69 kJ leaves it, so the netamount of heat flowing into the system is |Qi | = |QH | − |QL | = 2.01 kJ − 1.69 kJ = 0.32 kJ.

(c) From the First Law W = ∆U − Q = 0 − 0.32 kJ = −0.32 kJ, meaning that 0.32 kJ ofwork is done by the system per cycle: |Wo | = 0.32 kJ.

(d) By definition

e =|Wo ||Qi |

=0.32 kJ

2.01 kJ= 0.16 = 16% .

14.62

(a) The engine operates between TH = 700 K and TL = 20◦C = (20 + 273) K = 293 K.


(b) From Eq. (14.11) the maximum efficiency is

ec = 1 − TL

TH

= 1 − 293 K

700 K= 0.581 = 58.1% .

(c) The efficiency of the engine, according to the data from the designer, would be e =|Wo |/|QH | = 1.00 kJ/1.66 kJ = 0.602, which is greater than the maximum theoretical value of0.581. So this is impossible.

14.63The heat taken from the high-temperature reservoir per cycle is |QH | = 1.99 kJ, and the heatdischarged to the low-temperature reservoir is |QL | = 1.50 kJ. Since the engine is lossless theFirst Law gives the work-out per cycle as |Wo | = |QH | − |QL | = 1.99 kJ − 1.50 kJ = 0.49 kJ,and so by definition

e =|Wo ||QH |

=0.49 kJ

1.99 kJ= 0.246 = 24.6% .

14.64Denote the additional energy loss per cycle as |Qlost |, then the total amount of heat-out percycle is |QL | + |Qlost | = 1.50 kJ + 0.10 kJ = 1.60 kJ, and so First Law can be rewritten as|Wo | = |QH | − (|QL | + |Qlost |) = 1.99 kJ − 1.60 kJ = 0.39 kJ. Thus

e =|Wo ||QH |

=0.39 kJ

1.99 kJ= 0.196 = 19.6% .

14.65The ideal efficiency of the heat engine in question is given by Eq. (14.13):

ec =TH − TL

TH

=(200 + 273)K − (100 + 273)K

(200 + 273)K= 0.211 4 .

Thus the actual efficiency is e = 20.0% ec = (0.200)(0.211 4) = 4.23%.

14.66Use Eq. (14.13) for the coefficient of performance, with QL = 75 kJ and QH = 100 kJ:

η =|QL |

|QH | − |QL |=

75 kJ

100 kJ − 75 kJ= 3.0 .


14.67The heat Q that must be removed from a quantity of water of mass m at 0◦C to cause it tofreeze is Q = mLf . If this is accomplished in a time interval t, then the corresponding thermalpower P is given by P = Q/t = mLf/t. Plug in m = (2000 lb)(0.453 6 kg/lb) = 907.2 kg,Lf = 333.7 kJ/kg, and t = 1 d = 86 400 s to obtain

P =mLf

t=

(907.2 kg)(333.7 kJ/kg)86 400 s

= 3.5 kW .

14.68Use Eq. (14.15) for the coefficient of performance of a Carnot refrigerator, with TL = 0◦C =(0 + 273)K = 273 K and TH = 85◦C = (85 + 273)K = 358 K:

ηc =TL

TH − TL

=273 K

358 K − 273 K= 3.2 .

14.69Use Eq. (14.15), ηc = |QL |/|Wi | = TL/(TH − TL), to solve for Wi . Here QL = 1.0 J, TH =27◦C = (27 + 273)K = 300 K, and TL = −3.0◦C = (−3.0 + 273)K = 270 K; so

Wi =QL(TH − TL)

TL

=(1.0 J)(300 K − 270 K)

270 K= 0.11 J .

14.70

(a) For a Carnot engine operating between TH and TL , ec = 1 − TL/TH —– see Eq. (14.11).

(b) Since TH = 505◦C = (505 + 273) K = 778.15 K and TL = −30.0◦C = (−30.0 + 273) K =243 K, the best possible efficiency is

ec = 1 − TL

TH

= 1 − 243 K

778 K= 0.688 = 68.8% .

(c) The quantity is |Wo |, the amount of work done by the engine per cycle: |Wo | = |QH |−|QL |.(d) |QH | − |QL | = |Wo | = 2012 J.

