31??R?T2?298???R?T2?298??，T2?202.6K25??3?U?nCV,m?T2?T1??R??202.6?298???1990J?W25?H?nCp,m?T2?T1??R??202.6?298???1983J2?T2??p1?-1????S?nCp,mln??nRln?5.36J?K?T??p??1??2??A??U??S2T2?S1T1???1190??118?202.6?112.6?298??8454JS1?112.6J?K-1，S2?S1??S?112.6?5.36?118J?K-1
7．解：
?G??H??S2T2?S1T1???1983??118?202.6?112.6?298??7661J
∵ 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?? = J ?2?Q1 = ΔH1 = 40670 J ，Q2 = - W2 =
W1 = -p(Vg-Vl) = -pVg = -RT = -3101 J， W2 = J
Q' = Q1 + Q2 = 40670 + = J，W' = W1 + W2 = = J
ΔU = Q'-W' = = 37569 J
ΔH2 = 0，ΔH = ΔH1 = 40670 J，向真空膨胀：W = 0，Q = ΔU = 37569 J 40670?1?＋Rln??3730.5?? = + = J·K-1 ΔS = ΔS1 + ΔS2 =
ΔG = ΔH - TΔS = 40670 - 373 × = J
ΔA = ΔU + TΔS = 37569 - 373 × = J
8．解：Pb + Cu(Ac)2 → Pb(Ac)2 + Cu ，液相反应，p、T、V均不变。
W' = ，Q = J，W(体积功) = 0，W = W'
ΔU = Q + W = = J
QR-1
ΔH = ΔU + Δ(pV) = ΔU = J ΔS = T = 213635/298 = J·K ΔA = ΔU - TΔS = J，ΔG = J
9．解：确定初始和终了的状态
VHe?VH2初态：
终态：关键是求终态温度，绝热，刚性，ΔU = 0
nRTHe2?8.314?283.2??0.04649m35p1.013?10nRTH22?8.314?293.2???0.02406m35p1.013?10
nHeCV,m?He???T2?THe??nH2CV,m?H2??T2?TH2?0??即： 2 × × = 1 × × ， ∴ T2 = K
V2 =
= + = m
33
He m， K ) → ( m， K ) H2 m， K ) → ( m， K ) ?T2?SHe?nCV,m?He?ln??T?1所以：
??V2???nRln??V??1????
3
3
VHe?VH23
?287.7??0.07055?-1?2?1.5?8.314?ln???2?8.314?ln???7.328J?K?283.2??0.04649?
同理：
?SH2?8.550J?K-1?SH2
∴ΔS = ?SHe +
10．解：
= + = J·K
ΔH = ΔH1 + ΔH2 + ΔH3 = × 10) - 6032 - × 10) = J
?273?6032?263?ln?ln???263273?-273+ × ?? = J·K-1 ΔS = ΔS1 + ΔS2 + ΔS3 = × ?
ΔG = ΔH-TΔS = - + 263 × = J ?pl??ln??p?ΔG≈ΔG2 = - RT × ?s?
?pl?pl?G214.82?ln???p??s? = RT = 263?8.314 = ，ps=
11．解： 用公式 ΔS = -R∑nilnxi = - × × + × + × = J·K 12．解：
-1
ΔCp = × × = J·K
ΔHT = ∫ΔCpdT + Const = + Const，∵T = 298K 时，ΔH298 = -30585Ｊ 代入，求得：Const = -29524，ΔHT = - 29524，代入吉-赫公式， ?G2?G1??823298积分，
-1
?2988233.65?29524T2dT?66.91
?G210836.6???66.91?30.55，?G2?25143J823298
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?? = × 373 × = J ΔG1 = 0 ；ΔG2 =
ΔH3 = Cp,m × (473 - 373) = × 100 = 3358 J
?-1-1??196.24?Rln2?202.0J?K?mol??298??p??p??473??Sm?473??S?298??Cp,mln??209.98J?K-1?mol-1??Rln????298??p? Sm?373??S?p373???298??Cp,mln??Rln???ΔG3 = ΔH3 - (S2T2 - S1T1) = 3358 - ×473 – ×373) = J ΔG = ΔG2 + ΔG3 = - = -22766 J = kJ
1???＝14．解：
?p1?7，T2?T1??p??5?2???298??0.1??27?576K
5Q = 0， W = ΔU = nCV,m(T2-T1) = 10 × 2R × (576 - 298) = kJ
7ΔU = kJ，ΔH = nCpm(T2-T1) = 10 × 2R × (576 - 298) = kJ
--3
ΔS = 0 ，ΔG = ΔH - Sm,298ΔT = - 10 × × (576 - 298) × 10 =
ΔA = ΔU - SΔT = 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?T??T??VdU = 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?
?1?11??aa?1????????a??????2a??V??V??V?VVV1??21?1? ?2?2V2??S?V2??p?V2R?V2?S?dV?Rln???dV???dV??VV1??V?V1??T?V1V?1TV???????
?1?V21????A??U?T?S??a???RTln?V??VV1??2?1?????1?V2?1????G??H?T?S??2a???RTln?V??V??V1??2?1?
再求过程量，此时考虑实际过程恒温可逆： W???V2V1pdV???V2V1?1?V2a?1??RT???2?dV??a???RTln????VV??V?V2V1??1????
????
?V2Q?T?S?RTln??V?1对于恒温可逆过程：
物理化学第二章热力学第二定律练习题及答案
