?T2??p1????S3?Cpln??Rln?T??p???1??2? (2) 由公式：
?T??T2??S1?Cpln??S?Cln?2V?T?1???T设 C 点温度为 T'，(在等压下)，
???(在恒容下)
?T??T2???S1??S2?Cpln??C?Rln??p?T??T??1????T2??T2???Cpln??Rln???T??T??1??T2??p1???Cpln??Rln?T??p?1??2∴ ΔS3 = ΔS1 + ΔS2
?????T2p2?等容??C????B：???Tp1??p2V2p1V1??或：T?；T?21?nRnR???
5．证明：
???A????A??????????n???????i?T,V,nj?i??T?V,n?????????ni?T????????T,V,nj?i????????i??????T?????V,n??S???n?i????T,V,nj?i????V,n
???i???Td???SdT?Vdpiii∵ ，∴ ???S???n?i????i??????T???T,V,nj?i???p???S?V??ii???T??V,n ?V,n则
???p??S?V???ii??T??V,n?V,n
6．解：(1)等温过程：ΔU = ΔH = 0，
?p1?Q??W?nRTln?W??3987.5J?p???1?8.314?298?ln5?3987.5J；?2?Q3987.5?S?R??13.38J?K-1，?A??G??3987.5JT298
?p2??Q??W?p?V2?V1??RT?1????8.314?298??1?5??1982Jp1??(2) ΔU = ΔH = 0， ?p1?-1?S?nRln?，?G??A??3987.5J?p???13.38J?K?2?
1???p1?5??，T2?T1??p??3?2?(3)
??298?5?25?156.8K
3R??156.8?298???1761J，Q?025?H?nCp,m?T?R??156.8?298???2934J，W??U??1761J2?p1?-1?S?0，S1?S2?S?298K??Rln??p???126?Rln5?112.6J?K?2??U?nCV,m?T??G??H?S?T2?T1???2934?112.6??156.8?298??12965J3Q?0，?U?W，R?T2?T1???p2?V2?V1?2(4)
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?T??p???5.36J?K?1??2?S1?112.6J?K-1，S2?S1??S?112.6?5.36?118J?K-1?A??U?S?T2?T1???1761?112.6??156.8?298??14318J
?A??U??S2T2?S1T1???1190??118?202.6?112.6?298??8454J
7．解：
∵ p = 0， ∴ W = 0，设计如图，按1，2途经计算：
?G??H??S2T2?S1T1???1983??118?202.6?112.6?298??7661J
?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·ΔS = ΔS1 + ΔS2 = K1
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
QR-ΔH = ΔU + Δ(pV) = ΔU = 121796.2 J ΔS = T = 213635/298 = 716.9 J·K1 ΔA = ΔU - TΔS = -91838.8 J，ΔG = -91838.8 J
9．解：确定初始和终了的状态
nRTHe2?8.314?283.23VHe???0.04649mp1.013?105nRTH22?8.314?293.2VH2???0.02406m35p1.013?10初态：
终态：关键是求终态温度，绝热，刚性，ΔU = 0 nHeCV,m?He???T2?THe??nH2CV,m?H2??T2?TH2?0
即： 2 × 1.5R × (T2-283.2) = 1 × 2.5R × (293.2-T2) ， ∴ T2 = 287.7 K
V?VH2V2 = He = 0.04649 + 0.02406 = 0.07055 m3 He (0.04649 m3，283.2 K ) ( 0.07055 m3，287.7 K ) H2 (0.02406 m3，293.2 K ) ( 0.07055 m3，287.7 K ) ?T2?SHe?nCV,m?He?ln??T?1所以：
??V2???nRln??V??1????
???287.7??0.07055?-1?2?1.5?8.314?ln???2?8.314?ln???7.328J?K?283.2??0.04649?
-1?S?8.550J?KH2同理：
?SH2∴ΔS = ?SHe + = 7.328 + 8.550 = 15.88 J·K10．解：
ΔH = ΔH1 + ΔH2 + ΔH3 = (75.4 × 10) - 6032 - (37.7 × 10) = -5648.4 J
?273?6032?263?ln?ln???263273???? = -20.66 J·273ΔS = ΔS1 + ΔS2 + ΔS3 = 75.4 × -+ 37.7 × K-1
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·K1
12．解：
-ΔCp = 65.69-2 × 26.78-0.5 × 31.38 = -3.56 J·K1
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，
8233.65?29524?G2?G1??dT?66.912298823298T积分，
??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? = 8.314 ×ΔG1 = 0 ；ΔG2 = 373 × ln0.5 = -2149.5 J
Sm?373??SSm?473??S?p373??298??Cp,mln????Rln???298??-1-1??196.24?Rln2?202.0J?K?mol??p??-1-1??209.98J?K?mol??p?
?p473???298??Cp,mln??Rln????298?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???＝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) = 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?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?11???dV??a??V?V1V1V2V1? ?2所以：
又 ?H??U???pV???U??p2V2?p1V1?
?U??V2??U???dV??V??T?V2a?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???V2V1a?RT?2?V?V?1?V21??????RTln?dV??a??V??VV?1??2?1????
?V2Q?T?S?RTln??V?1对于恒温可逆过程： ????