Mean free path

Why does a smell spread so slowly?

Open a bottle of perfume at one end of a room. The molecules in the air move at hundreds of metres a second. Yet it takes many seconds before you smell it at the other end. Why so slow?

A molecule does not fly straight across the room. It hits another molecule after a very short distance and changes direction. Then it hits another, and another. Its path is a zigzag, and it makes little progress in any one direction.

The average length of the straight pieces between collisions is the mean free path, $\lambda$ (the Greek letter lambda).

A tube that the molecule sweeps

Picture the molecules as balls of diameter $d$. Two molecules touch when their centres are $d$ apart. So a moving molecule hits every molecule whose centre lies within $d$ of its own path.

In a time $t$, a molecule with speed $v$ sweeps out a tube. The tube has length $vt$ and radius $d$.

The volume of the tube is $\pi d^2 vt$. If there are $n$ molecules in each cubic metre, the number of centres inside the tube is $n\pi d^2 vt$. Each one is a collision.

The molecule travels $vt$ and makes $n\pi d^2 vt$ collisions. So the distance for each collision is

We assumed the other molecules stand still. They move too, so the collisions are a little more frequent. A careful average adds a factor $\sqrt{2}$:

Pressure and temperature

The ideal gas equation gives $n = \dfrac{p}{k_BT}$. Put it in:

• Double the pressure at a fixed temperature, and $\lambda$ halves. The molecules are packed closer.
• Heat a gas in a sealed rigid box. Both $p$ and $T$ grow, but $n$ stays the same, so $\lambda$ does not change.

Watch out. The second formula hides this. $\lambda$ depends on $n$ and $d$ only.

How big is it?

Take argon in a light bulb at $300\ \text{K}$ and $1.0 \times 10^5\ \text{Pa}$. An argon atom has $d = 3.4 \times 10^{-10}\ \text{m}$.

1. Number density. $n = \dfrac{1.0 \times 10^5\ \text{Pa}}{1.38 \times 10^{-23}\ \text{J/K} \times 300\ \text{K}} = 2.42 \times 10^{25}\ \text{m}^{-3}$.
2. Mean free path. First the bottom line: $\sqrt{2}\,\pi \times (3.4 \times 10^{-10}\ \text{m})^2 \times 2.42 \times 10^{25}\ \text{m}^{-3} = 1.24 \times 10^{7}\ \text{m}^{-1}$. Then $\lambda = \dfrac{1}{1.24 \times 10^{7}\ \text{m}^{-1}} = 8.1 \times 10^{-8}\ \text{m}$.
3. Compare. That is about $240$ atom diameters. The atoms are far apart compared with their size.

Collision frequency

How many collisions does one molecule make each second? In one second it travels $\bar v$, its average speed. Each collision takes $\lambda$ of that. So the collision frequency is

and the average time between collisions is $\tau = \lambda/\bar v$. Argon atoms at $300\ \text{K}$ have an average speed of about $400\ \text{m/s}$. With $\lambda = 8.1 \times 10^{-8}\ \text{m}$, each atom makes about $5 \times 10^9$ collisions every second.

Very low pressures

High in the atmosphere, about $100\ \text{km}$ up, the pressure is about three million times smaller than at sea level. There the mean free path is about $15\ \text{cm}$. In a good vacuum chamber it can be tens of metres, longer than the chamber. Then a molecule hits the walls far more often than it hits another molecule.

Halve the pressure. What happens to the mean free path? Now make the molecules bigger. Which change has the larger effect, and why?

Practise with new numbers

Try it in Python

The program finds the mean free path of air in three places: at sea level, high in the atmosphere and in a vacuum chamber.

Common mistakes

• Using the radius for $d$. Two molecules touch when their centres are one diameter apart, so $d$ is the diameter.
• Forgetting $\sqrt{2}$. It comes from the motion of the other molecules.
• Saying $\lambda$ grows when a sealed gas is heated. In a rigid box, $n$ does not change, so $\lambda$ does not change either.
• Mixing up $\lambda$ and the distance between neighbours. The mean free path is far longer than the spacing of the molecules.

Remember these

• $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n} = \dfrac{k_BT}{\sqrt{2}\,\pi d^2 p}$.
• $\lambda$ falls as $\dfrac{1}{p}$ at a fixed temperature, and does not change at a fixed $n$.
• Collision frequency: $\nu = \dfrac{\bar v}{\lambda}$. Time between collisions: $\tau = \dfrac{\lambda}{\bar v}$.
• In air at sea level, $\lambda$ is about a ten thousandth of a millimetre.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The pressure of a gas is made four times larger, at the same temperature. What happens to $\lambda$?

2. Gas A has molecules twice as wide as gas B. Both have the same $n$. Compare their mean free paths.

3. A molecule has a mean free path of $50\ \text{nm}$ and an average speed of $500\ \text{m/s}$. How many collisions does it make each second?

4. Why does a smell take seconds to cross a room, when the molecules move at hundreds of metres a second?