Visual walkthrough — Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms

Before anything, meet the cast of characters. We will not use a letter until you have seen it in a picture.

Why split the arrow into three? Because a wall can only feel the motion heading straight at it. Motion sliding sideways along a wall never pushes into it. So we must separate the diagonal velocity into its three edge-directions before we can talk about any one wall. That single decision drives the whole derivation.

---

Step 1 — One collision: momentum flips sign

WHAT. Watch one molecule approach the right-hand wall (the wall facing the $x$-direction). It comes in with $x$-velocity $+v_x$ and leaves with $-v_x$. Its $y$ and $z$ motion are untouched.

WHY. We start with the smallest possible event — one bounce — because pressure is built out of billions of these, and we can only add up things we first understand alone. Postulate 3 (elastic collision) is what lets us cleanly flip only the $x$-component: no speed is lost, the wall just mirrors it.

PICTURE.

Look at the two arrows on the molecule. Before (blue) points right, magnitude $v_x$. After (pink) points left, same magnitude. The vertical position is unchanged — that is the $y,z$ motion sliding harmlessly along the wall.

$$\underset{\text{change in momentum}}{\underbrace{\Delta p}}=\underset{\text{after: leftward}}{\underbrace{m(-v_x)}}-\underset{\text{before: rightward}}{\underbrace{m(+v_x)}}=-2m{v}_x$$

Here momentum $=m\times \text{velocity}$ — literally "mass carrying its motion". The molecule's own momentum drops by $2m{v}_x$; by Newton's Laws — Impulse and Momentum (action = reaction), the wall gains $+2m{v}_x$.

---

Step 2 — How often does this same wall get hit?

WHAT. After bouncing off the right wall, the molecule flies to the left wall and comes all the way back before it can hit the right wall again. That round-trip is a distance of $2L$.

WHY. Force is not just "how hard" — it is "how hard and how often". A hard hit once an hour is a weak force. So we must find the time between two hits on the same wall.

PICTURE.

The dashed yellow path traces the journey: right wall → left wall → right wall, total length $2L$. Since it covers that at $x$-speed $v_x$:

$$\underset{\text{time per round-trip}}{\underbrace{\Delta t}}=\frac{\stackrel{\text{distance there-and-back}}{\overbrace{2L}}}{\underset{\text{x-speed}}{\underbrace{v_x}}}$$

Between-wall bumps with other molecules? Postulate 4 (no forces) plus averaging means we can ignore them — on average the molecule still crosses the box in the same time. Faster molecule ($v_x$ large) → smaller $\Delta t$ → more hits per second.

---

Step 3 — Force from this one molecule on this one wall

WHAT. Combine Step 1 (each hit delivers $2m{v}_x$) with Step 2 (a hit happens every $2L/v_x$ seconds).

WHY. Force is the rate at which momentum is delivered — momentum per second. That is exactly (amount per hit) ÷ (time per hit).

PICTURE.

$$\underset{\begin{array}{c}\text{force from}\\ \text{one molecule}\end{array}}{\underbrace{f_x}}=\frac{\stackrel{\text{momentum per hit}}{\overbrace{2m{v}_x}}}{\underset{\text{time per hit}}{\underbrace{2L/v_x}}}=\frac{2m{v}_x\cdot v_x}{2L}=\frac{m{v}_x^2}{L}$$

Notice the $v_x$ appears twice and becomes $v_x^2$: once because a fast molecule hits harder, once because it hits more often. That squared speed is the seed of everything — it is why the final answer uses $v_{rms}$ (a squared-speed average) and not the plain average speed.

---

Step 4 — Add up all $N$ molecules

WHAT. Every molecule has its own $x$-speed. Sum all their pushes on this wall.

WHY. Total force on the wall is simply all the individual forces stacked together — pressure is a crowd effect.

PICTURE.

$$\underset{\text{total wall force}}{\underbrace{F_x}}=\frac{m}{L}\left(\underset{\text{one term per molecule}}{\underbrace{v_{x1}^2+v_{x2}^2+\cdots +v_{xN}^2}}\right)=\frac{mN}{L}\underset{\begin{array}{c}\text{average of the}\\ \text{squared x-speeds}\end{array}}{\underbrace{\overline{v_x^2}}}$$

The bar in $\overline{v_x^2}$ means "average over all $N$ molecules". Summing $N$ terms and dividing by $N$ turns the messy sum into $N\times (\text{average})$. This is the moment we stop tracking individuals and start describing the whole gas.

---

Step 5 — Turn force into pressure

WHAT. Pressure is force spread over area. The wall is a square of side $L$, so its area is $L^2$.

WHY. Pressure is defined as force per unit area — that is what "push per patch of wall" means. Bigger wall, same force → gentler pressure.

