- For solving this problem, we have to take in account the degree of freedom of air molecules.
- As, molecules of air have five degrees of freedom (three translational and two rotational)
- For each molecule, the kinetic energy of each degree of freedom = 1/2 KT
- So, K.E of 1 molecule for 5 degrees of freedom = 5/2KT
So, for molecules of air K.E = 5/2 KT = 5/2 x 1 x 1.38 x 10⁻²³ x 308 = 1062.6 x 10⁻²³ J
1 mole of air contains 6.022 x 10²³ molecules
K.E. of 1 mole = 1062.6 x 10⁻²³ x 6.022 x 10²³ = 6400 J
