Question

1.22 g of a gas measured over water at `150^(@)C` and a pressure of 775 mm of mercury occupied 900 mL. Calculate the volume of dry gas at NTP. Vapour pressure of water at `15^(@)C` is 14 mm.

Answer 1

Pressure of dry gas = Pressure of moist gas `-` Aqueous tension
`= 775 - 14`
`= 761 mm`
`{:("Initial conditions","NTP Conditions",),(V_(1) = 900 mL,V_(2) = ?,),(P_(1) = 761 mm,P_(2) = 760 mm,),(T_(1) = (273 + 15) = 288 K,T_(2) = 273 K,):}`
Since, `(P_(1)V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))`
So, `V_(2) = (P_(1) V_(1) T_(2))/(T_(1) P_(2))`
`= (761 xx 900 xx 273)/(288 xx 760)`
`= 854.2 mL`