16. Mean Free Path from Distances Between Successive Collisions of a Molecule.
The mean free path is the average distance a molecule travels in-between collisions. In reality a molecule will travel different distances in-between collisions. Suppose a molecule of size 0.3 nm travels the following distances between collisions: 11 nm, 5 nm, 12 nm, 15 nm, 25 nm, 2 nm, 11 nm, 30 nm, 18 nm, 20 nm, 14 nm, 15 nm, 10 nm, 28 nm, 9 nm, 17 nm, 5 nm, 10 nm, 6 nm, 13 nm, 17 nm, 15 nm, 10 nm, 22 nm, 9 nm, 23 nm, 19nm, 4 nm, 15 nm, 12 nm, 10 nm, 16 nm, 11 nm.
(a) Find the mean free path from this data.
(b) What is the standard deviation of space between collisions from the mean free path?
(c) What is the most probable free length?
(d) What is the rms free length?
(e) Supposing, the temperature of the gas is 200 K and the pressure of the gas is 1 atm, what would be the diameter of molecules of the gas
Hint.
Use a data processing sofware such as Mathematica or Excel or Google Sheets.
Answer.
(a) - (d) See solution, (e) \(D = 0.66\text{ nm}\text{.}\)
Solution.
I typed up the data in Mathematica and used their statistical functions. (a) 14 nm, (b) 6.7 nm, (c) Most common function gave two most-commonly occurring number, 10 nm and 15 nm. When I drew the histogram I found that 10 nm has more data points clustered around it than 15 nm points. The most probable length will be 10 nm. (d) 15.4 nm.
(e) Using the formula for the mean free path we get
\begin{equation*}
D = \sqrt{ \frac{k_B T}{\sqrt{2}\pi \lambda p}}.
\end{equation*}
Now, you can put in the numbers given.
\begin{equation*}
T = 200\ \text{K},\ \ p = 1.013\times 10^5\ \text{Pa}.
\end{equation*}
This gives \(D = 0.66\text{ nm}\text{.}\)
