Mean Free Path Between Molecules [Thermal & Statistical Mechanics I Studied #15]

This posting is probably going to stir up a lot of controversy. I just couldn’t accept the book’s explanation, so I reconstructed it in my own way using ideas from the internet and the book. It might not even make sense, so please do point things out.

The content of Chapter 8 is — collisions between gases! But rather than acknowledging the effect of those collisions and then gent~ly~ gent~ly~ lifting the ideal gas assumption, the goal is to find “the time between collisions” along with “the free distance between collisions.”

Hand sketch: a spherical gas molecule flying to the right at speed v₁; its cross-section, hatched in red, is labeled area = σ

One spherical gas molecule with cross section $\sigma$ is flying along at speed $v_1$.

Hand sketch: the same molecule sweeps out a cylinder of length v₁·dt as it flies

During time $dt$ this guy.. will sweep through a volume of $\sigma v_1 dt$.

Hand sketch: the cylinder is full of small green dots, the other molecules, n of them per unit volume

Let’s say the number of molecules in unit volume (the dots) inside that cylinder is $n$. That’s the basic setting done.

The length of the cylinder that the gas sweeps through during those $dt$ seconds is $v_1 dt$. Let me reinterpret that length $v_1 dt$ in a different way.

I think we could write it like this.

Now now now now, but the other molecules are also moving around as they please, so this is a bit confusing.

So we introduce the concept of relative velocity: calling those other ones v_i , we think of them as standing still, and instead let the moving gas molecule move with $\vec v_r = \vec v_1 - \vec v_i$ . Then let’s count the ’number of collisions during $dt$ seconds’ again.

Hand sketch: a cylinder of length v_r·dt and base area σ; the collisions of a molecule moving at v_r in time dt equal the number of molecules inside it, nσv_r dt

So let’s look at the equation above again.

Putting nσv_r dt collisions into it: nσv_r dt × (distance between other molecules) = v₁dt

Ah but, the distances between molecules aren’t all the same, so let me rewrite it like this.

With averages: nσ⟨v_r⟩dt × (average distance between other molecules) = ⟨v₁⟩dt, so the average distance is ⟨v₁⟩/(nσ⟨v_r⟩)

Working out ⟨v_r⟩ from v_r = v₁ − v_i: the squares add and the cross term averages to zero (⟨cos θ⟩ = 0), so ⟨v_r²⟩ = 2⟨v²⟩ and the rms of v_r is √2 v_rms

Ah damn it, what we need is $\langle v_r\rangle$ — I mean, the mean of v_r — but what came out was the root-mean-square.

Earlier we had

The rms speed v_rms = √(3k_BT/m) and the mean speed ⟨v⟩ = √(8k_BT/πm) differ by a constant factor: v_rms = √(3π/8)⟨v⟩

Again,

The rms of v_r is √2 v_rms; rewriting both sides with mean speeds gives ⟨v_r⟩ = √2⟨v⟩

The average distance between other molecules is ⟨v₁⟩/(nσ⟨v_r⟩) = 1/(√2 nσ), and with p = nk_BT it becomes k_BT/(√2 pσ)

Now… let me reveal my trick.

The reason we were able to get here was, it just so happened that,

The two speeds again: v_rms = √(3k_BT/m) and ⟨v⟩ = √(8k_BT/πm)

because of this. It was because the rms and mean values just happened to be related by a constant factor.

The reason they were related by a constant factor like that is because the gases followed the Maxwell-Boltzmann distribution!!! If it had been a different distribution, that constant-factor relation might not have held~ So we might not have arrived at our conclusion~

That is, we juuust quietly, sneakily slipped in the assumption that they follow the Maxwell-Boltzmann distribution….hehe

Ex 8.1

At room temperature and atmospheric pressure, calculate the mean free path of $\mathrm{N}_2$ gas. (For $\mathrm{N}_2$ , use a molecular diameter of $d = 0.37\ \mathrm{nm}$.)

Example 8.1: the mean free path of N₂ at 273 K and 1 atm, λ = 1/(√2 nσ) = k_BT/(√2 pσ), about 61 nm

Prob 8.1

With the pressure at $10^{-10}\ \mathrm{mbar}$ and the temperature at room temperature, what is the mean free path of the nitrogen molecules inside an ultra-high vacuum chamber? What is the average collision time?

The diameter of the chamber is $0.5\ \mathrm{m}$. Compared to collisions between molecules, at what ratio do molecules collide with the chamber walls?

If the pressure were suddenly increased to $10^{-6}\ \mathrm{mbar}$ , how do these results change?

Unit conversion: 10⁻⁶ mbar = 10⁻⁹ bar and 1 bar = 10⁵ Pa, so the pressure is 10⁻⁴ Pa = 10⁻⁴ N/m²

To find the mean free path we need n and σ: σ = πd² = 0.43×10⁻¹⁸ m², and n = p/k_BT with p = 10⁻⁴ N/m² and T = 273 K

The picture works out λ = 1/(√2 nσ) as 0.119 m and remarks that this is crazy long, probably because it is an ultra-high vacuum chamber

Raising the pressure 1000 times makes λ 1/1000 as large; the picture gives λ = 0.0119 m

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the relative-speed picture, the cross term is $2\vec v_1\cdot\vec v_i$, not $2\vec v_1\cdot\vec v_r$: $v_r^2 = v_1^2 + v_i^2 - 2\vec v_1\cdot\vec v_i$, and its average is zero. The result $\langle v_r^2\rangle = 2\langle v^2\rangle$ is right.
In the vacuum-chamber problem, the number density at $10^{-4}\ \mathrm{Pa}$ ($10^{-6}\ \mathrm{mbar}$) is $n = p/(k_BT) \approx 2.65\times10^{16}\ \mathrm{m^{-3}}$, not $1.3807\times10^{19}$, and $\sigma = 0.43\times10^{-18}\ \mathrm{m^2}$ is an area. So $\lambda = 1/(\sqrt2\,n\sigma) \approx 62\ \mathrm{m}$ at $10^{-6}\ \mathrm{mbar}$ and about $6.2\times10^{5}\ \mathrm{m}$ at the problem’s $10^{-10}\ \mathrm{mbar}$ ($\lambda \propto 1/p$, a factor of $10^4$ between the two), instead of the pictures’ $0.119\ \mathrm{m}$ and $0.0119\ \mathrm{m}$. These use $T = 273\ \mathrm{K}$ as the pictures do; at room temperature they are about 7% larger.

Ex 8.1

Prob 8.1

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