Sample Problem #3:

A mixture of 50 g mol of liquid benzene and 50 g mol of water is boiling at 101.32 kPa pressure. Liquid benzene is immiscible in water. Determine the boiling point of the mixture and the composition of the vapor. Which component will first be removed completely from the still? Vapor-pressure data for the pure components are as follows:

Temperature Pwater Pbenzene

K C (mm Hg) (mm Hg)

308.5 35.3 43 150

325.9 52.7 106 300

345.8 72.6 261 600

353.3 80.1 356 760

Given:

Required: boiling temperature and which component is completely removed firstSolution:

Get Ptotal = Pwater + Pbenzene:

Temperature Pwater Pbenzene Ptotal

K C (mm Hg) (mm Hg) (mm Hg)

308.5 35.3 150 43 193

325.9 52.7 300 106 406

345.8 72.6 600 261 861

353.3 80.1 760 356 1116

50 gmol water

50 gmol benzene

Vapors

Ptotal = 760 mm Hg

Plot temperature vs vapor pressure:

From the plot, at Ptotal = 760 mm Hg:

boiling temperature = 68.5CPwater = 230 mm Hg

Pbenzene = 530 mm Hg Because Pbenzene (=530 mm Hg) > Pwater (=230 mm Hg), benzene will be the component that will be completely removed first.

Answers: t = 68.5C and benzene

t = 68.5C

760530230

totalbenzenewater