Question

A gaseous mixture containing 0.170 mol fraction of carbon tetrachloride (CCl4) and 0.830 mol fraction of dry air is initially at 61.0°C and 1037 mmHg.

If this mixture is cooled at a constant pressure, at what temperature does the CCl4 first start to condense?

At what temperature would 44.0% of the CCl4 condense?

Answer 1

moles of CCl4 vapor/ mole of dry air = partial pressure of CCl4/ partial pressure of dry air

mole of CCl4/moles of dry air = mole fraction CCL4*total pressure/ mole fraction air*total pressure =0.17/0.83=0.205

moles of CCl4= 0.205* moles of dry air

when 44% of CCl4 condenses,

moles of CCl4 remaining = 56%

Moles of CCl4 = 0.204*0.56* moles of dry air =0.114 moles of dry air

moles of CCl4/ moles of dry air =0.114

0.114 = partial pressure of CCl4 vapor/ partial pressure of dry air

let x= partial pressure of CCl4 after condensation

0.114= x/(1037-x)

x =0.114*(1037-x)

x = 118-0.114x

1.114x= 118 , x= 118/1.114 =106 mm Hg

Since at the temperature to be calculated, partial pressure = Vapor pressure

Vapor presure of CCl4= 106 mm Hg

Antoine equation is logP= 6.89410-1219.58/(t+227.170), P is in mmHg and t in deg.c

log 106= 6.89410- 1219.58/(t+227.170)

2.025= 6.89410- 1219.58/(t+227.170)

1219.58/(t+227.170)= 6.89410-2.025= 4.87

t+227.170= 1219.58/4.87= 250.427

t = 250.427-227.170= 23.257 deg.c