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Relative humidity of air is 60°C and the saturation vapour pressure of water vapour in air is 3.6 KPa. The amount of water vapours presents in 2 lit air at 300 K is,

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28 g

44 g

31.2 g

108 g

answer is A.

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Detailed Solution

$P V = \frac{W}{M} R T W_{(H_{2} O)} = \frac{P V M}{R T} = \frac{3.6 \times 10^{3} \times 2 \times 10^{- 3} \times 18}{8.314 \times 300} = 52 g$

Relative humidity = 60% of weight of $H_{2} O$

= 0.6 × 52 g

= 31.2 g

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