3. the second and third laws of thermodynamicsramu/chem311/assigned/chap03_probs_4e.pdf · 3. the...

3. THE SECOND AND THIRD LAWS OFTHERMODYNAMICS

n The Carnot Cycle

3.1. a. Efficiency = Th – Tc

Th =

1000 – 2001000 = 0.8 = 80%

b. Heat rejected = 150 × 1000200 = 30 kJ

c. Entropy increase = 150 000

1000 = 150 J K–1

d. Entropy decrease = 30 000

200 = 150 J K–1

e. ∆S = 0

f. ∆S = 150 J K–1; ∆H = 0

∆G = ∆H – T∆S = 0 – 1000 × 150 = −150 000 J

= –150 kJ

3.2. Efficiency = wqh

= 398 – 313

398 = 0.214

15000.214 = 7010 J must be withdrawn.

THE SECOND AND THIRD LAWS OF THERMODYNAMICS n 49

3.3. a.

BC

S

T

AD

b.qhTh

+ qcTc

= 0 = qhTh

+ qc

300 K = 0

Work performed by system, –w = qh + qc = 10 kJ

∆Sh = qhTh

= 100 J K–1

100 J K–1 + qc

300 K = 0

qc = –30 000 J = –30 kJ

qh = 40 kJ

Th = 4030 × 300 K = 400 K

3.4. a. Efficiency = (Th – Tc)/Th = 0.25 = 25%

b. Heat absorbed at 400 K = 800/0.25 = 3200 J = 3.2 kJ

c. Heat rejected at 300 K = 3200 – 800 = 2400 J = 2.4 kJ

d. Entropy change for A → B = 3200 J/400 K = 8.0 J K–1

e. Zero

f. ∆H = 0 and thus ∆G = –T∆S = –400 × 8.0 = –3200 J = –3.2 kJ

g. The efficiency is 25%, so to produce 2000 J of work,

2000 J0.25 = 8000 J = 8.0 kJ

must be absorbed.

3.5. The heat transferred from the water to the ice in melting it is

50 n CHAPTER 3

(1012 g/18 g mol–1) × 6.025 kJ mol–1 = 3.35 × 1011 kJ

The fraction of this that can be converted into work is (Eq. 3.21)

Th – TcTh

= 20

293.15 = 0.0682

The available work is therefore

0.0682 × 3.35 × 1011 = 2.29 × 1010 kJ

One day is 24 × 60 × 60 = 86 400 s. The power, or the rate of doing work, is thus

2.29 × 1010/86 400 = 2.65 × 105 kJ s–1

= 2.65 × 108 J s–1 = 2.65 × 108 W

= 265 MW

3.6. A thermodynamic equation of state (Eq. 3.128) is

From the ideal gas law, we have P = RT/Vm; therefore, (∂P/∂T)V = R/Vm. Substituting above, andrecognizing that RT/Vm = P, we obtain

n Entropy Changes

3.7. C6H6:

CHCl3:

H2O:

C2H5OH:

30 800/353 = 87.3 J K–1 mol–1

29 400/334 = 88.0 J K–1 mol–1

40 600/373 = 108.8 J K–1 mol–1

38 500/351 = 109.7 J K–1 mol–1 } Hydrogen-bonded structure in liquids H2O and C2H5OH.

3.8. a. The reaction is

C(graphite) + 2H2(g) + 12 O2(g) → CH3OH(l)

∆f H° = 126.8 – [5.74 + (2 × 130.68) + 12(205.14) ]

= –242.9 J K–1 mol–1

b. The reaction is C(graphite) + 2H2(g) + 12 O2(g) + N2(g) → H2NCONH2(s)

∆f H° = 104.60 – [5.74 + (2 × 130.68) + 12(205.14) + 191.61]

= –456.8 J K–1 mol–1

.VT T

PTP

V

U

∂∂+−=

∂∂

.0=

+−=

∂∂

mT V

RTP

V

U


3.9. a. ∆rS° = [2(192.77) – 191.61 – 3(130.68)] J K–1 mol–1

= –198.11 J K–1 mol–1.

b. ∆rS° = 2(240.1) – 304.2 J K–1 mol–1 = –176.0 J K–1 mol–1.

3.10. First, calculate the value of ∆rS° at 298.15 K. Then, using this value, calculate the increase inentropy for the increase in temperature. Assume CP is constant throughout the range.

For the dissociation H2(g) → 2H(g),

∆rS°(298.15 K) = [2(114.717) – 130.680] J K–1 mol–1 = 98.754 J K–1 mol–1.

The value of ∆rS° at different temperatures may be calculated from

∆rS°(T) = ∆rS°(198.15 K) + 1273.15

r

298.15

,PCdT

T

°∆∫

where ∆rCP° is ,i P iiCν °∑ for the reaction. If ∆rCP° is independent of temperature, the quantity

∆rCP° can be taken out of the integral. Using the CP° values at 298.15 K, we get1273.15

298.15

(1273.15 K) 98.754 [2(20.784) 28.824]rdT

ST

°∆ = + − ∫

= 98.754 + 12.744 ln(1273.15/298.15)

= 186.46 J K–1 mol–1.

