Chapter 5, Problem 5.152QP

Interpretation Introduction

Interpretation:

The composition of the gases that pass through the orifice has to be calculated.

Concept Introduction:

The root-mean square speed of a gas $\left({\text{u}}_{\text{rms}}\right)$ is

  ${\text{u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3RT}}{\text{M}}}$

The rate of diffusion will be directly proportional to the root-mean square speed.  Moving the gases at greater speed will diffuse faster.  For two different gases, the rate of diffusion is written as,

  $\begin{array}{l}{\text{r}}_{\text{1}}\text{=}\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{1}}}}\\ \\ {\text{r}}_{\text{2}}\text{=}\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{2}}}}\end{array}$

Divide ${\text{r}}_{\text{1}}\text{by}{\text{r}}_{\text{2}}\text{gives,}$

  $\frac{{\text{r}}_{\text{1}}}{\text{r}{\text{2}}_{}}\text{=}\frac{\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{1}}}}}{\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{2}}}}}$

Cancel $\text{3RT}$ from the equation gives,

  $\begin{array}{l}\frac{\text{r}{\text{1}}_{}}{{\text{r}}_{\text{2}}}\text{=}\sqrt{\frac{\frac{\text{1}}{{\text{M}}_{\text{1}}}}{\frac{\text{1}}{{\text{M}}_{\text{2}}}}}\\ \frac{\text{r}{\text{1}}_{}}{{\text{r}}_{\text{2}}}\text{=}\sqrt{\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}}\end{array}$

The equation for Graham’s law is shown below,

  $\frac{\text{r}{\text{1}}_{}}{{\text{r}}_{\text{2}}}\text{=}\sqrt{\frac{{\text{M}}_{\text{2}}}{{\text{M}}_{\text{1}}}}\text{.}$