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.

1.22g of a gas mesured over water at 15^(@)C and undr a pressure of 1.

1.	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.
Answer» Solution :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)`