17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =Kmicro.Solve: The number of atoms is

NM

m= =

×= ×−

0

6 64 103 01 1027

23.0020 kg

kg..

Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is

m = = ×( ) = ×− −4 6 64 10 27 u 4 1.661 10 kg kg27 .

The average kinetic energy of each atom is

K mvavg avg2 kg 700 m s= = ×( )( )−1

212

27 26 64 10. = 1.63 × 10−21 J

Thus the thermal energy of the gas is

E K NKth micro avg J= = = ×( ) ×( )−3 01 10 1 63 1023 21. . = 490 J

17.2. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules.Solve: Oxygen atoms have an atomic mass number A = 16, so the mass of each molecule is

m = 32 u = 32(1.661 × 10−27 kg) = 5.32 × 10−26 kg

The number of molecules in the gas is

NM

m= =

× −

0 008026

. kg

5.32 10 kg = 1.505 × 1023

The thermal energy is

E NK N mvth avg avg2 = = ( )1

2 ⇒ = = ( )×( ) ×( )−v

E

Nmavgth

23

1700 J

1.505 10 kg

2 2

5 32 10 26.= 650 m/s

17.3. Model: The work done on a gas is the negative of the area under the pV curve.Visualize: Please refer to Figure Ex17.3. The gas is compressing, so we expect the work to be positive.Solve: The work done on the gas is

W pdV pV= − = −( )= − −( )( )( ) = ×( ) ×( ) =

∫−

area under the curve

200 cm 200 kPa m Pa 40 J3 3200 10 2 0 106 5.

Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environmentdoes positive work on the gas to compress it.

17.4. Model: The work done on a gas is the negative of the area under the pV curve.Visualize: Please refer to Figure Ex17.4. The gas is expanding, so we expect the work to be negative.Solve: The area under the pV curve is the area of the rectangle and triangle. We have

200 10 200 10 200 10 200 106 3 12

6 3×( ) ×( ) + ×( ) ×( )− − m Pa m Pa3 3 = 60 J

Thus, the work done on the gas is W = −60 J.Assess: The environment does negative work on the gas as it expands.

17.5. Visualize: Please refer to Figure Ex17.5.Solve: The work done on gas in an isobaric process is

W p V p V V= − = − −( )∆ f i

Substituting into this equation,

80 J Pa= − ×( ) −( )200 10 331 1V V ⇒ = × =−Vi

3 3 m cm2 0 10 2004.

Assess: The work done to compress a gas is positive.

17.6. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes.Solve: (a) Since the pressure (p i = pf = p) is constant the work done is

W p V p V VnRT

VV Von gas f i

i

if i

3 3

30.10 mol 8.31 J mol K 573 K cm cm

cm J

= − = − − = − −( )

= −( )( )( )−( )

=

∆ ( )

1000 2000

2000238

(b) For compression at a constant temperature,

W nRT V Von gas f i

3

30.10 mol 8.31 J mol K 573 K m

m330 J

= − ( )

= −( )( )( ) ××

=−

−

ln

ln1000 10

2000 10

6

6

(c) For the isobaric case,

pnRT

V= = ×i

i

Pa2 38 105.

For the isothermal case, pi Pa= ×2 38 105. and the final pressure is

pnRT

Vff

f

Pa= = ×4 76 105.

17.7. Visualize:

Solve: Because W pdV= −∫ and this is an isochoric process, W = 0 J. The final point is on a higher isotherm

than the initial point, so Tf > Ti. Heat energy is thus transferred into the gas (Q > 0) and the thermal energy of thegas increases (Eth f > Eth i) as the temperature increases.

17.8. Visualize:

Solve: Because this is an isobaric process W pdV p V V= − = − −( )∫ f i . Since Vf is smaller than Vi, W is positive.

That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, Tf < Ti. In otherwords, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be largerthan the work done.

17.9. Visualize:

Solve: Because the process is isothermal, ∆Eth = Eth f – Eth i = 0 J. According to the first law of thermodynamics,∆Eth = W + Q. This can only be satisfied if W = –Q. W is positive because the gas is compressing, hence Q isnegative. That is, heat energy is removed from the gas.

17.10. Visualize:

Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and theenvironment. That is, Q = 0 J. According to the first law of thermodynamics, the work done on a gas in an adiabaticprocess goes entirely to changing the thermal energy of the gas. The work W is positive because the gas iscompressed.

17.11. Visualize:

Solve: This is a case of gas compression and therefore W pdV= −∫ is a positive quantity. The final point is on a

lower isotherm than the initial point, so Tf < Ti. Heat energy is transferred out of the gas (Q < 0 J) and the thermalenergy of the gas decreases as the temperature falls. That is, Eth f < Eth i. To bring about this process: (1) The lockingpin is removed so the piston can slide up and down. (2) The masses on the top of the piston are not changed. Thiskeeps the pressure constant. (3) The ice block is brought into contact with the bottom of the cylinder.

17.12. Visualize:

Solve: For the isothermal process ∆T = 0 K. This means the first law of thermodynamics can only be satisfied ifW = −Q. When the gas compresses, W > 0 J implying Q < 0 J. Heat energy is transferred out of the gas, but thetemperature of the gas does not change. For the isochoric process, ′ =W 0 J because the volume does not change.Since the final point is on a lower isotherm, the final termperature is lower. That is, heat energy is transferred out ofthe gas (Q < 0) and hence the thermal energy of the gas decreases. To bring about this process: (1) Place thecylinder on the ice. The gas will begin to contract as heat energy is transferred to the ice. (2) Add masses on the topof the piston to increase the pressure and keep pV constant as the volume decreases. (3) At the desired volume andpressure, insert a locking pin into the piston to fix the volume. (4) Keep the ice in contact with the bottom of thecylinder until the original pressure is obtained in the gas.

17.13. Solve: The first law of thermodynamics is

∆Eth = W + Q ⇒ −200 J = 500 J + Q ⇒ Q = −700 J

The negative sign means a transfer of energy from the system to the environment.Assess: Because W > 0 means a transfer of energy into the system, Q must be less than zero and larger inmagnitude than W so that Eth f < Eth i.

17.14. Solve: This is an isobaric process. W > 0 because the gas is compressed. This transfers energy into thesystem. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics is

∆ ∆E W Q p Vth = + = − + Q = –(4.0 × 105 Pa)(200 – 600) × 10−6 m3 – 100 J = 60 J

Thermal energy increases by 60 J.

17.15. Model: The removal of heat from the ice reduces its thermal energy and its temperature.Solve: The heat needed to change an object’s temperature is Q = Mc∆T. The mass of the ice cube is

M = ρiceV = (920 kg/m3)(0.06 × 0.06 × 0.06) m3 = 0.199 kg

The specific heat of ice from Table 17.2 is cice = 2090 J/kg K, so

Q = (0.199 kg)(2090 J/kg K)(243 K – 273 K) = –12,500 J

Thus, the energy removed from the ice block is 12,500 J.Assess: The negative sign with Q means loss of energy.

17.16. Model: The spinning paddle wheel does work and changes the water’s thermal energy and itstemperature.Solve: (a) The temperature change is ∆T = Tf – Ti = 25°C − 21°C = 4 K. The mass of the water is

M = (200 × 10−6 m3)(1000 kg/m3) = 0.20 kg

The work done is

W = ∆Eth = Mcwater∆T = (0.20 kg)(4190 J/kg K)(4 K) = 3350 J

(b) Q = 0. No energy is transferred between the system and the environment because of a difference in temperature.

17.17. Model: Heating the mercury at its boiling point changes its thermal energy without a change intemperature.Solve: The mass of the mercury is M = 20 g = 2.0 × 10−2 kg, the specific heat cmercury = 140 J/kg K, the boilingpoint Tb = 357°C, and the heat of vaporization LV = 2.96 × 105 J/kg. The heat required for the mercury to change tothe vapor phase is the sum of two steps. The first step is

Q1 = Mcmercury∆T = (2.0 × 10−2 kg)(140 J/kg K)(357°C − 20°C) = 944 J

The second step isQ2 = MLV = (2.0 × 10−2 kg)(2.96 × 105 J/kg) = 5920 J

The total heat needed is 6864 J.

