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(A) The average speed of a gas molecule is given by the formula $v_{av} = \sqrt{\frac{8RT}{\pi M}}$.Since $R$ and $T$ are constant for both gases,we h

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$He$ (molecular mass $4$)

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(A) The average speed of a gas molecule is given by the formula $v_{av} = \sqrt{\frac{8RT}{\pi M}}$.Since $R$ and $T$ are constant for both gases,we have $v_{av} \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the average speeds is $\frac{v_{Gas}}{v_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{Gas}}}$.Given that the average speed of the gas is $4$ times that of $SO_2$,we have $\frac{v_{Gas}}{v_{SO_2}} = 4$.Substituting the values: $4 = \sqrt{\frac{64}{M_{Gas}}}$.Squaring both sides: $16 = \frac{64}{M_{Gas}}$.Solving for $M_{Gas}$: $M_{Gas} = \frac{64}{16} = 4$.$A$ gas with a molecular mass of $4$ is Helium $(He)$.

The gas having an average speed four times that of $SO_2$ (molecular mass $64$) is

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