Calculate the ratio of KE of molecule of oxygen and neon  gas at 27° C

For oxygen gas number of degree of freedom is 5

$$\begin{gathered} (\mathrm{KE}{)}_{\text{oxygen }}=\frac{5}{2}kT \\ k is boltzmann cons\tan t;T is absolute temperature \end{gathered}$$

Neon is mono atomic, therefore it has  ‘3’ degree of freedom 

$\begin{array}{l}(KE{)}_{\text{neon }}=\frac{3}{2}\mathrm{KT}\\ \frac{(KE{)}_{\text{oxygen }}}{(KE{)}_{\text{neon }}}=\frac{\frac{5}{2}KT}{\frac{3}{2}KT}=\frac{5}{3}\end{array}$