Q. 475.0( 1 Vote )

# 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=1.04Pa

Saturation vapor pressure=3.2kPa=3.2Pa

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=0.63.2103=1.92103Pa

If all the water vapor is removed, then new pressure =pressure of room -pressure due to water vapours

=1.04105-1.92105

= 1.02105Pa

If all the water vapour is removed from the room without changing the temperature, the new pressure will be 102kPa.

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