The temperature and the relative humidity are `300 K and 20%` in a room of volume `50 m^(3)`. 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 vapour pressure at `300 K = 3.3 kPa`.

Explanation

`RH = (VP/SVP)`
` rArr 0.20 = VP/3.3 xx (10^3)`
` rArr VP = 0.20 xx 3.3 xx (10^3)`
= 660
rArr PV = nRT
` rArr P = nRT/V = m/M xx (R xx T/V)`
` = (500 xx 8.3 xx 300/ 18 xx 50)`
` = 1383.3`
Net P = 1383.3 + 660 = 2043.3
Now , RH = 2043.3/3300 = 0.619 = 62%.