If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

Question

Physics

Moody

25 October, 13:50

If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

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Answers (1)

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Boyer · Answer 1

From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;

n = PV/RT = (4.24*1000*450) / (8.314*303) = 757.4 moles

Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles ...

25% of 757.4 moles = 25/100*757.4 = 189.35 moles

Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg