Chapter 5, Problem 5.76P

Interpretation Introduction

Interpretation:

The ratio of effusion rate of ${\text{O}}_{\text{2}}$ and $\text{Kr}$ 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 effusion 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:

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

Here,

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

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