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)》之铭言： : 问题如下: : 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. : 我现在如果设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) : 可是我现在不知道Ptotal是多少? : 有没有大大可以帮帮我的? : 谢谢喔^^ --

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