Question

3. A closed container at 58.1˚C and 285 mm Hg contains a mixture of n-heptane and cyclohexane. Both liquid and vapor phases exist in equilibrium, and the vapor phase contains 32 mol % n-heptane.
a) Calculate the mole fractions of the liquid phases.
b) Now, the pressure in the container doubles, will the saturation pressure of n-heptane increase, decrease, or stay the same? Justify your answer.

Answer 1

32 % of n-Heptane is in vapour phase then (100 -32) % = 68 % of n-heptane is in liquid phase.

32% of n-heptane is in vapour phase then (100 - 32) %= 68 % if cyclohexane in vapour phase.

hence 68% of cyclohexane in liquid phase.

vapour phase contain,

mole fraction of n-hexane = 0.32/(0.32+0.68) = 0.32

mole fraction of cyclo hexane = 0.68/(0.32+0.68) = 0.68

Liquid phase contains,

Mole fraction of n-hexane = 0.68

mole fraction of cyclohexane = 0.32

If we increase the pressure the gas molecule dissolve more in the given solution. because association of molecules takes place due to reduced volume of the container.