Question

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

(a) : Given temperature, T = 58.1 C = 58.1 + 273 = 331.1 K

We can calculate the vapor pressure of pure components by applying Antoine equation.

For n-heptane: A = 4.02832, B = 1268.636, C = - 56.199

logP⁰_h = A - B / (C+T) = 4.02832 - 1268.636 / ( - 56.199 + 331.1K)

=> P⁰_h = 0.260 atm = 0.260 atm x ( 760 mm Hg / 1 atm) = 198 mm Hg

For cyclohexane: A = 4.13983, B = 1316.554, C = - 35.581

logP⁰_c = A - B / (C+T) = 4.13983 - 1316.554 / ( - 35.581 + 331.1K)

=> P⁰_c = 0.484 atm = 0.484 atm x ( 760 mm Hg / 1 atm) = 368 mm Hg

From Raoult's law in binary solution,

285 mm Hg = X_h x P⁰_h + (1 - X_h) x P⁰_c

=> 285 mm Hg = X_h x (198 mm Hg) + (1 - X_h) x (368 mmHg) ------- (1)

=> 170 x Xh = 83

=> Xh = 0.488

Hence mole fraction of heptane = 0.488 (answer)

mole fraction of cyclohexane = 0.512 (answer)

(b) If we double the pressure inside the container, the saturation pressure of vapor pressure will also increase as from Raoult's law the pressure inside the container is directly proportional to saturation pressure (vapor pressure)