Graham's Law of Effusion [Thermal & Statistical Mechanics I Studied #13]

Already Chapter 7. Molecular Effusion. We’ve already sprinted to chapter 7! This time we learn effusion speed!

But this is super super easy. Because we learned it in high school chemistry class! The chem teacher probably threw a proportionality formula at you.

Like this: the effusion speed is inversely proportional to the square root of mass!!! Or you might have learned it like this

Anyway, we’re going to find out why this proportionality formula makes sense.

In chemistry class you probably learned this law as “Graham’s law (of effusion)”. Actually, it’s understandable that the chemistry teacher would just throw the above proportionality at you without deriving it. Because, first of all, the calculation is hard for a high schooler to derive, and second, this law was originally asserted based on experiments.

But, since we’re physics majors, let’s work through the cause and effect one by one!!! No need to talk so grandly though. The materials needed to derive this are all things we’ve used before at some point… hehe, lots of copy-paste incoming? hehe

Let’s take a look. Somewhere, molecules are effusing… We need to count how many of these pass through per unit time~

To count “the ones passing through”, I’ll introduce the concept of flux. It’s really nothing. Flux is

Flux: a stream of molecules (red arrows) crossing a unit area (the black parallelogram)

When what’s passing through like this is an electric field, or a magnetic field — just as we derived Maxwell’s Equations with this in electromagnetism — I’ll calculate the passing “molecules” as a flux.

The number of things passing through a unit area per unit time is…?

Huh?? We just did this right before, didn’t we?

We did “the number of things hitting a unit area per unit time”.

If we say that hitting means the number passing through an empty unit area, wouldn’t it be same-same?

The number of molecules crossing a unit area per unit time with speed v at angle θ: v cosθ · n · (4/√π)(m/2k_BT)^(3/2) · v² e^(−mv²/2k_BT) · ½ sinθ dθ

We did this here. (link - Ideal Gas Equation of State and Pressure (Thermal & Statistical Mechanics I Studied #11))

If we count all molecules passing through for every $\theta$ and every $v$, likewise, we just need to sum them all up with an integral.

The flux as a double integral over v and θ; pulling n/2 out front lets the v integral and the θ integral (cosθ sinθ dθ) separate

Handwritten working: the v integral is written ⟨v⟩, and the θ integral of cosθ sinθ dθ from 0 to π/2 comes out to ½

The flux is ¼ n⟨v⟩ with ⟨v⟩ = √(8k_BT/πm), so Φ = ¼ n √(8k_BT/πm)

(link - Expected Speed, Expected Energy, and Statistical Analysis of Ideal Gas Speed (Thermal & Statistical Mechanics I Studied #10))

Writing n as p/k_BT: Φ = ¼ (p/k_BT) √(8k_BT/πm) = p/√(2πm k_BT)

We didn’t compute the effusion speed. We ended up computing the effusion rate.

We looked at Graham’s law, which says the effusion rate is inversely proportional to the square root of mass hehe

Comments