`38.0 mL` of moist nitrogen were collected at `27^(@)C` and `0.98` bar pressure. Calculate the volume of the gas `N.T.P.` Aqueous tension at `27^(@)C` is `0.035 "bar"`.

Answer

`32.26 mL`

Explanation

According to available data :
`V_(1) = 38 mL , V_(2) = ?`
`P_(2) = (0.98 - 0.035) , V_(2 ) = ?`
`T_(1) = 27 + 273 = 300 K` , `T_(2) = 273 K`
By 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)) = ((0.945 "bar") xx (38 mL) xx (273 K))/((300 K) xx (1.013 "bar")) = 32.26 mL`