Q140 / 160 · AP EAMCET 2024 (21 May Shift 1) · Physics
A poly atomic gas with n degrees of freedom has a mean kinetic energy per molecule given by (if $N$ is Avogadro's number)
- A$\frac{n K T}{N}$
- B$\frac{n K T}{2 N}$
- Correct answerC$\frac{n K T}{2}$
- D$\frac{3 K T}{2}$
Solution
The mean kinetic energy of a polyatomic gas per molecule per degree of freedom is $ \lt \mathrm{K}$.E $\gt$ /dof/molecule
$\begin{aligned}
& =\frac{1}{2} \mathrm{KT} \\
& \therefore \quad \lt \mathrm{K} . \mathrm{E}\gt/ \text { molecule }=\left(\frac{1}{2} \mathrm{KT}\right) \mathrm{f}=\frac{\mathrm{nKT}}{2}
\end{aligned}$
Hence, the correct answer is option: 'C'.