Question

In a diatomic molecule, the rotational energy at a given temperature

• A. equals the translational kinetic energy for each molecule.
• B. is ${\left(\frac{2}{3}\right)}^{rd}$ the translational kinetic energy for each molecule.
• C. obeys Maxwell's distribution.
• D. have the same value for all molecules.

Answer 1

Answer: (B), (C)

obeys Maxwell's distribution.

$\text{Law of equipartition of energy}.$

According to the law of equipartition of energy, for any dynamic system in thermal equilibrium, the total energy for the system is equally divided among the degree of freedom.

Translational kinetic energy

$K_T=\frac{3}{2}kT$ .....$\left(i\right)$

Rotational kinetic energy,

$K_R=2\left(\frac{1}{2}KT\right)$ .....$\left(ii\right)$

Therefore, translational kinetic energy and rotational kinetic energy both obey Maxwell’s distribution independent of each other.

On dividing these two equations, we get

$\frac{K_R}{K_T}=\frac{\left(KT\right)}{\frac{3}{2}KT}$

$K_R=\frac{2}{3}{K}_T$

Rotational energy at a given temperature is ${\left(\frac{2}{3}\right)}^{rd}$ of translational $KE$ of each molecule.

$\text{Final Answer}$: (a), (d).