A more accurate value could be obtained if values of S° or CP° were available at 1273.15 K.

3.11. a. ∆Sm = ∫353

298 CP,mdT

T = 52 R ln 353298

= 3.52 J K–1 mol–1

b. ∆Sm = ∫353

298 CV,mdT

T = 32 R ln 353298

= 2.11 J K–1 mol–1

3.12. For N2 and O2, the volume increases by a factor of 2.5, increase of entropy for each

∆S = R ln 2.5 = 7.62 J K–1

For H2, volume increases by a factor of 5:

∆S = 0.5 × 8.3145 ln 5 = 6.69 J K–1

Total ∆S = 2 × 7.62 + 6.69 = 21.9 J K–1

3.13. By Eq. 3.63,

52 n CHAPTER 3

∆S = 8.3145 ln 31 + 5 × 8.3145 ln

32

= 9.13 + 16.86 = 25.99 J K–1

3.14. The mole fractions are

xN2 = 0.79 xO2

= 0.20 xAr = 0.01

The entropy change per mole of mixture is thus, by Eq. 3.65,

∆S = –8.3145(0.79 ln 0.79 + 0.20 ln 0.20 + 0.01 ln 0.01)

= –8.3145(–0.186 – 0.322 – 0.046) = 4.61 J K–1 mol–1

3.15. The equation for the formation of ethanol from its elements is

2C(graphite) + 3H2(g) + 12 O2(g) → C2H5OH(l)

The standard molar entropy of formation is thus

∆f S°m = S°product –ΣS°reactants

= 160.7 – [(2 × 5.74) + (3 × 130.68) + 12(205.14) ]

= –345.4 J K–1 mol–1

3.16. a. ∆Sgas = R lnV2V1

= 8.3145 ln10.0 = 19.1 J K–1 mol–1

∆Ssurr = –19.1 J K–1 mol–1

b. For an adiabatic process, Eq. 2.90 applies:

ThVγ–1 = TcVγ–1 where γ =

53

Tc = 64.2 K

Because ∆Sgas must be calculated from a reversible process,

∆Sgas = 19.1 J K–1 mol–1; ∆Ssurr = 0 since q = 0

Net ∆S = 19.1 J K–1 mol–1

3.17. a. Taking into account the entropy change for heating liquid water and the entropy change duringvaporization, we obtain

b. We recognize that changing the pressure on the surface of a liquid dies not affect the entropy.Therefore, we take into account the entropy changes for vaporization, heating the vapor at

−

− ×=×1

35

1 101298 cTγ

373 , (l)1 1

273

40 670(373.0 K) /(J K mol )

373.0373 40 670

75.48ln 132.59.272 373.0

P mCS dT

T− −∆ ° = +

= + =

∫


constant pressure and, finally, compressing the vapor to the final pressure at constanttemperature (see Eq. 3.51). This yields

373 , (g)1 1 2

2731

44 920(373.0 K) /(J K mol ) 8.3145ln

273.0P mC V

S dTT V

− − ∆ ° = + +

∫

Substituting CP,m = 30.54 + 10.29 × 10–3T and replacing V2/V1 in the last term by P1/P2 =0.00602/1.00, we get

3731 1 3 1

2732

3

44 920 30.54(373.0 K) /(J K mol ) 10.29 10 8.3145ln

273.0

44 920 37330.54 ln 10.29 10 (373 272)

273.0 273

0.006028.3145ln

1.00

132.59.

PS dT

T P− − −

−

∆ ° = + + × + = + + × −

+ =

∫

Thus the entropy change for the system depends only on the initial and final states and not thepath taken to go from one to the another.

3.18. a. Positive (increase in number of molecules)

b. Positive (decrease in electrostriction)

c. Negative (increase in electrostriction)

d. Positive (decrease in electrostriction)

3.19. Heat absorbed when temperature rises = dq = CPdT

Corresponding entropy change:

∆S = ∫CPdTT

Entropy increase when the temperature rises from T1 to T2:

∫T2

T1 CPT dT = n ∫T2

T1 CP,m

T dT

If the gas is ideal, CP is constant and

∆S = CP ln T2T1

= nCP,m ln T2T1

3.20. By Eq. 3.57 the entropy change is

∆S = 5 × 12.5 ln 373300 + 5 × 8.3145 ln

105

= 13.6 + 28.8 = 42.4 J K–1

3.21. Let the value of the final Celsius temperature be Tu. Then

54 n CHAPTER 3

200 × 0.140 (100 – Tu) = 100 × 4.18 (Tu – 20)

