## Mean Free Path of Argon Molecule

Determine the mean free path of argon molecules under normal conditions knowing that the molecule diameter is 0.4 nm.

• #### Hint 1

The assignment indicates that we have to determine the mean free path of molecules under normal conditions. What does it mean?

• #### Hint 2

The mean free path $$\bar{\lambda}$$ indicates the average distance traveled by a molecule between two collisions. Try to figure out on what variables this depends.

Then use the well-known formula of statistical physics for the calculation.

• #### Numerical values

 d = 0.4 nm = 4·10−10 m the diameter of argon molecules $$\bar{\lambda}\,=\,?$$ the mean free path of argon molecules

From the Handbook of chemistry and physics:

 T = 273.15 K the temperature under normal conditions p = 101,325 Pa the pressure under normal conditions k = 1.38·10−23 JK−1 the Boltzmann constant
• #### Analysis

Mean free path in essence gives us the average distance traveled by a molecule between two collisions.

To calculate the mean free path we use the formula of statistical physics, which says that the mean free path is directly proportional to the temperature and inversely proportional to the cross section of the molecule and pressure.

• #### Solution

Mean free path $$\bar{\lambda}$$, in its principle, the average distance traveled by a molecule between two collisions.

It is clear that it is inversely proportional to the pressure p (low pressure indicates a low density and thus low number of mutual collisions)

$\bar{\lambda}\sim\frac{1}{p}.$

It is also inversely proportional to the cross section of the molecule, and thus the square of the molecule diameter d (if something is small, it avoids collisions more easily)

$\bar{\lambda}\sim\frac{1}{{\pi}d^{2}}.$

It is also directly proportional to the temperature T (higher temperature results in a faster movement, therefore the molecule travels longer distance in the same time)

$\bar{\lambda}\sim T.$

The exact formula for calculating the mean free path of molecules is taken from statistical physics. It looks like this:

$\bar{\lambda}=\frac{kT}{p{\pi}d^{2}\sqrt{2}},$

where k is the Boltzmann constant.

• #### Numerical insertion

$\bar{\lambda}=\frac{kT}{p{\pi}d^{2}\sqrt{2}}$ $\bar{\lambda}=\frac{1.38\cdot{10^{-23}}\cdot{273.15}}{101{,}325\cdot\pi\cdot{4^{2}}\cdot{10^{-20}}\cdot\sqrt{2}}\,\,\mathrm{m}$ $\bar{\lambda}\dot{=}5.2\cdot{10^{-8}}\,\,\mathrm{m}=52\,\,\mathrm{nm}.$