Question

The gas having average speed four times as that of $S{O}_2$ at same temperature is
(Molar mass of $S{O}_2=64g$)

• A. $He$
• B. $O_2$
• C. $H_2$
• D. $C{H}_4$

Answer 1

The correct option is A $He$

We know,
Average speed $\left(V_{avg}\right)=\sqrt{\frac{8RT}{\pi M}}$
Given: Temperature is constant.
$V_{avg}^{\prime }=4V_{avg,S{O}_2}$
$\Rightarrow \sqrt{\frac{8RT}{\pi M^{\prime }}}=4\sqrt{\frac{8RT}{\pi M_{S{O}_2}}}$
$\Rightarrow \sqrt{\frac{M_{S{O}_2}}{M^{\prime }}}=4$
$M^{\prime }=\frac{M_{S{O}_2}}{16}=\frac{64}{16}=4\text{g}$
Since, molar mass of $He=4\text{g}$
$\therefore$ option (a) is correct.