Where pi is the probability to find one molecule in the i-th cell. (13) and (14) the properties of natural logarithms and the Stirling formula for the mathematical expansion of n!, Boltzmann derived the expressions (14), it must be N = 1 when T = 0 K, in order to have S = 0-that is, at the absolute temperature T = 0K, a system’s state is always composed by only one microstate.īy applying to Eqs. For a reason, which will clearly appear in the next section, it is preferable to use the logarithms with base two.Īccording to Nerst’s theorem and to Eq. The logarithm base used is arbitrary, since the conversion from a base to another one would simply require the introduction of a multiplicative constant. Where k = 1.38 x 10-23 joule/K, the so-called Boltzmann’s constant. ПМ, where n is the number of molecules contained in the i-th cell and n = У)i n Boltzmann also assumed that the entropy of the state is linked to the number N of microstates by the logarithmic expression It is then easy to demonstrate that the number of distinct microstates N corresponding to the macrostate of volume V is In 1872 he published a milestone paper in which he introduced the basic equations and concepts of statistical thermodynamics.Īs the first step, Boltzmann assumed that all the n molecules constituting the gas had the same velocity and divided the volume V into a very large number ofĬells, M, each having the smallest experimentally measurable size. Ludwig Boltzmann (1844-1906) understood the strong link between the thermodynamic and the statistical interpretation of the entropy concept. This state is also the macrostate characterized by the largest number of equivalent microstates. We know from the previous section that an isolated perfect gas tends to and will sooner or later achieve the state characterized by the maximum entropy consistent with the energy U of the gas. Of course, the larger the number of possible equivalent microstates N, the higher the probability of the corresponding macrostate. In each microstate there will be different values of the position and the velocity of each molecule, but the global result of the behavior of all the molecules will be exactly that macrostate characterized by the measured values of P, V, and T. It may result from a given number N of microstates-that is, a given number N of different positions and velocities of the billion molecules which compose it. Let us now consider a macroscopic state of a perfect gas, defined by given values of the quantities P (pressure), V (volume), and T (temperature). 3C) since it can be achieved by means of eight different microstates, all perfectly equivalent. It is easy to understand now that a macroscopic configuration having three balls in a compartment and only one in the other is more probable than the two previously considered configurations (Fig. The macroscopic state having two balls per compartment has more ways to occur (more microstates), which means that it is more probable that in Fig. We have now six possible ways for distributing our balls, each of them equivalent from a statistical point of view (Fig. Things are different if we want to distribute the four balls in both compartments, two in each one of them. Therefore, the macroscopic state ‘‘four balls in a compartment and no balls in another’’ can be achieved by means of two microscopic configurations perfectly equivalent. The probability is the same for both compartments, which are therefore perfectly equivalent from a statistical point of view. In fact, if we try to extract one ball from the box, in the dark, we have a 50% probability of putting our hand in the empty part and a 50% probability of putting our hand in the compartment with the four balls. The two states are identical from the microscopic and statistical points of view. We may put four balls in its left compartment leaving the right compartment empty. There are several ways of putting four balls in it. We may state that the macroscopic configuration seven has six microscopic states accomplishing it, and is therefore more probable than the configuration twelve.Ī similar result can be achieved if we consider a box divided into two compartments (Fig. There are therefore six ways to obtain a seven, and only one (6 + 6) to obtain a twelve. This is because a seven can be obtained as 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and finally 6 + 1, where the first number is the value obtained from the first die, and the second number is the value of the second die. Dice players know very well that if we cast two dice it is much easier to get a seven than a twelve.