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

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

Vedclass · Accepted Answer

$3.16$

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(D) The root mean square (rms) speed 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 the temperature $T$ is the same for both gases in the mixture,we have $V_{rms} \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the rms speeds is given by $\frac{V_{rms}(	ext{helium})}{V_{rms}(	ext{argon})} = \sqrt{\frac{M(	ext{argon})}{M(	ext{helium})}}$.Substituting the given values,$M(	ext{helium}) = 4 \ g/mol$ and $M(	ext{argon}) = 40 \ g/mol$,we get:$\frac{V_{rms}(	ext{helium})}{V_{rms}(	ext{argon})} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

$A$ 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 $\left( \frac{V_{rms}(\text{helium})}{V_{rms}(\text{argon})} \right)$ is

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