A vessel contains gaseous argon at 300 K and 1 atm.
(a) Calculate the collision rate (collisions/s) for a typical argon atom within the vessel.
(b) Determine the mean free path for argon in this vessel.

Question

Accepted Answer

(a) Employing Eq. (16.21), we can determine the collision rate for any labeled argon atom by using appropriate parameters available from Example 16.1. Hence, from

[latex]Z^{\ast} = \frac{2Z}{n} = 4n\sigma^{2}\left(\frac{\pi kT}{m} ight)^{{1}/{2}}[/latex], (16.21)

[latex]Z^{\ast} = 4n\sigma^{2}\left(\frac{\pi kT}{m} ight)^{{1}/{2}}[/latex]

we obtain

[latex]Z^{\ast} = 4(2.45 imes10^{19}cm^{-3})(3.42 imes10^{-8}cm)^{2} imes\left[\frac{\pi(300)(1.38 imes10^{-16}g_\ ._\ cm^{2}/s^{2} )}{(6.63 imes10^{-23}g)} ight]^{{1}/{2}} = 5.08 imes10^{9}[/latex] collisions/s.

(b) The mean free path can be obtained from Eq. (16.22). Using parameters previously calculated in Example 16.1, we find that

[latex]\ell = \frac{\overline{V}}{Z^{\ast}} = \frac{1}{\sqrt{2}\pi n\sigma^{2}}[/latex]. (16.22)

[latex]\ell = \frac{1}{\sqrt{2}\pi n\sigma^{2}} = \frac{1}{\sqrt{2}\pi(2.45 imes 10^{19}cm^{-3})(3.42 imes10^{-8}cm)^{2}} = 7.85 imes10^{-6}[/latex] cm.

We note again the huge collision rate experienced by any single argon atom, approximately five collisions per nanosecond. For these typical conditions, the mean free path is remarkably only about 100 times greater than the size of the atom itself.

Question 16.2: A vessel contains gaseous argon at 300 K and 1 atm. (a) Calc...

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