答案；1
① p1V1?nRTT p2/p1?T2/T1 T2?p2T1/p1=250 K 1 p2V2?nR21??1??1??② T3p3??T2p2? T3?T2???p2???p?3???50??250???100??1?1.41.4?304.75K
③ Q?Q1?Q2?Q1?nCV,m(T2?T1)?4?2.5?8.314?(250?750)??41570J
?U?nCV,m(T3?T1)?4?2.5?8.314?(304.75?750)??37018J ?H?nCp,m(T3?T1)?4?3.5?8.314?(304.75?750)??51825J ?S??S1?nCV,mlnT2250?4?2.5?8.314?ln??91.34J·K?1 T1750④ ?G??H??(TS)??H?(T3S3?T1S1)
?S0?nCp,mlnpT1?nRln0?292.80?13.48?279.30J·K?1 T0p1?1 S1?S0??S0?4S???S?4?191.6?279.30?1045.12J·K0mS3?S1??S?954.38J·K?1
?G??H?(T3S3?T1S1)=－51825－(304.75×945.38－750×1045.12)=441168 J
2解：反应：C2H6(g)= C2H4(g)+ H2(g)
?mol (1) ?rHm=52.26+84.68=136.94kJ·?1?mol?1·K?1 =219.56+130.684－229.60=120.644J·?rSm???mol?1＞0，=?rHm－T?rSm=136940－298×120.644=100988J·不能自发进行 ?rGm???rGm(2) ?rGm??RTlnK K?exp???RT??????18? ?1.985?10??11
(3) 425 K时，?G????1rm=?rHm－T?rSm=136940－425×120.644=85666J·mol K??exp??85666??11??8.314?425???2.957?10
3解：(a) ??-1vHm=33.273 kJ·mol
Q=△H=33.273 kJ
W=－p△V=－RT=－8.314×383.15=－3185.5 J △U= Q+W=30.087 kJ·mol-1
△S=△H/T=33273/383.15=86.84 J·K-1 △G=0
(b) △U, △H, △S,△G均同(a)
4
?Hm(268)=?H=?H1+?H2-?H3=?Hm(278.5)+ [C??p,m(l)－Cp,m(s)][278.5－268]
= 9956+(127.3－123.6)×10.5=9994.85 J/mol
?Sm(268)=?S=?S1+?S2-?S3=
9956278.5+ [C?p,m(l)－C?278.5p,m(s)]ln268 = 35.85 J·K-1·mol-1
5解： 惰性气体为单原子,
T不变，?U=?H=0
W=nRTln
p2p=－5743 J 1Q=－W=5743 ?S=
Q=19.14 J.K-1 T?G=?A=－T?S=－Q=5743 J
6解：（1）p=nRT5?8V=.314?37310=1551Pa
12
（2）
?H=?H1+?H2=?H1=5×40.64kJ=203.2 kJ W=0
?U=?U1+?U2=?U1=?H－?(pV)= ?H－nRT=203.2－5×8.314×373×10-3=187.7kJ Q=?U=187.7kJ
7解： He为理想气体 T不变，?U=?H=0
W=nRTln
p2=2305 J p1Q=－W=－2305 J ?S=
Q=－5.76 J.K-1 T?G=?A=－T?S=－Q=2305 J
8解：（1）比较式： ln p0=－
?H+C lnp0=19.434-1568/T RT ?vapHm=1568R=13036 J.mol-1
（2）同理升华焓 ?subHm=1975R=16420 J.mol-1 （3）熔化焓?fusHm=?subHm－?vapHm=3384 J.mol-1
（4）三相点的温度下：固态的蒸汽压=液态的蒸汽压
19.434-1568/T=23.136-1975/T 解得温度：T=110K
9解：绝热恒压
（A）1mol水由30℃→100℃需吸热：Q1=1×76×(100－30)=5320 J 3mol水蒸气由150℃→100℃放热：Q2=3×33×50=4950 J Q2 5320?4950=9.107×10-3mol 40630最终体系的平衡相态为100℃，有水1.0091mol，有水蒸气2.9909mol， （B）体系的终态温度为100℃ （C）绝热恒压，?H=0 ?S= ?S1(1mol水30℃→100℃)+?S2(3mol水蒸气150℃→100℃) +?S3(9.107×10-3mol水蒸气→水) 13 ?H373373 =nC(水)ln303+ nC水气)lnvapmp,mp,m(423－?n373 =15.769-12.454-0.992=2.35 J.K -1 10解：(A) 1摩尔理想气体300K由30 L恒温可逆膨胀至100 L ?U=0 W=－nRT ln V2V=－3003 J 1 Q=－W =3003 J ?S= Q/T=10.01 J.K-1 (B) 问真空膨胀是不可逆过程；?U、?S同(A)，Q=0 ，W=0 11解：在101.325kPa下 ?S= ?S1+ ?S2+ ?S3 ?S3731=Cp,m,Lln 298=75.291 ln 373298=16.902 J.K-1 ?S?H406902= T=373=109.088 J.K-1 ?S4234233= Cp,m,gln 373=35.58 ln 373=4.476 J.K-1 ?S=130.466 J.K-1 W=－p?V=－pVg=－nRT2=－1×8.314×423=－3516.8 J ?H= ? H 1+ ? H 2+ ? H 3=75.291×(100－25)＋40690＋35.58×(150－100)=48116 J 12解：(1) lnp2?p?H??1?1??1R??T1T2?? 对苯(A)：lnp3135.06??H?1R??363?1?178.65?H?11?368??① ln135.06?R??363?373??② 式①/式②得：p?3?pA?155.6kPa 对氯苯(B)：lnp327.73??H?11?39.06?H?11?R??363?368??③ ln27.73?R??363?373??④ 式③/式④得：p?3?pB?33.0kPa 14 ?33.0xBpBpBxB(2) 此时 yB?0.3000?= ???pA?pBpA(1?xB)?pBxB155.6(1?xB)?33.0xB? 得： xB?0.67 p?p? (1?x)?pABBxB?73.8kPa?(3) 此时 xB?0.3000 p?p? 97kPaA(1?xB)?pBxB?10.yB?pB33.0?0.3??0.09 pA?pB109.713解：(a) 苯为溶剂A： pA=p?1?xB?=10.01×(1－0.0385)=9.6246 kPa A?pB=yB=1-0.095 pB=91.687 kPa pA?pBp=pA+pB=9.6246+91.687 =101.31 kPa (b) pB=kBxB kB=pB/xB=2381 kPa 14解：CS2(B)和CH3COCH3(A)体系， aA= aApApyA= =，对B的公式类推(B代替A即可) ?A**xApApAp2?VHm= Rp1?11????TT?? 2??1-1 纯组分的蒸气压用克-克方程计算：ln对CH3COCH3(A)：T1=329.7K p1=101.325kPa T2=317.5K ?VHm=32370 J.mol 得 p2=pA=0.6352×101.325kPa 对CS2(B)：T1=319.7K p1=101.325kPa T2=317.5K ?VHm=27780 J.mol 得 p2=pB=0.9301×101.325kPa *-1 *aA= pyA0.620.9761==0.9751 ==1.220 ?A0.63521?0.2p*ApyB0.380.4086?==0.4086 ==2.043 A*0.2pB0.930111 ， nB=－1=0.111mol 0.91?nBaB= 15解：(A) 在1molA中加入nBB，xA(α相)=0.9= (B) xA(β相)=0.1 ，xA(α相)=0.1= 1 ，nB=9mol 1?nB 15