Homework 3
Chapter 20 - 6, 17, 19, 25, 28
Chapter 21 - 3, 12, 13, 16, 42
Q1
20-6. Calculate qrev and ΔS for a reversible cooling of one mole of an ideal gas at a constant volume V1 from P1,V1,T1 to P2,V1,T2 followed by a reversible expansion at constant pressure P2 from P2,V1,T2 to P2,V2,T1 (the final state for all the processes shown in Figure 20.3). Compare your result for ΔS 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 V1 下从 P1,V1,T1 可逆冷却到 P2,V1,T2,随后在恒定压力 P2 下从 P2,V1,T2 可逆膨胀到 P2,V2,T1(图 20.3 中所示所有过程的最终状态)时的 qrev 和 ΔS。将你得到的 ΔS 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
ΔS≥Tq
for an isothermal process. What does this equation say about the sign of ΔS? Can ΔS 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 dm3 to 50.0 dm3 at 300 K.
对于一个等温过程。这个方程说明了 ΔS 的符号有什么特征?在一个可逆等温过程中,ΔS 能否减小?计算一摩尔理想气体在 300 K 下从 100 dm3 的体积可逆等温压缩到 50.0 dm3 时的熵变。
Q3
20-19. Melting at the normal melting point (Tfus) 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 Tfus, then the substance will either melt or freeze. Calculate the change in entropy when two moles of water melt at 0∘C. The value of ΔHfus is 6.01 kJ⋅mol−1. Compare your answer with the one you obtained in Problem 20–18. Why is ΔSvap much larger than ΔSfus?
20-19. 物质在其标准熔点(Tfus)(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 Tfus 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 0∘C 时熔化的熵变。ΔHfus 的值为 6.01 kJ⋅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 远大于 ΔSfus?
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 against y1 for the mixing of two ideal gases. At what value of y1 is ΔmixSˉ 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 随 y1 变化的图像。在 y1 为何值时 ΔmixSˉ 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 K−1)T–(2.392×10−5 K−2)T2+(4.807×10−9 K−3)T3
over the temperature range 300 K ≤T≤ 1500 K. Calculate ΔS if one mole of butane is heated from 300 K to 1000 K at constant pressure.
在 300 K ≤T≤ 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。
Q7
21–12. Why is ΔvapSˉ>ΔfusSˉ?
21–12. 为什么 ΔvapSˉ>ΔfusSˉ?
Q8
21–13. Show that if CP(T)→Tα as T→0, where α is a positive constant, then S(T)→0 as T→0.
21–13. 证明如果当 T→0 时有 CP(T)→Tα,其中 α 为一正常数,那么当 T→0 时有 S(T)→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 K−1)T–(1.179×10−3 K−2)T2+(3.441×10−6 K−3)T3
15 K ≤T≤ 172.12 K
CˉP[Cl2(l)]/R=7.689+(5.582×10−3 K−1)T–(1.954×10−5 K−2)T2
172.12 K ≤T≤ 239.0 K
CˉP[Cl2(g)]/R=3.812+(1.220×10−3 K−1)T–(4.856×10−7 K−2)T2
239.0 K ≤T≤ 1000 K
Use the above molar heat capacities and Tfus=172.12 K, ΔfusHˉ=6.406 kJ⋅mol−1, Tvap=
239.0 K, ΔvapHˉ=20.40 kJ⋅mol−1, ΘD=116 K and the correction for nonideality =
0.502 J⋅K−1⋅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 K,ΔfusHˉ=6.406 kJ⋅mol−1,Tvap=
239.0 K,ΔvapHˉ=20.40 kJ⋅mol−1,ΘD=116 K 以及非理想性校正值 =
0.502 J⋅K−1⋅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∘ (do not consult any references).
a. S(s) + O2(g) → SO2(g)
b. H2(g) + O2(g) → H2O2(l)
c. CO(g) + 3 H2(g) → CH4(g) + H2O(l)
d. C(s) + H2O(g) → CO(g) + H2(g)
21–42. 根据 ΔrS∘ 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(g) → SO2(g)
b. H2(g) + O2(g) → H2O2(l)
c. CO(g) + 3 H2(g) → CH4(g) + H2O(l)
d. C(s) + H2O(g) → CO(g) + H2(g)