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(C) The average kinetic energy per molecule of an ideal gas is given by the formula $\langle KE \rangle = \frac{f}{2} K_B T$, where $f$ is the degrees

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$1: 1$

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(C) The average kinetic energy per molecule of an ideal gas is given by the formula $\langle KE \rangle = \frac{f}{2} K_B T$, where $f$ is the degrees of freedom, $K_B$ is the Boltzmann constant, and $T$ is the absolute temperature.Since both hydrogen and nitrogen gases are in the same vessel at thermal equilibrium, they are at the same temperature $T = 30^{\circ} C$.For both diatomic gases (hydrogen and nitrogen), the number of degrees of freedom $f$ is $5$ (at moderate temperatures).Therefore, the average kinetic energy per molecule for hydrogen is $\langle KE \rangle_{H_2} = \frac{5}{2} K_B T$.Similarly, the average kinetic energy per molecule for nitrogen is $\langle KE \rangle_{N_2} = \frac{5}{2} K_B T$.The ratio of the average kinetic energies per molecule is $\frac{\langle KE \rangle_{H_2}}{\langle KE \rangle_{N_2}} = \frac{\frac{5}{2} K_B T}{\frac{5}{2} K_B T} = 1: 1$.The mass ratio is irrelevant because the average kinetic energy per molecule depends only on the temperature.

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