6.70 – 0.0670 Tu = Tu – 20

Tu = 26.701.0670 = 25.02; T = 25.02 °C

a. ∆Smercury = 200 × 0.140 ∫298.17

373.15 dTT

= 200 × 0.140 ln 298.17373.15

= –6.28 J K–1

b. ∆Swater + vessel = 100 × 4.18 ∫298.17

293.15 dTT

= 100 × 4.18 ln 298.17293.15

= 7.10 J K–1

c. Net ∆S = –6.28 + 7.10 = 0.82 J K–1

3.22. Heat required to melt 20 g of ice is

2018 × 6020 = 6689 J

Heat required to heat 20 g of water from 0 °C to T °C is

(20 × 4.184 T) J

Heat required to cool 70 g of water from 30 °C to T °C is

70 × 4.184 (30 – T) J

Heat balance equation:

6689 + 20 × 4.184 T = 70 × 4.184 (30 – T)

T = 5.57 °C

Reversible processes:

a. Cool 70 g of water to 0 °C:

∆Ssystem = 70 × 4.184 ln 273.15303.15 = –30.52 J K–1

∆Ssurr = 30.52 J K–1

b. Melt 20 g of ice at 0 °C:

∆Ssystem = 20 6020

18 27315

×× .

= 24.49 J K–1

∆Ssurr = –24.49 J K–1

c. Heat 90 g of water to 5.57 °C:


∆Ssystem = 90 × 4.184 ln 278.72273.15 = 7.60 J K–1

∆Ssurr = –7.60 J K–1

Net ∆Ssystem = 1.57 J K–1; ∆Ssurr = –1.57 J K–1

3.23. ∆Sm = ∫1000

300 CP,m

T dT

∆Sm/J K–1 mol–1 = ∫1000

300 28.58T/K + 0.00376 –

50 000(T/K)3 d(T/K)

= 28.58 ln 1000300 + 0.00376 (1000 – 300) + 25 000

1

10002 –

13002

= 34.41 + 2.63 – 0.25 = 36.79

∆Sm = 36.8 J K–1 mol–1

3.24. In an isothermal reversible expansion

∆E = 0 = q + w and qrev = ∆S

TIf the process were reversible, the work done by the system would be

–wrev = T∆S = 300 × 50 = 15 000 J = 15 kJ

Since the actual work was less than this, the process is irreversible.

Degree of irreversibility = 615 = 0.4

3.25. It is necessary to devise a process in which the freezing occurs reversibly:

Step 1: Heat the supercooled water reversibly from –3 °C to 0 °C:

∆S1 = 75.3 ln 273.15270.15 = 0.83 J K–1 mol–1

Step 2: Freeze the water at 0 °C:

∆S2 = –6020273.15 = –22.04 J K–1 mol–1

Step 3: Cool the ice reversibly from 0 °C to –3 °C:

∆S3 = 37.7 ln 270.15273.15 = –0.42 J K–1 mol–1

The net entropy change in the system is therefore

∆S = 0.83 – 22.04 – 0.42 = –21.63 J K–1 mol–1

To calculate the entropy change in the environment, we calculate the heat that has been gained bythe environment in the three steps:

Step 1: –3 × 75.3 = –225.9 J mol–1

Step 2: 6020 J mol–1

Step 3: 3 × 37.7 = 113.1 J mol–1

56 n CHAPTER 3

The net heat gained by the environment is thus

–225.9 + 6020 + 113.1 = 5907.2 J mol–1

This heat was gained by the environment at –3 °C, and the entropy change is therefore

5907.2270.15 = 21.87 J K–1 mol–1

The net entropy change in the system and environment is thus

–21.63 + 21.87 = 0.24 J K–1 mol–1

3.26. The solution contains 0.1 mol, and the volume ratio is 1000/200 = 5. Then, by Eq. 3.51,

n = 0.5 × 0.2 = 0.1

∆S = 0.1 × 8.3145 × ln 5

= 1.338 J K–1 = 1.34 J K–1

3.27. 0.1 mol of substance A is present and the volume increases by a factor of 4:

∆S (A) = 0.1 × 8.3145 × ln 4

= 1.153 J K–1

0.15 mol of B is present and the volume increases by a factor of 4/3:

∆S (B) = 0.15 × 8.3145 × ln (4/3)

= 0.359 J K–1

Net ∆S = 1.51 J K–1

3.28. The final temperature is 40 °C. The water at 60 °C can be cooled reversibly to 40 °C, and the waterat 20°C can be heated reversibly to 40 °C.

∆S1 = 10 × 75.3 ln 313.15333.15 = –46.62 J K–1

∆S2 = 10 × 75.3 ln 313.15293.15 = 49.70 J K–1

The net entropy change is thus 49.70 – 46.62 = 3.08 J K–1.

3.29. For O2: ∆S = 5 × 8.3145 × ln 2

= 28.8 J K–1

The entropy change for the expansion of the N2 is the same, and the net entropy change is thus

57.6 J K–1

3.30. The process can be imagined as occurring by the following reversible processes:

Step 1: The water freezes to ice at 0 °C:

∆S2 = –6020

273.15 = –22.04 J K–1


Step 2: The ice is cooled reversibly to –12 °C:

∆S3 = 37.7 ln 261.15273.15 = –1.69 J K–1

Net ∆S(system) = –22.04 – 1.69 = –23.73 J K–1

The heat gained by the freezer = 6020 + 12 × 37.7 = 6472.4 J

This was gained at –12 °C, and therefore

∆S(surroundings) = 6472.4261.15 = 24.78 J K–1

The net entropy change is ∆S(system) + ∆S(surroundings) = 1.05 J K–1.

