`0.068 dm^(3)` of a sample of nitrogen is collected over water at `20^(@)C` and `0.92` bar of Hg. What is the volume of dry nitrogen N.T.P. ? (Aqueous tension of water at `20^(@)C = 0.023 "bar"` of Hg)

Explanation

Pressure of Hg nitrogen `=` Pressure of moist nitrogen - Aqueous tension
`= 0.92 - 0.23 = 0.897` bar ltbr From the available data : `V_(1) = 0.068 dm^(3)` , V_(2) = ?`
`P_(1) = 0.897` bar , `P_(2) = 1.013` bar
`T_(1) = 20 + 273 = 293 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) =- ((0.0897 "bar") xx (0.068 dm^(3)) xx (273 K))/((297 K) xx (1.013 "bar")) = 0.056 dm^(3)`