Mean free path
Why smells spread slowly
Light an agarbatti in one corner of a still room. Air molecules move at hundreds of metres per second, yet the smell takes many seconds, even minutes, to cross the room. Why so slow?
Because each molecule keeps bumping into others. It moves in a straight line, collides, changes direction, and does it again, tracing a zig-zag path that wanders only slowly across the room.
The average distance a molecule travels between two successive collisions is called its mean free path, written l.
Model each molecule as a hard sphere of diameter d. Our molecule hits any other molecule whose centre comes within a distance d of its own centre.
So as it moves, it sweeps out a cylinder of radius d, with cross-section π d squared. In a time Δt, moving at average speed v̄, the cylinder has volume π d squared v̄ Δt.
With n molecules per unit volume, the cylinder contains n π d squared v̄ Δt molecules, and our molecule hits every one. That is the number of collisions in the time Δt.
So the average time between collisions is τ equals 1 over n π v̄ d squared, and the mean free path, v̄ τ, is 1 over n π d squared.
But we pretended the other molecules stand still. They do not. Averaging over their motion brings in a factor of root two: l equals 1 over root 2 n π d squared.
So the mean free path is shorter when the molecules are more crowded, larger n, or bigger, larger d. Double the diameter and l drops to a quarter.
How far is that?
Let us estimate the mean free path of a water molecule in steam at 373 kelvin and atmospheric pressure. Take its diameter as about 2 × 10⁻¹⁰ metre.
First, the number density, from P equals n k B T. n equals 1.01 × 10⁵ divided by 1.38 × 10⁻²³ times 373: about 2 × 10²⁵ molecules per cubic metre.
Then l equals 1 over root 2, times 2 × 10²⁵, times π, times 2 × 10⁻¹⁰ squared. That works out to about 2.8 × 10⁻⁷ metre.
That is about 1400 molecular diameters. It is also about 75 times the average gap between neighbouring molecules, which is n to the power minus one-third, about 3.7 × 10⁻⁹ metre.
At about 700 metres per second, a molecule covers one free path in about 4 × 10⁻¹⁰ seconds. So each molecule collides more than a billion times every second.
Lower the pressure and n falls, so the mean free path grows. In a good vacuum, at a billionth of an atmosphere, it becomes a billion times longer: hundreds of metres, much bigger than the container.
Since n equals P over k B T, at a fixed temperature a lower pressure gives a longer mean free path, and at a fixed pressure a hotter gas gives a longer one.
Now we see why a smell spreads slowly. Each molecule takes billions of tiny random steps every second, and random steps carry it away only slowly. This slow spreading is called diffusion.
Try it: follow one molecule
- The amber molecule leaves a trail; each orange dot on it is a collision.
- The app adds up how far all the molecules travel and divides by the number of collisions (each involves two molecules). Compare with 1/(√2 n π d²).
- Double the number of molecules: l halves. Double d: l drops to a quarter.
- Turn collisions off, and the molecules fly from wall to wall in straight lines.
Bigger but fewer
The molecules of gas A have twice the diameter of those of gas B, but gas A has only half as many molecules per unit volume. How does the mean free path in A compare with that in B?
- l_A = 4 l_B
- l_A = l_B
- l_A = ½ l_B
- l_A = ¼ l_B
Answer
C. l = 1/(√2 n π d²), so l ∝ 1/(n d²). For gas A, n d² is ½ × 2² = 2 times that of B, so its mean free path is half as long. The diameter matters more than the number density, because it is squared.
Nitrogen at STP
Air at STP has about 2.7 × 10²⁵ molecules per cubic metre. Taking the diameter of a nitrogen molecule as 3 × 10⁻¹⁰ m, estimate its mean free path.
- 3.3 × 10⁻⁹ m
- 9.3 × 10⁻⁸ m
- 1.3 × 10⁻⁷ m
- 3.7 × 10⁻⁷ m
Answer
B. πd² = 3.14 × 9 × 10⁻²⁰ = 2.83 × 10⁻¹⁹ m². Then l = 1/(1.414 × 2.7 × 10²⁵ × 2.83 × 10⁻¹⁹) = 1/(1.08 × 10⁷) ≈ 9.3 × 10⁻⁸ m, about 0.1 μm. Leaving out √2 gives 1.3 × 10⁻⁷ m; using the radius instead of the diameter gives 3.7 × 10⁻⁷ m. 3.3 × 10⁻⁹ m is n^(−1/3), the average gap between molecules, not the distance between collisions.
Summary
- Mean free path:
l = 1 / (√2 n π d²) - Number density:
n = N/V = P / (k_B T) - Time between collisions:
τ = l / v̄
- The mean free path is the average distance a molecule travels between successive collisions.
- A molecule sweeps a cylinder of cross-section πd²; the √2 comes from the motion of the other molecules.
- l is shorter for a denser gas and for bigger molecules; it grows at low pressure.
- In air at room conditions l is about 0.1 μm: hundreds of molecular diameters, but tiny compared with a room. Molecules collide billions of times a second, so gases mix slowly by diffusion.