Chapter 5, Problem 5.75P

Interpretation Introduction

Interpretation:

For the lightest gas ${\text{H}}_{\text{2}}$ and heaviest gas, ${\text{UF}}_{\text{6}}$ the ratio of effusion rate 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.