(e) |QL | = |QH | − |Wo | = 4980 J − 2012 J = 2968 J.

(f)

e =|Wo ||QH |

=2012 J

4980 J= 0.404 = 40.4% < ec .


14.71(a) For a Carnot engine operating between TH and TL the efficiency is given by Eq. (14.11),ec = 1 − TL/TH .

(b) Since TH = 301◦C = (301 + 273) K = 574 K and TL = 149◦C = (149 + 273) K = 422 K,the maximum theoretical efficiency is

ec = 1 − TL

TH

= 1 − 422 K

574 K= 0.265 = 26.5% .

(c) The efficiency increases as TH increases or TL drops. To increase the efficiency withoutchanging TL , we can raise TH .

(d) The new efficiency is e′c

= 35.0% + 26.5% = 61.5%. Suppose that TH should be raised toT ′

H, then e′

c= 1 − TL/T ′

H, which gives

T ′H

=TL

1 − e′c

=422.15 K

1 − 0.615= 1096 K ≈ 1.10 × 103 K .

14.72The efficiency of a reversible engine is given by Eq.(14.11). Here TH = 325◦C = (325+273)K =598 K, and TC = 25◦C = (25 + 273)K = 298 K; so

e = 1 − TL

TH

= 1 − 288 K

598 K= 0.501 6 ≈ 50% .

The work done per cycle then follows from Eq. (14.10):

|Wo | = e|Qi | = (0.501 6)(610 kcal) = 3.1 × 102 kcal = 0.31 MJ ,

or 1.3 MJ. The heat exhausted per cycle is then

|QL | = |QH | − |Wo | = 610 kcal − 310 kcal = 300 kcal ,

or, to two significant figures, 0.30 Mcal. This is equivalent to 1.3 MJ. The time t it takes tocomplete one cycle is t = 1

5 s, so the power delivered by the engine is

P =Wo

t=

1.28 MJ15

s= 6.4 MW .

14.73The engine in question operates between TH and TL , where TH = 144 ◦F = (5/9)(144 −32)◦C = 62.2◦C = (62.2 + 273)K = 335 K and TL = 5 ◦F = (5/9)(5 − 32)◦C = −15◦C =(−15+273)K = 258 K. Such an engine, under ideal circumstances, would have an efficiency ofec = (TH − TL)/TH = (335 K − 258 K)/335 K = 0.230 = 23.0%. But even at this maximumpossible efficiency a heat intake of |Qi | = 102 kJ would still only produce a useful work outputof |Wo | = ec |Qi | = (0.230)(102 kJ) = 23.5 kJ, which is less than 26 kJ, the claimed value of|Wo |. So the engine is not physically possible.


When the engine is an ideal one (with e = ec), the heat intake |Qi | it needs to do an amountof work |Wo | is at its minimum: ec = |Wo |/|Qi(min)|. For ec = 0.230 and |Wo | = 100 kJ,

|Qi(min)| =|Wo |ec

=100 kJ

0.230= 435 kJ .

14.74The efficiency of an ideal heat engine operating between TH = 200◦C = (273 + 200)K = 473 Kand TL = 20.0◦C = (273 + 20.0)K = 293 K is ec = (TH − TL)/TH . Equate this to |Wo |/|Qi |,where |Wo | = 1000 J, to solve for the heat |Qi | that needs to be supplied:

|Qi | =|Wo |ec

=|Wo |TH

TH − TL

=(1000 J)(473 K)473 K − 293 K

= 2.63 × 103 J = 2.63 kJ .

