Question

A vessel containing water is put in a dry sealed room of volume 56 m³ at a temperature of 25° C. The saturation vapour pressure of water at 25° C is 20 mm of mercury. How much water will evaporate before the water is in equilibrium with the vapour?

1. 0.780 kg
2. 1.086 kg
3. 0.568 kg
4. 0.2665 kg

Answer 1

Correct Answer - Option 2 : 1.086 kg

Water will be in equilibrium with its vapour when the vapour gets saturated.

In this case, the pressure of vapour = saturation vapour pressure = 20 mm of mercury = 20 × 10^-3 × 13600 × 9.8 = 2665.6 Pa

Now from the gas law 

\(pV = \frac{m}{M}RT \Rightarrow m = \frac{{MpV}}{{RT}} = \frac{{18 \times 2665.6 \times 56}}{{8.3 \times 298}} = 1086g = 1.086\;kg\)