A dry gas occupies `127 mL` at `N.T.P.` If the same mass of the gas is collected over water at `23^(@)C` and a total pressure of `0.98 "bar"`, what volume will it occupy ? The vapour pressure of water at `23^(@)C` is `0.028 "bar"`.

Explanation

`{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=127mL,,V_(2)=?),(P_(1)=1.013"bar",,P_(2)=(0.98-0.028) = 0.952 "bar"),(T_(1)=273K,,T_(2)=23+273=296K):}`
Applying 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)) = ((1.013 "bar") xx (127 mL) xx (296 K))/((273 K) xx (0.952 "bar")) = 146.52 mL`