14.75The heat supplied to the engine during a time interval of one hour is |Qi | = 100 MJ. sincethe engine operates between TH = 200◦C = (300 + 273)K = 573 K and TL = 20.0◦C =(20.0 + 273)K = 293 K, its maximum possible efficiency is ec = (TH − TL)/TH , at which thework output Wo per hour would be

|Wo | = ec |Qi | =|Qi |(TH − TL)

TH

=(100 MJ)(573 K − 293 K)

573 K= 48.9 MJ .

14.76The efficiency is defined in Eq. (14.10): e = |Wo |/|Qi |. In this case |Wo | = (60 W)(6 h)×(3600 s/h) = 1.296 MJ and |Qi | = 330 kcal + 225 kcal + (1.0 kg)(650 kcal/kg) + 2(145 kcal) =1495 kcal = (1495 kcal)(4.186 kJ/kcal) = 6.258 MJ; so

e =|Wo ||Qi |

=1.296 MJ

6.258 MJ= 0.2 .

14.77

(a) e = |Wo |/|Qi | = 7.20 kJ/10.00 kJ = 0.720 = 72.0%.

(b) Of the 10.00-kJ of energy the engine receives from the high-temperature reservoir, 7.20 kJbecomes work-out and the remaining part, 10.00 kJ − 7.20 kJ = 2.80 kJ, is lost. So the per-centage of energy lost is 2.80 kJ/10.00 kJ = 28.0%.

(c) The energy lost could be exhausted to the low-temperature reservoir, or dissisipated intothe environment due to friction or thermal loss (by conduction, radiation or convection).


(d) A total of 2.80 kJ is lost per cycle, of which 2.00 kJ goes to the low-temperature reservoir.So the amount of energy lost by all other means must be 2.80 kJ − 2.00 kJ = 0.80 kJ.

14.78(a) For a Carnot engine operating between a TH and TL , ec = 1 − TL/TH .

(b) In our case TH = 1000 K and TL = 250 K, so ec = 1 − 250 K/1000 K = 0.750 = 75.0%.

(c) By definition e = |Wo |/|Qi |. Here e = ec = 0.750 and |Qi | = |QH | = 24.5 kJ, and so thework-our per cycle is |Wo | = ec |QH | = 0.750(24.5 kJ) = 18.4 kJ.

(d) The engine delivers |Wo | = 18.4 kJ of work per cycle, which lasts ∆t = 1.49 s. So theaverage mechanical power is

P =|Wo |∆t

=18.4 kJ

1.49 s= 12.3 kJ/s = 12.3 kW .

(e) Of the 24.5-kJ of energy the engine receives from the high-temperature reservoir per cycle,18.4 kJ becomes work-out and the remaining part, 24.5 kJ − 18.4 kJ = 6.1 kJ must exhaust tothe low-temperature reservoir.

14.79First, find TL from Eq. (14.11), with ec = 42.2% and TH = 473 K:

TL = TH(1 − ec) = (473 K)(1 − 0.422) = 273.4 K .

If the efficiency is to be increased to e′c

= 50% while TL remains the same, then the newtemperature T ′

Hof the high-temperature reservoir must satisfy e′

c= 1 − TL/T ′

H, or

T ′H

=TL

1 − e′c

=273.4 K

1 − 50%= 547 K .

14.80The heat engine in question operates between TH and TL , where TH = (273 + 67.0)K = 340 Kand TL = (273 + 17.0)K = 290 K. The corresponding Carnot efficiency is ec = (TH − TL)/TH ,while the actual efficiency of the engine is e = 90.0%ec = 0.900(TH −TL)/TH . At this efficiencythe work output |Wo | for a heat intake of |Qi | = 1.00 kJ is

|Wo | = e|Qi | =0.900Qi(TH − TL)

TH

=(0.900)(1.00 kJ)(340 K − 290 K)

340 K= 132 J .

14.81During each cycle of its operation the engine in question takes in an amount of heat |QH | =1200 J and dumps |QL | = 450 J to a cool reservoir. From Eq.(14.10) the efficiency of the engineis

e = 1 − |QL ||QH |

= 1 − 450 J

1200 J= 0.625 = 62.5% .


Now consider the engine operating backwards. During each cycle it takes out an amount ofheat QL from the low-temperature reservoir, receives an amount of work |Wi | from outside,and dumps an amount of heat |QH | to the high-temperature reservoir. Conservation of energyrequires that |QL |+ |Wi | = |QH |, so |QL | = |QH | − |Wi | = 1200 J− 1000 J = 200 J. This tellsus that, while |QH | is unchanged (at 1200 J), the value of |QL | is different (450 J for forwardoperation and 200 J for backward operation). Thus the engine must be irreversible. (And,consequently, its efficiency must be lower than that of a Carnot engine, which is reversible.)

