A vessel contains a mixture of one mole of ox | vedclass

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(A) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom per molecule is $\frac{1}{2} k T$,where $

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$1 : 1$

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(A) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom per molecule is $\frac{1}{2} k T$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.Both oxygen $(O_2)$ and nitrogen $(N_2)$ are diatomic gases.For a diatomic molecule,the number of rotational degrees of freedom is $2$.Therefore,the average rotational kinetic energy per molecule for any diatomic gas is given by $E_{rot} = 2 	imes \frac{1}{2} k T = k T$.Since both gases are at the same temperature $(300 \ K)$,the average rotational kinetic energy per molecule for $O_2$ is $k(300)$ and for $N_2$ is $k(300)$.Thus,the ratio of the average rotational kinetic energy per $O_2$ molecule to that per $N_2$ molecule is $k(300) : k(300) = 1 : 1$.

$A$ 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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