In this chapter, we are concentrating on the macroscopic properties of gases (pressure, volume, temperature). These can all be readily measured in the laboratory. The equation:

$\frac{{pV}}{T} = constant$

is an empirical relationship. In other words, it has been deduced from the results of experiments. It gives a good description of gases in many different situations. However, an empirical equation does not explain why gases behave in this way. An explanation requires us to think about the underlying nature of a gas and how this gives rise to our observations.
A gas is made of particles (atoms or molecules). Its pressure arises from collisions of the particles with the walls of the container; more frequent or harder collisions give rise to greater pressure. Its temperature indicates the average kinetic energy of its particles; the faster they move, the greater their average kinetic energy and the higher the temperature.
The kinetic theory of gases is a theory that links these microscopic properties of particles (atoms or molecules) to the macroscopic properties of a gas. Table 20.1 shows the assumptions on which the theory is based.
On the basis of these assumptions, it is possible to use Newtonian mechanics to show that pressure is inversely proportional to volume (Boyle’s law), volume is directly proportional to thermodynamic (kelvin) temperature (Charles’s law), and so on. The theory also shows that the particles of a gas have a range of speeds – some move faster than others. 
Learn the four assumptions of the kinetic theory shown in Table 20.1.

The kinetic theory has proved to be a very powerful model. It convinced many physicists of the existence of particles long before it was ever possible to visualise them.

Molecules in a box

We can use the kinetic model to deduce an equation that relates the macroscopic properties of a gas (pressure, volume) to the microscopic properties of its molecules (mass and speed). We start by picturing a single molecule in a cube-shaped box of side l (Figure 20.8). This molecule has mass m, and is moving with speed c parallel to one side of the box (c is not the speed of light in this case). It rattles back and forth, colliding at regular intervals with the ends of the box and thereby contributing to the pressure of the gas. We are going to work out the pressure this one molecule exerts on one end of the box and then deduce the total pressure produced by all the molecules..

You need to read through the proof carefully as you will need to be able to derive the final equation yourself.
The stages involved in this calculation are:
1. Find the change in momentum as a single molecule hits a wall at ${90^ \circ }$.
2. Calculate the number of collisions per second by the molecule on a wall.
3. Find the change in momentum per second.
4. Find the pressure on the wall.
5. Consider the effect of having three directions in which the molecule can move.
As you go through the proof, see for yourself where each stage starts and finishes.
Consider a collision in which the molecule strikes side ABCD of the cube. It rebounds elastically in the opposite direction, so that its velocity is −c. Its momentum changes from mc to −mc. The change in momentum arising from this single collision is thus:

$\begin{array}{l}
change\,in\,momentum =  - mc - (+mc) 
 =  - mc - mc =  - 2mc
\end{array}$

Between consecutive collisions with side ABCD, the molecule travels a distance of $2l$ at speed c. Hence:

$time\,between\,collisons\,with\,side\,ABCD = \frac{{2l}}{c}$

Now we can find the force that this one molecule exerts on side ABCD, using Newton’s second law of motion. This says that the force produced is equal to the rate of change of momentum:

$\begin{array}{l}
 = \frac{{2mc}}{{\left( {\frac{{2l}}{c}} \right)}}\\
 = \frac{{m{c^2}}}{l}
\end{array}$

(We use $+2mc$ because now we are considering the force of the molecule on side ABCD, which is in the opposite direction to the change in momentum of the molecule.)
The area of side ABCD is ${l^2}$. From the definition of pressure, we have:

$\begin{array}{l}
force = \frac{{change\,in\,momentum}}{{time\,taken}}\\
pressure\,p = \frac{{force}}{{area}}\\
 = \frac{{\left( {m{c^2}} \right)}}{l}\\
 = \frac{{m{c^2}}}{{{l^2}}}
\end{array}$

This is for one molecule, but there is a large number N of molecules in the box. Each has a different velocity, and each contributes to the pressure. We write the average value of ${c^2}$ as $ \lt  {c^2} \gt  $, and multiply by N to find the total pressure:

$p = \frac{{Nm \lt  {c^2} \gt  }}{{{t^2}}}$

But this assumes that all the molecules are travelling in the same direction and colliding with the same pair of opposite faces of the cube. In fact, they will be moving in all three dimensions equally.
If there are components ${c_x},{c_y}$ and ${c_z}$ of the velocity in the ${x^ - },{y^ - }$ and ${z^ - }$ directions then ${c^2} = {c_x}^2 + {c_y}^2 + {c_z}^2$.  There is nothing special about any particular direction and $ \lt  \,{c_x}^2 \gt  \, = \, \lt  {c_y}^2\, \gt  \, = \, \lt  \,{c_z}^2 \gt  $ and $ \lt  \,{c_x}^2 \gt  \, = \,\frac{1}{3} \lt  {c^2}\, \gt  \,$.
The equation for pressure worked out above involved just the component of the velocity in the x-direction and if c is the actual speed of the particle then we need to divide by 3 to find the pressure exerted.

$p = \frac{1}{3}\,\left( {\frac{{Nm \lt  {c^2} \gt  }}{{{l^2}}}} \right)$

Here, ${l^3}$ is equal to the volume V of the cube, so we can write:

$p = \frac{1}{3}\left( {\frac{{Nm}}{V}} \right)\, \lt  \,{c^2}\, \gt  $ or $pV = \frac{1}{3}\,Nm\, \lt  \,{c^2}\, \gt  $

(Notice that, in the second form of the equation, we have the macroscopic properties of the gas – pressure and volume – on one side of the equation and the microscopic properties of the molecules on the other side.)

Finally, the quantity Nm is the mass of all the molecules of the gas, and this is simply equal to the mass M of the gas. So $\frac{{Nm}}{V}$ is equal to the density ρ of the gas, and we can write:

$p = \frac{1}{3}\rho \, \lt  \,{c^2}\, \gt  $

So the pressure of a gas depends only on its density and the mean square speed of its molecules.

A plausible equation?

It is worth thinking a little about whether the equation $p = \frac{1}{3}\left( {\frac{{Nm}}{v}} \right)\, \lt  \,{c^2}\, \gt  $ seems to make sense. It should be clear to you that the pressure is proportional to the number of molecules, N. More molecules mean greater pressure. Also, the greater the mass of each molecule, the greater the force it will exert during a collision.
The equation also suggests that pressure p is proportional to the average value of the speed squared. This is because, if a molecule is moving faster, not only does it strike the container harder, but it also strikes the container more often.
The equation suggests that the pressure p is inversely proportional to the volume occupied by the gas.
Here, we have deduced Boyle’s law. If we think in terms of the kinetic model, we can see that if a mass of gas occupies a larger volume, the frequency of collision between the molecules and unit area of wall decreases. The equation shows us not just that pressure will be lower but that it is inversely proportional to volume.
These arguments should serve to convince you that the equation is plausible; this sort of argument cannot prove the equation.