First slide

Law of equipartition of energy

Question

The ratio of average translational KE to rotational KE of a linear polyatomic molecule at temperature T is

Moderate

A

3 : 1
B

5 : 1
C

3 : 2
D

7 : 5

Solution

For poly atomic linear molecule, number of degrees of freedom for translational motion is 3 and that for rotational motion is 2.
$∴$ The ratio = $\frac{3 x \frac{KT}{2}}{2 x \frac{KT}{2}} = \frac{3}{2}$

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