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： ????