17.18. Model: Heating the mercury changes its thermal energy and its temperature.Solve: (a) The heat needed to change the mercury’s temperature is

Q = McHg∆T ⇒ = = ( )( ) = = °∆TQ

McHg

100 J

0.020 kg 140 J kg K35.7 K 35.7 C

(b) The amount of heat required to raise the temperature of the same amount of water by the same number ofdegrees is

Q = Mcwater∆T = (0.020 kg)(4190 J/kg K)(35.7 K) = 2990 JAssess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat formercury. This explains why Qwater > Qmercury.

17.19. Model: Changing ethyl alcohol at 20°C to solid ethyl alcohol at its melting point requires two steps:lowering its temperature from 20°C to −114°C, then changing the ethyl alcohol to its solid phase at −114°C.Solve: The change in temperature is −114°C − 20°C = −134°C = −134 K. The mass is

M V= = ( ) ×( ) =−ρ 789 200 10 0 15786 kg / m m kg3 3 .

The heat needed for the two steps is

Q1 = Mcalcohol∆T = (0.1578 kg)(2400 J/kg K)(−134 K)= −5.075 × 104 J

Q2 = −MLf = − (0.1578 kg)(1.09 × 105 J/kg) = −1.720 × 104 J

The total heat required isQ = Q1 + Q2 = −6.79 × 104 J

Thus, the minimum amount of energy that must be removed is 6.79 × 104 J.Assess: The negative sign with Q indicates that 6.79 × 104 J will be removed from the system.

17.20. Model: Changing solid lead at 20°C to liquid lead at its melting point (Tm = 328°C) requires two steps:raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm.Solve: The equation for the total heat is

Q = Q1 + Q2 ⇒ 1000 J = Mclead(Tf – Ti) + MLf

⇒1000 J = M(128 J/kg K)(328 – 20) K + M(0.25 × 105 J/kg) ⇒ = ( ) =M1000 J

64,424 J kg15.5 g

The maximum mass of lead you can melt with 1000 J of heat is 15.5 g.

17.21. Model: We have a thermal interaction between the copper pellets and the water.Solve: The conservation of energy equation Qc + Ww = 0 is

M c T M c Tc c f w w fC) C) J( (− ° + − ° =300 20 0

Solving this equation for the final temperature Tf gives

TM c M c

M c M cfc c w w

c c w w

C) C)

kg)(385 J / kg K)(300 C) (0.10 kg)(4190 J / kg K)(20 C)

kg)(385 J / kg K) (0.10 kg)(4190 J / kg K) C

= ° + °+

= ° + °+

= °

( (

( .

( ..

300 20

0 030

0 03027 5

The final temperature of the water and the copper is 27.5°C.

17.22. Model: We have a thermal interaction between the copper block and water.Solve: The conservation of energy equation Qcopper + Qwater = 0 J is

Mcopperccopper (Tf – Ti copper) + Mwatercwater(Tf – Ti water) = 0 J

Both the copper and the water reach the common final temperature Tf = 25.5°C. Thus

Mcopper(385 J/kg K)(25.5°C – 300°C) + (1.0 × 10−3 m3)(1000 kg/m3)(4190 J/kg K)(25.5°C – 20°C) = 0 J

⇒ =M copper 0.218 kg

17.23. Model: We have a thermal interaction between the thermometer and the water.Solve: The conservation of energy equation Qthermo + Qwater = 0 J is

Mthermocthermo(Tf – Ti thermo) + Mwatercwater(Tf – Ti water) = 0 J

The thermometer slightly cools the water until both have the same final temperature Tf = 71.2°C. Thus

( . kg)( J / kg K)( . C . C) ( m )( kg / m )( J / kg K)( . C )

J (J / K)( . C ) J

3 3i water

i water

0 050 750 71 2 20 0 200 10 1000 4190 71 2

1920 838 71 2 0

6° − ° + × ° −= + ° − =

− T

T⇒ = °Ti water 73.5 C

Assess: The thermometer reads 71.2°C for a real temperature of 73.5°C. This is reasonable.

17.24. Model: We have a thermal interaction between the aluminum pan and the water.Solve: The conservation of energy equation QAl + Qwater = 0 J is

MAl cAl(Tf – Ti Al) + Mwater cwater(Tf – Ti water)The pan and water reach a common final temperature Tf = 24.0°C

(0.750 kg)(900 J / kg K)(24.0 C ) (10 10 m )(1000 kg / m )(4190 J / kg K)(24.0 C 20.0 C )

(675.0 J / K)(24.0 C ) 167,600 J 0 Ji Al

3 3 3

i Al

° − + × ° − °= ° − + =

−T

T

⇒ = °Ti Al 272 C = [(272) (9/5) + 32]°F = 522°F

17.25. Model: We have a thermal interaction between the metal sphere and the mercury.Solve: The conservation of energy equation Qmetal + QHg = 0 J is

Mmetalcmetal(Tf – Ti metal) + MHgcHg(Tf – Ti Hg) = 0 J

The metal and mercury reach a common final temperature Tf = 99.0°C. Thus

(0.500 kg)cmetal(99°C – 300°C) + (300 × 10−6 m3)(13,600 kg/m3)(140 J/kg K)(99°C – 20°C) = 0 J

We find that cmetal 449 J kg K= . The metal is iron.

17.26. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on theKelvin scale is the same as the change in temperature on the Celsius scale.Solve: (a) The atomic mass number of argon is 40. That is, Mmol = 40 g/mol. The number of moles of argon gas inthe container is

nM

M= = =

mol

1.0 g

40 g mol0.025 mol

The amount of heat is

Q = nCV∆T = (0.025 mol)(12.5 J/mol K)(100°C) = 31.25 J

(b) For the isobaric process Q = nCP∆T becomes

31.25 J = (0.025 mol)(20.8 J/mol K)∆T ⇒ = °∆T 60 C

17.27. Model: The heating processes are isobaric and isochoric. O2 is a diatomic ideal gas.Solve: (a) The number of moles of oxygen is

nM

M= = =

mol

1.0 g

32 g mol0.03125 mol

For the isobaric process,

Q = nCP∆T = (0.03125 mol)(29.2 J/mol K)(100°C) = 91.2 J

(b) For the isochoric process,

Q = nCV∆T = 91.2 J = (0.03125 mol)(20.9 J/mol K)∆T ⇒ = °∆T 140 C

17.28. Model: The heating is an isochoric process.Solve: The number of moles of helium is

nM

M= = =

mol

2.0 g

4 g mol0.50 mol

For the isochoric processes,

Q nC T THe V 0.50 mol 12.5 J mol K= = ( )( )∆ ∆ Q nC TM

TO V2 32 g mol20.9 J mol K= =

( )∆ ∆

Because Q QHe O2= ,

0.50 mol 12.5 J mol K32 g mol

20.9 J mol K( )( ) =

( )M ⇒ =M 9.57 g

17.29. Model: The O2 gas has γ = 1.40 and is an ideal gas.Solve: (a) For an adiabatic process, pVγ remains a constant. That is,

p V p Vi i fγ γ= f ⇒ =

= ( )

p pV

V

V

Vf ii

f

i

i

3.0 atm

γ

2

1 40.

= ( ) =3.0 atm 1.14 atm

1

2

1 40.

(b) Using the ideal-gas law, the final temperature of the gas is calculated as follows:

p V

T

p V

Ti i

i

f f

f

= ⇒ = = ( )

=T Tp

p

V

V

V

Vf if

i

f

i

i

i

423 K1.14 atm

3.0 atm321.5 K

2= 48.5°C

17.30. Model: We assume the gas is an ideal gas and γ = 1.40 for a diatomic gas.Solve: Using the ideal-gas law,

VnRT

pii

i

30.10 mol 8.31 J mol K 423 K

Pa m= = ( )( )( )

× ×( ) = × −

3 1 013 101 157 10

53

..

For an adiabatic process,

p V p Vi i f fγ γ=

⇒ =

= ×( )

= ×− −V Vp

p

p

pf ii

f

3 i

i

3 m

m

1

3

1 1 40

31 157 100 5

1 90 10

γ

..

.

.

