IB Physics HLThe Behaviour of GasesPaper 1 & 2Derivation~10 min read
Deriving the Kinetic Theory Equation
Now for something genuinely satisfying. We take our bouncing-particle picture, follow just one molecule around a box using nothing but Newton’s laws, and out drops a formula for the pressure of the entire gas. You won’t be asked to reproduce every line in an exam — but seeing how the pieces fit makes the final equation stick, and the logic is beautiful.
📘 What you need to know
The kinetic theory equation is P = ⅓ρv̄2, where v̄2 is the mean square speed (the average of the squared speeds)
Each elastic wall collision reverses a molecule’s momentum: Δp = 2mv
A molecule hits the same wall every 2l/v seconds (round trip of the box)
Force = rate of change of momentum leads to a pressure from one molecule, then from all N
Because motion is random in 3D, only one third is along each axis — that’s where the ⅓ comes from
The root-mean-square (rms) speed is vrms = √(v̄2)
The setup: one molecule in a box
Picture a single molecule of mass m in a cube-shaped box of side l, moving straight at one wall with speed v. It hits the wall, bounces back, crosses the box, bounces off the far wall, and returns — over and over. Each hit gives the wall a tiny shove. If we can find the average force from this one molecule, we can scale up to all of them.
The molecule bounces between the walls. Each elastic hit reverses its momentum from mv to −mv — a change of 2mv. One full round trip covers a distance 2l.
Following the collisions
Step 1 — momentum change per bounce. The collision is elastic, so the molecule rebounds at the same speed. Its momentum flips from mv to −mv, a change of:
Step 2 — time between hits. To hit the same wall again, the molecule must travel to the far wall and back — a distance 2l — at speed v:
Time between collisions
Δt = distance / speed = 2l / v
Step 3 — force from one molecule. Newton’s second law says force is the rate of change of momentum. Dividing the momentum change by the time between hits:
F = Δp / Δt = 2mv ÷ (2l/v) = mv2 / l
Step 4 — pressure from all N molecules. Pressure is force over area, and one wall has area l2. For one molecule, P = F/l2 = mv2/l3. Scale up to N molecules (now using the average of v2):
P = Nmv2 / l3
Accounting for 3D: where the ⅓ comes from
So far we pretended every molecule flies straight along one axis. In reality they move in all three dimensions. By Pythagoras in 3D, a molecule’s speed splits into components:
v2 = vx2 + vy2 + vz2
No direction is special, so on average the three shares are equal — each is one third of the total:
Because the three directions share the speed equally, vx2 = ⅓v̄2. Only the x-component drives collisions with our chosen wall, so a factor of ⅓ appears.
Replacing v2 with just its x-share, and writing the box volume as V = l3, the pressure becomes:
P = Nmv̄2 / (3V)
Bringing in density: the final equation
The last touch is neat. The total mass of gas is Nm (each of N molecules has mass m), so the density is ρ = Nm/V. Substituting that straight in gives the kinetic theory equation:
Kinetic theory of gases equationP = ⅓ρv̄2
where P is the pressure (Pa), ρ is the density (kg m−3), and v̄2 is the mean square speed (m2 s−2). Take the square root of the mean square speed and you get the root-mean-square (rms) speed — a useful “typical” molecular speed.
Δp = 2mv
÷ time 2l/v
F = mv²/l
÷ area, × N, × ⅓
P = ⅓ρv̄²
WE 1
Air has a density of 1.2 kg m−3 at a pressure of 1.0 × 105 Pa. Find the root-mean-square speed of its molecules.
Step 1 — rearrange P = ⅓ρv̄² for the mean square speed
v̄² = 3P / ρ
v̄² = (3 × 1.0×10⁵) ÷ 1.2 = 2.5 × 10⁵ m² s⁻²Step 2 — take the square root for the rms speedvrms = √(2.5×10⁵)vrms = 500 m s⁻¹Sensible — air molecules really do whizz around at hundreds of metres per second.
WE 2
A gas has density 0.90 kg m−3, and its molecules have an rms speed of 400 m s−1. Find the pressure of the gas.
Step 1 — use P = ⅓ρv̄² (with v̄² = vrms²)P = ⅓ × 0.90 × (400)²P = ⅓ × 0.90 × 160 000P = 4.8 × 10⁴ PaThe mean square speed is just the rms speed squared — no need to re-derive anything.
🛠️ The derivation in six steps
One bounce: momentum change Δp = 2mv (elastic reversal).
Timing: hits the same wall every Δt = 2l/v.
Force:F = Δp/Δt = mv2/l.
Pressure, all N:P = Nmv2/l3.
Go 3D: only ⅓ is along each axis → P = Nmv̄2/3V.
Density:ρ = Nm/V → P = ⅓ρv̄2.
💡 Top tips
You won’t recall the full derivation under exam pressure — but know the logic and be able to use the final equation.
The ⅓ is the 3D factor — explain it via vx2 = ⅓v̄2.
2l is the round trip — there and back, not just the box width.
rms speed = √(mean square speed) — don’t stop at v̄2 if a speed is wanted.
Density links to Nm/V — that’s the step that swaps N, m and V for ρ.
⚠ Common mistakes
Using Δp = mv instead of 2mv — the momentum reverses, it doesn’t just stop
Using l instead of 2l for the distance between hits on the same wall
Forgetting the ⅓ factor from the three dimensions
Confusing the mean square speed v̄2 with the rms speed (one is the square of the other)
Mixing up area (l2) and volume (l3) when going from force to pressure
Quick recap: Track one molecule: each elastic bounce changes its momentum by 2mv, and it hits the same wall every 2l/v. That gives a force, then a pressure; scaling to N molecules and averaging over 3D (the ⅓) and swapping in density ρ = Nm/V yields P = ⅓ρv̄2.
Look closely at that equation and something jumps out: pressure depends on the molecules’ speed squared — in other words, on their kinetic energy. That’s the doorway to the final idea in this topic. Next we connect molecular motion directly to temperature in Average Kinetic Energy of a Molecule, where ½mv̄2 meets ³⁄₂kBT.
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