Air at 30oC 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 15oC. Calculate the percent relative humidity of the original air. The vapor pressures of water at 15oC and 30oC are 1.7051 kPa and 4.246 kPa respectively.
Calculations:
At 15oC, and 200 kPa water is at 100% humidity.
i.e., Percentage humidity or percentage absolute humidity = 100 x [pA(p - pS)]/[ pS(p - pA)] = 100
Therefore,
pA/(p - pA) = pS/(p - pS)
where pA = partial pressure of water vapor;
pS = 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 30oC and 150 kPa is found as follows:
pA/(p - pA) = 0.0086
pA/(150 - pA) = 0.0086
pA + 0.0086 pA = 1.29
pA = 1.29/1.0086 = 1.279 kPa
Percentage relative humidity of the original air:
percentage relative humidity
= 100 x (pA/pS) = 100 x 1.279/4.246 = 30.12%Last Modified on: 01-May-2024
Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in