14.82The total work input in a time duration t for the air conditioner running at a power P is|Wi | = Pt. If the heat it removes from the room in the mean time is QL , then from Eq. (14.13)we find its efficiency to be η = |QL |/|Wi | = |QL |/Pt. Plug in P = 2070 W, t = 1.0 h, and|QL | = 18 000 Btu to obtain

η =|QL |Pt

=(18 000 Btu)(1.055 × 103 J/Btu)

(2070 W) [(1.0 h)(3600 s/h)]= 2.5 .

14.83Suppose that the refrigerator pumps out an amount of heat |QL | from its interior at a tem-perature TL while it is being supplied with an amount of work |Wi |. Then according toEq. (14.15) at its very best the coefficient of performance of such a refrigerator can attainis ηc = |QL |/|Wi | = TL/(TH −TL), where TH is the temperature of the environment into whichthe heat from the refrigerator is dumped.

Solve for |Wi | from the expression above: |Wi | = |QL |(TH − TL)/TL . Here |QL | is the heatthat must be removed from 2000 lb of water to freeze it into ice: |QL | = mLf , where m =(2000 lb)(0.453 6 kg/lb) = 907.2 kg and Lf = 333.7 kJ/kg (the heat of fusion of ice). Also,TH = 88 ◦F = [ 59 (88 − 32) + 273]K = 304 K and TL = 32 ◦F = [ 59 (32 − 32) + 273]K = 273 K.Substitute these numerical values into the expression for Wi obtained above to find

|Wi | = |QL |(

TH − TL

TL

)= (mLf)

(TH − TL

TL

)

= (907.2 kg)(333.7 kJ/kg)(

304 K − 273 K

273 K

)

= 3.45 × 107 J .

Since this much work is to be delivered in t = 1 d = 86 400 s, the corresponding power thatmust be supplied to the refrigerator is

P =|Wi |

t=

3.45 × 107 J

86 400 s= 399 W .


14.84The lowest possible temperature TL in the chamber is reached when the refrigerator is an idealone, operating at its maximum possible efficiency of ηc = |QL |/|Wi | = TL/(TH − TL). In atime interval t = 1.00 s, |QL | = (360 J/s)(1.00 s) = 360 J and |Wi | = (150 W)(1.00 s) = 150 J.Also, TH = 23◦C = (23 + 273)K = 300 K. Plug these data into the expression for ηc aboveand solve for TL :

TL = TH

( |QL ||QL | + |Wi |

)= (300 K)

(360 J

360 J + 150 J

)= 212 K ,

which is equivalent to −61◦C.

14.85Use Eq. (14.13) for the definition of coefficient of performance: η = |QL |/|Wi | = 5.5. Also,from conservation of energy |QH | = |QL | + |Wi |. Combine these two expressions to obtain theratio in question:

|QH ||Wi |

=|QL | + |Wi |

|Wi |=

|QL ||Wi |

+ 1 = η + 1 = 5.5 + 1 = 6.5 .

Thus for every 1 Joule of work you pay you get 6.5 Joules of heat into the house. This is 6.5times as efficient as heating the house directly, in which case you get only 1 Joule of heat withevery Joule of work you pay for.

14.86Ideally, the refrigerator should have a coefficient of performance given by Eq. (14.15): ηc =|QL |/|Wi | = TL/(TH − TL). Here TH = 80 ◦F = [ 59 (80 − 32) + 273]K = 299.7 K and TL =40 ◦F = [ 59 (40 − 32) + 273]K = 277.4 K. Thus

|QL ||Wi |

=TL

TH − TL

=277.4 K

299.7 K − 277.4 K= 12 ,

which means that 12 J of heat can be removed from the food for every Joule of work done bythe compressor.