To find the final temperature, we use the ideal-gas law once again as follows:

T Tp

p

V

V

p

pf if

i

f

i

i

i

3

3 3423 K m

1.157 10 m= = ( )

××

−

−

0 5 1 90 10 3. . = 346.9 K = 73.9°C

17.31. Model: γ is 1.40 for a diatomic gas and 1.67 for a monoatomic gas.Solve: (a) We will assume that air is a diatomic gas. For an adiabatic process,

T V T Vf f i iγ γ− −=1 1

Thus

V

V

T

Ti

f

f

i

1123 K

303 K

=

=

=

− −

1

11

1 40 126 4

γ ..

(b) For argon, a monatomic gas,

V

Vi

f

1123 K

303 K

=

=

−1

1 67 17 07

..

17.32. Model: Changing steam to ice requires four steps: removing heat to convert steam into water at 100°C,removing heat to lower the water temperature from 100°C to 0°C, converting water at 0°C to ice at 0°C, andlowering the temperature of ice to −20°C.Solve: Number of moles of steam is

npV

RT= =

× ×( ) ×( )( )( )

=−2 1 013 10 1530 10

0 105 6.

. Pa m

8.31 J mol K 373 K mol

3

The mass of steam is

M nM= = ( )( ) = =mol 0.10 mol 18 g mol 1.8 g 0.0018 kg

The heat needed for each step is

Q ML

Q Mc T

Q ML

Q Mc T

15

2

35

4

22 6 10 4068

0 754 2

3 33 10 599 4

20 0

= − = −( ) ×( ) = −

= = ( )( ) ° − °( ) = −

= − = −( ) ×( ) = −

= = ( )( ) − ° − °(

V

water water

f

ice ice

0.0018 kg J kg J

0.0018 kg 4190 J kg K C 100 C J

0.0018 kg J kg J

0.0018 kg 2090 J kg K C C

.

.

. .

∆

∆ )) = −75 24. J

The total heat required in this process is

Q Q Q Q Q= + + + = −1 2 3 4 5497 J

Assess: Approximately 75% of the heat is removed from steam at 100°C to make water at 100°C. This isconsistent with our experience that it takes much longer for a pan of water to boil away than it does to reachboiling.

17.33. Solve: The area of the garden pond is A = ( ) =π 2.5 m 19.635 m22 and its volume is V A= ( ) =0.30 m5.891 m3. The mass of water in the pond is

M V= = ( )( ) =ρ 1000 kg m 19.635 m 5 kg3 3 891

The water absorbs all the solar power which is

400 W m 19.635 m 7854 W2 2( )( ) =

This power is used to raise the temperature of the water. That is,

Q t Mc T= ( ) = = ( )( )( )7854 W 891 kg 4190 J kg K 10 Kwater∆ ∆ 5 ⇒ = =∆t 3 ,425 s 8.73 hr1

17.34. Model: The potential energy of the bowling ball is transferred into the thermal energy of the mixture.We assume the starting temperature of the bowling ball to be 0°C.Solve: The potential energy of the bowling ball is

U M gh h hg ball2 211 kg 9.8 m s 107.8 kg m s= = ( )( ) = ( )

This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is,

107.8 kg m s2th w f( ) = =h E M L∆ ⇒ =

( ) ×( )( ) =h

0.005 kg J kg

107.8 kg m s15.45 m

2

3 33 105.

17.35. Model: Heating the water and the kettle raises the temperature of the water to the boiling point andraises the temperature of the kettle to 100°C.Solve: The amount of heat energy from the electric stove’s output in 3 minutes is

Q = ( ) ×( ) = ×2000 J s 3 60 s J3 6 105.

This heat energy heats the kettle and brings the water to a boil. Thus,

Q M c T M c T= +water water kettle kettle∆ ∆

Substituting the given values into this equation,

3 6 10 100 20 0 750 100 205. .× = ( ) ° − °( ) + ( )( ) ° − °( ) J 4190 J kg K C C kg 449 J kg K C CwaterM

⇒ =Mwater 0 kg.994

The volume of water in the kettle is

VM= = = × =−water

water3

3 30.994 kg

kg m m 9 cm

ρ 10000 994 10 943.

Assess: 1 L = 103 cm3, so V ≈ 1 L. This is a reasonable volume of water.

17.36. Model: Each car’s kinetic energy is transformed into thermal energy.Solve: For each car,

K Mv E Mc T= = =12

2 ∆ ∆th car ⇒ =∆Tv

c

2

2 car

Assume ccar = ciron. The speed of the car is

v = × =80 km hr =80 1000 m

3600 s22.22 m s ⇒ = ( )

( ) = °∆T22.22 m s

449 J kg K0.55 C

2

2

17.37. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol.Solve: The aluminum, copper, and alcohol form a closed system, so Q = QAl + QCu + Qeth = 0 J. The mass of thealcohol is

M Veth3 3790 kg m m kg= = ( ) ×( ) =−ρ 50 10 0 03956 .

Expressed in terms of specific heats and using the fact that ∆T = Tf – Ti, the Q = 0 J condition is

M c T M c T M c TAl Al Al Cu Cu Cu eth eth eth∆ ∆ ∆+ + = 0 J

Substituting into this expression,

0.010 kg 900 J kg K 298 K 473 K 0.020 kg 385 J kg K 298 K

.0395 kg 2400 J kg K 298 K 288 K 1575 J 7.7 J K J 0 J

( )( ) −( ) + ( )( ) −( )+ ( )( ) −( ) = − + ( ) −( ) + =

T

T0 298 948

⇒ = = − °T 216.6 K C56 4.

17.38. Model: There are two interacting systems: aluminum and ice. The system comes to thermal equilibriumin four steps: (1) the ice temperature increases from −10°C to 0°C, (2) the ice becomes water at 0°C, (3) the watertemperature increases from 0°C to 20°C, and (4) the cup temperature decreases from 70°C to 20°C.Solve: The aluminum and ice form a closed system, so Q = Q1 + Q2 + Q3 + Q4 = 0 J. These quantities are

Q M c T

Q M L

Q M c T

Q M c T M M

1

25

3

4

3 33 10

= = ( )( )( ) =

= = ( ) ×( ) =

= = ( )( )( ) =

= = ( ) −( ) = −( )

ice ice

ice f

ice water

Al Al Al Al

0.100 kg 2090 J kg K 10 K 2090 J

0.100 kg J kg 33,300 J

0.100 kg 4190 J kg K 20 K 8380 J

900 J kg K 50 K 45,000 J kg

∆

∆

∆

.

The Q = 0 J equation now becomes

43,770 J – (45,000 J/kg)MAl = 0 JThe solution to this is MAl = 0.973 kg.

17.39. Model: There are three interacting systems: metal, aluminum, and water.Solve: The metal, aluminum container, and water form a closed system, so Qm + QAl + Qw = 0 J, where Qm is theheat transferred to the metal sample. This equation can be written:

Mmcm∆Tm + MAlcAl∆TAl + Mwcw∆T = 0 J

Substituting in the given values,

0.512 kg 351 K K 0.100 kg 900 J kg K 351 K 371 K

0.325 kg 4190 J kg K 351 K 371 K J 900 J kg K

m

m

( ) −( ) + ( )( ) −( )+ ( )( ) −( ) = ⇒ =

c

c

288

0

From Table 17.2, we see that this is the specific heat of aluminum.

17.40. Solve: For a monatomic gas, the molar specific heat at constant volume is CV 12.5 J mol K= . FromEquation 17.22,

12 5 625. J mol K1000 g kg

J kg Kmol= M ⇒ =Mmol 20 g mol

The gas is therefore neon.

17.41. Model: Heating the water raises its thermal energy and its temperature.Solve: A 5.0 kW heater has power P = 5000 W. That is, it supplies heat energy at the rate 5000 J/s. The heatsupplied in time ∆t is Q = 5000∆t J. The temperature increase is ∆TC = (5/9)∆TF = (5/9)(75°) = 41.67°C. Thus

Q t Mc T t= ∆ = ∆ = ° ⇒ ∆ =5000 150 5283 J kg)(4190 J / kg K)(41.67 C) s = 87.3 minw (

Assess: A time of ≈1.5 hours to heat 40 gallons of water is reasonable.

