**Air at 30 ^{o}C and 150 kPa in a closed container is compressed and cooled. It is found that the first droplet of water condenses at 200 kPa and 15^{o}C. Calculate the percent relative humidity of the original air. The vapor pressures of water at 15^{o}C and 30^{o}C are 1.7051 kPa and 4.246 kPa respectively**.

Calculations:

At 15^{o}C, and 200 kPa water is at 100% humidity.

i.e., Percentage humidity or percentage absolute humidity = 100 x [p_{A}(p - p_{S})]/[ p_{S}(p - p_{A})] = 100

Therefore,

p_{A}/(p - p_{A}) = p_{S}/(p - p_{S})

where p_{A} = partial pressure of water vapor;

p_{S} = vapor pressure of water vapor

p = total pressure of system

i.e., no of moles of water vapor per mole of dry air = 1.7051/(200 - 1.7051) = 0.0086

This ratio (moles of water vapor / mole of dry air) is not going to change for a closed system. Therefore, partial pressure of A at 30^{o}C and 150 kPa is found as follows:

p_{A}/(p - p_{A}) = 0.0086

p_{A}/(150 - p_{A}) = 0.0086

p_{A} + 0.0086 p_{A} = 1.29

p_{A} = 1.29/1.0086 = 1.279 kPa

Percentage relative humidity of the original air:

**percentage relative humidity** = 100 x (p_{A}/p_{S}) = 100 x 1.279/4.246 **= 30.12%**

Last Modified on: 11-Sep-2014

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