Question 3.1: Consider a system with N = 4, and E = 12, in which each part...
Consider a system with N = 4, and E = 12, in which each particle has rigid-rotor-like energy states:
\epsilon _{J}= J(J=1): J=0,1,2,3 (3.2)
Table 3.1 shows some of the possible distributions, and indicates those that are consistent with the system parameters, and that are therefore microstates.
Table 3.1 Distributions of Particles in Energy States
| \epsilon_{0}=0 | \epsilon_{1}=1 | \epsilon_{2}=6 | \epsilon_{3}=12 | E | Consistent with System? |
| 1 | 1 | 1 | 1 | 20 | No |
| 0 | 3 | 1 | 0 | 12 | Yes |
| 2 | 0 | 2 | 0 | 12 | Yes |
| 1 | 1 | 2 | 0 | 14 | No |
| 3 | 0 | 0 | 1 | 12 | Yes |
| … | … | … | … | … | … |
For Example 3.1, there are only three microstates. Mathematically, we write this procedure as
\Omega =\Sigma 1, for all possible distributions N_{i} such that
\sum\limits_{i}{N_{i} }=N and \sum\limits_{i}{N_{i} \epsilon _{i} }= E (3.3)
In this example, we assume that the particles are indistinguishable from each other. Thus, for the combination on line 5 of Table 3.1, we do not obtain additional microstates by reordering the three particles in J = 0. We have also assumed that more than one particle can occupy the same quantum state. It may be shown from quantum mechanics that this condition applies only to particles made up of an even number of elementary units: neutrons, protons, and electrons. Such particles are called bosons and are analyzed using Bose–Einstein statistics. Examples of bosons include hydrogen atoms, He^{4}, N^{2}, and photons. Particles made up of an odd number of elementary units are called fermions and are analyzed using Fermi–Dirac statistics. Examples of fermions include electrons, protons, and He^{3}. No two fermions may occupy the same quantum state (similar to the Pauli Exclusion Principle). If the particles in Example 3.1 are fermions, there are no microstates at all.
For real gas systems, the trial-and-error procedure is too slow for the large numbers of particles and energy states involved. We can use an alternative approach in the special case of translational energy where the spacing between adjacent energy levels is extremely small. Recall our result from Chapter 2 for the quantized translational energy:
\epsilon_{tr} = \frac{h^2}{8mL^2} (n^2_1 + n^2_2 + n^2_3) = \frac{h^2n^2}{8mL^2} (3.4)
where we saw that the spacing between adjacent levels \epsilon_{tr} ≈ 10^{−38}\text{ J for } N_2.
Consider the number of translational states with energy less than some value \epsilon ^* . By plotting all translational energy states on a Cartesian diagram with coordinates ( n_1, n_2, n_3), as illustrated in Fig. 3.1, it may be shown that all points lying inside the first octant of a sphere of radius \frac{2L}{ h} (2m\epsilon^∗)^{1/2} satisfy the condition that \epsilon < \epsilon^∗ . Although the combinations of ( n_1, n_2, n_3 ) are discrete points, since the spacing between adjacent states is very small, the number of states satisfying our condition is approximately given by the total volume of the sphere octant:
\Gamma = \frac{1}{8} \frac{4π}{3} \left[ \frac{2L}{h} (2m \epsilon^* )^{1/2} \right] ^3 = \frac{ 4π}{3} \frac{V}{h^3} (2m\epsilon^*)^{3/2} (3.5)
