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

The assumptions (WHAT we are allowed to use)

WHY these matter: elastic collisions let us reverse velocity cleanly; no forces means molecules travel in straight lines between hits; negligible size means they don't block each other. These simplifications make the maths exact for an ideal gas.

Setting up the problem (HOW we start)

Put $N$ molecules, each of mass $m$ , in a cube of side $L$ (so volume $V = L^3$ ).

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

Take one molecule with velocity components $v_x, v_y, v_z$ . We first find the force it exerts on one wall (the wall perpendicular to the $x$ -axis), then generalise.

Step 1 — Momentum change in one collision

The molecule hits the right wall moving with $x$ -velocity $+v_x$ and bounces back with $-v_x$ (elastic, only $x$ -component reverses).

$\Delta p = m(-v_x) - m(v_x) = -2mv_x$

Why this step? Only the $x$ -component matters for a wall facing the $x$ -direction — the $y,z$ motion slides along the wall and delivers no push into it. The wall receives $+2mv_x$ (Newton's third law).

Step 2 — How often does it hit?

Between two hits on the same wall, the molecule travels to the opposite wall and back: a distance $2L$ at speed $v_x$ .

$\text{time between collisions } \Delta t = \frac{2L}{v_x}$

Why this step? Rate of collisions = 1 per round-trip. More hits per second → more force.

Step 3 — Force from one molecule

Force = rate of momentum transfer:

$f_x = \frac{\text{momentum per hit}}{\text{time per hit}} = \frac{2mv_x}{\,2L/v_x\,} = \frac{m v_x^2}{L}$

Why this step? Force is how fast momentum is delivered. Notice $v_x^2$ appears: fast molecules hit harder and more often — two factors of $v_x$ .

Step 4 — All $N$ molecules on this wall

$F_x = \frac{m}{L}\left(v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2\right) = \frac{mN}{L}\,\overline{v_x^2}$

where $\overline{v_x^2}$ is the mean of the squared $x$ -velocities.

Why this step? Total force is just the sum of all individual pushes; factoring out $N$ turns the sum into $N \times$ average.

Step 5 — Pressure on that wall

$P = \frac{F_x}{\text{area}} = \frac{F_x}{L^2} = \frac{mN\,\overline{v_x^2}}{L^3} = \frac{mN\,\overline{v_x^2}}{V}$

Step 6 — Use isotropy (random directions) to remove $x$

Motion is random, so no direction is special:

$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$

Since $v^2 = v_x^2 + v_y^2 + v_z^2$ , taking averages:

$\overline{v^2} = 3\,\overline{v_x^2} \quad\Rightarrow\quad \overline{v_x^2} = \tfrac{1}{3}\overline{v^2}$

Why this step? This is the key trick. We measured only the $x$ -wall, but pressure is the same on all walls because motion has no preferred direction. So we replace $\overline{v_x^2}$ with the full 3-D speed.

Step 7 — Final result

$\boxed{PV = \tfrac{1}{3} N m\, \overline{v^2}} \quad\Longleftrightarrow\quad P = \tfrac{1}{3}\,\rho\, v_{rms}^2$

where $\rho = \dfrac{Nm}{V}$ is the density and $v_{rms} = \sqrt{\overline{v^2}}$ .

Worked examples

Common mistakes

Active recall

Recall Cloze check (reveal after answering)

Momentum change per wall collision = == $2mv_x$ ==.
Time between hits on the same wall = == $2L/v_x$ ==.
Force from one molecule on one wall = == $mv_x^2/L$ ==.
Key isotropy relation: $\overline{v_x^2} =$ == $\tfrac13\overline{v^2}$ ==.
Final: $P =$ == $\tfrac13\rho v_{rms}^2$ ==, and $v_{rms} =$ == $\sqrt{3RT/M}$ ==.

Recall Feynman: explain to a 12-year-old

Imagine a box full of super-bouncy tennis balls flying everywhere. Every time a ball hits the wall it bounces back and gives the wall a little shove. There are so many balls hitting so many times per second that all those tiny shoves feel like one steady push on the wall — that steady push is pressure. If the balls move faster (hotter box), they shove harder and more often, so the pressure goes up. Because the box is 3-D and the balls fly in all directions equally, each wall only feels one third of all the bouncing — that's where the $\tfrac13$ comes from!

