Assuming the vapor only consists of benzene and toluene
Equal partial pressures --> P(t) = P(b) = y(t)P = y(b)P
y(t) = y(b) = 0.5
Ideal gas, ideal solution: Raoult's Law
xiPi(sat) = yiP
x(t) = y(t)P/28
x(b) = y(b)P/95 = 1 - x(t)
and y(t) = y(b) = 0.5
combining the above 3 equations
x(t) = 1 - 0.5P/95 = 0.5P/28
Solve for P and substitute back to get
x(t) = 0.5P/28
x(b) = 1 - x(t)
the algebra is trivial
※ 引述《dougho (Doug)》之銘言：
: 問題如下:
: Benzene and toluene form an ideal solution. At 298 K, what is the mole fraction
: of benzene in the liquid that is in equilibrium with a vapor that has equal
: partial pressures of benzene and toluene? At 298 K, the vapor pressures of
: pure benzene and pure toluene are 95 and 28 torr, respectively.
: 我現在如果設x是mole fraction of benzene 然後1-x 就是mole fraction of toluene
: 所以Ptotal = (x)(95 torr) + (1-x)(28 torr)
: 可是我現在不知道Ptotal是多少?
: 有沒有大大可以幫幫我的?
: 謝謝喔^^
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