Physics - Heat & Thermo Question with Solution | TestHub

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