Nitrogen fills a vacuum chamber at 300 K…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Nitrogen fills a vacuum chamber at $300\ \text{K}$ . Take a nitrogen molecule as a sphere of diameter $3.7 \times 10^{-10}\ \text{m}$ , its molar mass as $0.028\ \text{kg/mol}$ , $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ and $R = 8.314\ \text{J/(mol K)}$ . Use $\lambda = \dfrac{1}{\sqrt2\,\pi d^2 n}$ for the mean free path and $\bar v = \sqrt{8RT/(\pi M)}$ for the mean speed.

At first the chamber is at $1.010 \times 10^{5}\ \text{Pa}$ . How many molecules are there in each cubic metre?
What is the mean free path, in nm?
How many collisions does one molecule make each second?
Compare the mean free path with the average distance between neighbouring molecules, $n^{-1/3}$ . How many times larger is it?
The chamber is now pumped out at the same temperature until the mean free path equals the size of the chamber, $0.40\ \text{m}$ . What is the pressure then?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Number density from $P = nk_BT$ , then the mean free path, the collision rate, and a scaling for the pumped chamber.

Step 1. Number density (a). $n = \dfrac{P}{k_BT} = \dfrac{1.010 \times 10^{5}\ \text{Pa}}{1.38 \times 10^{-23}\ \text{J/K} \times 300\ \text{K}} = 2.44 \times 10^{25}\ \text{m}^{-3}$ .

Step 2. Mean free path (b). $\pi d^2 = 4.301 \times 10^{-19}\ \text{m}^2$ . The bottom line is $\sqrt2 \times 4.301 \times 10^{-19}\ \text{m}^2 \times 2.440 \times 10^{25}\ \text{m}^{-3} = 1.484 \times 10^{7}\ \text{m}^{-1}$ , so

\displaystyle \lambda = \frac{1}{1.484 \times 10^{7}\ \text{m}^{-1}} = 6.739 \times 10^{-8}\ \text{m} = 67.4\ \text{nm}

Step 3. Collision rate (c). $\bar v = \sqrt{\dfrac{8RT}{\pi M}}$ , with $8RT = 8 \times 8.314\ \text{J/(mol K)} \times 300\ \text{K} = 19954\ \text{J/mol}$ and $\pi M = \pi \times 0.028\ \text{kg/mol} = 0.0880\ \text{kg/mol}$ . So $\bar v = 476\ \text{m/s}$ , so a molecule collides $\dfrac{\bar v}{\lambda} = 7.07 \times 10^{9}$ times each second.

Step 4. Spacing (d). $n^{-1/3} = 3.448 \times 10^{-9}\ \text{m} = 3.45\ \text{nm}$ . The mean free path is $19.5$ times larger. A molecule passes many neighbours before it hits one, because each neighbour is a very small target.

Step 5. Pumped chamber (e). At a fixed temperature $\lambda \propto 1/P$ :

\displaystyle P_2 = P_1\,\frac{\lambda_1}{\lambda_2} = 1.010 \times 10^{5}\ \text{Pa} \times \frac{6.739 \times 10^{-8}\ \text{m}}{0.40\ \text{m}} = 0.0170\ \text{Pa}

Below about this pressure, molecules cross the chamber from wall to wall without hitting each other.

Answer: (a) $2.44 \times 10^{25}\ \text{m}^{-3}$ . (b) $67.4\ \text{nm}$ . (c) $7.07 \times 10^{9}\ \text{s}^{-1}$ . (d) about $20$ times. (e) $0.0170\ \text{Pa}$ .

Check. The same $\lambda$ comes from $\dfrac{k_BT}{\sqrt2\,\pi d^2 P}$ in one step. And in (e) there are still $4.1 \times 10^{12}$ molecules in each cubic centimetre: a good vacuum is far from empty.

Watch out. Use the diameter, not the radius, in $\pi d^2$ . With the radius the mean free path comes out four times too large.

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