User:Inconspicuum/Physics (A Level)/Kinetic Theory

One formula which sums up a lot of the kinetic theory of an ideal gas is the following:

$$pV = \frac{1}{3}Nm\bar{c^2}$$,

where p is the pressure of the gas, V is its volume, N is the number of molecules, m is the mass of each molecule, and $$\bar{c^2}$$ is the mean square speed of the molecules. If you knew the speeds of all the molecules, you could calculate the mean square speed by squaring each speed, and then taking the mean average of all the squared speeds.

Derivation
This formula can be derived from first principles by modelling the gas as a lot of particles colliding. The particles have a momentum p = mc. If we put them in a box of volume V and length l, the change in momentum when they hit the side of the box is:

$$\Delta p = m(c - (-c)) = 2mc$$

Every time the particle travels the length of the box (l) and back (another l), it hits the wall, so:

$$c = \frac{2l}{t}$$,

where t is the time between collisions. Therefore:

$$t = \frac{2l}{c}$$

Each collision exerts a force on the wall. Force is the rate of change of momentum, so:

$$F = \frac{\Delta p}{\Delta t} = \frac{2mc}{\frac{2l}{c}} = \frac{2mc^2}{2l} = \frac{mc^2}{l}$$

However, we have got N particles all doing this, so the total force on the wall is given by:

$$F = \frac{Nm\bar{c^2}}{l}$$

The molecules all have different velocities, so we have to taken an average - the mean square speed. This force is the force in all three dimensions. The force in only one dimension is therefore:

$$F = \frac{Nm\bar{c^2}}{3l}$$

Pressure, by definition, is:

$$p = \frac{F}{A} = \frac{Nm\bar{c^2}}{3Al}$$

But area multiplied by length is volume, so:

$$p = \frac{Nm\bar{c^2}}{3V}$$

Therefore:

$$pV = \frac{1}{3}Nm\bar{c^2}$$