The kinetic molecular model explains gas pressure as the accumulated effect of molecular collisions with the container walls. That idea can be made quantitative using momentum transfer.

Consider a cubic container of side length $L$ and volume

$$V=L^3.$$

Take one molecule of mass $m$ with velocity component $v_x$ perpendicular to a wall.

Momentum transfer from one molecule

In an elastic collision with the wall, the molecule's $x$-momentum changes from $mv_x$ to $-mv_x$. The magnitude of the momentum transferred to the wall is therefore

$$\Delta p_x=2m|v_x|.$$

The molecule must travel to the opposite wall and back before striking the same wall again, a distance $2L$. The time between successive collisions with that wall is

$$\Delta t=\frac{2L}{|v_x|}.$$

Its average force on that wall is therefore

$$F_x=\frac{\Delta p_x}{\Delta t} =\frac{2m|v_x|}{2L/|v_x|} =\frac{mv_x^2}{L}.$$

Many molecules and gas pressure

For $N$ molecules, the total average force on a wall is

$$F=\frac{m}{L}\sum_{i=1}^N v_{x,i}^2.$$

The wall area is $A=L^2$, so

$$p=\frac{F}{A} =\frac{m}{L^3}\sum_i v_{x,i}^2.$$

Since $V=L^3$,

$$pV=m\sum_i v_{x,i}^2.$$

At equilibrium, molecular motion is isotropic: no direction is statistically preferred. Therefore

$$\langle v_x^2\rangle =\langle v_y^2\rangle =\langle v_z^2\rangle =\frac13\langle v^2\rangle.$$

Hence

$$\boxed{pV=\frac13Nm\langle v^2\rangle}.$$

The total translational kinetic energy of the molecules is

$$K_{\rm trans} =N\left\langle\frac12mv^2\right\rangle =\frac12Nm\langle v^2\rangle.$$

Therefore

$$\boxed{pV=\frac23K_{\rm trans}}.$$

This is a direct bridge from microscopic Newtonian mechanics to a macroscopic gas variable.

Temperature and mean translational kinetic energy

The ideal-gas law can be written in particle form as

$$pV=Nk_BT,$$

where $k_B$ is Boltzmann's constant. Comparing this with the kinetic-theory result gives

$$Nk_BT=\frac13Nm\langle v^2\rangle.$$

Thus

$$\boxed{\left\langle\frac12mv^2\right\rangle=\frac32k_BT}.$$

For a monatomic ideal gas, absolute temperature therefore sets the mean translational kinetic energy per particle.

Two ideal gases at the same temperature have the same mean translational kinetic energy per molecule even if their molecular masses are different.

Root-mean-square molecular speed

Define the root-mean-square speed by

$$v_{\rm rms}=\sqrt{\langle v^2\rangle}.$$

From the kinetic-energy relation,

$$\frac12m v_{\rm rms}^2=\frac32k_BT,$$

so

$$\boxed{v_{\rm rms}=\sqrt{\frac{3k_BT}{m}}}.$$

Using molar mass $M$ in $\mathrm{kg/mol}$ and the universal gas constant $R_u=N_Ak_B$ gives

$$\boxed{v_{\rm rms}=\sqrt{\frac{3R_uT}{M}}}.$$

This shows why lighter molecules move faster on average at the same temperature: equal thermal kinetic energy implies larger speed when the molecular mass is smaller.

Worked example: nitrogen at room temperature

For nitrogen gas with

$$M=0.0280,\mathrm{kg/mol}$$

at

$$T=300,\mathrm K,$$

$$v_{\rm rms} =\sqrt{\frac{3(8.314)(300)}{0.0280}} \approx5.17\times10^2,\mathrm{m/s}.$$

Thus

$$\boxed{v_{\rm rms}\approx517,\mathrm{m/s}}.$$

This does not mean every nitrogen molecule moves at $517,\mathrm{m/s}$. A gas contains a distribution of molecular velocities; $v_{\rm rms}$ is one statistical measure of that distribution. It is also not the same quantity as the arithmetic mean speed.

The central result is that macroscopic pressure and temperature emerge from statistical averages of microscopic momentum and kinetic energy.