Ratio of RMS Speeds of Helium and Argon Gas Mixture

The root-mean-square ($\text{rms}$) speed of molecules of an ideal gas at temperature $T$ is given by the Kinetic Theory of Gases formula: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ where:

• $R$ is the universal gas constant,
• $T$ is the absolute temperature of the gas mixture,
• $M$ is the molar mass (or atomic mass) of the gas species.

Since helium and argon are mixed together inside the same container at thermal equilibrium, both gases are at the same temperature $T = 300\text{ K}$.

The ratio of the $\text{rms}$ speed of helium to that of argon depends only on their respective atomic masses and is independent of the number of moles present: $\frac{v_{\text{rms}}(\text{helium})}{v_{\text{rms}}(\text{argon})} = \frac{\sqrt{\frac{3RT}{M_{\text{helium}}}}}{\sqrt{\frac{3RT}{M_{\text{argon}}}}} = \sqrt{\frac{M_{\text{argon}}}{M_{\text{helium}}}}$

Given:

• Atomic mass of helium, $M_{\text{helium}} = 4\text{ amu}$
• Atomic mass of argon, $M_{\text{argon}} = 40\text{ amu}$

Substituting these values into the ratio gives: $\frac{v_{\text{rms}}(\text{helium})}{v_{\text{rms}}(\text{argon})} = \sqrt{\frac{40}{4}} = \sqrt{10}$

Calculating the numerical value: $\sqrt{10} \approx 3.16$

Hence, the correct option is D.