第一章习题解答 第 1 页 共 6 页
第一章
1. 体系为隔离体系, ?U=0 2. ?V?nRTp2习题解答
W=Q=0
?T1??Rp?nRT1p?nRp?T2 ?W=p?V=R
3. (1) W=p?V=p(Vg–Vl)?pVg=nRT=1?8.314?373.15=3102 J 4. (1) 态1?态2 一般过程 (2) 态1?态2 绝热过程
W?RTlnV2V1?8.314?373.2ln0.10.025?4299 J(2) W=p?V=p(Vs–Vl)??11?11???M?101325?p?????0.018?0.16 J???33?0.92?101.0?10??l??s
Q 不确定 确定，Q=0 W 不确定 确定，W=–?U Q-W 确定值 确定值
?U 确定值 确定值 5. (1) 恒温可逆膨胀
(2) 真空膨胀
W = 0
W = p外(V2–V3) =
RTV2(3) 恒外压膨胀
?V2?V1??V1?? ?RT?1??V2???0.025???8.314?373.2??1??= 2327 J
0.1??
(4) 二次膨胀
12???W=W1 + W2 ?RT??1?V??RT?1?V?
2?3????V??V?
0.025?0.05????RT?1???RT?1???3103 J
0.05?0.1???6. (1) 理想气体为体系，等压过程， Q=?H>0
(2) 电热丝+理想气体为体系，等压过程，Q=0，?U=–W’>0，?H=?U+?(pV)>0
7. ?H=n??Hm,汽化=40670 J ?U=?H–?(pV)=?H–p(Vg-Vl)=40670–101325(30200–1880)?10–6 8. 过程 (1)理想气体自由膨胀 (2)理想气体恒温可逆膨胀 (3)理想气体节流膨胀 (4)理想气体绝热恒外压膨胀 (5)水蒸气通过蒸气机对外做功后复原，以水蒸气为体系 (6)水(p?，0?C)? 冰(p?，0?C) (7)刚性容器中石磨燃烧，整体为体系
– 0 +或?0 0 – 0 – + Q 0 + 0 0 + W 0 + 0 + + ?U 0 0 0 – 0 ?H 0 0 0 – 0 =40670–3058=37611 J
第一章习题解答 第 2 页 共 6 页
9. Cp,m=29.07–0.836?103T+2.01?10–6T2
(1) Qp=?H?n?TT2111??32?63?Cp,mdT??29.07??0.836?10T??2.01?10T?23??3001000
=20349–380+625=20.62 kJ
(2) QV=?U=?H–?(pV)=?H–(p2V2–p1V1) =?H–nR(T2–T1)
–3
=20.62–R(1000-300)?10=14.80 kJ 10．（1）等温可逆膨胀 ?U =?H = 0
Q =W ?nRTlnp1p2?p1V1lnp1p2?506.6?10?2?103?3ln51?1631 J
（2）等温恒外压膨胀
?U =?H = 0
Q = W = p2 (V2–V1) = p2V2–p2V1= p1V1–p2V1= (p1–p2)V1 =(506.6-101.3)?103?2?10–3 = 810 J
W=p(Vg–Vl)?pVg?RT=8.314?373=3.1 kJ ?U =Q=37.6 kJ
11. （1）常压蒸发：Q=?H=40.7 kJ 12.
T1?p1V1nR?273.2 K ?U=Q–W=37.6 kJ
(2) 真空蒸发：?H=40.7 kJ W=0
p1T1p2?136.5 K(1) p1T1=p2T2
T2?
V2?nRTp22?136.5R4?0.0028 m3
(2) ?U=nCV,m(T2–T1)= ?H=nCp,m(T2–T1)=
3252R(136.5?273.2)??1702 J
R(136.5?273.2)??2837 J(3) 以T为积分变量求算:
pT=C(常数) p?CT
V?nRTp?nRTc/T?nRTC2
dV?nRC?2TdT
W??pdV??C2nRT??dT?2nRTC?dT?2nR(T2?T1)??2270 J
也可以用p或V为积分变量进行求算。
13．?U=nCV,m(T2–T1)=20.92?(370–300)=1464 J
?H=nCp,m(T2–T1)=(20.92+R)?(370–300)=2046 J 始态体积 压力
V1?RT1p1?0.0246 m3 体积变化：V2
?V3?RT3p3?0.003026 m3
p2?RT2V2?821554 PaW=W1+W2=p2(V2–V1)+0=821554?(0.003026–0.0246)=–17724 J Q=?U+W=1464–17724=–16260 J
第一章习题解答 第 3 页 共 6 页
14. (1) ?H=Cp,m(T2–T1)?