17.42. Model: Heating the material increases its thermal energy.Visualize: Please refer to Figure P17.42. Heat raises the temperature of the substance from –40°C to –20°C, atwhich temperature a solid to liquid phase change occurs. From –20°C, heat raises the liquid’s temperature up to40°C. Boiling occurs at 40°C where all of the liquid is converted into the vapor phase.Solve: (a) In the solid phase,

∆ ∆Q Mc T= ⇒ =

=

=cQ

T M

∆∆

1 120 kJ

20 K 0.50 kg2000 J kg K

(b) In the liquid phase,

cM

Q

T=

=

=1 1∆

∆ 0.50 kg

80 kJ

60 K2667 J kg K

(c) The melting point Tm C = − °20 and the boiling point Tb C= + °40 .(d) The heat of fusion is

LQ

Mf

20,000 J

0.50 kg J kg= = = ×4 0 104.

The heat of vaporization is

LQ

Mv

60,000 J

0.50 kg J kg= = = ×1 2 105.

17.43. Model: Heating the material increases its thermal energy.Visualize: Please refer to Figure P17.43. The material melts at 300°C and undergoes a solid-liquid phase change.The material’s temperature increases from 300°C to 1500°C. Boiling occurs at 1500°C and the material undergoesa liquid-gas phase change.Solve: (a) In the liquid phase, the specific heat of the liquid can be obtained as follows:

∆ ∆Q Mc T= ⇒ =cM

Q

T

1 ∆∆

=

=1

83 30.200 kg

20 kJ

1200 K J kg K.

(b) The latent heat of vaporization is

LQ

Mv

40 kJ

0.200 kg J kg= = ( ) = ×2 0 105.

17.44. Model: The liquefaction of the nitrogen occurs in two steps: lowering nitrogen’s temperature from 20°Cto −196°C, and then liquefying it at −196°C. Assume the cooling occurs at a constant pressure of 1 atm.Solve: The mass of 1.0 L of liquid nitrogen is M V= = ( )( ) =−ρ 810 kg m m 0.810 kg3 310 3 . This mass corres-

ponds to

nM

M= = =

mol

810 g

28 g mol mols28 9.

At constant atmospheric pressure, the heat to be removed from 28.93 mols of nitrogen is

Q ML nC T= +

= −( ) ×( ) + ( )( ) −( ) = − ×v P

0.810 kg J kg 28.9 mols 29.1 J mol K 77 K K J

∆

1 99 10 293 3 43 105 5. .

17.45. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee temperaturefrom 90°C to 60°C requires four steps: (1) raise the temperature of ice from −20°C to 0°C, (2) change ice at 0°C towater at 0°C, (c) raise the water temperature from 0°C to 60°C, and (4) lower the coffee temperature from 90°C to60°C.Solve: For the closed coffee-ice system,

Q Q Q Q Q Q Q= + = + +( ) + ( ) =ice coffee J1 2 3 4 0

Q M c T M M1 = = ( )( ) = ( )ice ice ice ice2090 J kg K 20 K 41,800 J kg∆

Q M L M2 = = ( )ice f ice 330,000 J kg

Q M c T M M3 = = ( )( ) = ( )ice water ice ice4190 J kg K 60 K 251,400 J kg∆

Q M c T46300 10 37 000= = ×( )( )( ) −( ) = −−

coffee coffee3 3 m 1000 kg m 4190 J kg K 30 K J∆ ,

The Q = 0 J equation thus becomes

Mice J kg J 0 J41 800 330 000 251 400 37 710, , , ,+ +( ) − = ⇒ = =Mice 0.0605 kg 60.5 g

Visualize: 60.5 g is the mass of approximately 1 ice cube.

17.46. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by thenuclear reactor is used to raise the water temperature.Solve: For the closed reactor-water system, energy conservation per second requires

Q Q Q= + =reactor water 0 J

The heat from the reactor in ∆t = 1 s is Qreactor 2000 MJ J= − = − ×2 0 109. and the heat absorbed by the water is

Q m c T mwater water water water 4190 J kg K 12 K= = ( )( )∆⇒ − × + ( )( ) =2 0 10 4190 09. J J kg K 12 K Jwaterm ⇒ = ×mwater kg3 98 104.

Each second, 3.98 × 104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow perminute is

3 98 1060 14

3

. × × × × − kg

s

s

min

m

1000 kg

1 L

10 m3 3 = 2.39 × 106 L/min

17.47. Model: We have three interacting systems: the aluminum, the air, and the firecracker. The energyreleased by the firecracker raises the temperature of the aluminum and the air. We will assume that air is basical-ly N2.Solve: For the closed firecracker + air + aluminum system, energy conservation requires that

Q = Qfirecracker + QAl + Qair = 0 J QAl = mAlcAl∆T = (2.0 kg)(900 J/kg K)(3 K) = 5400 J

Q nC TpV

RTC Tair V V= =

∆ ∆

=×( ) ×( )

( )( ) ( )( )−1 01 10 20 10

29820 8 3

5 3 3.

.

Pa m

8.31 J / mol K K J / mol K K

= 51 J

Thus,

Qfirecracker = −QAl − Qair

= −5450 J

That is, 5450 J of energy are released on explosion of the firecracker.Assess: The negative sign with Qfirecracker means that the firecracker has lost energy.

17.48. Model: We have two interacting systems: the water and the gas. For the closed system comprised ofwater and gas to come to equilibrium, heat is transferred from one interacting system to the other.Solve: Energy conservation requires that

Qair + Qwater = 0 J ⇒ ngasCV(Tf – Ti gas) + mwaterc(Tf – Ti water) = 0 JUsing the ideal-gas law,

Tp V

n Ri gasgas gas

gas

5 310 1.013 10 Pa 10 m

mol J mol K= 1219 K= =

× ×( ) ×( )( )( )

−4000

0 40 8 31

6

. .

The energy conservation equation with Ti water = 293 K becomes

0.40 mol 12.5 J mol K K kg 4190 J kg K K J( )( ) −( ) + ×( )( ) −( ) =−T T1219 20 10 293 03 ⇒ =Tf 3 K45

We can now use the ideal-gas equation to find the final gas pressure.p V

T

p V

Ti i

i

f f

f

= ⇒ =p pT

Tf if

i

=

( ) =345 K

1219 K10 atm 2.83 atm

17.49. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, Eth =Kmicro. Also, the work W done on an expanding gas is negative.

Solve: (a) The thermal energy of N molecules is Eth = NKavg, where K m vavg avg= ( )12

2 is the average kinetic

energy per molecule. The mass of a hydrogen molecule is

m = = × ×( ) = ×− −2 3 32 10 27 u 2 1.661 10 kg kg27 .

The number of molecules in a 1 g = 0.001 kg sample is

NM

m= =

×= ×−

0.001 kg

kg3 32 103 01 1027

23

..

The average kinetic energy per molecule is

K m vavg avg kg 700 m s J= ( ) = ×( )( ) = ×− −12

212

27 2 223 32 10 8 13 10. .

Thus, the thermal energy in the 1-gram sample of the gas is

E NKth avg J 245 J= = ×( ) ×( ) =−3 01 10 8 13 1023 22. .

(b) The first law of thermodynamics tells us the change of thermal energy when work is done and heat is added:∆E W Qth 300 J 500 J 200 J= + = − + =

Here W is negative because energy is transferred from the system to the environment. The work and heat raise thethermal energy of the gas by 200 J to Eth = 445 J. Now the average kinetic energy is

K E N m vavg th12 avg J= = × = ( )−1 487 10 21 2

.

Solving for the new average speed gives

vK

mavgavg J

kgm s= =

×( )×

=−

−

2 2 1 478 10

3 32 10944

21

27

.

.

17.50. Model: These are isothermal and isobaric ideal-gas processes.Solve: (a) The work done at constant temperature is

W pdVnRT

VdV nRT V V nRT V V

V

V

V

V= − = − = − −( ) = − ( )= −( )( )( ) ( ) =

∫ ∫i

f

i

f

f i f i

2.0 mol 8.31 J mol K 303 K J

ln ln ln

ln 13 5530

(b) The work done at constant pressure is

W pdV p V V pV

V pV

nRT

V

V= − = − −( ) = − −

=

= = ( )( )( ) =

∫i

f

f ii

i i

2.0 mol 8.31 J mol K 303 K 3360 J

3

2

3

2

3

2

3

(c) For an isothermal process in which V Vf i= 13 , the pressure changes to p pf i= =3 4 5. atm.

