A mixture of 2 moles of helium gas (atomic ma | vedclass

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

Vedclass · Accepted Answer

$3.16$

Vedclass · Answer

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

$A$ mixture of $2$ moles of helium gas (atomic mass $= 4$ $amu$) and $1$ mole of argon gas (atomic mass $= 40$ $amu$) is kept at $300$ $K$ in a container. The ratio of the rms speeds $(\frac{V_{rms}(helium)}{V_{rms}(argon)})$ is

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