When we consider collisions between different gas molecules of the same substance, we can denote the relative velocity and the expected value of the relative velocity as \(v_{11}\) and \(\left\langle v_{11}\right\rangle\), respectively. By the argument we make above, we can find the number of collisions between any one of these molecules and all of the others. Letting this collision frequency be \({\widetilde{\nu }}_{11}\), we find

\[\widetilde{\nu }_{11}=N_1\pi {\sigma }^2_{11}\left\langle v_{11}\right\rangle,\]

where \({\sigma }_{11}=2{\sigma }_1\). Since we have

\[\left\langle v_{11}\right\rangle =\sqrt{2}\left\langle v_1\right\rangle,\]

while

\[\left\langle v_1\right\rangle =\sqrt{{8kT}/{\pi }m_1},\]

we have \(\left\langle v_{11}\right\rangle =4\sqrt{{kT}/{\pi }m_1}\). The frequency of collisions between molecules of the same substance becomes

\[{\widetilde{\nu }}_{11}=N_1\pi {\sigma }^2_{11}\left\langle v_{11}\right\rangle =4N_1{\sigma }^2_{11}{\left(\frac{\pi kT}{m_1}\right)}^{1/2}\]

The mean time between collisions, \({\tau }_{11}\), is

\[{\tau }_{11}={1}/{\widetilde{\nu}_{11}}\]

and the mean free path, \({\lambda }_{11}\),

\[{\lambda }_{11}=\left\langle v_1\right\rangle {\tau }_{11}=\frac{1}{\sqrt{2}N}_1\pi {\sigma }^2_{11}\]

When we consider the rate of collisions between all of the molecules of type \(1\) in a container, \({\rho }_{11}\), there is a minor complication. If we multiply the collision frequency per molecule, \({\widetilde{\nu }}_{11}\), by the number of molecules available to undergo such collisions, \(N_1\), we count each collision twice, because each such collision involves two type \(1\) molecules. To find the collision rate among like molecules, we must divide this product by 2. That is,

\[{\rho }_{11}=\frac{N_1{\widetilde{\nu }}_{11}}{2}=2N^2_1{\sigma }^2_{11}{\left(\frac{\pi kT}{m_1}\right)}^{1/2}\]