Pressure from molecular collisions

Where does the pressure of a gas come from?

Blow up a balloon. The rubber stretches and stays stretched. Something pushes on it from the inside, all the time and in every direction. What is it?

The kinetic theory gives a simple answer. A gas is made of a very large number of molecules. They fly about at high speed. Each time a molecule hits the rubber, it bounces back and gives the rubber a small push. Billions of billions of small pushes every second add up to a steady force. That force on each square metre is the pressure.

In this lesson we turn that picture into a formula.

The model of an ideal gas

We make a few simple assumptions. Together they describe an ideal gas.

• A gas is made of a very large number of identical molecules.
• The molecules are tiny. Their own volume is very small compared with the volume of the gas.
• They move at random, in every direction, with many different speeds.
• They exert no forces on each other, except during a collision.
• Every collision, with another molecule or with a wall, is elastic. Kinetic energy is not lost.
• A collision lasts a very short time compared with the time between collisions.

A real gas at low pressure and a high temperature behaves very nearly like this.

One molecule hitting one wall

Take a cube of side $L$. One molecule of mass $m$ moves inside it with velocity $(v_x, v_y, v_z)$.

The molecule hits wall A, the wall at the right, and bounces back elastically. Only $v_x$ changes. It becomes $-v_x$. The parts $v_y$ and $v_z$ do not change.

So the momentum of the molecule along x changes from $mv_x$ to $-mv_x$. The change is $-2mv_x$. By Newton's third law, the wall gets a momentum of $+2mv_x$ in each hit.

How often does the molecule hit wall A? It must cross the box and come back, a distance $2L$, at the speed $v_x$ along x. So the time between hits is $\dfrac{2L}{v_x}$, and the number of hits each second is $\dfrac{v_x}{2L}$.

The average force is the momentum given each second:

Double the speed along x. The hits come twice as often, and each hit gives twice the momentum. So the force becomes four times as large. Now double the length of the box. What happens to the force?

All the molecules together

Now the cube holds $N$ molecules. Each one adds its own $\dfrac{mv_x^2}{L}$. The total force on wall A is

Here $\bar{v_x^2}$ is the average of $v_x^2$ over all the molecules. Pressure is force divided by the area $L^2$, and $L^3$ is the volume $V$:

The molecules move at random, so no direction is special. The average of $v_x^2$ is the same as the average of $v_y^2$ and of $v_z^2$. Since $v^2 = v_x^2 + v_y^2 + v_z^2$, each one is a third of $\bar{v^2}$:

Write $n = N/V$ for the number of molecules in each cubic metre. Then

This is the main result of the kinetic theory of gases. The product $nm$ is the mass in each cubic metre, the density $\rho$. So we can also write $p = \tfrac13\rho\bar{v^2}$.

What about collisions between the molecules? They are elastic, so they only share out momentum and energy among the molecules. On average, the motion stays the same, and so does the pressure.

The square root of $\bar{v^2}$ is called the root mean square speed, $v_{rms}$. We meet it again in the third lesson.

Pressure and kinetic energy

The total translational kinetic energy of the molecules is $E = N \times \tfrac12 m\bar{v^2}$. Put this into the result above:

So the pressure of a gas is two thirds of its translational kinetic energy in each cubic metre. Take the air in a room at $1.0 \times 10^5$ Pa. Each cubic metre of it holds $\tfrac32 \times 1.0 \times 10^5\ \text{Pa} \times 1\ \text{m}^3 = 1.5 \times 10^5$ J of kinetic energy in the random motion of its molecules.

Practise with new numbers

Try it in Python

The program fills a cube with 20000 molecules, with random velocities. It adds up the push of every molecule on one wall. Then it compares that pressure with $\tfrac13 nm\bar{v^2}$.

Common mistakes

• Taking the change of momentum as $mv_x$. The molecule reverses, so the change is $2mv_x$.
• Forgetting the one third. Only one third of $\bar{v^2}$ belongs to the x direction.
• Using the average speed squared. The formula needs the average of $v^2$, which is not the same as the square of the average of $v$.
• Mixing $n$ and $N$. $N$ is the number of molecules. $n = N/V$ is the number in each cubic metre.

Remember these

• Pressure comes from molecules hitting the walls.
• One molecule gives a wall the momentum $2mv_x$ in each hit, and hits it $\dfrac{v_x}{2L}$ times each second.
• $p = \tfrac13\, nm\bar{v^2} = \tfrac13\rho\bar{v^2}$.
• $p = \tfrac23\,\dfrac{E}{V}$, where $E$ is the total translational kinetic energy.

Check yourself

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

1. A molecule hits a wall head on and bounces back with the same speed. Why does the wall get $2mv$ of momentum and not $mv$?

2. Every molecule in a closed box somehow moves twice as fast. What happens to the pressure?

3. A gas has a density of $1.5\ \text{kg/m}^3$ and a pressure of $1.2 \times 10^5\ \text{Pa}$. Find the rms speed of its molecules.

4. Molecules also hit each other on their way across the box. Why does the formula still work?