Derivation of the Kinetic Theory of Gases Equation

When molecules rebound from a wall in a container, the change in momentum gives rise to a force exerted by the particle on the wall
Many molecules moving in random motion exert forces on the walls which create an average overall pressure, since pressure is the force per unit area

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

Kinetic theory of gases equation derivation setup

Single molecule in box, downloadable AS & A Level Physics revision notes

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

Derivation

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
If they rebound in the opposite direction to their initial velocity, their final velocity is -c
The change in momentum is therefore:

$∆ p = - m c - (+ m c) = - m c - m c = - 2 m c$

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

The time between collisions of the molecule travelling to one wall and back is calculated by travelling a distance of 2l with speed c:

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

Note: c is not taken as the speed of light in this scenario

3. Find the change in momentum per second

The force the molecule exerts on one wall is found using Newton’s second law of motion:

Force = rate of change of momentum = $\frac{∆ p}{∆ t} = \frac{2 m c}{\frac{2 l}{c}} = \frac{m c^{2}}{l}$

The change in momentum is +2mc since the force on the molecule from the wall is in the opposite direction to its change in momentum

4. Calculate the total pressure from N molecules

The area of one wall is l²
The pressure, p is defined using the force and area:

$p = \frac{f o r c e}{a r e a} = \frac{\frac{m c^{2}}{l}}{l^{2}} = \frac{m c^{2}}{l^{3}}$

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

$p = \frac{N m c^{2}}{l^{3}}$

Each molecule has a different velocity and they all contribute to the pressure
The mean squared speed of c² is written with left and right-angled brackets <c²>
The pressure is now defined as:

$p = \frac{N m < c^{2} >}{l^{3}}$

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

The pressure equation still assumes all the molecules are travelling in the same direction and colliding with the same pair of opposite faces of the cube
In reality, all molecules will be moving in three dimensions equally
Splitting the velocity into its components c_x, c_y and c_z to denote the amount in the x, y and z directions, c² can be defined using pythagoras’ theorem in 3D:

$c^{2} = {c_{x}}^{2} + {c_{y}}^{2} + {c_{z}}^{2}$

Since there is nothing special about any particular direction, it can be determined that:

$< c_{x} >^{2} = < c_{y} >^{2} = < c_{z} >^{2}$

Therefore, <c_x²> can be defined as:

$< c_{x} >^{2} = \frac{1}{3} < c^{2} >$

6. Re-write the pressure equation

The box is a cube and all the sides are of length l
- This means l³ is equal to the volume of the cube, V
Substituting the new values for <c²> and l³ back into the pressure equation obtains the final equation:

$p V = \frac{1}{3} N m < c^{2} >$

Where:
- p = pressure (Pa)
- V = volume (m³)
- N = number of molecules
- m = mass of one molecule (kg)
- <c²> = mean square speed of the molecules (m s^–1)

This is known as the Kinetic Theory of Gases equation

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

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

Rearranging the pressure equation for p and substituting the density ρ:

$p = \frac{1}{3} ρ < c^{2} >$

Exam Tip

Make sure to revise and understand each step for the whole of the derivation, as you may be asked to derive all, or part, of the equation in an exam question.

Root-Mean-Square Speed

The pressure of an ideal gas equation includes the mean square speed of the particles:

$< c^{2} >$

Where
- c = average speed of the gas particles
- <c²> has the units m² s^-2

Since particles travel in all directions in 3D space and velocity is a vector, some particles will have a negative direction and others a positive direction
When there are a large number of particles, the total positive and negative velocity values will cancel out, giving a net zero value overall
In order to find the pressure of the gas, the velocities must be squared
- This is a more useful method, since a negative or positive number squared is always positive
To calculate the average speed of the particles in a gas, take the square root of the mean square speed:

$\sqrt{< c^{2} >} = c_{r . m . s}$

c_r.m.s is known as the root-mean-square speed and still has the units of m s^-1
The mean square speed is not the same as the mean speed

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 root-mean-square (r.m.s.) speed of the gas atoms at 504 K.

Answer:

Step 1: Write out the equation for the pressure of an ideal gas with density

$p = \frac{1}{3} ρ < c^{2} >$

Step 2: Rearrange for mean square speed

$< c^{2} > = \frac{3 p}{ρ}$

Step 3: Substitute in values

$< c^{2} > = \frac{3 \times (9.3 \times 10^{5})}{4.5} = 6.2 \times 10^{5} m^{2} s^{- 2}$

Step 4: To find the r.m.s value, take the square root of the mean square speed

$c_{r . m . s} = \sqrt{< c^{2} >} = \sqrt{6.2 \times 10^{5}} = 787.4 = 790 m s^{- 1} (2 s . f .)$

Derivation of the Kinetic Theory of Gases Equation (CIE A Level Physics)

Revision Note