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(D) The average kinetic energy (per molecule) of an ideal gas is given by the formula: $K.E. = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constan

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$1: 1$

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(D) The average kinetic energy (per molecule) of an ideal gas is given by the formula: $K.E. = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.Since both helium $(He)$ and argon $(Ar)$ are monatomic gases,they have the same number of degrees of freedom $(f = 3)$.Given that both gases are at the same temperature $(T = 300 \ K)$,the average kinetic energy per molecule depends only on the temperature.Therefore,the ratio of the average kinetic energies is: $\frac{K.E._{He}}{K.E._{Ar}} = \frac{\frac{3}{2} k_B T}{\frac{3}{2} k_B T} = 1: 1$.

Helium and argon are placed in a flask at the same room temperature $(300 \ K)$. What is the ratio of the average kinetic energies (per molecule) of helium and argon?

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