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

P_soln = P⁰_n-heptane x mole fraction of n-heptane + P_Cyclohexane x mole fraction of Cyclohexane

Vapour pressure of n-heptane = 285 x 0.54 = 154.89

Vapour pressure of cyclohexanol = 285 x 0.45 = 130.10

P_soln =( 154.89 x 0.54) + ( 130.10 x 0.45 )

= 142.18 mmHg

Mole fraction in liquid phase =

No. of moles = 58 / 84 = 0.69 (100 - 32)

= 0.69

b) Saturation pressure decreases bcoz as pressure increases the molecule sare converted to liquid phase from solid pha se.