Pressure from Molecular Bombardment

In Lesson 8 we set up the kinetic theory of gases with its six assumptions. We now use those assumptions to derive, from first principles, the microscopic origin of gas pressure: molecular bombardment of the container walls. The result is the celebrated formula

pV = (1/3) N m <c²>

which relates the macroscopic quantities p and V to the microscopic quantities N (number of molecules), m (mass of one molecule), and <c²> (the mean-square speed of the molecules). Derivation of this formula is a specification point in Module 5.1.4 of the OCR A-Level Physics A specification (H556), and the full derivation is a classic piece of physics that is worth learning thoroughly.

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The Strategy

We will consider a single molecule in a cubic box, compute the momentum change per unit time it imposes on one wall (which by Newton's second law gives the force), and then multiply up to get the total force due to all molecules. Dividing by wall area gives pressure. Finally, we multiply by volume to get pV on the left-hand side.

The key assumptions we rely on, from Lesson 8:

• Random motion in all directions (assumption 2).
• Elastic collisions with the walls (assumption 4).
• Short collision time compared to the time between collisions (assumption 5).
• No intermolecular forces between collisions (assumption 6).
• Many molecules, so averages are meaningful (assumption 1).

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Setting Up the Problem

Imagine a cubic container of side length L, so the volume is V = L³. Inside are N identical molecules, each of mass m. One of these molecules moves with velocity (u₁, v₁, w₁) — i.e. with velocity components u₁ in the x-direction, v₁ in the y-direction, w₁ in the z-direction.

We will focus on one particular wall — let us say the right-hand wall, perpendicular to the x-axis. This wall has area A = L².

graph TD
    M[Molecule with velocity u along x]
    L1[Left wall] -- u --- M
    M -- u --- R[Right wall<br/>area L²]
    M -. bounces elastically .- R

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Step 1: Momentum Change per Collision

The molecule travels across the box in the x-direction. When it hits the right-hand wall, the x-component of its velocity reverses (because the collision is elastic and the wall is much more massive than the molecule). Its new velocity is (-u₁, v₁, w₁): the y and z components are unchanged, but u₁ → -u₁.

The change in the molecule's x-momentum is therefore

Δp_x = m(-u₁) - m(u₁) = -2 m u₁

By Newton's third law, the wall receives an equal and opposite impulse: the wall's x-momentum increases by +2 m u₁ for each collision with this molecule.

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Step 2: Time Between Successive Collisions

How often does this molecule hit the right-hand wall?

After bouncing off the right wall, the molecule travels at speed u₁ across the box, hits the left wall at distance L away, bounces back, and returns to the right wall after travelling a total distance 2L. (We ignore collisions with the top, bottom, front and back walls, because those affect only the y- and z-components of velocity, not the x-component.)

The time between successive collisions of this molecule with the right-hand wall is therefore

Δt = 2L / u₁

And the number of collisions per unit time is u₁/(2L).

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Step 3: Force on the Wall from One Molecule

The average force on the right-hand wall due to this single molecule is the momentum transfer per collision times the number of collisions per unit time:

F₁ = (2 m u₁) × (u₁ / (2L))
   = m u₁² / L

So a single molecule with x-velocity u₁ exerts a (time-averaged) force of m u₁² / L on the right-hand wall.

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Step 4: Total Force from All Molecules

Now sum over all N molecules. Each molecule i has its own x-velocity component u_i, and contributes a force m u_i² / L to the right wall. The total force is

F = Σ (m u_i² / L)
  = (m/L) Σ u_i²

We now introduce the mean-square velocity component in the x-direction, <u²>, defined as

<u²> = (1/N) Σ u_i²

In words: <u²> is the average value of u² over all N molecules. Then Σ u_i² = N <u²>, and

F = (m/L) × N <u²> = N m <u²> / L

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Step 5: Pressure from the Force

The pressure on the right-hand wall is the force per unit area:

p = F/A = F/L²
  = (N m <u²> / L) / L²
  = N m <u²> / L³

Since L³ = V (the volume of the cube), we can write

p V = N m <u²>

That is our result so far: the pressure on one wall, expressed in terms of the mean square of the x-component of velocity.

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Step 6: From <u²> to <c²>

We now invoke the randomness of the motion. Each molecule has a speed c_i with components (u_i, v_i, w_i), so by Pythagoras,

c_i² = u_i² + v_i² + w_i²

Averaging over all molecules,

<c²> = <u²> + <v²> + <w²>

where <c²> is the mean-square speed, <v²> is the mean-square y-velocity component, and so on.

Now comes the crucial assumption of isotropy: because the motion is random, there is no preferred direction. On average, the x, y, and z components of velocity-squared are the same:

<u²> = <v²> = <w²>

Therefore

<c²> = 3 <u²>

or equivalently

<u²> = (1/3) <c²>

Substituting back into pV = Nm<u²>:

pV = (1/3) N m <c²>

This is the pressure equation of the kinetic theory of gases. It holds for any ideal gas in any container — the cubical shape was just a convenient choice for the derivation.

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The Result Unpacked

pV = (1/3) N m <c²>

Let us be clear about every symbol:

Symbol	Meaning	Units
p	Pressure of the gas	Pa (= N m⁻²)
V	Volume of the container	m³
N	Total number of molecules	(dimensionless)
m	Mass of one molecule	kg
<c²>	Mean-square speed of the molecules	m² s⁻²

Note the factor of 1/3. It comes from the randomness of motion in three dimensions, specifically from the step <u²> = (1/3) <c²>. In two dimensions (if you were looking at a gas confined to a plane) the factor would be 1/2. In one dimension it would be 1. The factor of 1/3 is a signature of three-dimensional isotropic motion.

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Alternative Forms of the Pressure Equation

Several equivalent forms of the same equation appear in textbooks and on data sheets:

Form 1 (with total mass):

pV = (1/3) N m <c²>

This is the form given on the OCR data sheet.

Form 2 (with density):

p = (1/3) ρ <c²>

where ρ = Nm/V is the mass density of the gas. This is useful because density is often more directly measurable than N.

Form 3 (with total gas mass):

pV = (1/3) M_total <c²>

where M_total = Nm is the total mass of the gas in the container.

All three are the same equation in different guises. You should be able to convert between them fluently.

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Root-Mean-Square Speed