Question

$2\text{moles}$ of gas $A$ effuses from a container in $10\text{min}$. How many moles of gas $B$ would effuse through the same container in same time under similar conditions. Ratio of specific gas constant for gas $A$ to the gas $B$ is $1:16$.

• A. $2$
• B. $1$
• C. $4$
• D. $8$

Answer 1

The correct option is D $8$

We know,
Specific gas constant, $R_{\text{specific}}=\frac{R}{M}$
where $M$ is molar mass.
$\Rightarrow \frac{R_{\text{specific (A)}}}{R_{\text{specific (B)}}}=\frac{M_B}{M_A}=\frac{1}{16}$
From Graham's law of effusion,
$\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}$
where $r_A$ and $r_B$ are the rate of effusion for gas $A$ and $B$.
$\therefore \frac{r_A}{r_B}=\sqrt{\frac{1}{16}}=\frac{1}{4}$
Also
$\Rightarrow \frac{\frac{V_A}{t}}{\frac{V_B}{t}}=\frac{r_A}{r_B}=\frac{1}{4}$
Where
$V_A$ and $V_B$ are the volume of gas effused and $t$ is the time taken for effusion.
We know,
For same pressure and temperature,
$V\propto n$ (no. of moles)
$\Rightarrow \frac{n_A}{n_B}=\frac{r_A}{r_B}=\frac{1}{4}$
$\Rightarrow n_B=4n_A=4\times 2=8\text{moles}$