3.31. For the freezing of 1 mol of water at 0 °C: ∆S = –6020

273.15 = –22.04 J K–1

For the reversible cooling of the ice from 0 °C to –10 °C:

∆S = 37.7 ln 263.15273.15 = –1.41 J K–1

Net ∆S(system) = –22.04 – 1.41 = –23.45 J K–1

The heat gained by the freezer = 6020 + (10 × 37.7) = 6397 J

This was gained at –10 °C, so that the entropy change in the freezer is

6397263.15 = 24.31 J K–1

The net entropy change is 0.86 J K–1.

3.32. The final temperature T is calculated in terms of the heat balance:

2 × 75.3 × (60 – T) = 4 × 75.3 × (T – 20)

T = 33.33 °C = 306.48 K

∆S = 2 × 75.3 ln 306.48333.15 + 4 × 75.3 ln

306.48293.15

= –12.6 + 13.4 = 0.8 J K–1

3.33. There are obviously several reversible paths that can be constructed between the initial and finalstates in this case. Let us consider four of them.

1. Isothermal expansion to the final volume followed by constant volume cooling to the finaltemperature.

3 1, 2

5 1, 2

3 1 1, 2

0 ( ) (253.2 298.0) 558.7 J mol

0 ( ) (253.2 298.0) 931.2 J mol

10.526 253.2ln ln ln ln 9.994 J K mol

2.478 298.0

V m f i

P m f i

f fV m

i i

U C T T R

H C T T R

V TS R C R R

V T

−

−

− −

∆ = + − = − = −

∆ = + − = − = −

∆ = + = + =

58 n CHAPTER 3

2. Isothermal expansion to the final pressure followed by constant pressure cooling to the finaltemperature.

3 1, 2

5 1, 2

3 1 1, 2

0 ( ) (253.2 298.0) 558.7 J mol

0 ( ) (253.2 298.0) 931.2 J mol

10.0 253.2ln ln ln ln 9.995 J K mol

2.0 298.0

−

−

− −

∆ = + − = − = −

∆ = + − = − = −

∆ = + = + =

V m f i

P m f i

fiP m

f i

U C T T R

H C T T R

TPS R C R R

P T

3. Isothermal expansion to (P0, V0) followed by adiabatic expansion to the final state.

Note that we need to find the intersection of the isotherm that passes through the initial state andthe adiabat that passes through the final state. This intersection is (P0, V0), at Ti. Using therelationships for adiabatic processs (Eq. 2.90),

V0 Ti(3/2) = Vf Tf

(3/2); Therefore,

3/ 23 3

0

3 1, 2

5 1, 2

1 10

253.210.526 dm 8.244 dm .

298.0

0 ( ) (253.2 298.0) 558.7 J mol

0 ( ) (253.2 298.0) 931.2 J mol

8.244ln 0 ln 9.994 J K mol

2.478

−

−

− −

= × = ∆ = + − = − = −

∆ = + − = − = −

∆ = + = =

V m f i

P m f i

i

V

U C T T R

H C T T R

VS R R

V

4. Constant Pressure heating to the final volume followed by constant volume cooling to the finalpressure.

The gas will have to be heated to T0 = 1266.0 K in order for it to reach the volume of 10.526

dm3 at 10.0 bar pressure. Therefore,

Yet another path we can try is constant volume cooling to the final pressure followed byconstant pressure heating to the final temperature.

In each of these cases, we have verified that ∆U, ∆H and ∆S are the same, thus proving that theyare independent of the path taken, as any state property should be. We now have to find theentropy change of the surroundings.

Entropy change of the surroundings.

The actual process is the expansion of the gas against a constant external pressure of 2 bar. Forthis process, according to the first law,

∆U = qact – Pext(Vf – Vi); Therefore,

qact = ∆U + Pext(Vf – Vi) = –558.7 + 2.0 ×(10.526 – 2.478) × (8.3145/0.083145)

3 1, 0 , 0 2

5 1, 0 , 0 2

50, , 2

0

( ) ( ) (1266.0 298.0 253.2 1266.0) 558.7 J mol

( ) ( ) (1266.0 298.0 253.2 1266.0) 931.2 J mol

1266.0ln ln ln

298.0

V m i V m f

P m i P m f

fP m V m

i

U C T T C T T R

H C T T C T T R

TTS C C R

T T

−

−

∆ = − + − = − + − = −

∆ = − + − = − + − = −

∆ = + = 3 1 12

253.2ln 9.995 J K mol

1266.0R − − + =


= 1050.9 J mol−1.

∆Ssurr = – qact /Tsurr = –1050.9 J mol−1/298 K = –3.526 J K−1mol−1

∆Suniv = ∆S + ∆Ssurr = 6.468 J K−1mol−1.

