(1) ≠ (2) ，所以 δQ 不具有全微分，Q不是状态函数。 2．证明：U = f(T，p) ，∴
又 ∵ dU = TdS-pdV
??U???U???dT????p??dp?TdS?pdV??T?p??T??U???U?dU???dT????p??dp0?T??p??T
，∴
在等温下： dT = 0 ， ∴ 等温下两边同除以 dp ： 又∵
??S???V?????????p??T??p??T??U???p???U???p????dp?TdS?pdV0?T
??U???p????S???V??????T?p???p???p???T??T??T ，∴
dS?T???V???V?????T?p?????p???T??p?T??TCp??S????T??T?p
3．证明：等压下，∴
CpdT，∴。
??S???S???V?Cp?T???T??????T?p??V?p??T?p ，
??T???V????????p??S?p??S?由麦克斯韦关系式：代入，得：
，则
??p???S???????T?V?p??S?
??p???V???V???p?Cp?T??????T?????T?p??T?S??T?S??T?p 。
4．证明：(1) 由于内能是状态函数，所以： ΔU3 = ΔU1 + ΔU2
∵ΔU = Q + W， ∴ Q3 + W3 = (Q1 + W1) + (Q2 + W2)，
即 Q3-(Q1 + Q2) = (W1 + W2)-W3
由图上可知：(W1 + W2)-W3 = W1-W3 = ΔABC的面积 ≠ 0
∴ Q3-( Q1 + Q2) ≠ 0，Q3 ≠ Q1 + Q2
(2) 由公式：
?T2??p1????S3?Cpln??Rln?T??p???1??2?
(在等压下)，
设 C 点温度为 T'，
?T??S1?Cpln??T???1??Sln??T2?2?CV?T??(在恒容下)
?S?T?1??S2?Cpln?????C?R?ln??T2??T1??p?T???C?T?pln??2???Rln??T2???T1???T???C?等容???B：T2p2???T?p1???C?T?pln??2?T??p??Rln??1?p??或：p2V2p1V1??1??2???T2?nR；T1?nR??∴ ΔS
3 = ΔS1 + ΔS2
5．证明：
????A???????S?????A??????????n?????????T?V,n??i?T,V,nj?i???ni???T,V,n?????????????i?j?i??ni????T????V,n??????TT,V,nj?i????V,n?∵ d???i??SidT?V????i?idp ，∴
??T???Si?Vi???p?V,n??T??V,n
?则
??S????????ni????????i???Si?V??p?i?T,V,nj?i??TV,n??T??V,n
6．解：(1)等温过程：ΔU = ΔH = 0， Q??W?nRTln??p?1?p???1?8.314?298?ln5?3987.5J；W??3987.5J?2??S?QRT?3987.5298?13.38J?K-1，?A??G??3987.5J(2) ΔU = ΔH
=
Q??W?p?V?p2?V1??RT??1?2????8.314?298??1?5?p??1982J1?
?S?nRln??p?1???13.38J?K-1，?G??A??3987?p.5J2??
0，
1??(3)
?p1?5??，T2?T1??p??3?2???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??A??U?S?T2?T1???1761?112.6??156.8?298??14318J?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?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?G??H??S2T2?S1T1???1983??118?202.6?112.6?298??7661J
7．解：
∵ p = 0， ∴ W = 0，设计如图，按1，2途经计算：
Q1 = ΔH1 = 40670 J ，Q2 = - W2 =
?p1??1??nRTln?＝8.314?373?ln???p?0.5???2?
= 2149.5 J
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
ΔS = ΔS1 + ΔS2 = = 109.03 + 5.76 = 114.8 J·K-1
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 ΔH = ΔU + Δ(pV) = ΔU = 121796.2 J ΔS = = 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
QRT40670?1?＋Rln??373?0.5?