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Question
The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m3. The saturation vapour pressure at 300 K 3.3 kPa. Calculate the mass of the water vapour present in the room.
Use R = 8.3 J K-1 mol-1
Solution
Here,
T = 300K
SVP = 3300 Pa at 300K
RH = 20%
⇒ `(P)/(SVP) = 0.2`
⇒ `P = 0.2 xx SVP = 0.2 xx 3300 = 660`
V = 50 `"m"^3`
M = 18 g
Now ,
PV = nRT
⇒ `PV = m/MRT`
⇒ `660 xx 50 = m/18 xx 8.3 xx 300`
⇒ m = 238.55 g≈238 g
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