Question

The gas having an average speed four times as that of ${SO}_2$ (molecular mass $64$) at the same temperature can be :

• A. $He$ (molecular mass $4$)
• B. $O_2$ (molecular mass $32$)
• C. $H_2$ (molecular mass $2$)
• D. ${CH}_4$ (molecular mass $16$)

Answer 1

The correct option is A $He$ (molecular mass $4$)

Average speed $(v_{av}$) of gas molecules is inversely proportional to the square root of the absolute temperature $\left(T\right)$ of the gas
$v_{av}=\sqrt{\frac{8RT}{\pi M}}$
where $R$ is gas constant and $M$ the molecular weight
Given $v_1=v$, $M_1=64,v_2=4v$
$\therefore$ $\frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}}$
$\frac{v}{4v}=\sqrt{\frac{M_2}{64}}$
$\Rightarrow M_2=\frac{64}{16}=4$
hence, the gas is helium (molecular mass $4$)