17.51. Model: This is an isothermal process. The work done is positive for a compression.Solve: For an isothermal process,

W nRT V V= − ( )ln f i

For the first process,

W nRT= = − ( )500 J ln 12 ⇒ =nRT 721.35 J

For the second process,

W nRT W= − ( ) ⇒ = −( ) ( )ln ln110

110721.35 J = 1660 J

17.52. Visualize:

Solve: (a) The gas exerts a force on the piston of magnitude

F p Agas on piston gas= = ×( ) ( )[ ]3 atm 101,300 Pa atm 0.080 mπ 2 = 6110 N

This force is directed toward the right.(b) The piston is in static equilibrium, so the environment must exert a force on the piston of equal magnitudeFenviron on piston 6110 N= but in the opposite direction, toward the left.

(c) The work done by the environment is

W F r F xenviron environ on piston environ on piston 6110 N 0.10 m J= ⋅ = − = −( )( ) = −r r r

∆ ∆ 611

The work is negative because the force and the displacement are in opposite directions.(d) The work done by the gas is

W F r F xgas gas on piston gas on piston 6110 N 0.10 m 611 J= ⋅ = + = ( )( ) =r r

∆ ∆

This work is positive because the force and the displacement are in the same direction.(e) The first law of thermodynamics is W + Q = ∆Eth where W is Wenviron. So here W = −611 J and we find

Q E W= − = ( ) − −( ) =∆ th 196 J 611 J 807 J

Thus, 807 J of heat is added to the gas.

17.53. Model: This is an isobaric process.Visualize:

Solve: (a) The initial conditions are p1 = 10 atm = 1.013 × 106 Pa, T1 = 50°C = 323 K, V1 = πr2L1 = π(0.050 m)2

(0.20 m) = 1.57 × 10−3 m3. The gas is heated at a constant pressure, so heat and temperature change are related byQ = nCP∆T. From the ideal gas law, the number of moles of gas is

np V

RT= =

×( ) ×( )( )( )

=−

1 1

1

6 31 013 10 1 57 10. . Pa m

8.31 J mol K 323 K0.593 mol

3

The temperature change due to the addition of Q = 2500 J of heat is thus

∆TQ

nC= = ( )( ) =

P

2500 J

0.593 mol 20.8 J mol K203 K

The final temperature is T2 = T1 + ∆T = 526 K = 253°C.(b) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal gas relationship for anisobaric process is

V

T

V

T

L

T

L

TL

T

TL2

2

1

1

2

2

1

12

2

11 20= ⇒ = ⇒ = = ( ) =526 K

323 K cm 32.6 cm

17.54. Model: The process in part (a) is isochoric and the process in part (b) is isobaric.Solve: (a) Initially V1 = (0.20 m)3 = 0.0080 m3 = 8.0 L and T1 = 293 K. Helium has an atomic mass number A = 4,so 3 g of helium is n = M/Mmol = 0.75 mole of helium. We can find the initial pressure from the ideal-gas law:

pnRT

V11

1

= =( )( )( )

= =0.75 mol 8.31 J mol K 293 K

0.0080 m228 kPa 2.25 atm3

Heating the gas will raise its temperature. A constant volume process has Q = nCV∆T, so

∆TQ

nC= = ( )( ) =

V

1000 J

0.75 mol 12.5 J mol K107 K

This raises the final temperature to T2 = T1 + ∆T = 400 K. Because the process is isochoric,p

T

p

Tp

T

Tp2

2

1

12

2

11 2 25= ⇒ = = ( ) =400 K

293 K atm 3.07 atm.

(b) The initial conditions are the same as part a, but now Q = nCP∆T. Thus,

∆TQ

nC= = ( )( ) =

P

1000 J

0.75 mol 20.8 J mol K64.1 K

Now the final temperature is T2 = T1 + ∆T = 357 K. Because the process is isobaric,V

T

V

TV

T

TV2

2

1

12

2

11= ⇒ = = ( ) = =357 K

293 K0.0080 m 0.00975 m 9.75 L3 3

(c)

17.55. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules.Solve: At T = 0 K, atoms of the gas have no thermal energy ((Eth)i = 0) and no velocity. Consequently, thepressure due to the atoms of the gas is zero. If we start the gas in a volume of

V VnRT

pi ff

f5

32.0 mol 8.31 J mol K 310 K

1.013 10 Pa0.05086 m= = =

( )( )( )×( ) =

and increase the temperature from Ti = 0 K to Tf = 310 K at constant volume, then the pressure will rise with T asheat Q is added. The amount of heat added at constant volume is

Q nC T= = ( )( )( ) =V 2.0 mol 12.5 J mol K 310 K J∆ 7750

No work is done, so the first law is ∆Eth = Q + W = Q = 7750 J. The change in thermal energy is ∆Eth =( ) ( ) .E Eth f th i− But (Eth)i = 0, so the thermal energy of the gas is (Eth)f = 7750 J.

17.56. Model: This is an isothermal process.Solve: (a) The final temperature is T2 = T1 because the process is isothermal.(b) The work done on the gas is

W pdVnRT

VdV nRT

V

VV

V

V

V

= − = − = −∫ ∫1

2

1

2

11

2

1

ln = −nRT1ln 2

(c) From the first law of thermodynamics ∆Eth = W + Q = 0 J because ∆T = 0 K. Thus, the heat energy transferredto the gas is Q W nRT= − = 1 2ln .

17.57. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process.Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa and T1 = 293 K. Nitrogen has a molar mass Mmol =28 g/mol, so 5 g of nitrogen gas has n = M/Mmol = 0.1786 mol. From this, we can find the initial volume:

VnRT

p11

1

31 430 10= =( )( )( )

= × =−0.1786 mol 8.31 J mol K 293 K

304,000 Pa m 1430 cm3 3.

The volume triples, so V2 = 3V1 = 4290 cm3. The expansion is isobaric (p2 = p1 = 3.0 atm), so

V

T

V

TT

V

VT2

2

1

12

2

11= ⇒ = = ( ) = °3 293 K 879 K = 606 C

(b) The process is isobaric, so

Q nC T= = ( )( ) −( ) =P 0.1786 mol 29.1 J mol K 879 K 293 K 3050 J∆

(c) The pressure is decreased at constant volume (V3 = V2 = 4290 cm3) until the original temperature is reached(T3 = T1 = 293 K). For an isochoric process,

p

T

p

Tp

T

Tp3

3

2

23

3

22 3 0= ⇒ = = ( ) =293 K

879 K atm 1.0 atm.

(d) The process is isochoric, so

Q nC T= = ( )( ) −( ) = −V 0.1786 mol 20.8 J mol K 293 K 879 K J∆ 2180

So, 2180 J of heat was removed to decrease the pressure.(e)

17.58. Model: The gas is an ideal gas.Visualize: Please refer to Figure P17.58. Call the upper right corner (on process A) 2 and the lower left corner(on process B) 3.Solve: The change in thermal energy is the same for any gas process that has the same ∆T. Processes A and Bhave the same ∆T, since they start and end at the same points, so (∆Eth)A = (∆Eth)B. The first law is then

(∆Eth)A = QA + WA = (∆Eth)B = QB + WB ⇒ QA – QB = WB – WA

In process B, work W = –p∆V = –pi(2Vi – Vi) = –piVi is done during the isobaric process i → 3. No work is doneduring the isochoric process 3 → f. Thus WB = –piVi. Similarly, no work is done during the isochoric process i → 2of process A, but W = –p∆V = –2pi(2Vi – Vi) = –2piVi is done during the isobaric process 2→ f. Thus WA = –2piVi.Combining these,

QA – QB = WB – WA = –piVi – (–2piVi) = piVi

17.59. Model: The two processes are isochoric and isobaric.Visualize: Please refer to Figure P17.59.Solve: Process A is isochoric which means

T T p p T T p p T Tf i f i f i f i i i1 atm 3 atm= ⇒ = ( ) = ( ) = 13

From the ideal-gas equation,

Tp V

nRii i= =

× ×( ) ×( )( )( ) =

−3 1 013 10 2000 105 6. Pa m

0.10 mol 8.31 J mol K731.4 K

3

⇒ = =T Tf i 243.8 K13

⇒ − = −T Tf i 487.6 K

Thus, the heat required for process A isQ nC TA V 0.10 mol 20.8 J mol K 487.6 K J= = ( )( ) −( ) = −∆ 1010

Process B is isobaric which means

T V T V T T V V T Tf f i i f i f i i3 3

i3000 cm 1000 cm= ⇒ = ( ) = ( ) = 3

From the ideal-gas equation,

Tp V

nRii i= =

× ×( ) ×( )( )( ) =

−2 1 013 10 1000 10

8 31

5 6.

