## Section27.8Kinetic Theory Bootcamp

### Subsection27.8.6Miscellaneous

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.

(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 $D = 0.66\text{ nm}\text{.}$

What should be the maximum temperature of a sun-sized star if Helium atoms cannot escape the gravitational pull of the star? Hint: Equate gravitational escape speed with the rms speed.

Hint

Solution

The escape speed $v_e$ of a body of mass $m$ from a body of mass $M$ and radius $R$ is given by equating the kinetic energy to the negative of the potential energy.

\begin{equation*} \frac{1}{2} m v_e^2 = G_N \frac{M m}{R}. \end{equation*}

This gives

\begin{equation*} v_e = \sqrt{\frac{2G_N M }{R} }. \end{equation*}

The root-mean square speed of a molecule when in the environment of temperature $T$ is

\begin{equation*} v_\text{rms} = \sqrt{\frac{3k_B T}{m}} \end{equation*}

Equating the two speeds will give the condition we seek in this problem.

\begin{equation*} \frac{3k_B T}{m} = \frac{2G_N M }{R} \end{equation*}

The temperature will be

\begin{equation*} T =\frac{2G_N Mm }{3 k_B R}. \end{equation*}

Now, we put in the following numbers.

\begin{align*} \amp M = 2\times 10^{30}\ \text{kg}\ \ \text{(Mass of Sun)}\\ \amp m = 4\times 10^{-3}\ \text{kg}/6.022\times 10^{23} \ \ \text{(Mass of one He atom)}\\ \amp R = 7.0\times 10^9 \ \text{m}\ \ \text{(Radius of Sun)}. \end{align*}

These give the following for $T\text{.}$

\begin{equation*} T = 9 \times 10^6 \ \text{K}. \end{equation*}

This temperature is much larger than the surface temperature of the Sun.