Helium collected over Water measures 350 ml at 20 degree centigrade. If atmospheric pressure is 752. 5 mm of Hg and vapour pressure of water at 20 degree centigrade be 17.5 mm Hg, what is the percentage of water vapour in moist helium gas?

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Solution

$V a p o u r p r e s s u r e o f t h e m o i s t g a s = 752.5 - 17.5 = 635 m m H g A s p e r t h e c o m b i n e d g a s l a w, \frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}} \frac{760 \times 22400}{273} = \frac{635 \times V_{2}}{293} V_{2} = 28773.5 m l V o l u m e o f w a t e r v a p o u r i n t h e g a s = 28773 - 350 = 28423 m l % o f w a t e r v a p o u r = \frac{28423}{28773.5} \times 100 = 98.76 %$

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Helium collected over Water measures 350 ml at 20 degree centigrade. If atmospheric pressure is 752. 5 mm of Hg and vapour pressure of water at 20 degree centigrade be 17.5 mm Hg, what is the percentage of water vapour in moist helium gas?

Vapour pressure of the moist gas = 752.5 - 17.5 = 635 mm HgAs per the combined gas law, p1V1 T1= p2V2T2760 × 22400273 = 635 × V2293V2 =28773.5 mlVolume of water vapour in the gas = 28773-350 = 28423 ml% of water vapour = 2842328773.5 × 100 = 98.76 %