14.87According to Eq. (14.15), ηc = |QL |/|Wi | = TL/(TH − TL), the work input |Wi | needed for anideal refrigerator to expel an amount of heat QL from the cold chamber is |Wi | = |QL |(TH −TL)/TL . In the present case TH = 27◦C = (27 + 273)K = 300 K is the temperature of theenvironment and TL = −4.0◦C = (−4.0 + 273)K = 269 K is that of the cold chamber. Nowconsider a time interval t = 1.0 min = 60 s, during which the heat removed from the cold


chamber is |QL | = 360 J. Divide both sides of the equation for |Wi | by t to find the electricpower P needed:

P =|Wi |

t=

(QL

t

)(TH − TL

TL

)=

(360 J

60 s

)(300 K − 269 K

269 K

)= 0.69 W .

14.88Use Eq. (14.17), with Q = −5000 J and T = 22◦C = (22 + 273)K = 295 K. Thus the changein entropy of the room for every 5000 J of heat removed is

∆S =Q

T=

−5000 J

295 K= −16.9 J/K ,

where ∆S < 0 since the entropy of the room decreases.

14.89Use Eq.(14.17) for entropy change due to a certain amount of heat exchange Q which takes placeat a constant temperature T . Here Q = +1.00 kJ and T = 100◦C = (100+273)K = 373 K, so

∆S =Q

T=

+1.00 × 103 J

373 K= +2.68 J/K .

14.90The temperature T of the furnace remains at T = 200◦C, or 473 K. When an amount of heatQ (= −1.2 kJ) leaves the furnace, its entropy changes by

∆S =Q

T=

−1.20 kJ

473 K= −2.54 J/K .

14.91Similar to the previous problem, the temperature T of the chamber remains a constant, atT = −10.0◦C = (−10.0 + 273)K = 263 K. When an amount of heat Q (= 5.00 kJ) is addedto it, the entropy of the chamber increases by

∆S =Q

T=

5.00 kJ

263 K= +19.0 J/K .

14.92

(a) The rate of energy loss is 8000 W, or 8000 J/s. In one minute the total energy lost isQ = −(8000 J/s)(60 s) = −480 kJ. (here the negative sign indicates that heat is transferredout of the house.)


(b) If the amount of heat leaving the house (h) per minute is Q and the absolute temperatureinside the house is Th , then according to Eq. (14.17) the change in entropy of the house perminute is

∆Sh =Q

Th

=−480 kJ

(23 + 273.15) K= −1.62 kJ/K .

(c) The amount of heat transferred into the environment per minute is −Q = +480 kJ. if theabsolute temperature of the environment (e) is Te , then the change in entropy of the environmentper minute is

∆Se =−Q

Te

=+480 kJ

(0.0 + 273.15) K= +1.76 kJ/K .

(Note that, since the overall change in entropy of both the house and the environment is∆S = ∆h + ∆Se > 0, this process is spontaneous.)

14.93

(a) According to Eq.(14.17), if an amount of heat Q enters a system at an absolute temperatureT , then the entropy of the system changes by ∆S = Q/T .

(b) The heat needed is Q = Lfm = (333.7 kJ/kg)(4.00 kg) = 1334.8 kJ ≈ 1.33 MJ.

(c) T = 0.000◦C = (0.000 + 273.15) K = 273.15 K.

(d)

∆S =Q

T=

1334.8 kJ

273.15 K= 4.89 kJ/K .

14.94The temperature of the water-ice mixture remains at T = 0◦C = 273 K while water freezesinto ice. The heat that must be removed from a quantity of water of mass m at 0◦C to causeit to freeze is Q = mLf . Thus the entropy change for this much water as it freezes into ice is∆S = −Q/T = −mLf/T , where the minus sign is because heat has to be removed from, notadded to, the water. Plug into this expression m = 10 g, Lf = 333.7 J/g, and T = 273 K toobtain

∆S = −mLf

T= − (10 g)(333.7 J/g)

273 K= −1.2 J/K .

The entropy of the water decreases as it freezes into ice.

14.95A Carnot cycle consists of two isothermal processes at TL and TH , respectively, plus two adia-batic processes for which ∆S = 0 (since Q = 0). In a T -S diagram, an isothermal process isrepresented by a straight line parallel to the T -axis, while an adiabatic process corresponds toa straight line parallel to the S-axis (since S does not change). Therefore in a T -S diagram a

QL

S

T

A B

CD

isothermal

isothermal

adiabaticadiabatic

QH

TL

TH


Carnot cycle assumes the shape of a rectangle, as shown below. Here A → B and C → D areisothermal, while B → C and D → A are adiabatic. The system takes in an amount of heatQH from A to B, and disposes of an amount of heat QL from C to D.