This is, therefore, a spontaneous process.

3.34. ∆S(system) = 5 × 75.3 ln 276.15323.15 = –59.18 J K–1

The heat accepted by the refrigerator is

5 × 75.3 × (50 – 3) = 17 696 J

∆S(refrigerator) = 17 696/276.15 = 64.08 J K–1

Net ∆S = 64.08 – 59.18 = 4.90 J K–1

3.35. a. In this case all of the ice melts and the final temperature is 12 °C. (See the solution to Problem2.32.) The entropy changes are

1. Reversible melting of the ice at 0 °C:

∆S1 = 6025 J mol–1 × (100/18) mol/273.15 K

= 122.5 J K–1

2. Reversible heating of 100 g of water from 0 °C to 12 °C:

∆S2 = 75.3 J K–1 mol–1 × (100/18) mol ln (285.15/273.15)

= 18.0 J K–1

3. Reversible cooling of 1 kg of water from 20 °C to 12 °C:

∆S3 = (1000/18) mol × 75.3 J K–1 mol–1 ln (285.15/293.15)

= –115.7 J K–1

The net entropy change is therefore

∆S = 122.5 + 18.0 – 115.7 = 24.8 J K–1

b. In this case only 250 g of the ice melts, and the final temperature of the water is 0 °C. (See thesolution to Problem 2.32.) The entropy changes are now

1. For the reversible melting of 250 g of ice at 0 °C:

∆S1 = (250/18) mol × 6025 J mol–1/273.15 K

= 306.4 J K–1

2. For the cooling of 1 kg of water from 20 °C to 0 °C:

∆S2 = (1000/18) mol × 75.3 J K–1 mol–1 ln (273.15/293.15)

= – 295.6 J K–1

The net entropy change is

∆S = 306.4 – 295.6 = 10.8 J K–1.

60 n CHAPTER 3

3.36. Using the expression for CP,m given in Table 2.1,

= 217.3 J K–1 mol–1

n Gibbs and Helmholtz Energies

3.37. From Eq. 3.91,

∆G° = 2(–181.64) + 2(–394.36) – (–914.54)

= –237.46 kJ mol–1

∆H° = 2(–288.3) + 2(–393.51) – (–1263.07)

= –100.55 kJ mol–1

∆S° = ∆ ∆H G

T

°− °=

–100 550 + 237 460298.15

= 459.2 J K–1 mol–1

3.38. Work done by system = P∆V = 30.19 dm3 atm mol–1

= 3059 J mol–1

∆H = 40 600 J mol–1

∆U = ∆H – ∆ (PV)

= 40 600 – 3 059 = 37 540 J mol–1

∆G = 0 ∆S = 40 600373.15 = 108.8 J K–1 mol–1

3.39. From Example 3.6, ∆H and ∆S are available.

a. At 0°C, ∆H = –6020 J mol–1

∆S = –22.04 J K–1 mol–1

∆G = –6020 + 22.04 × 273.15 = 0

b. At –10°C, ∆H = –5644 J mol–1

∆S = –20.64 J K–1 mol–1

∆G = –5644 + 20.64 × 263.15 = –213 J mol–1

3.40. ∆U = 0

dTT

TTS ∫

×−×++=°−

−− 0.800

32.77

24311 /100.51076.358.28

94.151 )molJ K/() K0.800(


∆H = ∆U + ∆ (PV) = 0

∆S = qrevT =

1T ∫V2

V1 PdV = R ln

V2V1

= 8.3145 ln 10

= 19.14 J K–1 mol–1

∆A = ∆U – T∆S = –298.15 × 19.14 = –5.706 kJ mol–1

∆G = –5.706 kJ mol–1

The quantities are all state functions, and the preceding values therefore do not depend on how theprocess is carried out.

3.41. a. (a) ∆G = –85 200 + (300 × 170.2) J mol–1

= –34.14 kJ mol–1

(b) ∆G = –85 200 + (600 × 170.2) J mol–1

= 16.92 kJ mol–1

(c) ∆G = –85 200 + (1000 × 170.2) J mol–1

= 85.00 kJ mol–1

b. At T = 85 200/170.2 = 500.6 K

3.42. From Eq. (3.168), we write

Where T is the mid-point of the temperature range (T1,T2). In the limit ∆T → 0, this will yield Eq.(3.168). Substituting, we get

Therefore, ∆cH° = –(323.0)2 × 8.567×10–3 = –894.73 kJ mol–1.

