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A mixture of $2$ moles of helium gas (atomic mass $= 4 u$ ), and $1$ mole of argon gas (atomic mass $= 40 u$ ) is kept at $300 K$ in a container. The ratio of their $r m s$ speeds $[\frac{v_{r m s} (h e l i u m)}{v_{r m s} (a r g o n)}]$ is close to

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3.16

0.32

0.45

2.24

answer is D.

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Detailed Solution

$v_{r m s} = \sqrt{\frac{3 R T}{m}} \Rightarrow \frac{{(v_{r m s})}_{H e}}{{(v_{r m s})}_{A r}} = \sqrt{\frac{40}{4}} = \sqrt{10} = 3.16$

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