Homework 3

Chapter 20 - 6, 17, 19, 25, 28

Chapter 21 - 3, 12, 13, 16, 42

Q1

20-6. Calculate qrevq_{rev} and ΔS\Delta S for a reversible cooling of one mole of an ideal gas at a constant volume V1V_1 from P1,V1,T1P_1, V_1, T_1 to P2,V1,T2P_2, V_1, T_2 followed by a reversible expansion at constant pressure P2P_2 from P2,V1,T2P_2, V_1, T_2 to P2,V2,T1P_2, V_2, T_1 (the final state for all the processes shown in Figure 20.3). Compare your result for ΔS\Delta S with those for paths A, B + C, and D + E in Figure 20.3.

20-6. 计算一摩尔理想气体在恒定体积 V1V_1 下从 P1,V1,T1P_1, V_1, T_1 可逆冷却到 P2,V1,T2P_2, V_1, T_2,随后在恒定压力 P2P_2 下从 P2,V1,T2P_2, V_1, T_2 可逆膨胀到 P2,V2,T1P_2, V_2, T_1(图 20.3 中所示所有过程的最终状态)时的 qrevq_{rev}ΔS\Delta S。将你得到的 ΔS\Delta S 结果与图 20.3 中路径 A、B + C 以及 D + E 的结果进行比较。

Q2

20-17. Show that

20-17. 证明

ΔSqT\Delta S \ge \frac{q}{T}

for an isothermal process. What does this equation say about the sign of ΔS\Delta S? Can ΔS\Delta S decrease in a reversible isothermal process? Calculate the entropy change when one mole of an ideal gas is compressed reversibly and isothermally from a volume of 100 dm3100\text{ dm}^3 to 50.0 dm350.0\text{ dm}^3 at 300 K.

对于一个等温过程。这个方程说明了 ΔS\Delta S 的符号有什么特征?在一个可逆等温过程中,ΔS\Delta S 能否减小?计算一摩尔理想气体在 300 K 下从 100 dm3100\text{ dm}^3 的体积可逆等温压缩到 50.0 dm350.0\text{ dm}^3 时的熵变。

Q3

20-19. Melting at the normal melting point (TfusT_{fus}) of a substance (the melting point at one atm) can be regarded as a reversible process because if the temperature is changed infinitesimally from exactly TfusT_{fus}, then the substance will either melt or freeze. Calculate the change in entropy when two moles of water melt at 00^\circC. The value of ΔHfus\Delta H_{fus} is 6.01 kJmol16.01\text{ kJ}\cdot\text{mol}^{-1}. Compare your answer with the one you obtained in Problem 20–18. Why is ΔSvap\Delta S_{vap} much larger than ΔSfus\Delta S_{fus}?

20-19. 物质在其标准熔点(TfusT_{fus})(一个大气压下的熔点)的熔化过程可被视为一个可逆过程,因为如果温度从恰好为 TfusT_{fus} 的值发生无限小的变化,物质便会熔化或凝固。计算两摩尔水在 00^\circC 时熔化的熵变。ΔHfus\Delta H_{fus} 的值为 6.01 kJmol16.01\text{ kJ}\cdot\text{mol}^{-1}。将你的答案与你在问题 20-18 中得到的答案进行比较。为什么 ΔSvap\Delta S_{vap} 远大于 ΔSfus\Delta S_{fus}

Q4

20-25. Calculate the change in entropy of the system and of the surroundings and the total change in entropy if one mole of an ideal gas is expanded isothermally and reversibly from a pressure of 10.0 bar to 2.00 bar at 300 K.

20-25. 如果一摩尔理想气体在 300 K 下从 10.0 bar 的压力等温可逆膨胀到 2.00 bar,计算系统熵变、环境熵变以及总熵变。

Q5

20-28. Plot ΔmixSˉ\Delta_{mix}\bar{S} against y1y_1 for the mixing of two ideal gases. At what value of y1y_1 is ΔmixSˉ\Delta_{mix}\bar{S} a maximum? Can you give a physical interpretation of this result?

20-28. 绘制两种理想气体混合时 ΔmixSˉ\Delta_{mix}\bar{S}y1y_1 变化的图像。在 y1y_1 为何值时 ΔmixSˉ\Delta_{mix}\bar{S} 达到最大值?你能对这一结果给出物理解释吗?

Q6

21–3. The molar heat capacity of butane can be expressed by

21–3. 丁烷的摩尔热容可以表示为

CˉP/R=0.05641+(0.04631 K1)T(2.392×105 K2)T2+(4.807×109 K3)T3\bar{C}_P/R = 0.05641 + (0.04631 \text{ K}^{-1})T – (2.392 \times 10^{-5} \text{ K}^{-2})T^2 + (4.807 \times 10^{-9} \text{ K}^{-3})T^3

over the temperature range 300 K T\le T \le 1500 K. Calculate ΔS\Delta S if one mole of butane is heated from 300 K to 1000 K at constant pressure.

