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`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. |
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Answer» 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)` |
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