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`
(a) For the two gases, ratio of kinetic energies ` = ((K.E.)_(1))/((K.E.)_(2)) = (n_(1)T_(1))/(n_(2)T_(2)) = (7)/(28) xx 400 xx (32)/(16) xx (1)/(300) = 2 : 3`
(b) 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`
Average kinetic energy is only dependent on temperature