A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O₂ molecule to that per N₂ molecule is

The correct option is 1:1.

According to the law of equipartition of energy, for a thermodynamic system at absolute temperature $T$, the average kinetic energy associated with each degree of freedom per molecule is given by:
$E=\frac{1}{2}{k}_{B}T$
where $k_B$ is the Boltzmann constant.

Both oxygen (O₂) and nitrogen (N₂) are linear, diatomic molecules. At moderate temperatures such as 300 K, a diatomic molecule has:
- 3 translational degrees of freedom
- 2 rotational degrees of freedom

Therefore, the average rotational kinetic energy per molecule for any diatomic gas is:
$E_{\text{rot}}=2\times \left(\frac{1}{2}{k}_{B}T\right)=k_{B}T$

Since both gases in the mixture are at the same temperature ($T=300\text{K}$), the average rotational kinetic energy per molecule is the same for both oxygen and nitrogen:
$E_{\text{rot},{\text{O}}_2}=k_{B}T$
and
$E_{\text{rot},{\text{N}}_2}=k_{B}T$

Hence, the ratio of the average rotational kinetic energy per O₂ molecule to that per N₂ molecule is:
$\text{Ratio}=\frac{k_{B}T}{k_{B}T}=1$
which is written as 1:1.