The condition of air in a closed room is described as follows. Temperature 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapor is removed from the room without changing the temperature, what will be the new pressure? The saturation vapor pressure at 25°C = 3.2 kPa.
Given
Temperature T = 25
=298K
Relative humidity =60%
Initial pressure of room=104kPa=1.04
Pa
Saturation vapor pressure=3.2kPa=3.2
Pa
Formula for relative humidity is given as
Where both actual vapor pressure and saturation vapor pressure are at same temperature.
So,
Therefore, actual vapor pressure of water vapor=0.6 
saturation vapor pressure
Vapor pressure of water vapor=0.6
3.2
103=1.92
103Pa
If all the water vapor is removed, then new pressure =pressure of room -pressure due to water vapours
=1.04
105-1.92
105
= 1.02
105Pa
If all the water vapour is removed from the room without changing the temperature, the new pressure will be 102kPa.