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