Question

$question$
Let two gases , methane and helium each of 1 mole kept in a container. A
small hole is made in the container. Find relative rate of effussion.

Pls solve this...​

Answer 1

Explanation:

It takes 243 s for 4.46 × 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10−5 mol Ne to effuse?

Solution

It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:

\text{rate of effusion} = \frac{\text{amount of gas transferred}}{\text{time}}

and combine it with Graham’s law:

\frac{\text{rate of effusion of gas Xe}}{\text{rate of effusion of gas Ne}} = \frac{\sqrt{\mathcal{M}_\text{Ne}}}{\sqrt{\mathcal{M}_\text{Xe}}}

To get:

\frac{\frac{\text{amount of Xe transferred}}{{\text{time for Xe}}}}{\frac{\text{amount of Ne transferred}}{\text{time for Ne}}} = \frac{\sqrt{\mathcal{M}_\text{Ne}}}{\sqrt{\mathcal{M}_\text{Xe}}}

Noting that amount of A = amount of B, and solving for time for Ne:

\frac{\frac{\rule[0.25ex]{3.8em}{0.1ex}\hspace{-3.8em}\text{amount of Xe}}{\text{time for Xe}}}{\frac{\rule[0.25ex]{3.8em}{0.1ex}\hspace{-3.8em}\text{amount of Ne}}{\text{time for Ne}}} = \frac{\text{time for Ne}}{\text{time for Xe}} = \frac{\sqrt{\mathcal{M}_\text{Ne}}}{\sqrt{\mathcal{M}_\text{Xe}}} = \frac{\sqrt{\mathcal{M}_\text{Ne}}}{\sqrt{\mathcal{M}_\text{Xe}}}

and substitute values:

\frac{\text{time for Ne}}{243 \;\text{s}} = \sqrt{\frac{20.2 \;\rule[0.5ex]{1.8em}{0.1ex}\hspace{-1.8em}\text{g mol}}{131.3 \;\rule[0.5ex]{1.8em}{0.1ex}\hspace{-1.8em}\text{g mol}}} = 0.392

Finally, solve for the desired quantity:

\text{time for Ne} = 0.392 \times 243 \;\text{s} = 95.3 \;\text{s}

Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.