CCl4 Removal by Humidification

Carbon tetra chloride is to be removed from a polymer solution by bubbling dry air through it at 297 K. The resulting mixture has % relative humidity of 70. It is required to remove 90% of carbon tetra chloride present by compressing to a suitable pressure and cooling to 283 K. What this pressure should be?

Data: Vapor pressure of CCl4 at 297 K = 12.2 kN/m2 and at 283 K = 6 kN/m2

Calculations:

Initial conditions:

Temperature = 297 K

Total pressure = 101 kN/m2

Relative humidity = pA/pS = 0.70

pA = 0.7 x 12.2 = 8.54 kN/m2

moles of CCl4 / moles of dry air = 8.54 / (101.3 - 8.54) = 0.09207

CCl4 present per mole of dry air = 0.09207 mole

Final conditions:

Temperature = 283 K

Total pressure = to be determined

Relative humidity = 100 %

CCl4 to be present in the exit air per mole of dry air = 0.09207 x 0.1 = 0.009207 mole

If the partial pressure of a vapor is greater than its vapor pressure, it will condense till its partial pressure is equal to vapor pressure.

i.e., 6/(pT - 6) = 0.009207

pT = 657.71 kN/m2

that is, the air mixture has to be compressed to 657.71 kN/m2 and cooled to 283 K to remove 90% of carbon tetra chloride.