PICTURE.

$$\underset{\text{pressure}}{\underbrace{P}}=\frac{F_x}{\underset{\text{wall area}}{\underbrace{L^2}}}=\frac{mN\overline{v_x^2}}{L\cdot L^2}=\frac{mN\overline{v_x^2}}{\underset{=V}{\underbrace{L^3}}}=\frac{mN\overline{v_x^2}}{V}$$

The three $L$'s (one from the force, two from the area) collapse into $L^3$, which is exactly the volume $V$. Neat: the box size vanishes into a single physical quantity.

---

Step 6 — The key trick: three directions are equal

WHAT. We only studied the $x$-wall. But molecules fly randomly, so on average no direction is special. Therefore the average squared speeds in all three directions are equal.

WHY. This is postulate 1 (random motion) doing its job. If $x$ were somehow favoured, the gas would drift — but it does not. Randomness forces symmetry.

PICTURE.

The full speed splits by the 3-D Pythagoras rule (each edge-shadow squared, added):

$$\underset{\text{full speed squared}}{\underbrace{v^2}}=v_x^2+v_y^2+v_z^2$$

Take the average of every term, then use the equality of the three:

$$\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\overline{v_x^2}⟹\boxed{\overline{v_x^2}=\frac{1}{3}\overline{v^2}}$$

This is where the famous $\frac{1}{3}$ is born — not from any collision, but from geometry: three equal directions share the total motion, so each wall feels exactly one-third.

---

Step 7 — Assemble the final result

WHAT. Put $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$ back into the pressure from Step 5.

WHY. We measured the $x$-wall but we want a statement about the whole 3-D gas. This substitution promotes our one-wall result to a universal one.

PICTURE.

$$P=\frac{mN}{V}\cdot \frac{1}{3}\overline{v^2}⟹\boxed{PV=\frac{1}{3}Nm\overline{v^2}}$$

Now define density $\rho =\frac{Nm}{V}$ (total mass ÷ volume) and the root-mean-square speed $v_{rms}=\sqrt{\overline{v^2}}$ (so $\overline{v^2}=v_{rms}^2$):

$$\boxed{P=\frac{1}{3}\rho v_{rms}^2}$$

---

Step 8 — Edge and degenerate cases (never skipped)

WHAT. Check the corners: a molecule sliding along the wall, a molecule with zero $x$-speed, and heavy vs light gases.

WHY. A derivation you can trust must survive its extremes. If a strange input broke it, the whole story would be suspect.

PICTURE.

• Molecule moving purely along the wall ($v_x=0$). From Step 3, $f_x=m(0{)}^2/L=0$. It never reaches the far wall in the $x$-sense — it grazes forever. Zero push, exactly as the picture shows the arrow parallel to the wall. Correct.
• Head-on molecule ($v_x=v$, so $v_y=v_z=0$). It alone contributes its full $v^2$ to the $x$-wall, but such special molecules are rare; averaged over the random crowd, symmetry restores $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$. No single molecule breaks the law; the average does the work.
• Light vs heavy gas (same $T$). Using $v_{rms}=\sqrt{3RT/M}$: at equal temperature, equal $K{E}_{avg}$, so a lighter $M$ must move faster. For $H_2$ vs $O_2$: $\sqrt{32/2}=4$ — hydrogen is four times faster (this powers Graham's Law of Diffusion and shapes the Maxwell-Boltzmann Speed Distribution).
• Real gases. At high pressure, molecule volume and attractions stop being negligible (postulates 2 and 4 fail) — the tidy $\frac{1}{3}\rho v_{rms}^2$ needs correcting: see Real Gases and van der Waals Equation.

---

The one-picture summary

This single figure chains every step: one bounce ($2m{v}_x$) → its rate ($v_x/2L$) → one force ($m{v}_x^2/L$) → all molecules ($mN\overline{v_x^2}/L$) → pressure ($/L^2$) → the $\frac{1}{3}$ from three directions → $P=\frac{1}{3}\rho v_{rms}^2$.

Recall Feynman retelling — the whole walkthrough in plain words

Picture one bouncy ball in a box. It hits the right wall and bounces straight back, so its motion doesn't just stop — it flips, giving the wall a double-strength shove of $2m{v}_x$. Then it races across to the far wall and back — a trip of $2L$ — before it can shove the right wall again. Shove ÷ trip-time = the steady force from that one ball, and because a faster ball shoves harder and more often, the speed shows up squared. Add up all the balls' shoves, then spread that force over the wall's area to get pressure. Finally: the balls fly in all three directions equally, so each wall only feels one third of the total zooming — and that's the $\frac{1}{3}$. Bundle the mass-per-volume into density $\rho$ and the squared-speed-average into $v_{rms}^2$, and out pops $P=\frac{1}{3}\rho v_{rms}^2$.

Recall Quick cloze check

• Momentum delivered per hit ::: $2m{v}_x$
• Time between hits on the same wall ::: $2L/v_x$
• Force from one molecule ::: $m{v}_x^2/L$
• Why the $\frac{1}{3}$ ::: three equal directions, so $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$
• Final law ::: $P=\frac{1}{3}\rho v_{rms}^2$