14.96(a) An amount of heat Q (= 10.0 kcal = 4.186×104 J) is transferred from the warmer reservoir(at TH = 600 K) to the colder one (at TL = 300 K), so the entropy change is

∆SH = − Q

TH

=4.186 × 104 J

600 K= +69.8 J/K

for the warmer reservoir and

∆SL = +Q

TL

= +4.186 × 104 J

300 K= +140 J/K

for the colder reservoir.

(b) Heat enters from one end of the bar and leaves at the other end, resulting in no net heatintake or loss for the bar. Its entropy is therefore a constant: ∆Sbar = 0.

(c) The entropy change for the Universe is the net entropy change of the two reservoirs (sincethe rest of the Universe is unaffected):

∆SUniverse = ∆SH + ∆SL = −69.77 J/K + 139.53 J/K = +69.8 J/K .

14.97The heat it takes to melt a quantity of copper of mass m is Q = mLf , where in this casem = 500 g and Lf = 205 J/g. Thus the entropy change for the copper in the process (whichtakes place at T = 1083◦C = 1356 K) is

∆S =Q

T=

mLf

T=

(500 g)(205 J/g)1356 K

= +75.6 J/K .


Here ∆S > 0 since the copper absorbs heat in the melting process.

14.98This problem is similar to the previous one. The heat it takes to convert a quantity of water ofmass m into steam is Q = mLv , where in this case m = 1.00 kg, and Lv = 2259 kJ/kg. Thusthe entropy change for the water in the process (which takes place at T = 100◦C = 373 K) is

∆S =Q

T=

mLf

T=

(1.00 kg)(2259 kJ/kg)373 K

= +6.06 kJ/K .

Here ∆S > 0, since the water absorbs heat in the process.

14.99According to the previous problem, the entropy change for a quantity of ice of mass m as itmelts into water is ∆S = mLf/T . Plug in m = 1.00 kg, Lf = 333.7 kJ/kg, and T = 273 K toobtain

∆S =mLf

T=

(1.00 kg)(333.7 kJ/kg)273 K

= +1.22 kJ/K .

From the previous problem we learned that the entropy change for the same amount of water asit vaporizes into steam is ∆S = +6.06 kJ/K, which is considerably greater than 1.22 kJ/K. Thisis largely due to the relative increase in volume. 1 kg of water occupies a volume of 1000 cm3, or10−3 m3, which represents a volume change of less than 10% from the solid state (ice); whereasthe volume of 1 kg of steam at atmospheric pressure is about 1.7 m3, which corresponds to avolume increase by a factor of 1700, or 170 000%, from the liquid state. The more the volumeavailable for the molecules to fly around the more disordered they become, hence the moreentropy.

14.100As the ice (I) of mass m melts its temperature remains at TI = 0◦C = 273 K, so its entropychanges by ∆SI = mLf/TI (see the previous problem). Meanwhile the large marble table(M) maintains a temperature of TM = 27◦C = (27 + 273)K = 300 K so its entropy changeis ∆SM = −mLf/TM , where we noted that an amount of heat Q = mLf must be transferredfrom the table to the ice for it to melt. Plug m = 100 g and Lf = 333.7 J/g into the formulasabove to find the total entropy change:

∆Stotal = ∆SI + ∆SM =mLf

TI

− mLf

TM

g

= (100 g)(333.7 J/g)(

1273 K

− 1300 K

)

= +11 J/K .


