The tempareture and the relative humidity are 300K and 20% in a room of volume `50m^(3)`. `The floor is washed with water , soog of water sticking on the floor. Assuming mo communication with the surrouding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at `300K=3.3kPa`.

Explanation

Given , T= 300K,
Relative humidity = 20%,
`V= 50(m^3)`
SVP at 300K = 3.3 K Pa
VP = Relative humidity xx SVP
` = (0.2 xx 3.3 xx (10^3)) `
` PV = mRT/ M `
` rArr 0.2 xx 3.3 xx (10^3) xx 50 `
` (= (m xx 8.3 xx 300) / 18)`
` rArr m = (0.2 xx 3.3 xx 50 xx 18 xx (10^3)/ 8.3 xx 300)`
` = 238.55 g ~~ 238 g` .