Chapter 5.5, Problem 5.14AFP

Interpretation Introduction

Interpretation:

If it takes $1.25\min$ for $0.010\text{ mol}$ of $\text{He}$ to effuse, the time it will take for the same amount of ethane to effuse is to be calculated.

Concept introduction:

According to Graham’s law, the rate of effusion of any gas is inversely proportional to the square root of its molecular weight.

$r_{\text{A}}\propto \frac{\text{1}}{\sqrt{{\text{M}}_{\text{A}}}}$

Here,

$r_{\text{A}}$ is the rate of gas A.

${\text{M}}_{\text{A}}$ is the molar mass of gas A.

For two different gas this equation can be modified as follows:

$\frac{r_A}{r_B}=\sqrt{\frac{M_{\text{B}}}{M_{\text{A}}}}$ (1)

Here,

$r_B$ is the rate of gas B.

$M_{\text{B}}$ is the molar mass of gas B.

The rate can be expressed as $\frac{1}{t}$.

Here, $t$ is the time.