.

Pa m

0.10 mol J mol K243.8 K

3

⇒ = =T Tf i 731.4 K3 ⇒ − =T Tf i 487.6 K

Thus, heat required for process B isQ nC TB P 0.10 mol 29.1 J mol K 487.6 K= = ( )( )( )∆ = 1419 J

Assess: Heat is transferred out of the gas in process A, but transferred into the gas in process B.

17.60. Model: We have an adiabatic and an isothermal process.Visualize: Please refer to P17.60.Solve: For the adiabatic process, no heat is added or removed. That is Q = 0 J.

Isothermal processes occur at a fixed temperature, so ∆T = 0 K. Thus ∆Eth = 0 J, and the first law ofthermodynamics gives

Q W= − = nRT V Vln f i( )The temperature T can be obtained from the ideal gas equation as follows:

p V nRT Tp V

nRi ii i

3 Pa m

0.10 mol 8.31 J mol K= ⇒ = =

×( ) ×( )( )( )

−1 013 10 3000 105 6.= 366 K

Substituting into the equation for Q we get

Q = ( )( )( ) ××

= −−

−0 10 8 31 3661000 10

3000 10

6

6. . ln mol J mol K K m

m334 J

3

3

That is, 334 J of heat energy is removed from the gas.

17.61. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes.Visualize: Please refer to Figure P17.61.Solve: (a) For this isobaric process, p1 = 4.0 atm, V1 = 800 × 10−6 m3, p2 = 4.0 atm, and V2 = 1600 × 10−6 m3. Thetemperature T1 of the gas is obtained from the ideal-gas equation as:

Tp V

nR11 1 390= = K

where n = 0.10 mol. Also,

T TV

VT T2 1

2

11

6

1

1600 102= = ×

×

=−

−

m

800 10 m

3

6 3 = 780 K

Thus, the heat required for the process 1 → 2 is

Q nC T T= −( ) = ( )( )( )P 0.10 mol 29.1 J mol K 390 K2 1 = 1135 J

This is heat transferred to the gas.(b) For the isochoric process, V2 = V3 = 1600 × 10−6 m3, p2 = 4.0 atm, p3 = 2.0 atm, and T2 = 780 K. T3 can be obtainedfrom the ideal-gas equation as follows:

p V

T

p V

TT T p p2 2

2

3 3

33 2 3 2= ⇒ = ( ) = ( )

=780 K

2.0 atm

4.0 atm390 K

The heat required for the process 2 → 3 is

Q nC T T= −( ) = ( )( ) −( ) = −V 0.10 mol J / mol K 390 K 780 K 811 J3 2 20 8.

Because of the negative sign, this is the amount of heat removed from the gas.(c) The change in the thermal energy of the gas is

∆E Q Q W W Wth J J + J= +( ) + +( ) = − +→ → → → →1 2 2 3 1 2 2 3 1 21135 811 0 = −324 J p V∆

= 324 J – (4.0 × 1.013 × 105 Pa)(1600 × 10−6 m3 – 800 × 10−6 m3) = 0 J

Assess: This result was expected since T3 = T1.

17.62. Model: Assume that the gas is an ideal gas.Visualize:

The volume of container A is a constant. On the other hand, heating container B causes the volume to change, butthe pressure remains the same.Solve: (a) For the heating of the gas in container A, ∆T Q nCA A V= / . Similarly, for the gas in container B,∆T Q nCB B P= / . Because QA = QB and CP > CV, we see that ∆ ∆T TA B> . The gases started at the same temperature,so TA > TB.(b)

(c) The pressure in container B exerted by the gas is equal to the pressure on the gas by the piston. That is,

p pw

AB atmospiston

piston

= + = × +( )( )

× −1 013 105. Pa10 kg 9.8 m s

1.0 10 m

2

4 2 = 1.081 × 106 Pa

Container A has the same volume, temperature, and number of moles of gas as container B, so P PA B= =1 081 106. × Pa.(d) The heating of container B is isobaric, so

V

T

V

TV V

T

Tf

f

i

if i

f

i

= ⇒ =

We have Ti = 293 K, and Tf can be obtained from

Q P t nC T T= = −( )∆ P f i

The number of moles of gas is n PV RT= =i i i mol./ .0 355 Thus

25 W 15 s 0.355 mol 20.8 J mol K 293 Kf( )( ) = ( )( ) −( )T

⇒ =Tf 344 K ⇒ = ×( )( )−Vf3 m 344 K 293 K8 0 10 4. = 9.39 × 10−4 m3

= 939 cm3

17.63. Model: Assume that the gas is an ideal gas. A diatomic gas has 1.40.Solve: (a) For container A,

VnRT

piAiA

iA

30.10 mol 8.31 J mol K 300 K

Pa m= =

( )( )( )× ×

= × −

3 0 1 013 108 20 105

4

. ..

For an isothermal process pfAVfA = piAViA. This means T TfA iA K= = 300 and

V V p pfA iA iA fA3 3 m 3.0 atm 1.0 atm m= ( ) = ×( )( ) = ×− −8 20 10 2 46 104 3. .

The gas in container B starts with the same initial volume. For an adiabatic process,

p V p VfB fB iB iBγ γ= ⇒ =

= ×( ) = ×− −V V

p

pfB iBiB

fB

3 3m3.0 atm

1.0 atm m

1

4

11 40

38 20 10 1 80 10γ

. ..

The final temperature TfB can now be obtained by using the ideal-gas equation:

T Tp

p

V

VfB iBfB

iB

fB

iB

3

4 3300 K1.0 atm

3.0 atm

m

8.20 10 m= = ( )

××

−

−

1 80 10 3. = 220 K

(b)

17.64. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes.Visualize: Please refer to Figure P17.64.Solve: (a) The initial conditions are p1 = 3.0 atm = 304,000 Pa, V1 = 100 cm3 = 1.0 × 10−4 m3, and T1 = 100°C =373 K. The number of moles of gas is

np V

RT= =

( ) ×( )( ) = ×

−−1 1

1

4

31 0 10

3739 81 10

304,000 Pa m

8.31 J mol K K mol

3.

( ).

At point 2 we have p2 = p1 = 3.0 atm and V2 = 300 cm3 = 3V1. This is an isobaric process, so

V

T

V

TT

V

VT2

2

1

12

2

11= ⇒ = = =3(373 K) 1119 K

The gas is heated to raise the temperature from T1 to T2. The amount of heat required is

Q nC T= = ×( )( ) −( ) =−P mol 20.8 J mol K 1119 K 373 K J∆ 9 81 10 1523.

This amount of heat is added during process 1 → 2.(b) Point 3 returns to T3 = 100°C = 373 K. This is an isochoric process, so

Q nC T= = ×( )( ) −( ) = −−V mol 12.5 J mol K 373 K 1119 K 91.5 J∆ 9 81 10 3.

This amount of heat is removed during process 2 → 3.

17.65. Model: Assume the gas to be an ideal gas.Visualize: Please refer to Figure P17.65.Solve: (a) The work done on the gas is the negative of the area under the p-versus-V graph, that is

W = − = −area under curve J 50 7.

(b) The change in thermal energy is

∆ ∆E nC T nC T Tth V V f i= = −( )Using the ideal-gas law to calculate the initial and final temperatures,

Tp V

nRii i= =

× ×( ) ×( )( )( ) =

−4 0 1 013 10 100 10

0 015

5 6. .

.

Pa m

mol 8.31 J mol K325 K

3

Tp V

nRff f

3 Pa m

0.015 mol 8.31 J mol K244 K= =

×( ) ×( )( )( ) =

−1 013 10 300 105 6.

⇒ = ( )( ) −( ) = −∆Eth 0.015 mol 12.5 J mol K 244 K 325 K J15 2.

(c) From the first law of thermodynamics,

∆Eth = Q + W ⇒ = − = − − −( ) =Q E W∆ th 15.2 J 50.7 J 35.5 J

That is, 35.5 J of heat energy is transferred to the gas.

