Collection of Solved Problems in Physics

Mean Free Path of Argon Molecule

Task number: 1273

• Hint 1

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

• Normal conditions

  This means that we have to determine the mean free path of molecules at the temperature 273.15 K a and the pressure 101,325 Pa.

• 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.

• The equation for the mean free path

  The mean free path \(\bar{\lambda}\) of molecules can be calculated by using the formula

  \[ \bar{\lambda}=\frac{kT}{\pi\sqrt{2}\,pd^{2}}, \]

  where k is the Boltzmann constant, T the temperature, p the pressure and  d is the diameter of the molecule.

• 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

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

• Answer

  The mean free path of the argon molecules under normal conditions is about 52 nm.