14.101Since the two samples are equal in mass the final temperature of the mixture is Tf = 1

2 (Ti1+Ti2),where Ti1 = 40.0◦C and Ti2 = 32.0◦C. It follows that Tf = 36.0◦C. The average temperatureof the first sample of mass m (= 20 kg) is then (T1)av = 1

2 (40.0◦C + 36.0◦C) = 38.0◦C =(38.0 + 273)K = 311 K, while that of the second one is (T2)av = 1

2 (32.0◦C + 36.0◦C) =34.0◦C = (34.0 + 273)K = 307 K. The heat transferred from the first sample to the secondone is Q = cWm(Ti1 − Tf) = cWm(Tf − Ti2). The corresponding entropy changes for the twosamples are then approximately ∆S1 ≈ −Q/(T1)av and ∆S2 ≈ +Q/(T2)av , respectively. Notethat ∆S1 < 0 and ∆S2 > 0 as sample 1 loses heat to sample 2. The net entropy change of thesystem is then

∆S = ∆S1 + ∆S2 ≈ − Q

(T1)av

+Q

(T2)av

= cWm(Ti1 − Tf)[− 1

(T1)av

+1

(T2)av

]

= (4.186 kJ/kg·K)(20 kg) [(40.0 − 36.0)K](− 1

311 K+ − 1

307 K

)

= +14 J/K .

14.102

(a)

ec = 1 − TL

TH

= 1 − 300 K

500 K= 0.400 = 40.0% .

(b) Since ec = |Wo |/|QH |, |Wo | = ec |QH | = 0.400(1000 J) = 400 J.

(c) |QH | = 1000 J, meaning that 1000 J of heat flows out of the high-temperature reservoir percycle.

(d) The high-temperature reservoir loses |QH | = 1000 J of heat per cycle, so

∆SH =−|QH |

TH

=−1000 J

500 K= −2.00 J/K .

(e) Decrease, as heat is transferred out of the high-temperature source.

(f) Use |QH | − |QL | = |Wo | to find |QL | = |QH | − |Wo | = 1000 J − 400 J = 600 J.

(g) The low-temperature reservoir absorbs |QL | = 600 J of heat per cycle, so

∆SL =|QL |TL

=−600 J

300 K= +2.00 J/K .

(h) The entropy, like the internal energy, is a function of the state. Since the engine returnsto its initial state at the end of each cycle, its entropy also returns to its initial value, meaningthat ∆S = 0 for the engine per cycle.


(i) The entire system consists of the engine and the high- and low-temperature reservoirs. Forthe engine itself ∆S = 0, and for the two reservoirs ∆SH + ∆SL = −2.00 J/K + 2.00 J/K = 0as well [which is guaranteed by Eq. (14.14)]; so there is no net change in the entire system percycle.

14.103

(a) Since the process is isothermal ∆T = 0.

(b) Start with the Ideal Law: PiVi = nRT = PfVf , which gives PiVi = PfVf , or Pf/Pi =Vi/Vf = Vi/2Vi = 1

2 , i.e., Pf = 12Pi .

(c) For an ideal gas U ∝ T . Since T is constant so is U , i.e., ∆U = 0.

(d) The relationship is the First Law of Thermodynamics: ∆U = Q + W , which in our casereduces to Q = −W , since ∆U = 0.

(e) W = −nRT ln(Vf/Vi) = −nRT ln 2 (as Vf = 2Vi).

(f) Since Q = −W = nRT ln 2, by definition

∆S =Q

T=

nRT ln 2T

= nR ln 2 ≈ 0.693 nR .

14.104The work W done on n moles of ideal gas at temperature T as it expands isothermally fromVi to Vf is given by Eq. (14.5): W = −nRT ln(Vf/Vi). In an isothermal process ∆U = 0 (forideal gases), so the heat Q absorbed by the gas is Q = ∆U −W = −W . Thus from Eq. (14.17)the entropy change for the gas is

∆S =Q

T=

−W

T=

−[−nRT ln(Vf/Vi)]T

= nR lnVf

Vi

.

14.105According to Eq.(14.17) the heat Q absorbed by a system at a constant temperature T is relatedto its entropy change ∆S by ∆S = Q/T , or Q = T∆S. From Fig. P105 the temperature ofthe system is TH = 300 K as its entropy changes from 10 kJ/K to 40 kJ/K, so the heat intake is

QH = TH∆SH = (300 K)(30 kJ/K) = 9.0 × 103 kJ = 9.0 MJ .

Similarly, as the system decreases its entropy by the same amount (30 kJ/K) at TL = 100 K,the heat that passes through the system is

QL = TL∆SL = (100 K)(−30 kJ/K) = −3.0 × 103 kJ = −3.0 MJ ,

where the minus sign indicates that heat is removed from, rather than added to, the system.


