发布时间 : 星期六 文章物理化学第二章热力学第二定律练习题及答案更新完毕开始阅读c227ca200540be1e650e52ea551810a6f524c818
6．解：(1)等温过程：ΔU = ΔH = 0，
?p2??Q??W?p?V2?V1??RT?1????8.314?298??1?5??1982Jp1??(2) ΔU = ΔH = 0，
1??25?p1?5??，T2?T1??p??32??(3)
??298?5??156.8K
3Q?0，?U?W，R?T2?T1???p2?V2?V1?2(4)
7．解：
∵ p = 0， ∴ W = 0，设计如图，按1，2途经计算：
?p1??1??nRTln?＝8.314?373?ln???p?0.5?? = 2149.5 J ?2?Q1 = ΔH1 = 40670 J ，Q2 = - W2 =
W1 = -p(Vg-Vl) = -pVg = -RT = -3101 J， W2 = -2149.5 J
Q' = Q1 + Q2 = 40670 + 2149.5 = 42819.5 J，W' = W1 + W2 = -3101-2149.5 = -5250.5 J
ΔU = Q'-W' = 42819.5-5250.5 = 37569 J
ΔH2 = 0，ΔH = ΔH1 = 40670 J，向真空膨胀：W = 0，Q = ΔU = 37569 J
40670?1?＋Rln??3730.5?? = 109.03 + 5.76 = 114.8 J·K-1 ΔS = ΔS1 + ΔS2 =
ΔG = ΔH - TΔS = 40670 - 373 × 114.8 = -2150.4 J ΔA = ΔU + TΔS = 37569 - 373 × 114.8 = -5251.4 J
8．解：Pb + Cu(Ac)2 → Pb(Ac)2 + Cu ，液相反应，p、T、V均不变。
W' = -91838.8J，Q = 213635.0 J，W(体积功) = 0，W = W' ΔU = Q + W = 213635-91838.8 = 121796.2 J
QRΔH = ΔU + Δ(pV) = ΔU = 121796.2 J ΔS = T = 213635/298 = 716.9 J·K-1
ΔA = ΔU - TΔS = -91838.8 J，ΔG = -91838.8 J
9．解：确定初始和终了的状态
VHe?VH2初态：
nRTHe2?8.314?283.23??0.04649mp1.013?105nRTH22?8.314?293.23???0.02406mp1.013?105
终态：关键是求终态温度，绝热，刚性，ΔU = 0
即： 2 × 1.5R × (T2-283.2) = 1 × 2.5R × (293.2-T2) ， ∴ T2 = 287.7 K
V2 = VHe?VH = 0.04649 + 0.02406 = 0.07055 m3
2He (0.04649 m3，283.2 K ) H2 (0.02406 m3，293.2 K )
→ ( 0.07055 m3，287.7 K ) → ( 0.07055 m3，287.7 K )
?T2?SHe?nCV,m?He?ln??T?1所以：
??V2???nRln??V??1????
同理：
?SH2?8.550J?K-1?SH2
∴ΔS = ?SHe + 10．解：
= 7.328 + 8.550 = 15.88 J·K
ΔH = ΔH1 + ΔH2 + ΔH3 = (75.4 × 10) - 6032 - (37.7 × 10) = -5648.4 J
?273?6032?263?ln?ln???263273???? = -20.66 J·K-1 ΔS = ΔS1 + ΔS2 + ΔS3 = 75.4 × -273+ 37.7 ×
ΔG = ΔH-TΔS = - 5648.4 + 263 × (-20.66) = -214.82 J
?pl??ln??p?ΔG≈ΔG2 = - RT × ?s?
?pl?pl?G214.82?ln???p??s? = RT = 263?8.314 = 0.09824，ps= 1.1032
11．解： 用公式 ΔS = -R∑nilnxi
= - 8.314 × (0.2 × ln0.2 + 0.5 × ln0.5 + 0.3 × ln0.3) = 8.561 J·K-1
12．解：
ΔCp = 65.69-2 × 26.78-0.5 × 31.38 = -3.56 J·K-1
ΔHT = ∫ΔCpdT + Const = -3.65T + Const，∵T = 298K 时，ΔH298 = -30585Ｊ 代入，求得：Const = -29524，ΔHT = -3.65T - 29524，代入吉-赫公式，
?G2?G1??823298积分，
?2988233.65?29524T2dT?66.91
恒温恒压下，ΔG > 0，反应不能自发进行，因此不是形成 Ag2O 所致。
13．解：
?p2?RTln??p??1?? = 8.314 × 373 × ln0.5 = -2149.5 J ΔG1 = 0 ；ΔG2 =
ΔH3 = Cp,m × (473 - 373) = 33.58 × 100 = 3358 J
ΔG3 = ΔH3 - (S2T2 - S1T1) = 3358 - (209.98×473 – 202.00×373) = -20616.59 J ΔG = ΔG2 + ΔG3 = -20616.5 - 2149.5 = -22766 J = -22.77 kJ
1???p1?7??＝，T2?T1???5p2??14．解：
??298??0.1??27?576K
5Q = 0， W = ΔU = nCV,m(T2-T1) = 10 × 2R × (576 - 298) = 57.8 kJ
7ΔU = 57.8 kJ，ΔH = nCpm(T2-T1) = 10 × 2R × (576 - 298) = 80.8 kJ
ΔS = 0 ，ΔG = ΔH - Sm,298ΔT = 80.9 - 10 × 130.59 × (576 - 298) × 10--3 = -282.1kJ
ΔA = ΔU - SΔT = -305.2 kJ
15．解：这类题目非常典型，计算时可把热力学量分为两类：一类是状态函数的变化， 包括ΔU、ΔH、ΔS、ΔA、ΔG，计算时无需考虑实际过程；另一类是过程量，包括
Q、W，不同的过程有不同的数值。
先求状态函数的变化，状态变化为 ：(p1,V1,T1)
→ (p2,V2,T2)
??p???U???S???p???T???p?T??V?V?T??????VTTdU = TdS - pdV ，
RRT2a??p???p?a??，???????2?3?p?2?V?RTVV ??V?TV?对状态方程 ? 而言：??T?VVRa??U????T??p?2VV ∴ ??V?T?U?所以：
?V2V1??U???dV???V?T?V2V1?11???dV??a???2V?V2V1? a又 ?H??U???pV???U??p2V2?p1V1? 再求过程量，此时考虑实际过程恒温可逆：
?V2Q?T?S?RTln??V?1对于恒温可逆过程： ????