1.2.5 Mean free path
If a perfume bottle is opened in the corner of a room it is a very long time before the aromatic gaseous substances can be detected in the opposite corner of the room. This experience seems to contradict the mean gas velocities described in the previous chapter. The reason for this lies in the great number of collisions that a gas particle sustains along its way. The mean free path is the average distance that a particle can travel between two successive collisions with other particles.
Figure 1.4: Mean free path between two collisions
For collisions of identical particles, the following applies for the mean free path:
\[\bar l = \frac{k\cdot T}{\sqrt 2\cdot\pi\cdot p\cdot d_m^2}\]
Formula 1-11: Mean free path [9]
| $\bar l$ | Mean free path | [m] |
| $d_m$ | Molecular diameter | [m] |
| $m$ | Mass | [kg] |
From Formula 1-11 it can be seen that the mean free path displays linear proportionality to the temperature and inverse proportionality to the pressure and molecular diameter. At this point we will disregard the further variants of this equation discussed in academic literature which examine issues such as collisions between different gas particles, collisions of gas particles with ions or electrons, and temperature effects.
To demonstrate the temperature dependence of the mean free path, Formula 1-11 is often written with the temperature as the only variable on the right-hand side of the equation:
\[\bar l\cdot p = \frac{k\cdot T}{\sqrt 2\cdot\pi\cdot d_m^2}\]
Formula 1-12: Mean free path II
Table 1.5 shows the $\bar l\cdot p$ values for a number of selected gases at 0°C.
| Gas | Chemical Symbol | $\bar l\cdot p$ [m hPa] |
$\bar l\cdot p$ [m Pa] |
|---|---|---|---|
| Hydrogen | H2 | 11.5·10-5 | 11.5·10-3 |
| Nitrogen | N2 | 5.9·10-5 | 5.9·10-3 |
| Oxygen | O2 | 6.5·10-5 | 6.5·10-3 |
| Helium | He | 17.5·10-5 | 17.5·10-3 |
| Neon | Ne | 12.7·10-5 | 12.7·10-3 |
| Argon | Ar | 6.4·10-5 | 6.4·10-3 |
| Air | 6.7·10-5 | 6.7·10-3 | |
| Krypton | Kr | 4.9·10-5 | 4.9·10-3 |
| Xenon | Xe | 3.6·10-5 | 3.6·10-3 |
| Mercury | Hg | 3.1·10-5 | 3.1·10-3 |
| Water vapor | H2O | 6.8·10-5 | 6.8·10-3 |
| Carbon monoxide | CO | 6.0·10-5 | 6.0·10-3 |
| Carbon dioxide | CO2 | 4.0·10-5 | 4.0·10-3 |
| Hydrogen chloride | HCl | 3.3·10-5 | 3.3·10-3 |
| Ammonia | NH3 | 3.2·10-5 | 3.2·10-3 |
| Chlorine | Cl2 | 2.1·10-5 | 2.1·10-3 |
Table 1.5: Mean free path of selected gases at 273.15K [10]
Using the values from Table 1.5 we now estimate the mean free path of a nitrogen molecule at various pressures:
| Pressure [Pa] | Pressure [hPa] | Mean free path [m] |
|---|---|---|
| 1·105 | 1·103 | 5.9·10-8 |
| 1·104 | 1·102 | 5.9·10-7 |
| 1·103 | 1·101 | 5.9·10-6 |
| 1·102 | 1·100 | 5.9·10-5 |
| 1·101 | 1·10-1 | 5.9·10-4 |
| 1·100 | 1·10-2 | 5.9·10-3 |
| 1·10-1 | 1·10-3 | 5.9·10-2 |
| 1·10-2 | 1·10-4 | 5.9·10-1 |
| 1·10-3 | 1·10-5 | 5.9·100 |
| 1·10-4 | 1·10-6 | 5.9·101 |
| 1·10-5 | 1·10-7 | 5.9·102 |
| 1·10-6 | 1·10-8 | 5.9·103 |
| 1·10-7 | 1·10-9 | 5.9·104 |
| 1·10-8 | 1·10-10 | 5.9·105 |
| 1·10-9 | 1·10-11 | 5.9·106 |
| 1·10-10 | 1·10-12 | 5.9·107 |
Table 1.6: Mean free path of a nitrogen molecule at 273.15K (0°C)
At atmospheric pressure a nitrogen molecule therefore travels a distance of 59 nm between two collisions, while at ultra-high vacuum at pressures below 10-8 hPa it travels a distance of several kilometers.
The relation between molecular number density and the mean free path is shown in a graph in Figure 1.5.
Figure 1.5: Molecular number density (red, right-hand y axis) and mean free path (blue, left-hand y axis) for nitrogen at a temperature of 273.15 K
