The temperature and the relative humidity are 300K and 20% in a room of volume 50 cm3. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapor pressure at 300 K = 3.3 kPa.

Given


Temperature T=300K


Relative humidity=20%


Volume of room= 50m3


Mass of water=500g


Molar mass of H2O M=2+16=18g


Saturation vapor pressure(SVP) at 300 K = 3.3 kPa=3300Pa


Let vapor pressure inside the room be P1




P1=0.23.3103=660Pa


Since the floor has dried that means water on the floor has been evaporated.


Let P2 be the partial pressure of evaporated water


We know ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant =8.3JK-1mol-1


T=temperature


n=number of moles of gas


P=pressure of gas.


And Number of moles n=




P2=1385Pa


Total pressure of room= partial pressure of evaporated water+ pressure of air inside the room


P=P2+P1


P=1385+660=2045Pa



The relative humidity when the floor dries is 61.9%.


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