Ideal Gas Law
In the last lecture, we discussed the Maxwell Boltzmann velocity and speed distribution functions for an ideal gas. Remember that, in an ideal gas, there are no interactions between the particles, hence, the particles do not exert forces on each other. However, particles do experience a force when they collide with the walls of the container. Let us assume that each collision with a wall is elastic. Let us assume that the gas is in a cubic box of length (a\ and that two of the walls are located at (x 0\ and at (x a\ Thus, a particle moving along the (x\ direction will eventually collide with one of these walls and will exert a force on the wall when it strikes it, which we will denote as (F_x\ Since every action has an equal and opposite reaction, the wall exerts a force F_x\ on the particle. According to Newton's second law, the force F_x\ on the particle in this direction gives rise to an acceleration via F_x ma_x m\dfrac{\Delta v_x}\Delta t} \label{2.1} Here, (t\ represents the time interval between collisions with the same wall of the box. In an elastic collision, all that happens to the velocity is that it changes sign. Thus, if (v_x\ is the velocity in the (x\ direction before the collision, then v_x\ is the velocity after, and \Delta v_x v_x v_x 2v_x\ so that F_x 2m\dfrac{v_x}\Delta t} \label{2.2} In order to find \Delta t\ we recall that the particles move at constant speed. Thus, a collision between a particle and, say, the wall at (x 0\ will not change the particle's speed. Before it strikes this wall again, it will proceed to the wall at (x a\ first, bounce off that wall, and then return to the wall at (x 0\ The total distance in the (x\ direction traversed is (2a\ and since the speed in the (x\ direction is always (v_x\ the interval \Delta t \dfrac{2a}{v_x} Consequently, the force is F_x \dfrac{mv_x^2}{a} \label{2.3} Thus, the force that the particle exerts on the wall is [F_x \dfrac{mv_x^2}{a} \label{2.4} The mechanical definition of pressure is [P \dfrac{\langle F \rangle}{A} \label{2.5} where \langle F \rangle\ is the average force exerted by all (N\ particles on a wall of the box of area (A\ Here (A a^2\ If we use the wall at (x 0\ we have been considering, then [P \dfrac{N \langle F_x \rangle}{a^2} \label{2.6} because we have (N\ particles hitting the wall. Hence, [P \dfrac{N m \langle v_x^2 \rangle}{a^3} \label{2.7} from our study of the Maxwell Boltzmann distribution, we found that