p2V2p1V1?nRT252R(T2?273.2)?2092p2105 T2=373.8 K p2=0.684?105 Pa J
nRT1
32?2?373.8273.2
?U= CV,m(T2–T1)?R(373.8?273.2)?1255W=Q–?U=1674–1255=419 J
恒温，W恒容，W＝01(2) 状态1???????中间状态?????2???状态2
状态函数变化同上 ?U=1255 J ?H=2092 J W=W1?nRTln
15. 双原子分子
?p1???T1??p??2??V2V1=8.314?273.2?ln2=1574 J
Q=?U+W=1255+1574=2829 J
52CV,m?1??R
???Cp,mCV.m1.41??1???1.4
T1p1??T2p2?
1?1.4
T2?p?273.2??0.5p???????224.1 K
?3W=–?U=–nCV,m(T2–T1)?Cp,mCV.m28.828.8?R?52R?224.1?273.2??1020 J16. (1) ??
??1.406 n?p1V1RT1??3p?1.4?10298R1.406??0.1755 mol
?p1V1?p?2V2
p2?V1???p1??V??2??1.43??3p???2.86???114.7 kPa
T2?p2V1RT2?224.9 K
(2) ?U=nCV.m(T2–T1)=n (28.8–R)?(224.9–298)= –263 J ?H=nCp.m(T2–T1)=n ?28.8?(224.9–298)= –369 J
17. 设物质量为1mol
??Cp,mCV,m?3.5R2.5R?1.4
始态温度：T1???12V2p1V1nR??1?2?105?0.011?8.3141.4?1?240.6 K
求终态温度：
?U?nC?H?nC??1T1V1?T
?V1??T2?T1??V??2??10??240.6??30???155.0 K
V,m(T2?T1)?5272R(155.0?240.6)??1779R(155.0?240.6)??2490 J J
p,m(T2?T1)?Q=0 W=–?U=1779 J
若设物质量为n mol，可如下计算： 始态温度T1?p1V1nR?240.6n， 终态温度T2?V1???T1??V??2???1?155.0n
第一章习题解答 第 4 页 共 6 页
在计算?U、?H、W时n可消去，得相同的结果。
18. 证明 U=H–pV
19. 证明
???H???p????H???U???H?Cp?CV????????????????V??T?T?T?T?T??p??V??p??V??V??????U???H???V???V???????p???Cp?p????T?p??T?p??T?p??T?p
（1）
H=f(T,p)
??H??H?dH???dT????p??T?p???dp??T
???p??????T??T?VV不变，对T求导
Cp?CV??H?????p???H??H???H???????????T?V??T?p??p 代入(1)
???p???p???p????V??????????T?V??T?V?T??T?V???H???????p????V???T??