17.66. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected bythe first law of thermodynamics.Visualize: Please refer to Figure P17.66.Solve: (a) For point 1, V1 = 1000 cm3 = 1.0 × 10−3 m3, T1 = 133°C = 406 K, and the number of moles is

nM

M= = ×

=−

mol

g

4 g / mol0.030 mol

120 10 3

Thus, the pressure p1 is

pnRT

V11

1

51 012 10 1 0= = ×. . Pa = atm

The process 1 → 2 is isochoric (V2 = V1) and p2 = 5p1 = 5.0 atm. Thus,

T T p p2 1 2 1 5= ( ) = ( )( ) = = °406 K 2030 K 1757 C

The process 2 → 3 is isothermal (T2 = T3), so

V3 = V2(p2/p3) = V2(p2/p1) = 5V2 = 5000 cm3

p (atm) T (°C) V (cm3)

Point 1 1.0 133 1000Point 2 5.0 1757 1000Point 3 1.0 1757 5000

(b) The work W1 2 0→ = J because it is an isochoric process. The work in process 2 → 3 can be found usingEquation 17.16 as follows:

W nRT V V2 3 2 3 2→ = − ( )ln = −( )( )( ) ( ) = −0.030 mol 8.31 J mol K 2030 K Jln 5 815

The work in the isobaric process 3 → 1 is

W p V V3 1→ = − −( )f i = − ×( ) × − ×( )− −1 012 10 1 0 10 5 0 105 3 3. . . Pa m m3 3 = 405 J

(c) The heat transferred in process 1 → 2 is

Q nC T1 2→ = = ( )( ) −( )V 0.030 mol 12.5 J mol K 2030 K 406 K∆ = 609 J

The heat transferred in the isothermal process 2 → 3 is Q W2 3 2 3 815→ →= − = J . The heat transferred in the isobaricprocess 3 → 1 is

Q nC T3 1 1013→ = = ( )( ) −( ) = −P 0.030 mol 20.8 J mol K 406 K 2030 K J∆

17.67. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.Solve: The air admitted into the cylinder at T0 = 30°C = 303 K and p0 = 1 atm = 1.013 × 105 Pa has a volume V0 =600 × 10−6 m3 and contains

np V

RT= =0 0

0

0.024 mol

Using Equation 17.36 and the fact that Q = 0 J for an adiabatic process,

∆ ∆E Q W nC Tth V= + = ⇒ W = nCVT

⇒400 J = (0.024 mol)(20.8 J/mol K)(Tf – 303 K) ⇒ =Tf 1100 K

For an adiabatic process Equation 17.40 is

T V T Vf fγ γ− −=1

0 01 ⇒ =

= ×( ) = × =

−− − −V V

T

Tf3 3 m

K

1100 K m cm0

0

1

14

1

1 4 1 5 36 0 10303

2 39 10 23 9f

.. . .

γ

Assess: Note that W is positive because the environment does work on the gas.

17.68. Model: The gas is an ideal gas that is subjected to an adiabatic process.Solve: (a) For an adiabatic process,

p V p Vf f i iγ γ= ⇒ =

p

p

V

Vf

i

i

f

γ

⇒ = ( )2 5 2 0. . γ ⇒ = =γ ln( . )

ln( . ).

2 5

2 01 32

(b) Equation 17.40 for an adiabatic process is

T V T Vf f i iγ γ− −=1 1 ⇒ =

= ( ) = ( ) =−

−T

T

V

Vf

i

i

f

γ 1

1 32 1 0 322 0 2 0 1 25. . .. .

17.69. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.Solve: (a) Equation 17.40 for an adiabatic process is

T V T Vf f i iγ γ− −=1 1 ⇒ =

−V

V

T

Tf

i

i

f

1

1γ

For the temperature to increase from Ti = 20°C = 293 K to Tf = 1000°C = 1273 K, the compression ratio will be

V

Vf

i

293 K

1273 K=

=

−1

1 4 10 02542

.. ⇒ = =V

Vmax

min ..

1

0 0254239 3

(b) From the Equation 17.39,

p V p Vp

p

V

Vf f i if

i

i

f

γ γγ

= ⇒ =

= ( ) =39 3 1711 4. .

17.70. Model: The helium gas is assumed to be an ideal gas that is subjected to an isobaric process.Solve: (a) The number of moles in 2.0 g of helium is

nM

M= = =

mol

2.0 g

4.0 g mol0.50 mol

At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume

VnRT

pii

i

30.0153 m L= = = 15 3.

For an isobaric process (pf = pi) that doubles the volume Vf = 2Vi,

T T V Vf i f i( ) = ( ) = 2 ⇒ = = ( ) = = °T Tf i 373 K 746 K 473 C2 2

(b) The work done by the environment on the gas is

W p V V p V= − −( ) = − −( )i f i i i 2 1 = − ×( )( )1 013 105. Pa 0.0153 m3 = −1550 J

(c) The heat input to the gas is

Q nC T T= −( ) = ( )( ) −( )P f i 0.50 mol 20.8 J mol K 746 K 373 K = 3880 J

(d) The change in the thermal energy of the gas is

∆E Q Wth 3880 J 1550 J 2330 J= + = − =

(e)

Assess: The internal energy can also be calculated as follows:

∆ ∆E nC Tth V 0.5 mol 12.5 J mol K 746 K 373 K 2330 J= = ( )( ) −( ) =

This is the same result as we got in part (d).

17.71. Model: The helium gas is assumed to be an ideal gas that is subjected to an isothermal process.Solve: (a) The number of moles in 2.0 g of helium gas is

nM

M= = =

mol

2.0 g

4.0 g mol0.50 mol

At Ti = 100°C = 373 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume

VnRT

pii

i

30.0153 m L= = = 15 3.

For an isothermal process (Tf = Ti) that doubles the volume Vf = 2Vi,

p V p V p p V Vf f i i f i i f atm atm= ⇒ = ( ) = ( )( ) =1 0 0 5012. .

(b) The work done by the environment on the gas is

W nRT V V= − ( )i f iln = −( )( )( ) ( )0.50 mol 8.31 J mol K 373 K ln 2 = −1074 J

(c) Because ∆Eth = Q + W = 0 J for an isothermal process, the heat input to the gas is Q = −W = 1074 J.(d) The change in internal energy ∆Eth = 0 J.(e)

17.72. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process.Solve: (a) The number of moles in 14.0 g of N2 gas is

nM

M= = =

mol

14.0 g

28 g mol0.50 mol

At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume

VnRT

pii

i

30.0112 m L= = = 11 2.

For an adiabatic process that compresses to a pressure pf = 20 atm, we can use Equation 17.39 and Equation 17.40as follows:

T V T VT

T

V

Vp V p V

p

p

V

Vf f i if

i

i

ff f i i

f

i

i

f

γ γγ

γ γγ

− −

−

= ⇒

=

= ⇒

=1 1

11

Combining the above two equations yields

T T p pf i f i( ) = ( )−γγ

1

⇒ = ( ) = ( ) =−

T T Tf i i 643 K20 2 35351 4 1 0

1 4

. .

. .

(b) The work done on the gas is

W E nC T T= = −( ) = ( )( ) −( )∆ th V f i 0.50 mol 20.8 J mol K 643 K 273 K = 3850 J

(c) The heat input to the gas is Q = 0 J.(d) From the above equation,

V

V

p

p

V

Vi

f

f

i

max

min

=

= ( ) = =

1

1

1 420 8 50γ

. .

(e)

17.73. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process.Solve: (a) The number of moles in 14.0 g of N2 gas is

nM

M= = =

mol

14.0 g

28 g mol0.50 mol

At Ti = 273 K and pi = 1.0 atm = 1.013 × 105 Pa, the gas has a volume

VnRT

pii

i

30.0112 m L= = = 11 2.

For an isochoric process (Vi = Vf),

T

T

p

pf

i

f

i

20 atm

1 atm= = = 20 ⇒ = ( ) =Tf 273 K 5460 K20

(b) The work done on the gas is W p V= − =∆ 0 J.(c) The heat input to the gas is

Q nC T T= −( ) = ( )( ) −( ) = ×V f i 0.50 mol 20.8 J mol K 5460 K 273 K J5 39 104.

