Humidity

Air at 30^oC 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^oC. Calculate the percent relative humidity of the original air. The vapor pressures of water at 15^oC and 30^oC are 1.7051 kPa and 4.246 kPa respectively.

At 15^oC, 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^oC 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%