A mixture of 2, moles of helium gas (atomic m | vedclass

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(A) 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 const

Vedclass · Accepted Answer

$3.16$

Vedclass · Answer

(A) 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 both gases are in the same container at the same temperature $T = 300\, K$, the ratio of their rms speeds is:
$\frac{V_{rms}(	ext{helium})}{V_{rms}(	ext{argon})} = \frac{\sqrt{\frac{3RT}{M_{He}}}}{\sqrt{\frac{3RT}{M_{Ar}}}} = \sqrt{\frac{M_{Ar}}{M_{He}}}$
Given $M_{He} = 4\, u$ and $M_{Ar} = 40\, u$, we substitute these values:
$\frac{V_{rms}(	ext{helium})}{V_{rms}(	ext{argon})} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

$A$ mixture of $2\, moles$ of helium gas (atomic mass $= 4\, u$) and $1\, mole$ of argon gas (atomic mass $= 40\, u$) is kept at $300\, K$ in a container. The ratio of their rms speeds $\left[ \frac{V_{rms}(\text{helium})}{V_{rms}(\text{argon})} \right]$ is close to:

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