Physics with Calculus/Thermodynamics/The Ideal Gas

Kinetic Theory

An ideal gas is one which has molecules of negligible size, and with negligible interactions, excluding elastic collisions with each other and perhaps the walls of the container. Furthermore, when dealing with ideal gases, we assume that they have a uniform pressure, temperature, and that they fill up the container (although there are times when we will drop this assumption, but we will always assume it is true for a sufficiently small volume). We will also neglect gravitational potential energy. While it seems that we are placing too many restrictions on the gas, it is gives very accurate results when applied to gases of relatively high temperature and relatively low pressure -- that is, the molecules are far apart and moving quickly. For example, it would be a good approximation of neon at atmospheric pressure and 20 degrees Celsius.

From our postulates, we will derive some useful information about ideal gases. First, let's look at pressure. If we have a piston, and molecules are colliding with it (i.e. there is a pressure on it), how much force do we need to keep it in place? If during each second, the piston were to gain some momentum from the molecules, it is the forces job to impart the same momentum the other way. In other words, the momentum per second delivered by the molecules is the force we need to apply. We know this is equal to the pressure times the area (we assumed the pressure is uniform). If the molecules make perfectly elastic collisions, then the total momentum delivered is $2mv_{x}$. That is, nothing heats up; when the piston is in thermal equilibrium with the environment, then the collisions are essentially elastic, because nothing is getting hotter.

Now, we will find how many molecules hit the piston in a given time dt. If the molecules are traveling in the x direction with velocity $v_{x}$, and there are n molecules per unit volume, then there are $1/2nv_{x}Adt$ collisions. You can easily see this by imagining a rectangular prism of area A and thickness v_x dt, and since half (because half are going the other way) of these molecules are going to hit the wall, you get the desired result. Therefore, the collisions per second is $1/2nv_{x}A$. By multiplying the momentum per collision by the collisions per second, we get momentum per second, $nAmv_{x}^{2}$. The pressure is therefore $nmv_{x}^{2}$. However, this is only true if all the molecules have the same x-velocity component. If we average the force over all the different speeds the molecules could have, we get $nm<v_{x}^{2}>$. This is not the same as $nm<v_{x}>^{2}$ because the second one is zero! It is important that we find the pressure, then average, because the average momentum per collision times the average collisions per second is not the average momentum per second.

In any case, the pressure is

$P=nm<v_{x}^{2}>$.

Since the gas is uniform, $<v_{x}^{2}>=<v_{y}^{2}>=<v_{z}^{2}>$. Therefore,

$P=1/3nm<v^{2}>=2/3n(1/2m<v^{2}>)=2/3nU$

where U is the total internal energy of the gas. Actually, this is only true if there is no machinery inside the molecule, like vibrations and the sort. Replacing n by N/V, we have

$PV=2/3U$.