A system of propane and n-butane exists in two-phase vapor/l

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A system of propane and n-butane exists in two-phase vapor/liquid equilibrium at 10.0
bar and 323 K.The mole fraction of propane is about 0.67 in the vapor phase and about
0.40 in the liquid phase.Additional pure propane is added to the system,which is brought
again to equilibrium at the same T and P,with both liquid and vapor phases still present.
What is the effect of the addition of propane on the mole fractions of propane in the
vapor and liquid phases?

这一题考查的是“杠杆原理（lever rule）”.从二元气液平衡相图上可以看出来,加入纯丙烷后,在同温同压下达到新的相平衡,此时丙烷在气相中的摩尔分数仍然是0.67,在液相中的摩尔分数仍然是0.40.
设加入纯丙烷以前,体系的总物质的量为n,气相总物质的量为n(g),液相总物质的量为n(l),丙烷在气相中摩尔分数为x2=0.67,丙烷在液相中摩尔分数为x1=0.40,体系中丙烷的总摩尔分数为xp
那么加入纯丙烷后,体系的总物质的量为n',气相总物质的量为n'(g),液相总物质的量为n'(l),丙烷在气相中摩尔分数为x2=0.67,丙烷在液相中摩尔分数为x1=0.40,体系中丙烷的总摩尔分数为xp'
n'>n,xp'>xp
在原来的平衡中,n·xp = n(l)·x1 + n(g)·x2
n = n(l) + n(g)
所以 n(l)·(xp - x1) = n(g)·(x2 - xp)
写成比例形式 n(g) / n(l) = (xp - x1) / (x2 - xp)
同样的,加入了纯丙烷后的新平衡中
n'(l)·(xp' - x1) = n'(g)·(x2 - xp')
n'(g) / n'(l) = (xp' - x1) / (x2 - xp')
由于xp' > xp,所以 n'(g) / n'(l) > n(g) / n(l)
也就是说,加入了纯丙烷达到了新的平衡后,体系中气相所占的比例升高了,液相所占的比例下降了.
呼……终于敲完了……

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