(d) The pressure ratio is

p

p

p

pmax

min

f

i

20 atm

1 atm= = = 20

(e)

17.74. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchangeheat with its surroundings, the compression is an adiabatic process.Solve: The initial pressure of air in the mountains behind Los Angeles is pi = 60 × 103 Pa at Ti = 273 K. Thepressure of this air when it is carried down to the elevation near sea level is pf = 100 × 103 Pa. The adiabaticcompression of a gas leads to an increase in temperature according to Equation 17.39 and Equation 17.40, which are

T V T VT

T

V

Vp V p V

p

p

V

Vf f i if

i

i

ff f i i

f

i

i

f

γ γγ

γ γγ

− −

−

= ⇒

=

= ⇒

=1 1

11

Combining these two equations,

T T p p T Tf i f i f i

Pa

Pa273 K( ) = ( ) ⇒ = ×

×

= ( )

−−

γγ

1 3

3

1 4 1

1 4 0 286100 10

60 10

5

3

.

. .

= 316 K = 43°C = 109°F

17.75. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a constanttemperature of 77°C. How many moles of the gas are in the sample?(b) The number of moles is

n =−( )( )( ) =50 J

8.31 J mol K 350 K0.0156 mol

ln 13

17.76. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15°C. If the mercurytemperature rises to 90°C, what was the initial temperature of the iron slug?(b) The initial temperature was 217°C.

17.77. Solve: (a) 2.0 × 105 J must be removed from a sample of nitrogen gas at 20°C to convert it to liquidnitrogen. What is the mass of the sample?(b) The mass is M = 0.472 kg.

17.78. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What isthe compression ratio Vmax/Vmin?

(b) The ratio is V Vmax min = =10 5 181

1 4. . .

17.79. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, andadiabatic processes.Visualize: Please refer to Figure CP17.79. The gas at point 1 has volume V1 = 1000 cm3 = 1.0 × 10−3 m3 andpressure p1 = 3.0 atm. At point 2, V2 = 3000 cm3 = 3.0 × 10−3 m3 and p2 = 1.0 atm. These values mean that T2 = T1,so process 1 → 2 is an isothermal process. The process 2 → 3 occurs at constant volume and is thus an isochoricprocess. Finally, because temperature T3 is lower than T1 or T2, the process 3 → 1 is an adiabatic process.Solve: (a) The number of moles of gas is

n = =0.120 g

4 g mol0.030 mols

The temperature T1 can be calculated to be

Tp V

nR11 1

5 33 0 1 013 10 1 0 10

8 311219= =

× ×( ) ×( )( )( ) =

−. . .

.

Pa m

0.030 mol J mol K K

3

= 946°C

For the isothermal process 1 → 2, T2 = 1219 K. For the adiabatic process 3 → 1,

p p V V T T V V3 1 1 3

1 67

3 1 1 3

1 13

0 67= ( ) = ( )

= = ( ) = ( )( ) = °−γ γ3.0 atm

1000 cm

3000 cm0.48 atm 1219 K 583 K = 310 C

3

3

..

Point p (atm) T (°C) V (cm3)

1 3.0 946 10002 1.0 946 30003 0.48 310 3000

Note that the values obtained above are consistent with the isochoric process 2 → 3, for which

p

T

p

T2

2

3

3

= ⇒ = ( ) = ( )( ) =p T T p2 2 3 3 1219 K 583 K 0.48 atm 1 atm

(b) From Equation 17.15,

W nRTV

V1 2 12

1→ = −

ln = −( )( )( ) ( ) = −0.030 mol 8.31 J mol K 1219 K Jln 3 334

The work done in the ischochoric process is W2 3→ = 0 J . The work done in the adiabatic process is

W nC T T3 1 1 3 239→ = −( ) = ( )( ) −( ) =V 0.030 mol 12.5 J mol K 1219 K 583 K J

(c) For the process 1 → 2, ∆ ∆T E Q W= ⇒ = ⇒ = − =0 0 K J 334 Jth

For the process 2 → 3, W = 0 J,

∆E Q nC T Tth V 239 J= = −( ) = −3 2

For the process 3 → 1, Q = 0 J.

17.80. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the movingmolecules.Visualize: Please refer to Figure P17.80.Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the piston’s mass mustexactly balance the upward force of the gas, F pAgas = where A = πr2 is the area of the face of the piston. Since the

upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus,

M g pA pM g

A

V g

Aghpiston

piston Cu pistonCu

3 kg m m s m Pa= ⇒ = = = = ( )( )( ) =ρ

ρ 8920 9 80 0 040 35002. .

(b) The gas volume is V r L12 2 4 30 030 0 20 5 65 10= = = × −π π ( . ) ( . ) . . m m The number of moles is

np V

RT= =

( ) ×( )( )( )

= ×−

−1 1

1

4

45 65 10

8 12 103500 Pa m

8.31 J mol K 293 K mol

3..

The number of molecules is

N = nNA = (8.12 × 10−4 mol)(6.02 × 1023 mol−1) = 4.89 × 1020

(c) The thermal energy is

E NK N m vth avg avg= = ( )12

2

Molecular nitrogen N2 has an atomic mass number of A = 28, so the mass of one molecule is

m = 28 × (1.661 × 10−27 kg) = 4.65 × 10−26 kg

If the average speed is vavg = 550 m/s, then the thermal energy is

Eth = (4.89 × 1020)(0.5)(4.65 × 10−26 kg)(550 m/s)2 = 3.44 J

(d) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is added,so the heating takes place at constant pressure with Q = nCP∆T. The temperature increase is

∆TQ

nC= =

×( )( )=−

P4

2.0 J

8.12 10 mol 29.1 J mol K85 K

This raises the gas temperature to T2 = T1 + ∆T = 378 K = 105°C.(e) Noting that the volume of a cylinder is V = πr2L and that r doesn’t change, the ideal gas relationship for anisobaric process is

V

T

V

T

L

T

L

TL

T

TL2

2

1

1

2

2

1

12

2

11= ⇒ = ⇒ = =

=378 K

293 K20 cm 25.8 cm

(f) The work done by the gas is W F ygas gas= ∆ . The force exerted on the piston by the gas is

F pA p rgas 9.90 N= = =π 2

This force is applied through ∆y = 5.8 cm = 0.058 m, so the work done is

Wgas = (9.90 N)(0.058 m) = 0.574 JThus, 0.574 J is the work done by the gas on the piston. The work done on the gas is −0.574 J.

17.81. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature(20°C), and the liquid nitrogen. The boiling point of liquid nitrogen is –196°C = 77 K.Solve: The piece of iron has mass Miron = 197 g = 0.197 kg and volume Viron = Miron/ρiron = (0.197 kg)/(7870 kg/m3) =25 × 10−6 m3 = 25 cm3 = 25 mL. The heat lost by the iron is

Qiron = Mironciron∆T = (0.197 kg)(449 J/kg K)(77 K – 293 K) = –1.911 × 104 J

This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires

Q Q Q ML MQ

Liron N2 iron firon

f5

J

1.99 10 J / kg kg+ = + = ⇒ = − = ×

×=0

1 911 100 0960

4..

The volume of liquid nitrogen boiled away is thus

VM

boilN2

33 kg

810 kg / m m mL= = = × =−

ρ0 0960

1 19 10 1194..

Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of ironexcludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is V1 = 1475 mL.The pressure is p1 = 1.0 atm and the temperature is T1 = 77 K. Thus the number of moles of nitrogen gas is

np V

RT11 1

1

3101 300

8 310 234= = × =

−( , )

( ..

Pa)(1.475 10 m

J / mol K)(77 K) mol

3

119 mL of liquid boils away, so the gas volume increases to V2 = V1 + Vboil = 1475 mL + 119 mL = 1594 mL. Thetemperature is still T2 = 77 K, but the number of moles of gas has been increased by the liquid that boiled. Thenumber of moles that boiled away is

nboil

6.0 g

28 mol / g mol= =9

3 429.

Thus the number of moles of nitrogen gas increases to n2 = n1 + nboil = 0.234 mol + 3.429 mol = 3.663 mol.Consequently, the gas pressure increases to

pn RT

V22 2

23

63 6631 470 10= =

×= × =−

( ..

mol)(8.31 J / mol K)(77 K)

1.594 10 m Pa 14.5 atm3

Assess: Don’t try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading toserious injuries.