在 300 K T\le T \le 1500 K 的温度范围内。如果一摩尔丁烷在恒定压力下从 300 K 加热到 1000 K,计算 ΔS\Delta S

Q7

21–12. Why is ΔvapSˉ>ΔfusSˉ\Delta_{vap}\bar{S} > \Delta_{fus}\bar{S}?

21–12. 为什么 ΔvapSˉ>ΔfusSˉ\Delta_{vap}\bar{S} > \Delta_{fus}\bar{S}

Q8

21–13. Show that if CP(T)TαC_P(T) \to T^\alpha as T0T \to 0, where α\alpha is a positive constant, then S(T)0S(T) \to 0 as T0T \to 0.

21–13. 证明如果当 T0T \to 0 时有 CP(T)TαC_P(T) \to T^\alpha,其中 α\alpha 为一正常数,那么当 T0T \to 0 时有 S(T)0S(T) \to 0

Q9

21–16. The molar heat capacities of solid, liquid, and gaseous chlorine can be expressed as

21–16. 固态、液态和气态氯的摩尔热容可以表示为

CˉP[Cl2(s)]/R=1.545+(0.1502 K1)T(1.179×103 K2)T2+(3.441×106 K3)T3\bar{C}_P[\text{Cl}_2(\text{s})]/R = -1.545 + (0.1502 \text{ K}^{-1})T – (1.179 \times 10^{-3} \text{ K}^{-2})T^2 + (3.441 \times 10^{-6} \text{ K}^{-3})T^3

15 K T\le T \le 172.12 K

CˉP[Cl2(l)]/R=7.689+(5.582×103 K1)T(1.954×105 K2)T2\bar{C}_P[\text{Cl}_2(\text{l})]/R = 7.689 + (5.582 \times 10^{-3} \text{ K}^{-1})T – (1.954 \times 10^{-5} \text{ K}^{-2})T^2

172.12 K T\le T \le 239.0 K

CˉP[Cl2(g)]/R=3.812+(1.220×103 K1)T(4.856×107 K2)T2\bar{C}_P[\text{Cl}_2(\text{g})]/R = 3.812 + (1.220 \times 10^{-3} \text{ K}^{-1})T – (4.856 \times 10^{-7} \text{ K}^{-2})T^2

239.0 K T\le T \le 1000 K

Use the above molar heat capacities and Tfus=172.12 KT_{fus} = 172.12 \text{ K}, ΔfusHˉ=6.406 kJmol1\Delta_{fus}\bar{H} = 6.406 \text{ kJ} \cdot \text{mol}^{-1}, Tvap=T_{vap} =

239.0 K, ΔvapHˉ=20.40 kJmol1\Delta_{vap}\bar{H} = 20.40 \text{ kJ} \cdot \text{mol}^{-1}, ΘD=116\Theta_D = 116 K and the correction for nonideality =

0.502  JK1mol1\text{ J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} to calculate the standard molar entropy of chlorine at 298.15 K. Compare your result with the value given in Table 21.2.

使用上述摩尔热容以及 Tfus=172.12 KT_{fus} = 172.12 \text{ K}ΔfusHˉ=6.406 kJmol1\Delta_{fus}\bar{H} = 6.406 \text{ kJ} \cdot \text{mol}^{-1}Tvap=T_{vap} =

239.0 K,ΔvapHˉ=20.40 kJmol1\Delta_{vap}\bar{H} = 20.40 \text{ kJ} \cdot \text{mol}^{-1}ΘD=116\Theta_D = 116 K 以及非理想性校正值 =

0.502  JK1mol1\text{ J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} 来计算氯在 298.15 K 时的标准摩尔熵。将你的结果与表 21.2 中给出的值进行比较。

Q10

21–42. Arrange the following reactions according to increasing values of ΔrS\Delta_r S^\circ (do not consult any references).

a. S(s) + O2_2(g) \to SO2_2(g)

b. H2_2(g) + O2_2(g) \to H2_2O2_2(l)

c. CO(g) + 3 H2_2(g) \to CH4_4(g) + H2_2O(l)

d. C(s) + H2_2O(g) \to CO(g) + H2_2(g)

21–42. 根据 ΔrS\Delta_r S^\circ 的递增值对下列反应进行排序(不要查阅任何参考文献)。

a. S(s) + O2_2(g) \to SO2_2(g)

b. H2_2(g) + O2_2(g) \to H2_2O2_2(l)

c. CO(g) + 3 H2_2(g) \to CH4_4(g) + H2_2O(l)

d. C(s) + H2_2O(g) \to CO(g) + H2_2(g)