The net work done by the system per cycle is |Wo | = |QH |−|QL | = 9.0 MJ−3.0 MJ = 6.0 MJ.

The efficiency of the engine follows from Eq. (14.10):

e =|Wo ||Qi |

=6.0 MJ

9.0 MJ= 0.67 = 67% .

14.106If the heat removed per cycle from the cold region (at temperature TL) is |QL |, then thecorresponding change in the entropy of that region is ∆SL = −|QL |/TL . The value of |QL | canbe obtained from the coefficient of performance of the Carnot refrigerator [see Eqs. (14.15) and(14.17)]: |QL | = ηc |Wi | = |Wi | [TL/(TH − TL)]. Thus

∆SL = −|QL |TL

= −|Wi | [TL/(TH − TL)]TL

= − |Wi |TH − TL

= − 2.00 J

(27 − 0)K= −0.074 J/K .

Here we noted that the numerical value of TH − TL is the same both in kelvin and Celsiusdegrees.

14.107For isothermal processes of an ideal gas ∆U = Q + W = 0, so Q = −W = −800 J. HereW > 0 because the 800 J of work is done on the gas. The temperature T of the gas remains atT = 22.0◦C = (22.0 + 273)K = 295 K, so its change in entropy is

∆S =Q

T=

−800 J

295 K= −2.71 J/K .

14.108By definition, the change in entropy ∆S = Sf −Si is a function of the initial state (i) and finalstate (f) of a system and is independent of the path which the system takes to reach the finalstate from the initial state. Thus we may take any path that connects i and f to find ∆S. Sincein a free expansion states i and f must have the same temperature we may choose a reversible,isothermal path that leads from i to f. The entropy change for such a path has already beencalculated in Problem (14.104) to be

∆S = nR lnVf

Vi

.

Since this result is path-independent ∆S should be the same for a free expansion from Vi to Vf .

Note: you might be tempted to use Eq. (14.17), ∆S = Q/T , which would give you ∆S = 0since Q = 0 in a free expansion. This is incorrect, however, since Eq. (14.17) is applicable onlyif the process which results in ∆S is reversible. A free-expansion is an irreversible process.


As an alternative, you may use Eq.(14.18) directly and note that the total number of microstatesavailable for an N -particle system with volume V satisfies W = CV N , with C a constantindependent of V ; so

Si = kB lnWi = kB ln(CVi)N = NkB ln(CVi) = n(NAkB) ln(CVi) = nR ln(CVi) .

Similarly, Sf = nR ln(CVf). It follows that

∆S = Sf − Si = nR ln(CVf) − nR ln(CVi) = nR lnCVf

CVi

= nR lnVf

Vi

.

14.109With the previous problem in mind, we plug n = 1.0 mol and Vf = 2Vi into the result for ∆Sobtained above to find

∆S = nR lnVf

Vi

= nR ln2Vi

Vi

= (1.0 mol)(8.314 4 J/mol·K) ln 2 = +5.8 J/K .

14.110The amount of the boulder’s mechanical energy that is lost during the tumble is mgh, wherem (= 200 kg) is its mass and h (= 200 m) is the vertical distance through which it falls.Immediately after the tumble the boulder will heat up a little bit, but eventually its temperaturedrops back to that of its environment, at T = 7.0◦C (or 280 K). This means that eventuallythe mechanical energy lost by the boulder is entirely absorbed by the Universe in terms of heat.The resulting change in entropy for the Universe is then

∆SUniverse =Q

T=

mgh

T=

(200 kg)(9.81 m/s2)(200 m)280 K

= +1.4 kJ/K .

Here we assumed that Eq.(14.17) is applicable. This is justified by imagining that the initial andfinal states of the Universe is connected by a reversible path (rather than the actual irreversibleone), and noticing that ∆S is path-independent.

· 484 CHAPTER 14 THERMODYNAMICS 14 Thermodynamics Answers to Discussion Questions •14.1 Air is...

Documents

Transcript of · 484 CHAPTER 14 THERMODYNAMICS 14 Thermodynamics Answers to Discussion Questions •14.1 Air is...