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 is given by the formula $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,

Vedclass · Accepted Answer

$3.16$

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(D) The root mean square (rms) speed 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 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:$\frac{v_{	ext{rms}} 	ext{ (helium)}}{v_{	ext{rms}} 	ext{ (argon)}} = \frac{\sqrt{\frac{3RT}{M_{	ext{He}}}}}{\sqrt{\frac{3RT}{M_{	ext{Ar}}}}} = \sqrt{\frac{M_{	ext{Ar}}}{M_{	ext{He}}}}$Given $M_{	ext{He}} = 4 \ amu$ and $M_{	ext{Ar}} = 40 \ amu$,we have:$\frac{v_{	ext{rms}} 	ext{ (helium)}}{v_{	ext{rms}} 	ext{ (argon)}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$.

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

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