则T的估算值太大，应减小T的值。通过这样的试差过程可解出共沸温度为
Taz?330.144K。在共沸点处各组分的活度系数为
ln?1?lnp?lnp1s?lnp?A1?B1/(Taz?C1)?0.01069
sln?2?lnp?lnp2?lnp?A2?B2/(Taz?C2)?0.3114
将式（4）除以式（5），得
2/A12x12 （11） ln?1/ln?2?A21x2将x2用x1表示，并整理得
n [1?A12l?12A/(21?ln2x?)1]x?2? 1 0 （12） 1将各组分的活度系数数值及van Laar 方程参数值代入式（12），并整理得 (x1az)2?2.06889x1az?1.03444?0 （13）
解此一元二次方程便可得到乙酸甲酯-甲醇体系101.325kPa下的共沸组成为
x1az?0.8457
5-28 丙酮（1）-己烷（2）系统在101.325kPa下液相可用Wilson方程表示其非理想性，方程参数(g12?g22)/R?582.075K，(g21?g11)/R?132.219K，试求液相组成x1=0.25时的沸点及气相组成。已知丙酮和己烷的摩尔体积分别为73.52cm3?mol-1和130.77cm3?mol-1，其饱和蒸气压可用Antoine方程Antoine方程常数分别为 表示，logpis?Ai?Bi/T(?Ci丙酮 A1 = 6.24204，B1 = 1210.595，C1 =－43.486 己烷 A2 = 6.03548，B2 = 1189.640，C2 =－46.870
解：在101.325kpa下丙酮-已烷系统的气相可当做理想气体处理，于是汽液平衡
方程为
py1?ps1x?1,1py?2psx? (a) 22 p?p1sx??psx2?2 (b) 11 其中活度系数用Wilson方程计算
21
??12?21? (c)
ln?1??ln(x1??12x2)?x2???x??xx??x?11222211???12?21?ln?2??ln(x2??21x1)?x1??? (d)
?x1??12x2x2??21x1? 式（c）和式（d）中参数?12和?21为
V2L?g?g??12?Lexp??1222??1.7787exp??582.075/T? (e)
V1RT??V1L?g?g??21?Lexp??2111??0.56221exp??132.219/T? (f)
V2RT?? 饱和蒸汽压可用Antoine方程表示，即
lgp1s?6.24204?1210.595/(T?43.486) (g)
slgp2?6.03548?1189.640/(T?46.870) (i)
采用试差法解出泡点温度T，具体方法是给定一个初值T?0?，由式(c)和式(d)计算
出?2，由式(g)和式(i)得到饱和蒸汽压值，再由式(b)计算出总压 p。若计算出的总压 p 值大于101.33kPa，说明TkPa，则说明T?0?值太大；反之，若计算出的p值小于101.33
?0?太小。试算两个T后，可用线性插入得到后面的试算值，即
i?T 以Ti?1??T???101.3?2p5??i??T???T??ii?1 (j) p???p??ii?1?0??331K和T?1??330K开始计算，计算过程各变量值如下：
i 0 1 2 3 4 T?i? ?1i 2.40686 2.41065 2.42870 2.43025 2.43036 ?2i 1.12894 1.12945 1.13193 1.13214 1.13216 p1S?i? 107.518 103.940 88.372 87.137 87.047 S?i?p2 p?? i331.00 330.00 325.30 324.90 324.87 70.554 68.192 57.917 57.103 57.143 124.43 120.40 102.83 101.43 101.33 22
因此丙酮-已烷体系在液相浓度为x1组成可由式(a)求得
?0.25时的泡点T=324.87 K，其汽相
y1?p1S?1x1/p?87.047?2.43036?0.25/101.33?0.5220
?5-29 已知乙酸乙酯-水（1）（2）二元系统在70 ℃时的互溶度数据为x1?0.0109，
x1??0.7756，试确定NRTL方程中的参数。设NRTL方程的第三参数选定为， ?12
＝0.2。
?解：对于乙酸乙酯（1）-水（2）二元系统，已知x1?0.0109、x1??0.7756，有
??x2?1?x1?1?0.0109?0.9891
?x2?1?x1??1?0.7756?0.2244
?x1?x2x1??x2?x2?x1?x2x1?0.0109?0.011020.98910.7756??3.4560.2244
0.9891??90.740.01090.2244??0.28940.7756?将这些数据代入到NRTL方程中，并与相平衡关系式联立求解
??x1?1?x1??1? ????x2?2?x2?2??x1?x2?1
?x1??x2?1ln?1??x2??2???2?G?G21211212?? ?22??x??x?G??x??x?G??2212112?1?ln?2??x1??2???2?G?G12122121?? ?22??x??x?G??x??x?G??1121221?2? 23
ln?1??x2??2???2?G?G21211212?? ?22??x??x?G??x??x?G??2212112?1???2?G?G12122121?? ?22??????x?xG??x?xG??1121221?2??ln?2??x1??2G21?exp???12?21? G12?exp???12?12?
????x1???G21ln????21?x1??x1?G21???x??2?????x??G12ln2???12??x2??x2?G12???x??12?????????G???G211212????2??21???x?x1????G12????x??G21??1?1???x2????22?????G? 1212??2???x?????1?1?G12????x2??2?????????21G21???G12????2??12????x?x2???G21????x??G12??1?2???x1????12?????21G21? ??2???x???G??1?2???x1?21???2????exp?0.2??exp?0.2???????211212ln71.15???21???2??0.01102?exp??0.2?21??????1?0.01102exp??0.2?12???????
2????exp??0.2?21??12exp??0.2?12??????21???2?3.456?exp?0.2????21?????1?3.456exp??0.2?12???????2????exp??0.2?12??21exp??0.2?21???ln0.2269???12???2??90.74?exp??0.2?12??????1?90.74exp??0.2?21???????
2????exp??0.2?12??21exp??0.2?21??????12???2?0.2894?exp?0.2????12?????1?0.2894exp??0.2?21???????解得?12?0.03
?21?4.52
5-30 A-B是一个形成简单最低共熔点的系统，液相是理想溶液，并已知下列数据
24
