A bucket full of water is placed in a room at `15^(@)C` with initial relative humidity 40%. The volume of the room is `50m^(3)`.(a) How much water will evaporated? The saturation vapour pressure of water at `15^(@)C`and `20^(@)C`are 1.6kPa and 2.4 kPa respectively.

Explanation

(a) Relative humidity
` =(VP/(SVP at (15^@C)))`
` rArr 0.4 = VP/(1.6 xx (10^3))`
` rArr VP = 0.4 xx 1.6 xx (10^3)`
The evaporation occurs as long as the
atmosphere does not saturated
Net pressure change
`= 1.6 xx (10^3) - 0.4 xx 1.6 xx (10^3)`
` = (1.6 - 0.4 xx 1.6 )(10^3)`
` 0.96 xx (10^3).`
Mass of water evaporated =m
` rArr 0.96 xx (10^3) xx 50 = ((m xx 8.3 xx 288) /18)`
` rArr m = (0.96 xx 50 xx 18 xx (10^3)/ 8.3 xx 288)`
` = 361.45 ~~ 361 g .`
` (b) At `20^@C`, SVP = 2.4 K Pa,
` At 15^@C, SVP = 1.6 K Pa.`
Net pressure change
`= ( 2.4 xx 1.6) xx (10^3) Pa`
`= 0.8 xx (10^3) Pa`
Mass of water evaporated
` m= (m' xx 8.3 xx 293 / 18)`
` rArr m' = (0.8 xx 50 xx 18 xx (10^3)/ 8.3 xx 293)`
` = 296.06 ~~296 grams .`