Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Explanation

According to kinetic theory of gases, the pressure .P. exerted by one mole of an ideal gas is
`P = 1/3 rho C^(2)` where `rho` is density of the gas.
`:. P = 1/3 M/V C^(2) implies PV = 1/3 MC^(2)`
But `PV = RT` for one mole of ideal gas.
`:. 1/3 MC^(2) = RT implies 1/3 M/(N_A) C^(2) = R/(N_A) T = K_(B)T`
`( :. K_(B) = R/(N_A))`
(or) `1/2 mC^(2) = 3/2 K_(B)T implies 1/2 mC^(2) prop T`
But, 1/2 mC^(2)` is average translational kinetic energy per molecule of a gas.
Therefore, average kinetic energy of molecule of an ideal gas is directly proportion to the absolute temperature of the gas.