7 grams of nitrogen is present at `127^@C` and 16 grams of oxygen at `27^@C`. Calculate (a) ratio of kinetic energy and (b) ratio of average kinetic energy of nitrogen and oxygen.

Explanation

`(K.E) = 3/2 nRT `
Average kinetic energy means, kinetic energy per one mole.
` ((A.K.E)_1)/((A.K.E)_2) = (T_1)/(T_2) = (400 )/(300) = 4:3`