Ideal gas law PV = nRT — derivation from kinetic theory: Visual walkthrough

Step 0 — The stage: a box of balls

Before any maths, meet the cast.

A cube with each side of length $L$ (so its volume is $V = L^{3}$ — length times width times height, all equal).
Inside: $N$ tiny balls (molecules), each with mass $m$ (how heavy one ball is).
Each ball zooms around with a velocity — an arrow saying how fast and in which direction it moves.

We slice that velocity arrow into three pieces along the three edges of the box: $v_{x}$ (how fast it moves left–right), $v_{y}$ (front–back), $v_{z}$ (up–down). These are the velocity components.

Figure — Ideal gas law PV = nRT — derivation from kinetic theory

We will study the wall on the right, the one the $x$ -axis pokes through. Only $v_{x}$ matters for pushing that wall.

Step 1 — One bounce off the wall

WHAT: track a single ball smacking the right wall and bouncing straight back.

WHY: a wall cannot feel "temperature". It only feels pushes — forces. Newton told us a force is nothing but momentum changing over time: $F = \frac{Δ p}{Δ t}$ where $p = m v$ is momentum (mass × velocity, "how much oomph a moving thing carries") and $Δ$ means "the change in". So to find any push, we must first find how much momentum a ball hands over per bounce. That is this step. See Pressure and Newton's Second Law.

PICTURE: the ball arrives moving right with $x$ -momentum $+ m v_{x}$ , and leaves moving left with $x$ -momentum $- m v_{x}$ . The collision is perfectly elastic — the ball keeps its speed, it just flips direction.

$change in x -momentum Δ p_{x} = after: moving left (- m v_{x}) - before: moving right (+ m v_{x}) = - 2 m v_{x}$

The ball's momentum changed by $- 2 m v_{x}$ . By Newton's third law the ball pushes the wall the opposite way, so the wall receives a kick of size $2 m v_{x}$ . The minus sign just says "leftward"; its size is what pushes the wall.

Step 2 — How often does it come back?

WHAT: figure out the time between two hits of the same ball on the same wall.

WHY: one kick isn't a steady push. A wall feels pressure because kicks arrive again and again. So we need the rate of kicks, which means the time gap between them.

PICTURE: after bouncing off the right wall, the ball flies all the way to the left wall (distance $L$ ) and comes back (another $L$ ). Total round trip $= 2 L$ . Travelling at speed $v_{x}$ in the $x$ -direction, the time is distance ÷ speed.

$Δ t = \frac{2 L round-trip distance}{x -speed v _{x}}$

So a faster ball ( $v_{x}$ big) returns sooner — it hits more often. Hold that thought: $v_{x}$ has now shown up in two places — the size of each kick and how often kicks come.

Step 3 — Average push from one ball

WHAT: combine Step 1 (kick size) and Step 2 (kick rate) into a steady average force.

WHY: because a wall being hit 2 000 times a second with tiny taps feels the same as one smooth push equal to (tap size) × (taps per second). That product is the average force, straight from $F = Δ p /Δ t$ .

$F_{1} = \frac{∣Δ p _{x} ∣}{Δ t} = \frac{2 m v _{x} kick per hit (Step 1)}{time per hit (Step 2) 2 L / v _{x}} = \frac{m v _{x}^{2}}{L}$

PICTURE: watch $v_{x}$ appear twice and multiply into $v_{x}^{2}$ . This is why the final law contains speed-squared, not speed.

Step 4 — Add up all $N$ balls

WHAT: sum the one-ball force over every ball in the box.

WHY: the wall feels all the balls at once. Each contributes independently, so total force is the sum. And a sum of $N$ things equals $N$ times their average — that is literally what "average" means.

$F = \frac{m}{L} \sum_{i = 1}^{N} v_{x, i}^{2} = \frac{m}{L} N mean of v_{x}^{2} over all balls ⟨ v_{x}^{2} ⟩$

The angle brackets $⟨ ⟩$ mean "average over all $N$ balls". We traded a giant messy sum of $N$ different numbers for one tidy quantity: the mean square $x$ -speed.

Step 5 — Randomness makes the $\frac{1}{3}$

WHAT: replace the wall-specific $⟨ v_{x}^{2} ⟩$ with the full speed $⟨ v^{2} ⟩$ .

WHY: we picked the $x$ -wall arbitrarily. In a real gas, no direction is special — this is called isotropy (Greek: "same in all directions"). So the average sideways-ness must equal the average front-ness must equal the average up-ness: $⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩$

Since $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ , averaging both sides gives $⟨ v^{2} ⟩ = 3 ⟨ v_{x}^{2} ⟩$ , so:

$⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩$

PICTURE: the speed-squared is a "pie" split into three equal slices, one per direction. Each wall only feels its own slice — one third.

