Derivation of the Kinetic Theory of Gases Equation

Gas pressure arises due to the collisions of the gas particles with the walls of the container that holds it
- When a gas particle collides with a wall in the container, it exerts a force on the wall
- The random motion of a large number of molecules exerting a force on the walls creates an overall pressure
- This is because pressure = force per unit area

Derivation

Picture a single molecule in a cube-shaped box with sides of equal length $l$
The molecule has a mass m and moves with speed c, parallel to one side of the box
It collides at regular intervals with the ends of the box, exerting a force and contributing to the pressure of the gas
By calculating the pressure this one molecule exerts on one end of the box, the total pressure produced by all the molecules can be deduced

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A single molecule in a box collides with the walls and exerts a pressure

1. Find the change in momentum as a single molecule hits a wall perpendicularly

One assumption of the kinetic theory is that molecules rebound elastically
- This means there is no kinetic energy lost in the collision, so initial and final velocities are equal in magnitude
When a gas particle of mass m, travels at an average speed v, and hits a wall of the container, it undergoes a change in momentum due to the force exerted by the wall on the particle

initial momentum = $m v$

final momentum = $- m v$

Therefore, the change in momentum $∆ p$ is:

$∆ p$ = final momentum – initial momentum

$∆ p = - m v - (m v) = - 2 m v$

Force is equal to the rate of change of momentum (Newton's Second Law), so the force exerted by the wall on a molecule is:

$F = \frac{∆ p}{∆ t} = - \frac{2 m v}{∆ t}$

According to Newton's Third Law, the particle exerts an equal and opposite force on the wall
Therefore force exerted on the wall by the particle is:

$F = \frac{2 m v}{∆ t}$

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A particle hitting a wall of the container in which the gas is held experiences a force from the wall and a change in momentum. The particle exerts an equal and opposite force on the wall

2. Calculate the number of collisions per second by the molecule on a wall

The time between collisions from one wall to the other, and then back again, over a distance of $2 l$ with speed v is:

time between collisions = $\frac{d i s t a n c e}{s p e e d} = \frac{2 l}{v}$

3. Find the change in momentum per second

The force the molecule exerts on one wall, F, from step 1 now becomes the following when substituting the time between collisions for Δt:

$F = \frac{2 m v}{∆ t} = \frac{2 m v}{\frac{2 l}{v}} = \frac{m v^{2}}{l}$

4. Calculate the total pressure from N molecules

The area of one wall is $l^{2}$
The pressure, P can be written as:

$P = \frac{F}{A} = \frac{\frac{m v^{2}}{l}}{l^{2}} = \frac{m v^{2}}{l^{3}}$

This is the pressure exerted from one molecule
To account for the large number of N molecules, the pressure can now be written as:

$P = \frac{N m v^{2}}{l^{3}}$

5. Consider the effect of the molecule moving in 3D space

The pressure equation written above still assumes all the molecules are travelling in the same x direction and colliding with the same pair of opposite faces of the cube
- To reflect this, it can be rewritten as:

$P = \frac{N m {v_{x}}^{2}}{l^{3}}$

Where v_x is the x component of the average velocity of all the particles
In reality, all molecules will be moving in three dimensions equally
Splitting the velocity into its components v_x, v_y and v_z to denote the amount in the x, y and z directions, v² (the square of the average speed) can be defined using pythagoras’ theorem in 3D:

$v^{2} = {v_{x}}^{2} + {v_{y}}^{2} + {v_{z}}^{2}$

Since there is nothing special about any particular direction, it can be determined that the square of the average speed in each direction is equal:

${v_{x}}^{2} = {v_{y}}^{2} = {v_{z}}^{2}$

$v^{2} = 3 {v_{x}}^{2}$

Therefore, by combining the two equations, v can be defined as:

${v_{x}}^{2} = \frac{1}{3} v^{2}$

6. Re-write the pressure equation

The box is a cube with a side length $l$
Therefore, volume of the cube $V = l^{3}$

$P = \frac{N m v^{2}}{3 V}$

This can also be written using the density ρ of the gas:

$ρ = \frac{m a s s}{v o l u m e} = \frac{N m}{V}$

- m is the mass of a single particle so Nm is the total mass of all the particles
Rearranging the pressure equation for p and substituting the density ρ:

$P = \frac{1}{3} ρ v^{2}$

Where:
- P = pressure of the gas (Pa)
- ρ = density of the gas (kg m^–3)
- v = mean square speed (m s^–1)

This is known as the Kinetic theory of gases equation

Worked example

An ideal gas has a density of 4.5 kg m^-3 at a pressure of 9.3 × 10⁵ Pa and a temperature of 504 K.

Determine the mean square speed of the gas molecules at 504 K.

Answer:

Step 1: List the known quantities

Density of the gas, ρ = 4.5 kg m^-3
Gas pressure, P = 9.3 × 10⁵ Pa
Temperature, T = 504 K

Step 2: Rearrange the pressure equation for the mean square speed

$P = \frac{1}{3} ρ v^{2}$

$v = \sqrt{\frac{3 P}{ρ}}$

Step 3: Substitute in the values

$v = \sqrt{3 \times \frac{9.3 \times 10^{5}}{4.5}} = 787.4 = 790 m s^{- 1}$

Exam Tip

Although you will not be asked to recall the derivation, you need to understand the physics that underlines it and how to use the equation in the final result.

The 'mean square' speed just means an average speed, as we're assuming that all the molecules travel at the same speed.

Derivation of the Kinetic Theory of Gases Equation (SL IB Physics)

Revision Note