Graham's law of effusion - diffusion (rate ∝ 1 - √M): Visual walkthrough

Before Step 1, let us agree on the cast of characters, in plain words:

Step 1 — Temperature is kicking energy (the great equalizer)

WHAT. We claim: two different gases sitting at the same temperature give every molecule the same average kicking energy, no matter how heavy the molecule is.

WHY this step. This is the single fact that makes light gases fast and heavy gases slow. If we did not have "same energy," we would have no way to compare two gases at all. It is the foundation stone — everything else is bookkeeping on top of it.

PICTURE. Look at s01. On the left, light violet balls; on the right, heavy magenta balls. Every ball carries the same size energy-arrow (same length). Same energy, but the light balls will have to move faster to spend that energy — that is the seed of the whole law.

Figure — Graham's law of effusion - diffusion (rate ∝ 1 - √M)

This idea comes straight from Kinetic Theory of Gases.

Step 2 — Energy is made of mass and speed

WHAT. We now write down what kicking energy actually is mechanically: it depends on how heavy a ball is ( $m$ ) and how fast it moves ( $v$ ).

WHY this step. Step 1 told us the energy value. But we want speed, because speed is what carries a molecule to the hole. So we need the bridge between energy and speed — that bridge is the kinetic-energy formula.

PICTURE. In s02, a single ball of mass $m$ moves right with speed $v$ . Its energy grows as the ball gets fatter (bigger $m$ ) or faster (bigger $v$ ) — but notice the $v$ is squared: doubling the speed quadruples the energy. Watch that squaring; it is where the square-root will eventually come from.

Step 3 — Set the two energies equal and solve for speed

WHAT. Step 1 and Step 2 are two descriptions of the same quantity $\overline{K E}$ . So we set them equal and rearrange to get speed alone.

WHY this step. This is the payoff: we combine "same energy for all" (Step 1) with "energy = ½mv²" (Step 2). The mass that was absent in Step 1 now reappears in Step 2 — and because it is tied to $v^{2}$ , it will control the speed.

PICTURE. s03 shows the two energy-expressions as two pans of a balance that must weigh the same. As we slide $m$ up, the speed-arrow must shrink to keep the balance level. Heavy ⇒ slow, drawn literally as a see-saw.

Take the square root of both sides to turn "speed squared" back into a plain speed:

Step 4 — Speed decides how often a ball finds the hole

WHAT. We connect molecular speed to the rate of effusion — the number of balls slipping through the pinhole per second.

WHY this step. Graham's law is about the hole, not directly about speed. We must argue that faster balls hit the hole more often. Reasoning: in one second, a fast ball sweeps out a longer path, so it arrives at the tiny opening more frequently. Double the speed ⇒ double the hits ⇒ double the escape rate.

PICTURE. s04: two chambers, each with a pinhole on the right wall. Fast violet balls stream out in a thick jet; slow magenta balls dribble out. The arrow-thickness is the rate.

Step 5 — Same temperature makes the constants vanish

WHAT. We compare two gases at the same $T$ . Everything inside the root except $M$ is identical for both, so it cancels.

WHY this step. We want a clean law with only masses in it. The pieces $3$ , $R$ , $T$ are the same number for both gases (same room, same temperature), so when we take the ratio they divide away.

PICTURE. s05: two rate-expressions side by side; the shared block $3 R T$ is greyed out and struck through on both, leaving only the $1/ M$ tails.

Step 6 — The degenerate & edge cases (never leave a gap)

WHAT. We check the situations that break naive intuition.

WHY this step. A law you cannot stress-test you do not really own. Let us push it to extremes.

PICTURE. s06 plots $r \propto 1/ M$ as a smooth curve, and marks four spots on it.

Step 7 — One worked number, fully drawn: H₂ vs O₂

WHAT. Plug in real masses to see the machine turn.

WHY. A number makes the abstract chain concrete and lets us check the whole page.

PICTURE. s07: a labelled bar showing H₂'s escape jet exactly 4× the O₂ jet.

The one-picture summary

Everything above, compressed: same $T$ → same energy → light balls faster → faster balls escape more → rate $\propto 1/ M$ → masses flip when comparing two gases.

Recall Feynman retelling — the whole walk in plain words

Picture a warm room where every ball, big or small, gets kicked with the same amount of oomph (that is what "same temperature" means — Step 1). Energy is heaviness times speed-squared (Step 2), so if two balls carry equal energy, the light one has to move faster to use up that energy — and because energy uses speed squared, making a ball four times heavier only makes it two times slower, not four (Step 3). Now poke a pinhole: fast balls sweep more ground each second, so they find the hole more often and leak out faster (Step 4). When you line up two gases in the same warm room, the temperature and constants are shared and cancel, leaving pure mass (Step 5) — and the masses flip, because faster means lighter. Push it to extremes and it still behaves: identical gases tie, monster-heavy gases crawl toward (never past) zero, and if the rooms differ in warmth the shortcut breaks (Step 6). Feed in H₂ and O₂ and out pops "four times faster" (Step 7). That whole story is the single formula $r_{1} / r_{2} = M_{2} / M_{1}$ .

Recall Rebuild the chain from memory

What single fact makes light gases faster? ::: At the same temperature every molecule has the same average kinetic energy (Step 1). Why a square root and not $1/ M$ ? ::: Because energy uses $v^{2}$ ; undoing the square gives (Steps 2–3). Why do the masses flip in $r_{1} / r_{2} = M_{2} / M_{1}$ ? ::: Rate rides on speed, and speed is inverse to $M$ ; the heavier gas's mass lands on top (Step 5). When does the simple form fail? ::: When temperatures (or pressures) differ, so $3 R T$ no longer cancels (Step 6).

Connections

Kinetic Theory of Gases — supplies the equal-energy fact of Step 1.
Root Mean Square Speed — the $v_{r m s}$ we built in Step 3.
Maxwell-Boltzmann Distribution — why "average" speed is the honest quantity to use.
Ideal Gas Equation — where $R$ and density $ρ \propto M$ come from.
Isotope Separation (Uranium Hexafluoride) — the tiny-mass-difference application of this exact chain.