Imagine a room full of tiny ping‑pong balls (the gas molecules) bouncing all around.
When a ball hits a wall, it pushes on that wall – that push is what we feel as pressure.
The faster and more often the balls hit, the higher the pressure.
Let’s look at just one direction, the x‑axis.
A molecule of mass \$m\$ moves with velocity \$c_x\$ and collides elastically with a wall perpendicular to x.
If the time between successive hits is \$\Delta t\$, the average force on the wall is
\$F = \dfrac{\Delta p}{\Delta t} = \dfrac{2 m c_x}{\Delta t}\$.
\$p = \dfrac{F}{A} = \dfrac{2 m c_x}{A \Delta t}\$.
If we have many molecules, each with its own \$c_x\$, we average over all of them.
The average of \$cx\$ over many collisions is zero (equal numbers go left and right), but the average of \$cx^2\$ is not.
So we replace \$cx\$ by its mean‑square value \$\langle cx^2 \rangle\$ in the pressure expression.
In reality, molecules move in all three directions.
Because the gas is isotropic, the mean‑square speeds in each direction are equal:
\$\langle cx^2 \rangle = \langle cy^2 \rangle = \langle c_z^2 \rangle.\$
The total mean‑square speed is
\$\langle c^2 \rangle = \langle cx^2 + cy^2 + cz^2 \rangle = 3 \langle cx^2 \rangle.\$
Substituting this into the 1‑D pressure formula and multiplying by the number of molecules \$N\$ gives the familiar relation:
\$pV = \frac{1}{3} N m \langle c^2 \rangle.\$
| Symbol | Meaning | Units |
|---|---|---|
| \$p\$ | Pressure | Pa (N m⁻²) |
| \$V\$ | Volume | m³ |
| \$N\$ | Number of molecules | dimensionless |
| \$m\$ | Mass of one molecule | kg |
| \$\langle c^2 \rangle\$ | Mean‑square speed | m² s⁻² |
\$pV = \frac{1}{3} N m \langle c^2 \rangle.\$