The volume of a certain mass of gas at `N.T.P.` is `27.5 cm^(3)`. What pressure will be necessary to keep the volume same at `27^(@)C`?

Answer

`1.113 "bar"`

Explanation

`V_(1) = 27.5 cm^(3) , V_(2) = 27.5 cm^(3)`
`P_(1) = 1.013 "bar" , P_(2) = ?`
`T_(1) = 273 K , T_(2) = 27 + 273 =300 K`
According to ideal gas equation,
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `P_(2) = (P_(1)V_(1)T_(2))/(T_(1)V_(2)) = ((1.013 "bar") xx (27.5 cm^(3)) xx (300 K))/((273 k) xx (27.5 cm^(3))) = 1.113 "bar"`