Kinetic Theory of Gases.
One of the great successes of the classical Newtonian mechanics was the kinetic theory of Gases. It treated a gas as a collection of very small particles moving at various velocities in a container. The particle masses were assumed very small compared to the mass of the walls. The particles obeyed Newtonian mechanics.
We model an idealized gas by starting with what we already experimentally knew about gases: (1) The ideal-gas law, PV = nRT, and (2) The relations for specific heats of gases. Textbooks usually start with the simplest case, a monatomic gas. We call it the "ideal-gas" or "idealized gas". In spite of the idealizations, many actual monatomic gases come very close to this ideal.
Certain assumptions are made:
Most derivations of the kinetic theory model begin by considering a collision of a a typical particle with one wall (we will focus our attention on the right wall). We see how much force it exerts on that wall during each collision, and express that in terms of its x component of velocity. The elastic collision does not change the size of the components of velocity, only their directions, as we showed in the calculation of the ball bouncing from a massive wall. Even if the particle takes a side trip to one of the side walls, that won't change the x component of its velocity. Each collision with the right wall changes the particle's momentum by 2mvx, and each collision exerts an impulse on the wall of 2mvx.
The number of collisions in a given time is found from distance = velocity × time, 2L = vxt where t is the time it takes a particle to go from one face to the other and back again.
Now we want the total impulse on the wall due to all collisions by this particle in a longer time, T, such that T >> t >> Δt. The total impulse on the wall in time T due to repeated collisions with the same particle is:
I = (vx/2L)T(2mvx) .
So, the average force on the wall during time T due to this one particle is <F> = I/T:
<F> = mvx2/L .
Then the average pressure on the wall from this particle is given by:
p = <F>/A = <F>/L2 = mvx2/L3 = mvx2/V .
So far we have considered only one particle. The pressure on a wall due to all particles is the total over all possible values of vx, which we may write:
P = (Nm/V)Σ(vx2)
<vx2> = <vy2> = <vz2> .
From the Pythagorean theorem we know that v2 = vx2 + vx2 + vx2 . Therefore:
<v2> = <vx2> + <vy2> + <vz2> = 3<vx2>.
This fortunate result allows us to re-express the pressure a particle exerts on the wall in terms of its speed (rather than in terms of a component of its speed):
p = (m/3V)Σv2
Multiply by N/N where N is the number of particles.
P = (mN/3V)(Σv2)/N
But (Σv2)/N = <v2> So:
PV = (1/3)Nm<v2> = (1/3)M<v2>
M = nM is the total mass of the gas.
We can rearrange this to look more like the experimental ideal-gas law PV = nRT.
PV = (2N/3)[(Σ(1/2)mv2)/N]
The quantity in square brackets is the average kinetic energy.
PV = (2N/3)<Ek>.
By comparing with the experimental result PV = nRT, we can associate the average kinetic energy with the temperature. Carrying this further, we can show that the sum of all the particle's kinetic energies represents the total thermal energy of the gas. These conclusions hold firm as more complicated situations are studied: diatomic gases, gases of molecules with two or three degrees of freedom, and gases with vibrational energies. This is one of the remarkable successes of modeling a gas as mechanical particles obeying Newton's laws of mechanics.
Carried further, we can use this model to correctly predict specific heats of such gases. It also shows us that the thermal energy content of a gas is simply the sum of the kinetic energies of its particles. The model gives us useful insight into what happens when work is done on an enclosed gas, by compressing the gas to smaller volume, and why a gas does work when it expands.
In this enetire process we were totally ignorant of the distrubtion of particle velocities over the entire range of velocities. It remained for Maxwell to make a hypothesis about the velocity distrubution, giving an equation for it. We can say a few things about this distrubution without going to all that trouble. It's possible that a particle may have zero velocity, but unlikely. The distrubution must have a maximum particle velocity, the unlikely case where all of the energy is in one particle. The most probable velocity must lie somewhere between these two extremes. So we expect the velocity distribution graph to be a smooth curve that's zero at v = 0, rises smoothly, has a peak or hump somewhere, and drops smoothly to zero at v = Vmax.