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CCl4 Removal by Humidification

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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 CCl_{4} at 297 K = 12.2 kN/m^{2} and at 283 K = 6 kN/m^{2}

Calculations:

Initial conditions:

Temperature = 297 K

Total pressure = 101 kN/m^{2}

Relative humidity = p_{A}/p_{S} = 0.70

p_{A} = 0.7 x 12.2 = 8.54 kN/m^{2 }

moles of CCl_{4} / moles of dry air = 8.54 / (101.3 - 8.54) = 0.09207

CCl_{4} present per mole of dry air = 0.09207 mole

*
*Final conditions:

Temperature = 283 K

Total pressure = to be determined

Relative humidity = 100 %

CCl_{4} 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/(p_{T} - 6) = 0.009207

p_{T} = 657.71 kN/m^{2}

that is, the air mixture has to be compressed to **657.71 kN/m**^{2} and cooled to 283 K to remove 90% of carbon tetra chloride.

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Last Modified on: 04-Feb-2022

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