??yP???4ye1y2 ? ?f ?f111113??ln??(1?y) 同理 ln?22dln?3223 ?y2?y2?(1?y2)(1?3y2)?1?3y2?2y2dy2323)(1?3y2?2y2)???e(1?3y22?2y2 ?? f?4ye222??2.568ΜPα f??3.29M 当y1?y2?0.5时：f 7Pa124-28解：当汽—液两相平衡时，须满足
?v?f?l fii 上标v和l分别指的是汽相和液相。
若汽相可视为理想气体，则系统的压力比较低，液相的标准态逸度fi??Pis，则
有
?Py??fx?Py??Psx ?iiiiiiiii 如果存在恒沸物，即yi?xi
?1P2s 对二元物系来说，则有?s
?2P1?APBs8?104 对本题 ?s??0.67 5?BPA1.2?10GE?0.5xAxB 已知RT?(nGE/RT)2 ?ln?A?[ ]T.P.nB?0.5xB?nA
?A?e20.5xB
同理?B?e ?0.5x2A?A?e0.5(x?BB?xA)?0.67
解得：xA?0.9056
0?xA?1，说明在353K时该系统有共沸物存在。 4-29解：
（1）Van Laar方程 ln?1?A12(A21x2A12x1)2 ， ln?2?A21()2
A12x1?A21x2A12x1?A21x2 式中A12和A21由恒沸点的数据求得。 在恒沸点，yi?xi ??i?PyiP ?ssPixiPi 则?1?P101.3P101.3，??1.0191????1.0414 2ss99.4097.27P1P2 A12?ln?1(1?x2ln?220.475ln1.04142)?ln1.0191(1?)?0.1635
x1ln?10.525ln1.0191x1ln?120.475ln1.01912)?ln1.0414(1?)?0.0932
x2ln?20.525ln1.0414A21?ln?2(1?全浓度范围内，苯和环己烷的活度系数为
20.0932x20.1635x22 ln?1?0.1635()?20.1635x1?0.0932x2(1.7543x1?x2)0.1635x10.0932x122 ln?2?0.0932()?20.1635x1?0.0932x2(x1?0.5700x2)（2）Statchard和Hildebrand方程
V1?89cm?mol ，V2?109cm?mol
2V1?22109x20.461x7892222?ln?1?(?1??2)??()(18.82?14.93)?2RT8.3145?350.889x1?109x2(0.816x51?x2)2V2?1289x20.565x51092222?ln?2?(?1??2)??()(18.82?14.93)?2RT8.3145?350.889x1?109x2(x1?1.224x72)3?13?1?1?18.82(J0.5?cm?1.5)，?2?14.93(J0.5?cm?1.5) ?1?x1V189x1x2V2109x1，?2? ??x1V1?x2V289x1?109x2x1V1?x2V289x1?109x2 在恒沸点，x1?0.525，x2?0.475
0.4617?0.4752 ?ln?1??0.1276 ?1?1.1361 2(0.8165?0.525?0.475)0.565?50.5225 ln?2? ?1?1.1357 ?0.127 32(0.525?51.224?70.47)5 活度系数比（1）中计算偏大。
（3）当x1?0.8时，x2?0.2
0.1635?0.22 用Van Laar方程：ln?1??0.002544 ?1?1.0025 2(1.7543?0.8?0.2)0.0932?0.82 ln?2??0.07140 ?2?1.074 0(0.8?0.5700?0.2)2 y1??1P1sx1P?1.0025?99.40?0.8?0.79
101.3 用Scatchard和Hildebrand方程：
0.4617?0.22 ln?1??0.02537 ?1?1.02569
(0.8165?0.8?0.2)20.5655?0.82 ln?2??0.3315 ?2?1.3930 2(0.8?1.2247?0.2) y1??1P1sx1P?1.02569?99.40?0.8?0.81
101.3
4-30解：根据Wilson方程 ln?i?1?ln(??ijxj)??jk?kixk ?kjxj 将上式应用于三元系统，并将已知的各参数代入，即可求得该三元系统各组分的
活度系数： ln?1?1?ln(x1?x2?12?x3?13)?x1
x1?x2?12?x3?13 ?x3?31x2?21 ?x?21?x2?x3?23x1?31?x2?32?x30.34
0.34?0.33?0.7189?0.33?0.5088 ?1?ln(0.34?0.33?0.5088)?
?
0.33?1.18160.33?0.9751??0.030.34?1.1816?0.33?0.33?0.52290.34?0.9751?0.33?0.5793?0.33故 r1?1.03
lnr2?1?ln(x1?21?x2?x3?23)?x3?32x1?12x2??x1?x2?12?x3?13x1?21?x2?x3?23x1?31?x2?32?x30.34?0.71890.34?0.33?0.7189?0.33?0.50880.330.33?0.5793? ?
0.34?1.1816?0.33?0.33?0.52290.34?0.9751?0.33?0.5793?0.33 ?0.183
?1?ln(0.34?1.1816?0.33?0.33?0.5229)? 故 r2?1.20
lnr3?1?ln(x1?31?x2?32?x3)?x1?13
x1?x2?12?x3?13 ?
x2?23x3 ?x1?21?x2?x3?23x1?31?x2?32?x30.34?0.50880.34?0.33?0.7189?0.33?0.50880.33?0.52290.33? ?
0.34?1.1816?0.33?0.33?0.52290.34?0.9751?0.33?0.5793?0.33 ?0.348
?1?ln(0.34?0.9751?0.33?0.5793?0.33)? 故 r3?1.42
在50℃下该三元体系的总压力P
000 P?r1x1P1?r2x2P2?r3x3P3
?(1.03?0.34?81.82?1.20?0.33?78.05?1.42?0.33?55.58)KPa ?85.59KPa 平衡时的气相组成：
r1P10x11.03?81.82?0.34y1???0.335P85.59r2P20x21.20?78.05?0.33??0.361 y2?P85.59r3P30x31.42?55.58?0.33y3???0.304P85.59