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(A) According to the law of equipartition of energy,the average kinetic energy associated with each degree of freedom is $\frac{1}{2} k_B T$,where $k_

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$1:1$

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(A) According to the law of equipartition of energy,the average kinetic energy associated with each degree of freedom is $\frac{1}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.Both oxygen $(O_2)$ and nitrogen $(N_2)$ are diatomic molecules.$A$ diatomic molecule has $2$ degrees of freedom associated with rotational motion.Therefore,the average rotational kinetic energy for each molecule is $2 	imes (\frac{1}{2} k_B T) = k_B T$.Since both gases are in the same vessel at the same temperature $(T = 300 K)$,the average rotational kinetic energy per molecule for both $O_2$ and $N_2$ is $k_B T$.Thus,the ratio of the average rotational kinetic energy per $O_2$ molecule to that per $N_2$ molecule is $k_B T : k_B T = 1:1$.

$A$ 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 $O_2$ molecule to that per $N_2$ molecule is

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