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Question
A bucket full of water is placed in a room at 15°C with initial relative humidity 40%. The volume of the room is 50 m3. (a) How much water will evaporate? (b) If the room temperature is increased by 5°C, how much more water will evaporate? The saturation vapour pressure of water at 15°C and 20°C are 1.6 kPa and 2.4 kPa respectively.
Use R = 8.3 J K-1 mol-1
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Solution
(a) Relative humidity is given by
`(VP)/("SVP at" 15^circC)`
⇒ 0.4 = `(VP)/(1.6 xx 10^3)`
⇒VP = `0.4 xx 1.6 xx 10^3`
Evaporation occurs as long as the atmosphere is 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`
Let the mass of water evaporated be m. Then,
⇒ `0.96 xx 10^3 xx 50 = (m xx 8.3 xx 288)/18`
⇒ `m = (0.96 xx 50 xx 18 xx 10^3)/(8.3 xx 288)`
=361.45 ≈ 361 g
(b) At `20^circC` , SVP = 2.4 KPa
At `15^circC` , SVP = 1.6 KPa
Net pressure change = `(2.4 - 1.6) xx 10^3 Pa`
= `0.8 xx 10^3 Pa`
Mass of water evaporated is given by
`m = (m^' xx 8.3 xx 293)/18`
⇒ `m^' = (0.8 xx 50 xx 18 xx 10^3)/(8.3 xx 293)`
= 296.06 ≈ 296 g
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