`1.22g` of a gas mesured over water at `15^(@)C` and undr a pressure of `1.02` bar of mercury occupied `0.9 dm^(3)`. Calculate the volume of the volume of the dry gas at `N.T.P.` Vapour pressure of water at `15^(@)C` is `0.018` bar.

Explanation

Pressure of dry gas = Pressure of moist gas - Aqueous tension
`= 1.02 - 0.018 = 1.002` bar
From the available data: `V_(1) = 0.9 dm^(3) , V_(2) = ?`
`P_(1) = 1.002` bar , `P_(2) = 1.013` bar
`T_(1) = 15+273 = 288 K , T_(2) = 273 K`
According the Gas equation , `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))`
Substituting the values, `V_(2) = ((1.022 "bar")xx (0.9 dm^(3)) xx (273 K))/((288 K) xx (1.013 "bar" )) = 0.844 dm^(3)`