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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.

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Solution

Here,T=300KSVP=3300 Pa at 300 KRH=20%PSVP=0.2P=0.2×SVP=0.2×3300=660V=50 m3M=18 g Now,PV = nRT PV = mMRT660×50=m18×8.3×300m=238.55 g≈238 g

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