What is mean free path?
Gas molecules move very fast, about 500 m/s in air. But they are not alone. Every molecule keeps hitting other molecules.
The straight distance a molecule covers between two hits is a free path. The average of many free paths is the mean free path, written λ (lambda).
For air at room temperature and pressure, λ ≈ 10⁻⁷ m, about 300 times the size of a molecule.
Deriving λ = 1/(√2 n π d²)
Step 1: the collision tube
Let each molecule be a sphere of diameter d. Two molecules touch when their centres are d apart. So our molecule hits every molecule whose centre is within d of its line of motion. It sweeps a tube of radius d, cross-section πd².
Step 2: volume swept
In time t, at speed v̄, it sweeps a volume πd² v̄ t.
Step 3: number of hits
If n molecules are in each m³, the number of hits = n πd² v̄ t.
Step 4: distance per hit
λ = distance / hits = v̄ t / (n πd² v̄ t) = 1/(nπd²).
Step 5: others move too
The other molecules also move, so the relative speed is larger, on average √2 times. This adds √2:
λ = 1 / (√2 n π d²)
Using n = P/(kBT): λ = kBT / (√2 π d² P).
What does mean free path depend on?
- Number density n: λ ∝ 1/n. A crowded gas means short free paths.
- Diameter d: λ ∝ 1/d². Bigger molecules collide more.
- Pressure P (T fixed): λ ∝ 1/P. Pump out air and λ grows.
- Temperature T (P fixed): λ ∝ T, because the gas spreads out.
- Collision frequency: ν = v̄/λ; time between collisions τ = λ/v̄.
Why does a smell spread slowly?
A perfume molecule moves at hundreds of m/s, but it travels only about 0.1 μm before being knocked into a new direction. Billions of random turns each second make it wander, not race. This slow wandering is called diffusion. Air currents usually carry smells faster than diffusion alone.
Try it: the 3D and at home
- In the 3D: set crowding to 1 and count hits for 10 seconds. Predict the count at crowding 2, then check. Now double d and predict λ before you look.
- At home: walk across an empty corridor, then a crowded one (like a railway platform). In the crowd your straight walks between bumps get much shorter: that is a shorter mean free path.
Board exam focus
- Define mean free path and write λ = 1/(√2 n π d²); the collision-tube derivation is a common 3-mark question.
- Know how λ changes with P, T, n and d.
- Convert units: d in m, n in m⁻³; answers are often in nm or μm.
Key formulas and definitions
- λ = 1 / (√2 n π d²)
- λ = k_B T / (√2 π d² P)
- n = P / (k_B T)
- Collision cross-section = π d²
- Collision frequency ν = v̄ / λ
- Time between collisions τ = λ / v̄
- λ ∝ 1/n, λ ∝ 1/d², λ ∝ T/P
Worked examples
1. Nitrogen at STP has n = 2.7 × 10²⁵ m⁻³ and d = 3.0 × 10⁻¹⁰ m. Find the mean free path.
Step 1: πd² = 3.14 × 9 × 10⁻²⁰ = 2.83 × 10⁻¹⁹ m². Step 2: √2 n πd² = 1.414 × 2.7 × 10²⁵ × 2.83 × 10⁻¹⁹ = 1.08 × 10⁷. Step 3: λ = 1/1.08 × 10⁷. Answer: λ ≈ 9.3 × 10⁻⁸ m (about 93 nm).
2. Find λ for a gas at 300 K and 1.0 × 10⁵ Pa with d = 2.0 × 10⁻¹⁰ m. (k = 1.38 × 10⁻²³ J/K)
Step 1: kT = 4.14 × 10⁻²¹ J. Step 2: √2 π d² P = 1.414 × 3.14 × 4 × 10⁻²⁰ × 10⁵ = 1.78 × 10⁻¹⁴. Step 3: λ = 4.14 × 10⁻²¹ / 1.78 × 10⁻¹⁴. Answer: λ ≈ 2.3 × 10⁻⁷ m.
3. The mean free path of a gas is 100 nm at 1 atm. What is it at 0.01 atm, same temperature?
Step 1: λ ∝ 1/P at fixed T. Step 2: pressure is 100 times smaller, so λ is 100 times larger. Answer: λ = 10⁴ nm = 10 μm.
4. Molecules with average speed 500 m/s have λ = 1.0 × 10⁻⁷ m. Find the collision frequency and the time between collisions.
Step 1: ν = v̄/λ = 500 / 10⁻⁷ = 5 × 10⁹ per second. Step 2: τ = 1/ν. Answer: ν = 5 × 10⁹ s⁻¹, τ = 2 × 10⁻¹⁰ s.
5. Gas A molecules are twice as wide as gas B molecules. Both have the same n. Compare their mean free paths.
Step 1: λ ∝ 1/d². Step 2: λ_A/λ_B = (d_B/d_A)² = (1/2)². Answer: λ_A = λ_B / 4.
6. Find the number of molecules per m³ in a gas at 300 K and 1.0 × 10⁵ Pa.
Step 1: n = P/(kT). Step 2: n = 10⁵ / 4.14 × 10⁻²¹. Answer: n ≈ 2.4 × 10²⁵ m⁻³.
7. At 300 K, down to what pressure must a gas with d = 3.0 × 10⁻¹⁰ m be pumped so that λ = 1 m?
Step 1: P = kT/(√2 π d² λ). Step 2: √2 π d² = 1.414 × 3.14 × 9 × 10⁻²⁰ = 4.0 × 10⁻¹⁹ m². Step 3: P = 4.14 × 10⁻²¹ / (4.0 × 10⁻¹⁹ × 1). Answer: P ≈ 1.0 × 10⁻² Pa (a good vacuum).
Common mistakes
- Using radius instead of diameter in λ = 1/(√2 n π d²). The collision tube's radius is the full diameter d.
- Forgetting the √2 that comes from the motion of the other molecules.
- Thinking λ changes with temperature when n is fixed. For fixed n, λ is fixed; λ ∝ T only at fixed pressure.
- Thinking a fast molecule should cross a room in a millisecond. Collisions make it zig-zag, so it spreads slowly.