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(A) The average speed $(v_{av})$ of gas molecules is given by the formula:$v_{av} = \sqrt{\frac{8RT}{\pi M}}$where $R$ is the universal gas constant,$

Vedclass · Accepted Answer

$He$ (molecular mass $4$)

Vedclass · Answer

(A) The average speed $(v_{av})$ of gas molecules is given by the formula:$v_{av} = \sqrt{\frac{8RT}{\pi M}}$where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.Assuming the temperature $T$ is constant for both gases,the average speed is inversely proportional to the square root of the molar mass:$v_{av} \propto \frac{1}{\sqrt{M}}$Let $v_{1}$ be the speed of $SO_{2}$ and $v_{2}$ be the speed of the unknown gas. Given $v_{2} = 4v_{1}$ and $M_{1} = 64$:$\frac{v_{2}}{v_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$$4 = \sqrt{\frac{64}{M_{2}}}$Squaring both sides:$16 = \frac{64}{M_{2}}$$M_{2} = \frac{64}{16} = 4$The gas with a molar mass of $4$ is Helium $(He)$.Therefore,the correct option is $A$.

The gas having average speed four times as that of $SO_{2}$ (molecular mass $64$) is

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