Chapter 5, Problem 5.119P

(a)

Interpretation Introduction

Interpretation:

The root-mean-square speed of ${\text{O}}_{\text{2}}$ at $\text{STP}$ is to be calculated.

Concept introduction:

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 kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

(b)

Interpretation Introduction

Interpretation:

The collision frequency is to be calculated.

Concept introduction:

The expression to calculate the collision frequency is as follows:

$\text{Collision frequency}=\left(\frac{\text{ root-mean-square speed}}{\text{mean free path}}\right)$ (3)

The kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

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

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

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

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