Question

$X\text{mL}$ of $H_2$ gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is:

• A. $\text{10 seconds}:He$
• B. $\text{20 seconds}:O_2$
• C. $\text{25 seconds}:CO$
• D. $\text{55 seconds}:C{O}_2$

Answer 1

The correct option is B $\text{20 seconds}:O_2$

According to Graham's law of diffusion for two gases undergoing diffusion under identical conditions through the same hole,
$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$
Also,
Rate of diffusion for a gas $\left(r\right)$ = $\frac{Volumeofgascomingout}{Timetaken}=\frac{V}{t}$

$\frac{r_1}{r_2}=\frac{t_2}{t_1}$
$\therefore \frac{t_2}{t_{H_2}}=\sqrt{\frac{M_2}{M_{H_2}}}$

a) For $He,t_2=\sqrt{\frac{4}{2}}\times \left(5s\right)=5\sqrt{2}s\ne 10s$

b) For $O_2,t_2=\sqrt{\frac{32}{2}}\times \left(5s\right)=20s$

c) For $CO,t_2=\sqrt{\frac{28}{2}}\times \left(5s\right)=18.71\ne 25s$

d) For $C{O}_2,t_2=\sqrt{\frac{44}{2}}\times \left(5s\right)=23.45\ne 55s$