Chapter 1, Problem 1B.6E

Interpretation Introduction

Interpretation:

Mean free path of ${\text{N}}_{\text{2}}$ molecules and the number of collisions per second has to be calculated.

Concept introduction:

Mean free path:

The average distance travelled by a molecule between the collisions is called the mean free path.  The mean free path is calculated by the formula,

    $\text{λ=}\frac{\text{kT}}{\text{σp}}$

    $\boxed{\begin{array}{l}\text{where,}\\ \text{λ}\text{=}\text{mean}\text{free}\text{path}\\ \text{P}\text{=}\text{Pressure}\\ \text{σ}\text{=}collisioncross-\sec tionofthemolecules\\ \text{T}\text{=}\text{Temperature}\end{array}}$

Collision frequency:

$\begin{array}{c}\text{Collision}\text{frequency,}\text{z}\text{=}\frac{\text{2σp}}{\text{kT}}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\text{1/2}}\\ \\ \text{where,}\\ \text{σ}\text{-}\text{collision cross-section of the molecules}\\ \text{p}\text{-}\text{pressure}\\ \text{k-}\text{Boltzmann constant}\text{.}\\ \text{T}\text{-}\text{Temperature}\\ \text{M}\text{-}\text{Molar}\text{mass}\end{array}$