Expression for Pressure Exerted by a Gas

Setting up the problem. Consider a monatomic gas of $N$ identical molecules, each of mass $m$, enclosed in a cubical container of side $l$ (Figure 9.1(a)). Every molecule moves with its own velocity, which can be resolved into components $(v_x, v_y, v_z)$ along the three edges of the cube.

One molecule's collision with a wall. Focus on a single molecule striking the right-hand wall (perpendicular to the x-axis) with velocity components $(v_x, v_y, v_z)$. Because the collision is perfectly elastic, the molecule rebounds with the same speed, its x-component reversed to $-v_x$ while $v_y$ and $v_z$ stay unchanged (Figure 9.1(b)). The molecule's momentum in the x-direction therefore changes from $mv_x$ to $-mv_x$, a change of $-2mv_x$; by the law of conservation of momentum, the wall itself gains momentum $+2mv_x$ from this one collision.

Counting how many molecules hit the wall. In a short time interval $\Delta t$, only those molecules that (i) lie within a distance $v_x\Delta t$ of the wall and (ii) happen to be moving toward it will actually strike it (Figure 9.2). This defines a thin slab of volume $Av_x\Delta t$, where $A$ is the wall's area. If $n=N/V$ is the number density of molecules, and (since the motion is completely random) only half of the molecules in this slab are moving toward the wall at any instant, the number of molecules hitting the wall in time $\Delta t$ is

Total momentum transferred and the resulting force. Each of these collisions transfers momentum $2mv_x$ to the wall, so the total momentum transferred in time $\Delta t$ is

By Newton's second law, force is the rate of change of momentum, so the force on the wall is

Pressure. Dividing by the wall's area gives the pressure,

Since all $N$ molecules do not move with the same speed or in the same direction, this $v_x^2$ must really be understood as the average value of $v_x^2$ across every molecule. Because the motion is isotropic (no preferred direction, gravity neglected), each molecule has, on average, the same mean-square speed along all three directions: $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$. Since the mean square speed is $\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\overline{v_x^2}$, it follows that $\overline{v_x^2}=\tfrac13\overline{v^2}$. Substituting this into the pressure expression gives the central result of this section:

$P = \frac13 nm\overline{v^2} = \frac13\frac{N}{V}m\overline{v^2}. \qquad (9.6)$ …