20．
nQV+C?T=0
0.5100QV?2.94?8175.5?0
QV=–4807200 J
?n =–4
–1
–1
C7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g)
?cHm = QV + ?nRT =–4807200–4R?298 = –4817100 J?mol =–4817.1 kJ?mol
21．(1) 2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方) ?n =–3 QV =–223.8 kJ ?rHm = QV + ?nRT = –223.8 + (–3)RT?10–3 = –231.2 kJ
22. (1) ?=4 mol
23． ?fH?m(kJ?mol–1)
(2) ?=2 mol
(3) ?=8 mol
(2) 2C(石墨) + O2(g) = 2CO(g) ?n = 1
?rHm = QV + ?nRT = –231.3 +RT?10–3 = –228.8 kJ (3) 2H2(g)+Cl2(g) = HCl (g) ?rHm = QV =–184 kJ
?n =0
QV =–231.3 kJ QV =–184 kJ
2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)
–411 –811.3 –1383 –92.3
?rH?m=?(n?fH?m)产物–?(n?fH?m)反应物= (–1383–2?92.3)–(–811.3–2?411) = 65.7 kJ ?rU?m =?rH?m–?nRT=65.7–2RT?10–3=60.7 kJ
7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l)
–286
–3230
24．生成反应
?cH?m(kJ?mol–1) –394
?rH?m=?(n?cH?m) 反应物–?(n?cH?m) 产物= [7?(–394) + 3?(–286)] – (–3230)= –386 kJ
25. 反应 C(石墨) ? C(金刚石) ?cH?m(kJ?mol–1) –393.5 –395.4
?rH?m=?cH?m，石墨–?cH?m，金刚石 =–393.5–(–395.4)=1.9 kJ
第一章习题解答 第 5 页 共 6 页
26. 3C2H2 (g) = C6H6 (l)
–1
?fH?m (kJ?mol) 226.73 49.04 ?cH?m (kJ?mol–1) –1300 –3268
–1
由生成热数据计算： ?rH?m=49.04–3?226.73=–631.15 kJ?mol ?rU?m=?rH?m–?nRT=–631.15+3?8.314?298.2?10–3=623.71 kJ?mol–1
由燃烧热数据计算： ?rH?m=3?(–1300)–(–3268) =–632 kJ?mol ?rU?m=?rH?m–?nRT=–632+3?8.314?298.2?10–3=624.6 kJ?mol–1
KCl(s) ? K(aq, ?) + Cl(aq, ?)
–435.87 ? –167.44
+
–
–1
27. 反应 ?fH?m(kJ?mol–1)
?rH?m=17.18 kJ
?rH?m=?(n?fH?m)产物–?(n?fH?m)反应物
17.18=[?fH?m (K+,aq, ?)–167.44]–(–435.87)
?fH?m (K+,aq, ?)=–251.25 kJ?mol–1
28. 生成反应
H2(g) + 0.5O2(g) = H2O(g) ?rH298=–285.8 kJ?mol–1
Cp,m(J?K–1?mol–1) 28.83
?rH373??rH298???CpdT29.16 75.31 ?Cp=75.31–(28.83+0.5?29.16)=31.9J?K–1 =–285.8+31.9?(373–298)?10–3=–283.4 kJ?mol–1
c?107 –3.376 20.12 –30.46 –117.904
2
–1
29. 反应N2(g) + 3H2(g) = 2NH3(g) ?rH298=–92.888 kJ?mol
N2(g) H2(g) NH3(g) ?
?rH398??rH298?a 26.98 29.07 25.89 –62.41
b?103 5.912 –0.837 33.00 62.599
??CpdT??rH298?2???a??bT??cT398?dT
??rH2981????aT??bT2????cT?3?29831
398??rH29811??32?73???(?62.41)T?(62.6?10)T?(?117.9?10)T?23??298
=–92880+[–6241+2178–144]= –97086 J
?rH291=49.455 kJ
30. 按图示过程计算：
H2(g) + I2(s) 2HI(g) Cp,m=55.64 J?K–1?mol–1 ?HI2 s?l T=386.7K ?熔H?m=16736 J ?HHI ?HH2 Cp,m=62.76 J?K–1?mol–1 Cp.m=7R/2 Cp.m=7R/2 l ?g T=457.5K ?蒸H?m=42677 J –1–1Cp,m=7R/2 J?K?mol H2(g) + I2(g) ?HH2=nCp,m?T=3.5R(473–291)=5296 J
?HHI=nCp,m?T=2?3.5R(473–291)=10592 J
?HI2=?H1(s,291?386.7K) + ?H2(s?l) + ?H3(l,1386.7?457.5K) + ?H4(l?g)
+ ?H5(g,457.5?473K)
?rH473=？ 2HI(g) 第一章习题解答 第 6 页 共 6 页
=55.64?(386.7–291)+16736+62.76?(457.5–386.7)+42677+3.5R?(473–457.5)
=69632 J
?HH2+?HI2+?rH473=?rH291+?HHI ?rH473=–14881 J
5296+69632+?rH473=49455+10592