# Vapour Pressure and Mass of Water Vapor

A room is filled with saturated moist air at 30oC and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg, the mass of water vapor is:

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Solution:

The water vapors in air behaves as ideal gas and therefore we may consider the moist air as a mixture of two ideal gases, where total pressure is the sum of the pressure exerted by the components individually. The saturated vapor pressure is a constant at a given temperature and is measured experimentally. We have tables for S V P at different temperatures. According to the table the S V P at 30C is 31.7 mm of mercury which is equal to 1333 Pa. So the pressure exerted by the air is 100 - 1.333 = 98.667 K Pa =98667 Pa.

Using ideal gas equation P V = n RT = (m/M) RT (m is the mass of gas and M is molar mass) the volume V of the air in the room or the volume of the room is given as V = (m/M) (RT/P) = (100000/29)(8.31 x 303)/98667 = 88 m^3.

So now the if the mass of water vapor is m' then using the gas equation P' V = (m'/M') RT or m'/M' = P'V / RT = 46.6 (' for vapor)

Therefore mass of vapor m' = 46.6 x 18 = 838.8 gm =0.8388 kg.

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