Connections

Ideal Gas Equation PV = nRT — combine to get $KE_{avg}=\tfrac32 k_BT$ .
Maxwell-Boltzmann Speed Distribution — gives $v_{mp}, \overline{v}, v_{rms}$ .
Graham's Law of Diffusion — rate $\propto v_{rms}\propto 1/\sqrt{M}$ .
Average Kinetic Energy and Temperature
Real Gases and van der Waals Equation — what breaks when postulates fail.
Newton's Laws — Impulse and Momentum — the physics engine of this derivation.

What is the momentum change of a molecule in one elastic collision with a wall (x-facing)?

2mv_x

(velocity reverses from

+v_x

-v_x

What is the time between successive collisions of one molecule with the same wall?

\Delta t = 2L/v_x

(round trip of length

2L

at speed

v_x

Force exerted by one molecule on one wall?

f = mv_x^2/L

Why does

v_x^2

(not

v_x

) appear in the force?

Faster molecules both hit harder (

\propto v_x

) and hit more often (

\propto v_x

), giving two factors.

State the isotropy relation used in the derivation.

\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}=\tfrac13\overline{v^2}

Final KMT pressure equation in terms of density?

P=\tfrac13\rho v_{rms}^2

Express

PV

in terms of

N,m,\overline{v^2}

PV=\tfrac13 Nm\overline{v^2}

Derive

v_{rms}

formula from

PV=\tfrac13Nm\overline{v^2}=Nk_BT

v_{rms}=\sqrt{3k_BT/m}=\sqrt{3RT/M}

Why is

v_{rms}

used and not average speed?

Pressure depends on

\overline{v^2}

;

v_{rms}=\sqrt{\overline{v^2}}\neq\overline{v}

Common error giving 2× too small pressure?

Using

\Delta p=mv_x

instead of

2mv_x

(forgetting the bounce reverses momentum).

At same T, ratio of

v_{rms}

for H₂ vs O₂?

\sqrt{32/2}=4

(lighter = faster).

Where does the factor 1/3 come from physically?

The 3 spatial dimensions share the speed equally; each wall feels one third.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, gas ke andar molecules bilkul random direction me udd rahe hote hain, empty space me. Pressure ka "push" kahan se aata hai? Jab yeh molecules wall se takraate hain aur wapas bounce karte hain, tab har collision wall ko ek chhota sa dhakka deta hai. Ek molecule ki velocity $+v_x$ se $-v_x$ ho jaati hai (elastic bounce), yaani momentum change $2mv_x$ — yahan factor 2 mat bhoolna, kyunki molecule rukta nahi, ulta bounce karta hai.

Ab ek molecule wall se dobara takraane me $2L$ distance travel karta hai speed $v_x$ se, to time lagta hai $2L/v_x$ . Force = momentum change / time = $mv_x^2/L$ . Saare $N$ molecules jodo, wall area $L^2$ se divide karo, aur $L^3 = V$ maan lo — mil jaata hai pressure. Ek final trick: motion har direction me barabar hai (isotropy), isliye $\overline{v_x^2} = \tfrac13 \overline{v^2}$ . Yahi se aata hai wo famous $\tfrac13$ — kyunki 3 dimensions speed ko barabar baant lete hain.

Result: $P = \tfrac13 \rho v_{rms}^2$ . Isko $PV = nRT$ ke saath compare karo to average kinetic energy $= \tfrac32 k_B T$ aur $v_{rms} = \sqrt{3RT/M}$ nikal aata hai. Yaad rakho — $M$ ko hamesha kg/mol me lena, warna answer galat aayega. Aur $v_{rms}$ use karna, average speed nahi, kyunki pressure $v^2$ pe depend karta hai. Yeh derivation important hai kyunki isne pehli baar temperature ko molecules ki speed se jod diya — pura ideal gas behaviour isi se samajh aata hai.

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

The assumptions (WHAT we are allowed to use)

Setting up the problem (HOW we start)

Step 1 — Momentum change in one collision

Step 2 — How often does it hit?

Step 3 — Force from one molecule

Step 4 — All $N$ molecules on this wall

Step 5 — Pressure on that wall

Step 6 — Use isotropy (random directions) to remove $x$

Step 7 — Final result

Worked examples

Common mistakes

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — States of Matter (Quantitative)

Connections