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Question
The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m3. 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.
Use R = 8.31 J K-1 mol-1
Solution
Here ,
M = 18g for water
m = 500g
V = 50 m3
T = 300K
SVP = 3300 Pa
RH = 20%
`(VP)/(SVP) = 0.2`
⇒ VP = `P_1` = `0.2 xx 3300 = 660 "Pa"`
Partial pressure `"P"_2` For evaporated water is given by
`P_2V = m/MRT`
⇒ `P_2 = 500/(18 xx 50) xx 8.31 xx 300`
⇒ `P_2 = 1385 "Pa"`
Total pressure , `P = P_1 + P_2 = 1385 + 660 = 2045"Pa"`
`"RH" = (P)/(SVP) xx 100 = 2045/3300 xx 100% = 61.9 ≈ 62%`
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