Now consider one molecule of mass m moving towards face 1 and the x- component of its velocity is c_x.

            Þ its momentum in the x direction is mc_x

At face 1 there is a perfectly elastic collision so that the momentum is reversed to -mc_x.

This results in a momentum change of:

                        mc_x - (-mc_x) = 2mc_x

After colliding with face 1, the molecule travels a distance 2l before it collides again with face 1 again.  This takes a time t, which is given by:

                        t = 2l/c_x

Stage 2 Impulse

So that we can get the pressure, we need to know the force.  This can be worked out by Newton II, considering the impulse, which is the change of momentum.

impulse = FDt

Rearranging gives us:

F = change in momentum

Dt

 Time interval is the time taken for the molecule to move up the box to the far end, bounce off and come back again. Time  = distance / speed = 2l / c_x

                             Þ F = 2mc_x  = m(c_x)²

                                   2l/c_x         l

Stage 3 Working out the pressure

Since pressure = force/area, we can work out the pressure on face 1.

The area of face 1 = l².

                        p = F = mc_x²/l = mc_x²

                             A       l²          l³

Suppose we have N molecules of gas. The pressure will be:

                        p = m ( c_x1² + c_x2² + c_x3²....+ c_xN²) = m(N_cx²)

                              l³                                                  l³

Now l³ is the volume of the gas, so we can rewrite the equation:

Stage 4 The Statistics

The term <c_{x 2}> ("cx-squared bar") is called the mean square speed of the molecules in the x direction.  The bar is written over the c², as in the yellow box.  (Notice the different way of writing it in the text.  You may see this in some books.  It is also almost impossible to write c-bar in this web-page editor!)  However the molecules are moving randomly in the container and very few would be moving exactly parallel to the x-axis.  However we can consider each molecule’s velocity to be the resultant of three components, c_x, c_y, and c_z.

As in two dimensions, the three components can be combined by Pythagoras to give the resultant velocity.

Similarly we can combine the mean square velocities:

Since there are a large number of molecules we can assume that there are equal numbers moving in each of the co-ordinate directions:

So we can write:

So our final equation becomes:

Stage 5 Density

However we can go further.  Since Nm is the total mass of the gas and V is the volume, we can express the pressure in terms of the density.

                                    density =   mass   =    Nm

                                                    volume       V

This gives us the equation:

Be careful of the difference between the square of the mean (<c>)² ["c-bar squared"] and the mean square <c²> ["c-squared bar"].  Consider 1 + 2 + 3 = 6.  The average is 2 and the square of the mean is 4.  (1² + 2² + 3² ) ÷ 3 = (1 + 4 + 9) ÷ 3 = 14 ÷ 3 = 4.67, so there is quite a difference.

The most probable speed is that at which the greatest number of molecules are moving. The mean speed is the average value of all the speeds. The root mean square speed is the square root of the mean square speed of the molecules. They are slightly but significantly different.

Internal Energy and Kinetic Energy of Molecules and Temperature

The internal energy of a material is the sum of its potential and kinetic energies.  In an ideal gas the molecules are so far apart that intermolecular forces can be ignored.  Therefore there is no potential energy.  So all the energy is kinetic.  The total kinetic energy is shared randomly throughout the molecules in the gas.  Therefore there is a random distribution of the speeds of the molecules, which we can show on a graph.

If no energy is being transferred as heat between an object and its surroundings, we say that it is in thermal equilibrium.  If a gas is in thermal equilibrium and it is not being compressed or expanded, the average kinetic energy will remain constant, so will the temperature.

We have seen that:

                                    pV = nRT

  and:

                                    pV = 1/3 Nmc²

It does not take a genius to see that:

                                    nRT = 1/3 Nmc²

We also know that kinetic energy, E_k = 1/2 mv² (or 1/2 mc²).

We can now rewrite our expression as (since 2/3 ×1/2 = 1/3):

nRT = 2/3 N[1/2mc²]

We can rearrange this to write:

                                    1/2mc² = 3/2nRT = 3/2RT

                                                        N         N/n

Since N is the total number of molecules and n is the number of moles, N/n will always be N_A, which is Avogadro's number, 6.02 ×10²³.

                                    Þ 1/2mc² = 3/2RT

                                                             N_A

Now R/N_A is called Boltzmann's constant and is given the code k.  We can easily work out the value for k.

                                                 k =     8.31    = 1.38 × 10^-23 J K^-1

                                                      6.02 × 10²³

So the equation now becomes: