# Question c71ad

Dec 4, 2016

$\textsf{P = 2.84 \times {10}^{7} \textcolor{w h i t e}{x} {\text{N/m}}^{2}}$

#### Explanation:

The graphic shows the effective collision cross section, or area:

From this we can estimate the diameter of the atoms:

$\textsf{A = \pi {d}^{2}}$

$\therefore$$\textsf{d = \sqrt{\frac{A}{\pi}} = \frac{0.36}{3.142} = = 0.3385 \textcolor{w h i t e}{x} n m}$

The mean free path $\textsf{\lambda}$ is the average distance between collisions.

The expression is:

sf(lambda=(1)/(sqrt(2)pid^(2)n_v)#

$\textsf{{n}_{v}}$ is the number of molecules per unit volume.

We can eliminate this using the Ideal Gas Expression:

$\textsf{P V = n R T}$

$\textsf{{n}_{v} = \frac{n {N}_{A}}{V}}$

Where $\textsf{{N}_{A}}$ is the Avogadro Constant.

Since $\textsf{V = \frac{n R T}{P}}$

We can write:

$\textsf{{n}_{v} = \frac{\cancel{n} {N}_{A}}{\frac{\cancel{n} R T}{P}} = \frac{{N}_{A} P}{R T}}$

Substituting this into the expression for $\textsf{\lambda \Rightarrow}$

$\textsf{\lambda = \frac{R T}{\sqrt{2} \pi {d}^{2} {N}_{A} P}}$

$\therefore$$\textsf{P = \frac{R T}{\lambda \sqrt{2} {d}^{2} {N}_{A}}}$

We can now set the condition that $\textsf{\lambda = d}$

$\therefore$$\textsf{P = \frac{R T}{\sqrt{2} \pi {d}^{3} {N}_{A}}}$

Putting in the numbers:

$\textsf{P = \frac{8.31 \times 298}{1.414 \times 3.142 \times {\left(0.3385 \times {10}^{- 9}\right)}^{3} \times 6.02 \times {10}^{23}} \textcolor{w h i t e}{x} {\text{N/m}}^{2}}$

$\textsf{P = 2.837 \times {10}^{7} \textcolor{w h i t e}{x} {\text{N/m}}^{2}}$