3.43. Conversion of water to vapor at 0.0313 atm:

∆H = 44.01 kJ mol–1

∆S = 44 010298.15 = 147.6 J K–1 mol–1

Reversible isothermal expansion from 0.0313 atm to 10–5 atm:

∆H = 0

∆S = 8.3145 ln 0.031310–5 = 66.9 J K–1 mol–1

Net ∆H = 44.01 kJ mol–1

o oc c 2 c 1 c

22 1 2 1

1,

P

G G G H

T T T T T T T

∂ ∆ ° ∆ ∆ ∆ ° ≈ − = − ∂ −

3 c2

1 802.57 815.048.576 10

348.0 298.0 348.0 298.0 323.0

H−− − ∆ ° − = × = − −

62 n CHAPTER 3

∆S = 147.6 + 66.9 = 214.5 J K–1 mol–1

∆G = 44 010 – 298.15 × 214.5 = –19 940 J mol–1

= –19.94 kJ mol–1

3.44. a. ∆U, ∆H e. None

b. ∆S f. ∆G

c. ∆H g. ∆U

d. ∆H h. None

3.45.

∂∂P

G

T = V

Therefore

∆G = ∫VdP

The molar volume of mercury is

Vm = 200.6 g mol–1

13.5 g cm–3 = 1.486 × 10–5 m3 mol–1

Then ∆Gm = 1.486 × 10–5 m3 × 999 bar × 105 Pa bar–1

= 1485 J mol–1 = 1.485 kJ mol–1

3.46.

∂∂

T

Gm

P= –Sm

= –(A + B ln T)

where A = 36.36 J K–1 mol–1

B = 20.79 J K–1 mol–1

∆Gm = –∫323.15

298.15 (A + B ln T) dT

= –[AT + B(T ln T – T)] 323.15298.15

= –(A – B) 50 – B (323.15 ln 323.15) – 298.15 ln 298.15)

= –778.5 – 3502.3 J mol–1 = –4.28 kJ mol–1.

3.47. The entropy at 300.0 K is calculated as

∫ ∫+++=° −− 64.197

0.15

08.263

64.197

11 20.87

64.197

740265.3226.1)molJ K/() K0.300( dT

TdT

TS

1-0.300

08.263

1-molJ K 86.247

88.39

08.263

937 24 ∫ =++ dTT


3.48. ∆U and ∆H are zero for the isothermal expansion of an ideal gas.

q = –w = 500 J mol–1

∆S = R ln 2 = 5.76 J K–1 mol–1

∆G = –T∆S = –1728 J mol–1 = –1.73 kJ mol–1

wrev = +1.73 kJ mol–1; qrev = –1.73 kJ mol–1

3.49. Since Pext = 0 (evacuated vessel), no work is done; w = 0.

∆U = q + w = q = 30 kJ mol–1.

For H2O(l) → H2O(g), ∆n = 1 (see Example 2.5)

∆H = ∆U + ∆ (PV) = ∆U + ∆nRT

= 30 000 + 3102 = 33 102 J mol–1 = 33.1 kJ mol–1

To obtain the entropy change, consider the reversible processes:

(1) H2O (l, 100 °C) → H2O (g, 100 °C, 1 atm)

∆S1 = 40 600373.15 = 108.8 J K–1

(2) H2O (g, 100 °C, 1 atm) → H2O (g, 100 °C, 0.5 atm)

∆S2 = R ln V2V1

= R ln 2 = 5.76 J K–1 mol–1

Net ∆S = 114.6 J K–1 mol–1

Net ∆G = ∆H – T∆S

= 33 102 – 42 760 = –9658 J mol–1

= –9.66 kJ mol–1

3.50. Allow the process to occur by the following reversible steps:

(1) H2O (g, 100 °C, 2 atm) → H2O (g, 100 °C, 1 atm)

∆H1 = 0

∆S1 = R ln V2/V1 = R ln 2 = 5.76 J K–1 mol–1

∆G1 = –T∆S1 = –2150 J mol–1

(2) H2O (g, 100 °C, 1 atm) → H2O (l, 100 °C, 1 atm)

∆H2 = –40 600 J mol–1

∆S2 = – 40 600373.15 = –108.8 J K–1 mol–1

∆G2 = 0 (reversible process at constants P and T)

64 n CHAPTER 3

(3) H2O (l, 100 °C, 1 atm) → H2O (l, 100 °C, 2 atm)

The ∆H, ∆S, and ∆G changes are negligible for this process.

