$\begin{aligned}
& \text { Degree of freedom }(f)=5+2(3 \mathrm{N}-5) \
& \qquad f=5+2(3 \times 2-1)=7 \
& \text { energy of one molecule }=\frac{f}{2} K_B T
\end{aligned}energyof10molecules=10\left(\frac{\mathrm{f}}{2} \mathrm{K}{\mathrm{B}} \mathrm{T}\right)=10\left(\frac{7}{2} \mathrm{~K}{\mathrm{B}} \mathrm{T}\right)=35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$