Pressure is Confined Kinetic Energy

According the kinetic theory of ideal gases [181], air pressure can be defined as the average momentum transfer per unit area per unit time due to molecular collisions between a confined gas and its boundary. Using Newton's second law, this pressure can be shown to be given by one third of the average kinetic energy of molecules in the gas.

$$p = \frac{1}{3}\rho \left<u^2\right>$$

Here, $\langle u^2\rangle$ denotes the average squared particle velocity in the gas. (The constant $1/3$ comes from the fact that we are interested only in the kinetic energy directed along one dimension in 3D space.)

Proof: This is a classical result from the kinetic theory of gases [181]. Let $M$ be the total mass of a gas confined to a rectangular volume $V = Aw$ , where $A$ is the area of one side and $w$ the distance to the opposite side. Let $\overline{u}_x$ denote the average molecule velocity in the $x$ direction. Then the total net molecular momentum in the $x$ direction is given by $M\vert\overline{u}_x\vert$ . Suppose the momentum $\overline{u}_x$ is directed against a face of area $A$ . A rigid-wall elastic collision by a mass $M$ traveling into the wall at velocity $\overline{u}_x$ imparts a momentum of magnitude $2M\overline{u}_x$ to the wall (because the momentum of the mass is changed from $+M\overline{u}_x$ to $-M\overline{u}_x$ , and momentum is conserved). The average momentum-transfer per unit area is therefore $2M\overline{u}_x/A$ at any instant in time. To obtain the definition of pressure, we need only multiply by the average collision rate, which is given by $\overline{u}_x/(2w)$ . That is, the average $x$ -velocity divided by the round-trip distance along the $x$ dimension gives the collision rate at either wall bounding the $x$ dimension. Thus, we obtain

$$p \isdefs \frac{2M\overline{u}_x}{A}\cdot \frac{\overline{u}_x}{2w} \eqsp \rho \overline{u}_x^2$$

where $\rho=M/V$ is the density of the gas in mass per unit volume. The quantity $\rho\overline{u}_x^2/2$ is the average kinetic energy density of molecules in the gas along the $x$ dimension. The total kinetic energy density is $\rho\overline{u}^2/2$ , where $\overline{u}=\sqrt{\overline{u}_x^2+\overline{u}_y^2+\overline{u}_z^2}$ is the average molecular velocity magnitude of the gas. Since the gas pressure must be the same in all directions, by symmetry, we must have $\overline{u}_x^2=\overline{u}_y^2=\overline{u}_z^2 = \overline{u}^2/3$ , so that

$$p = \frac{1}{3}\rho \overline{u}^2.$$