The overall changes are thus

∆H = –40 600 J mol–1 = –40.6 kJ mol–1

∆S = 5.76 – 108.8 = –103 J K–1 mol–1

∆G = –2.15 kJ mol–1

3.51. ∆U = ∫T2

T1 CV,mdT

CV,m = CP,m – R

= (20.27 + 1.76 × 10–2 T/K) J K–1 mol–1

q = 0; CV,mdT = dw = –P2dV

∆Um = 300 ∫T

TT K2

1

220 27 176 10( . . / )+ × − dT

= –P2 ∫V2

V1 dV = –P2

RT2

P2 –

300RP1

= –4 × 8.3145

T2

4 – 30010

20.27 (T2 – 300) + 0.0088 (T22 – 3002) = –4 × 8.3145

T2

4 – 30

0.0088 T22 + 28.58 T2 – 7871 = 0

T2 = –28.58 ± (816.816 + 277.06)1/2

0.0176

= –28.58 + 33.07

0.0176 = 255.3 K

∆Um = 20.27 (255.3 – 300) + 0.0088 (255.32 – 3002) = –1124.5 J mol–1

∆Hm = 28.58 (255.3 – 300) + 0.0088 (255.32 – 3002) = –1495.9 J mol–1

dS = dqT =

dU + PdVT

For 1 mol of ideal gas, PVm = RT

d(PVm) = RdT = PdVm + VmdP

PdVm = RdT – VmdP = RdT – RTP dP

dSm = dUm + RdT –

RTdPP

T = CP,mdT

T – R dPP


∆Sm/J K–1 mol–1 = ∫255.3

300 (28.58 + 0.0176 T) dT

T – 8.314 ln 410

= 28.58 ln 255.3300 + 0.0176 (255.3 – 300) + 7.618 = 2.22

∆Sm = 2.22 J K–1 mol–1

3.52. ∆H° = 2(–285.85) – 393.51 + 74.81 = –890.4 kJ mol–1

∆G° = 2(–237.13) – 394.36 + 50.72 = –817.9 kJ mol–1

∆S° = ∆ ∆H G

T

°− °=

( .4 . )

.

890 817 9 1000

29815

+ ×= –243.2 J K–1 mol–1

3.53. a. and b. True only for an ideal gas.

c. True only if the process is reversible.

d. True only for the total entropy.

e. True only for an isothermal process occurring at constant pressure.

3.54. First vaporize the water at 1 atm pressure:

∆S1 = 40 600373.15 = 108.8 J K–1

Then expand from 1 bar to 0.1 bar:

∆S2 = R ln 10 = 19.1 J K–1

The net entropy change is

∆S = ∆S1 + ∆S2 = 108.8 + 19.1

= 127.9 J K–1

The Gibbs energy change is

∆G = ∆H – T∆S = 40 600 – (373.15 × 127.9)

= –7125.9 J = –7.13 kJ

3.55. ∆H° = –205.0 + 104.6 = –100.4 kJ mol–1

∆G° = –108.74 + 32.2 = –76.5 kJ mol–1

∆S° = ∆ ∆H G

T

°− °=

–100 400 + 76 540298.15

= –80.0 J K–1 mol–1

66 n CHAPTER 3

n Energy Conversion

3.56. 100-atm engine: Th = 585 K; Tc = 303 K

Efficiency = 585 – 303

585 = 48.2%

5-atm engine: Th = 425; Tc = 303 K

Efficiency = 425 – 303

425 = 28.7%

3.57.wqc

= 293 – 269

269 = 0.089

w = 0.089 × 104 = 890 J min–1 = 14.8 J s–1

At 40% efficiency, power = 14.8/0.4 = 37.0 J s–1

= 37.0 W

3.58. Performance factor = Th

Th – Tc

a. 59.6%; b. 11.9%; c. 6.6%

3.59. Efficiency = 10732273 = 0.472

1 liter = 8.0 kg = 8000 g

114.2 g mol–1 = 70.05 mol

Energy produced = 70.05 mol × 5500 kJ mol–1

= 3.85 × 105 kJ

Work = 0.472 × 3.85 × 105 = 1.82 × 105 kJ

3.60. Let qh be the heat supplied to the building at 20 °C and qc be the heat taken in by the heat pump at10 °C:

qhqc

= 293.15283.15

Work supplied to heat pump:

w = qh – qc = qh

1 –

283.15293.15 = 0.034 qh

Let q′h be the heat produced by the fuel at 1000 °C and q′

c be the heat rejected at 20 °C. Workperformed by the heat engine and supplied to the heat pump is


w = q′h – q′

c = q′h

1 –

293.151273.15 = 0.770 q′

h

Thus

0.034 qh = 0.770 q′h

and the performance factor is

qh

q′h =

0.7700.034 = 22.6

3.61. The heat that must be removed from 1 kg (= 55.5 mol) of water in order to freeze it is

6.02 × 55.5 = 334 kJ = qc

This is the heat gained by the refrigerator.

a. If the efficiency were 100%,

ThTc

= 298.15273.15 = −

q

qh

c

= –qh

334 kJ

Thus, the heat discharged at 25 °C is

–qh = 365 kJ

Work required to be supplied to the refrigerator is

365 – 334 = 31 kJ

With an efficiency of 40%, the actual work will be

10040 × 31 = 78 kJ

b. The heat discharged at 25 °C will be

334 + 78 = 412 kJ.

n Thermodynamic Relationships

3.62. a. Using the given relationship and the definitions of α and κ, we have

Substituting into Eq. (3.128), we obtain

b. Using the chain rule of partial differentiation, we obtain

( )( ) .

