First slide

Law of equipartition of energy

Question

A rigid container contains a mixture of one mole of nitrogen to two moles of He at $0^{o} C$ . Then the ratio of rotational kinetic energy per $N_{2}$ molecule to the transational kinetic energy per He molecule is

Moderate

A

3 : 2
B

3 : 1
C

1 : 2
D

2 : 3

Solution

For $N_{2}$ molecule, number of degrees of freedom for rotational motion is 2 and for He molecule, number of degrees of freedom for translational motion is 3.
$∴$ The required ratio $= \frac{2. (\frac{K T}{2})}{3. (\frac{K T}{2})} = \frac{2}{3}$

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