A mixture of 2, text{moles} of helium gas (at | vedclass

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(D) The root mean square speed $(v_{	ext{rms}})$ of a gas is given by the formula: $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universa

Vedclass · Accepted Answer

$3.16$

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(D) The root mean square speed $(v_{	ext{rms}})$ of a gas is given by the formula: $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the temperature, and $M$ is the molar mass of the gas.Since both gases are in the same container at the same temperature $(T = 300\, 	ext{K})$, the ratio of their $r.m.s.$ speeds is:$\frac{v_{	ext{rms}}(	ext{He})}{v_{	ext{rms}}(	ext{Ar})} = \frac{\sqrt{\frac{3RT}{M_{	ext{He}}}}}{\sqrt{\frac{3RT}{M_{	ext{Ar}}}}} = \sqrt{\frac{M_{	ext{Ar}}}{M_{	ext{He}}}}$Given $M_{	ext{He}} = 4\, 	ext{g/mol}$ and $M_{	ext{Ar}} = 40\, 	ext{g/mol}$:Ratio $= \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

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