Physics Friday 69

Consider a particle of mass m moving freely in a rectangular box, and undergoing perfectly elastic collisions with the walls of the box. Let us consider the motion in the x direction. Let the x-component of the particle’s velocity be v_x, and the length of the box in the x direction be L_x. The time it takes to travel from one of the x-bounding walls to the other is , and so the number of collisions with a particular x-bounding wall per unit time is . When it collides with the wall, the x-component of the velocity changes from v_x to –v_x, and so, due to conservation of momentum, it imparts a momentum of 2m|v_x|. The product of the momentum imparted per collision and the rate of collisions per unit time gives the (time-)average force exerted on the wall by the particle:
.

Now, let us replace the particle with N identical non-interacting particles (an ideal gas). Then the force on one of the x-bounding walls is
, where the bar represents the average over the N particles. Dividing this by the area A of the wall gives us pressure:

where is the volume of the box.
If our gas is isometric, then , where is the root-mean-square speed. This means that the pressure is the same on all sides of the box, and
.

Now, the Maxwell speed distribution tells us that for ideal gas molecule, the average kinetic energy is given by . This allows us to rewrite the above equation for pressure as
, or multiplying both sides by the volume,
.

This is the ideal gas law. Often, one is introduced to the law in terms of the number of moles, n, instead of the number of molecules N. As , where N_A is Avogadro’s number, this gives us the form more commonly seen in introductory courses:
,
where is the ideal gas constant.

Tags: Boltzmann Constant, Friday Physics, Ideal Gas, Ideal Gas Constant, Ideal Gas Law, Maxwell Speed Distribution, physics, Statistical Mechanics