Chapter 5, Problem 5.72P

Interpretation Introduction

Interpretation:

The rate of effusion of a gas higher than, lower than, or equal to its rate of diffusion is to be explained. Also, the ratio of their effusion rate for two gases with molecules of approximately the same size higher than, lower than, or equal to its rate of diffusion is to be explained.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  $\frac{\text{Rate of A}}{\text{Rate of B}}=\frac{u_{rms\text{ of A}}}{u_{rms\text{ of B}}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}=\frac{\text{time B}}{\text{time A}}$

Here,

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

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

The expression to calculate the root-mean-square speed is as follows:

  $u_{rms}=\sqrt{\frac{3RT}{M}}$

Here, $M$ is the molar mass, $T$ is the temperature and $R$ is the gas constant.

The expression to calculate the average kinetic energy of gases is as follows:

  $\text{Average kinectic energy of gases}=\left(\frac{3}{2}\right)\left(RT\right)$

Here, $T$ is the temperature and $R$ is the gas constant.

The expression to calculate the kinetic energy is as follows:

  $E_{\text{k}}=\left(\frac{1}{2}\right)\left(\text{mass}\right){\left(\text{speed}\right)}^2$