Step 6 — From force to pressure

WHAT: turn the total force into pressure by dividing by the wall's area.

WHY: pressure is force spread over area — a big force on a tiny patch presses hard; the same force on a huge wall barely presses. The wall is a square of side $L$ , so its area is $L^{2}$ .

$P = \frac{F}{area} = \frac{F}{L ^{2}} = \frac{m N ⟨ v _{x}^{2} ⟩}{L ^{3}} = \frac{1}{3} \frac{m N ⟨ v ^{2} ⟩}{L ^{3}}$

Now the magic: $L^{3}$ is just the volume $V$ . Substituting $L^{3} = V$ :

Step 7 — The bridge: what is temperature?

WHAT: connect the mechanical result to temperature $T$ .

WHY: our formula speaks "speed"; thermometers speak "temperature". We need a translator. Nature provides one: for an ideal gas, temperature is defined as a measure of average translational kinetic energy (via the Equipartition Theorem). Each of the 3 movement directions stores $\frac{1}{2} k_{B} T$ of energy, so all three give:

$avg kinetic energy ⟨ E_{k} ⟩ = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$

Here $k_{B}$ is the Boltzmann constant — the tiny number ( $1.38 \times 1 0^{- 23} J/K$ ) that converts "degrees" into "joules per molecule". See Boltzmann Constant and Gas Constant and Temperature and Internal Energy.

PICTURE: a dial where "hotness" (T) and "average jiggle energy" ( $⟨ E_{k} ⟩$ ) are the same needle read on two scales.

Now nudge our Step 6 result into that shape by inserting a $\frac{1}{2}$ (multiply and divide by 2):

$P V = \frac{1}{3} N m ⟨ v^{2} ⟩ = \frac{2}{3} N = ⟨ E_{k} ⟩ = \frac{3}{2} k_{B} T (\frac{1}{2} m ⟨ v^{2} ⟩) = \frac{2}{3} N \cdot \frac{3}{2} k_{B} T = N k_{B} T$

The $\frac{2}{3}$ and $\frac{3}{2}$ cancel perfectly — a sign the definition was the right key.

$P V = N k_{B} T$

Step 8 — Counting in moles

WHAT: swap "number of balls $N$ " for "number of moles $n$ ".

WHY: chemists count gas in moles, not individual molecules. One mole is $N_{A} = 6.022 \times 1 0^{23}$ molecules (Avogadro's number). So $N = n N_{A}$ .

$P V = N k_{B} T = (n N_{A}) k_{B} T = n call it R (N_{A} k_{B}) T$

Bundling the two constants $N_{A} k_{B}$ into a single gas constant $R$ :

Real gases deviate when the "point particle, no forces" assumptions break — that's Real Gases and Van der Waals Equation. The spread of individual speeds behind $⟨ v^{2} ⟩$ is the Maxwell-Boltzmann Speed Distribution.

The one-picture summary

Here is the whole climb on one slide: bounce → kick → rate → force → sum → $\frac{1}{3}$ → pressure → temperature → moles.

Recall Feynman retelling — the walkthrough in plain words

Picture a box crammed with billions of tiny bouncy balls. Follow one ball. It smacks the right wall and bounces straight back — its sideways motion flips, and that flip hands the wall a kick worth $2 m v_{x}$ . Then it flies across, hits the far wall, comes back, and kicks again — every $2 L / v_{x}$ seconds. A kick this size arriving this often feels like a steady push of $m v_{x}^{2} / L$ . Add up the push from all $N$ balls and you get $N$ times the average of $v_{x}^{2}$ . But no direction is special, so the sideways jiggle is exactly one-third of the total jiggle — that's where the famous $\frac{1}{3}$ comes from. Divide the total force by the wall's area and out pops the pressure: $P = \frac{1}{3} N m ⟨ v^{2} ⟩ / V$ . So far it's pure Newton — no heat anywhere. The last trick is realising that "temperature" is just nature's word for "average jiggle energy": $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ . Slot that in and the fractions cancel to give $P V = N k_{B} T$ . Finally, count in moles instead of molecules, roll the two constants into $R$ , and you've built $P V = n R T$ from a single bouncing ball.

Recall Rapid self-test

Why does force scale with $v_{x}^{2}$ , not $v_{x}$ ? ::: Kick size $\propto v_{x}$ and hit rate $\propto v_{x}$ ; they multiply.
Where does $\frac{1}{3}$ come from? ::: Isotropy — $x$ is one of three equal directions, $⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩$ .
What physical input turns mechanics into thermodynamics? ::: $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ (temperature = average kinetic energy).
What is $R$ ? ::: $R = N_{A} k_{B}$ — Boltzmann's constant scaled from per-molecule to per-mole.

Parent: 1.7.08 Ideal gas law PV = nRT — derivation from kinetic theory (Hinglish)