Kinetic Theory Bootcamp

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.

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(a) Find the mean free path from this data.

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(b) What is the standard deviation of space between collisions from the mean free path?

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(d) What is the rms free length?

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(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

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Hint.

Use a data processing sofware such as Mathematica or Excel or Google Sheets.

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Answer.

(a) - (d) See solution, (e) \(D = 0.66\text{ nm}\text{.}\)

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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.

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

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17. Surface Temperature of a Star for Helium Atoms to Escape Gravitational Pull.

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.

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Hint.

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Answer.

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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.

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