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(B) The root mean square speed $(V_{rms})$ of a gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M_w}}$.Since the gases are in the same sample

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{1}$

Vedclass · Answer

(B) The root mean square speed $(V_{rms})$ of a gas is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M_w}}$.Since the gases are in the same sample,they are at the same temperature $(T)$.Therefore,$V_{rms} \propto \frac{1}{\sqrt{M_w}}$.The ratio of the root mean square speed of helium $(He)$ to oxygen $(O_2)$ is:$\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{M_{w, O_2}}{M_{w, He}}}$.Given the molar mass of helium $(M_{w, He} = 4 \ g/mol)$ and oxygen $(M_{w, O_2} = 32 \ g/mol)$:$\frac{V_{He}}{V_{O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}$.Thus,the ratio is $\frac{2\sqrt{2}}{1}$.

$A$ sample contains a mixture of helium and oxygen gas. The ratio of the root mean square speed of helium to oxygen in the sample is:

Similar Questions

At a temperature of $300 \ K$,the kinetic energy of $O_2$ is $6.21 \times 10^{-21} \ J$ and its $rms$ speed is $484 \ m/s$. What will be the kinetic energy and $rms$ speed at $600 \ K$?

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$Assertion :$ The root mean square and most probable speeds of the molecules in a gas are the same.
$Reason :$ The Maxwell distribution for the speed of molecules in a gas is symmetrical.

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