Question

Air containing 50,000ppm acetone at 1 atm is cooled from 150°F to 0°F. What percent of the acetone will condense when cooled?

Answer 1

Unit conversion:

deg F to deg C

C=5/9(F-32)

For 150 deg F ,C=5/9(150-32)=65.5 deg C

For 0 deg F, C=5/9(0-32)=-17.8 deg C

vapour pressure of acetone at 65.5 deg C using Antoine equation-

P=10^(A- B/C+T) where T in is deg C,P in mmHg

A=7.1327,B=1219.97,C=230.653 for -65 deg C <T<70 deg C

Vapour pressure of pure ethanol at 65.5 deg C=Pi=10^[7.1327-(1219.97/230.653+65.5)]=1031.1195 mmhg=1031.1195 mmhg*(1 atm/760 mmhg)=1.357 atm

Vapour pressure of pure ethanol at -17.8 deg C=Pf=10^[7.1327-(1219.97/230.653-17.8)]=25.1875 mmhg=25.1875mmhg*(1 atm/760 mmhg)=0.003314 atm

Now you have following values:

fraction of acetone=Xacetone=50,000 ppm=50,000*10^-6 =0.05

At T=65.5 deg C, Po=vapour pressure of pure ethanol=1.357 atm

So,At T=-17.8 deg C,Po=0.003314 atm ,

As fraction of acetone varies directly with its vapour pressure,p

,p=X*Po [Raoults law]

So, percent decrease in vapour pressure(p) is equal to percent decrease in its mole fraction (X)

percent decrease in vapour pressure =[(1.357 atm-0.003314 atm , )/1.357 atm]*100=99.756%

percent decrease in its mole fraction (X)=99.756% [answer]

new value for X=0.05-(99.756%*0.05)=0.05-0.99756*0.05=0.000122=122 *10^-6

or in units of ppm X=122ppm