Question

Oxygen gas diffuses through a hole in a container in $10$ seconds. The time taken (in seconds) for the effusion of the same volume of $S{O}_2$ gas under identical conditions is:

• A. $7.07$
• B. $20$
• C. $14.14$
• D. $10$

Answer 1

The correct option is C $14.14$

Given,

Time taken for oxygen diffusion=10 sec

Also pressure,volume,temperature is constant.

Molar mass of oxygen=32g/mole

Molar mass of sulphur dioxide=64g/mole

According to graham's law of diffusion/effusion

$r\propto \sqrt{\frac{1}{M}}$

r=rate of diffusion

M=molar mass

$\frac{r_{O_2}}{r_{{SO}_2}}=\sqrt{\frac{M_{{SO}_2}}{M_{O_2}}}$

since,

$r=V/T$,$r\propto \frac{1}{t}$

$\frac{t_{{SO}_2}}{t_{O_2}}=\sqrt{\frac{M_{{SO}_2}}{M_{O_2}}}$

$\frac{t_{{SO}_2}}{10}=\sqrt{\frac{64}{32}}$

$t_{{SO}_2}=14.14$