/

/

κα−=

∂∂∂∂

−=

∂∂

T

P

T PV

TV

T

P

.κ

κακα PT

TPV

U

T

−=

+−=

∂∂

).()( TPVVPT

P

V

V

U

P

U

TTT

ακκκ

κα −=−

−=

∂∂

∂∂=

∂∂

68 n CHAPTER 3

3.63. .T T

H ST V

P P

∂ ∂ = + ∂ ∂

Since ,T P

S V

P T

∂ ∂ − = ∂ ∂ we get

.T P

H VV T

P T

∂ ∂ = − ∂ ∂

Since 1

,P

V

V Tα ∂ = ∂

we get

( )1 .T

HV T

Pα∂ = − ∂

3.64. a. Vm = RT/P. Therefore, (∂Vm/∂T )P = R/P, and the cubic expansion coefficient is

1/( ) 1/ .m

mm P

VR PV T

V Tα ∂ = = = ∂

b. Since Vm = RT/P, (∂Vm/∂P )T = –RT/P2. Therefore,

2 21/( ) 1/ ,m

mm T

VRT P V P

V Pκ ∂ = − = = ∂

since (RT/P Vm) = 1.

3.65. From the first and second laws

dU = TdS – PdV

Therefore

∂∂V

UT

= T

∂∂V

ST – P

= T

∂∂T

PV – P

using the Maxwell equation, Eq. 3.124. From the van der Waals equation

∂∂T

PV =

PVm – b =

1T

P +

aV2

m

∂∂V

UT =

aV2

m

3.66. µ ≡

∂∂P

TH

= P

T

TH

PH

)/(

)/(

∂∂∂∂−

= –

∂∂

P

H

CP

1T

Since dH = TdS + VdP,


P

H

∂∂

T= T

∂∂P

ST + V

= –T

∂∂T

VP + V

from the Maxwell equation, Eq. 3.125. Thus

µ = P

P

C

VT

VT −

∂∂

= mP

mP

m

C

VT

VT

,

−

∂∂

for 1 mol

The equation P(Vm – b) = RT applies to 1 mol of gas, and it follows that

∂∂T

VP =

RP

Therefore

µ =

RTP – Vm

CP,m

3.67. a.

∂∂T

GP = –S (Eq. 3.119)

Therefore

∂∂

2

2

T

G

P = –

S

T

∂ ∂ P

dS = qrevT =

CPdTT

S

T

∂ ∂ P

= CPT

CP = –T

∂∂

2

2

T

G

P

b.

∂∂

P

CP

T =

∂∂

∂∂

PT

H

P T =

∂∂

∂∂

TP

H

T P

∂∂

P

HT = –T

∂∂T

VP + V (see Problem 3.66)

70 n CHAPTER 3

∂∂

∂∂

TP

H

T P = –T

∂∂

2

2

T

VP –

∂∂T

VP +

∂∂T

VP

∂∂

P

CPT = –T

∂∂

2

2

T

V

P

3.68. A = U – TS

dA = dU – TdS – SdT

At constant temperature

dA = dU – TdS = dq + dw – TdS

But dq = TdS, and therefore dA = dw.

3.69. dU = TdS – PdV (Eq. 3.105)

Dividing by dV at constant T:

V

U

∂∂

T = T

V

S

∂∂

T – P = 0

But

∂∂V

ST =

∂∂T

PV

(Eq. 3.124)

and therefore

∂∂T

PV =

PT

Integrating, ln P = ln T + const, or P ∝ T.

Thus

PV = const × T

3.70. By Euler’s reciprocity theorem (Appendix C),

∂∂V

SU

= –V

S

SU

VU

)/(

)/(

∂∂∂∂

∂∂V

US = –P (Eq. 3.116)

∂∂

S

UV = T (Eq. 3.116)

∴

∂∂V

SU

= PT

For an ideal gas, U depends only on T so that

∂∂V

SU

=

∂∂V

ST


For an isothermal process involving n mol of an ideal gas

dS = nRd ln V = nRdV

V = PdV

T

Thus, (∂S/∂V)T

= P/T and therefore (∂S/∂V)U

= P/T.

3.71. a. Eq. (3.159) gives the following relationship for the fugacity of a gas:

Dividing both sides by RT and using the definition of the compression factor Z, we get

Which is the desired result.

b. Dividing both sides of the given equation of state by RT, we get

Z = 1 + [b – A/(RT2/3)](P/RT) = 1 + [b/(RT) – A/( RT5/3)]P, which means

(Z – 1)/P = b/(RT) – A/( RT5/3).

Now, setting P1 = 0 and integrating, we obtain

3.72. Using the expression for fugacity derived in Problem (3.71), we get

which yields

Therefore, the fugacity of the gas f = P × exp(-0.622 84) = 268 bar.

∫ ∫

−

=

−=

2

1

2

1

.ln2

2P

P

P

P

mm dP

P

RTPVdP

P

RTV

P

fRT

∫

−=

2

1

,1

ln2

2P

PdP

P

Z

P

f

∫ ∫ ∫

+

−=

−=

P P P

PdPRT

bdP

RT

ab

RTdP

P

Z

P

f0 0 0

211

ln

∫

−=

−=

2

023/5

2

.1

lnP

PRT

A

RT

bdP

P

Z

P

f

.84 622.02

11ln 2

2

−=

+

−=

P

RT

bP

RT

ab

RTP

f

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