A vessel containing water is put in a dry sealed room of volume `76 m^(3)` at `15^(@)C` The saturation vapour pressure of water is 15 mm of Hg at `15^(@)C` How much water will evaportate before the water is in equilibrium with the vapour ? .

Explanation

Water vapour will attain equilibrium when vapour pressure room becomes 15 mm at `15^(@)C`
or ` w = (PVm)/(RT) = (15 xx 76xx 10^(3) xx 18)/(760 xx 0.0821 xx 288)`
`= 1141 .90g = 1.1.42 kg`
Thus water evaporated is `1.142 kg` .