In Lesson 19 we saw that the macroscopic properties of gases can be explained in terms of their microscopic, or molecular, behavior. For one example, recall that we were able to account for the pressure exerted by a gas on the walls of its container in terms of the momentum transferred to those walls by rebounding gas molecules. Such explanations are part of a study called statistical mechanics.
Here we shall focus our attention on a single problem, one involving the distribution of gas molecules between the two halves of an insulated box. This problem is reasonably simple to analyze, and it allows us to use statistical mechanics to calculate the entropy change for the free expansion of an ideal gas. You will see in Sample Problem 20-7 that statistical mechanics leads to the same entropy change we obtained in Sample Problem 20-1 using thermodynamics.
Figure 20-16 shows a box that contains six identical (and thus indistinguishable) molecules of a gas. At any instant, a given molecule will be in either the left or the right half of the box; because the two halves have equal volumes, the molecule has the same likelihood, or probability, of being in either half.
Table 20-1 shows the seven possible configurations of the six molecules, each configuration labeled with a Roman numeral. For example, in configuration I, all six molecules are in the left half of the box (n1 = 6) and none are in the right half (n2 = 0). We see that, in general, a given configuration can be achieved in a number of different ways. We call these different arrangements of the molecules microstates. Let us see how to calculate the number of microstates that correspond to a given configuration.
Fig. 20-16 An insulated box contains six gas molecules. Each molecule has the same probability of being in the left half of the box as in the right half. The arrangement in (a) corresponds to configuration III in Table 20-1, and that in (b) corresponds to configuration IV.
Suppose we have N molecules, distributed with n1 molecules in one half of the box and n2 in the other. (Thus n1 + n2 = N.) Let us imagine that we distribute the molecules “by hand,” one at a time. If N = 6, we can select the first molecule in six independent ways; that is, we can pick any one of the six molecules. We can pick the second molecule in five ways, by picking any one of the remaining five molecules; and so on. The total number of ways in which we can select all six molecules is the product of these independent ways, or 6 × 5 × 4 × 3 × 2 × 1 = 720. In mathematical shorthand we write this product as 6! = 720, where 6! is pronounced “six factorial.” Your hand calculator can probably calculate factorials. For later use you will need to know that 0! = 1. (Check this on your calculator.)
However, because the molecules are indistinguishable, these 720 arrangements are not all different. In the case that n1 = 4 and n2 = 2 (which is configuration III in Table 20-1), for example, the order in which you put four molecules in one half of the box does not matter, because after you have put all four in, there is no way that you can tell the order in which you did so. The number of ways in which you can order the four molecules is 4! = 24. Similarly, the number of ways in which you can order two molecules for the other half of the box is simply 2! = 2. To get the number of different arrangements that lead to the (4, 2) split of configuration III, we must divide 720 by 24 and also by 2. We call the resulting quantity, which is the number of microstates that correspond to a given configuration, the multiplicity W of that configuration. Thus, for configuration III,
Thus, Table 20-1 tells us there are 15 independent microstates that correspond to configuration III. Note that, as the table also tells us, the total number of microstates for six molecules distributed over the seven configurations is 64.
Extrapolating from six molecules to the general case of N molecules, we have
You should verify that Eq. 20-18 gives the multiplicities for all the configurations listed in Table 20-1.
The basic assumption of statistical mechanics is
All microstates are equally probable.
In other words, if we were to take a great many snapshots of the six molecules as they jostle around in the box of Fig. 20-16 and then count the number of times each microstate occurred, we would find that all 64 microstates would occur equally often. Thus the system will spend, on average, the same amount of time in each of the 64 microstates.
Fig. 20-17 For a large number of molecules in a box, a plot of the number of microstates that require various percentages of the molecules to be in the left half of the box. Nearly all the microstates correspond to an approximately equal sharing of the molecules between the two halves of the box; those microstates form the central configuration peak on the plot. For N ≈ 1022, the central configuration peak is much too narrow to be drawn on this plot.
Because all microstates are equally probable but different configurations have different numbers of microstates, the configurations are not all equally probable. In Table 20-1 configuration IV, with 20 microstates, is the most probable configuration, with a probability of 20/64 = 0.313. This result means that the system is in configuration IV 31.3% of the time. Configurations I and VII, in which all the molecules are in one half of the box, are the least probable, each with a probability of 1/64 = 0.016 or 1.6%. It is not surprising that the most probable configuration is the one in which the molecules are evenly divided between the two halves of the box, because that is what we expect at thermal equilibrium. However, it is surprising that there is any probability, however small, of finding all six molecules clustered in half of the box, with the other half empty.
For large values of N there are extremely large numbers of microstates, but nearly all the microstates belong to the configuration in which the molecules are divided equally between the two halves of the box, as Fig. 20-17 indicates. Even though the measured temperature and pressure of the gas remain constant, the gas is churning away endlessly as its molecules “visit” all probable microstates with equal probability. However, because so few microstates lie outside the very narrow central configuration peak of Fig. 20-17, we might as well assume that the gas molecules are always divided equally between the two halves of the box. As we shall see, this is the configuration with the greatest entropy.
Suppose that there are 100 indistinguishable molecules in the box of Fig. 20-16. How many microstates are associated with the configuration n1 = 50 and n2 = 50, and with the configuration n1 = 100 and n2 = 0? Interpret the results in terms of the relative probabilities of the two configurations.
Solution: The Key Idea here is that the multiplicity W of a configuration of indistinguishable molecules in a closed box is the number of independent microstates with that configuration, as given by Eq. 20-18. For the (n1, n2) configuration (50, 50), that equation yields
Similarly, for the configuration (100, 0), we have
Thus, a 50–50 distribution is more likely than a 100–0 distribution by the enormous factor of about 1 × 1029. If you could count, at one per nanosecond, the number of microstates that correspond to the 50–50 distribution, it would take you about 3 × 1012 years, which is about 200 times longer than the age of the universe. Even 100 molecules is still a very small number. Imagine what these calculated probabilities would be like for a mole of molecules, say about N = 1024. Thus, you need never worry about suddenly finding all the air molecules clustering in one corner of your room, with you gasping for air in another corner.
Probability and Entropy
In 1877, Austrian physicist Ludwig Boltzmann (the Boltzmann of Boltzmann’s constant k) derived a relationship between the entropy S of a configuration of a gas and the multiplicity W of that configuration. That relationship is
This famous formula is engraved on Boltzmann’s tombstone.
It is natural that S and W should be related by a logarithmic function. The total entropy of two systems is the sum of their separate entropies. The probability of occurrence of two independent systems is the product of their separate probabilities. Because ln ab = ln a + ln b, the logarithm seems the logical way to connect these quantities.
Table 20-1 displays the entropies of the configurations of the six-molecule system of Fig. 20-16, computed using Eq. 20-19. Configuration IV, which has the greatest multiplicity, also has the greatest entropy.
When you use Eq. 20-18 to calculate W, your calculator may signal “OVER-FLOW” if you try to find the factorial of a number greater than a few hundred. Fortunately, there is a very good approximation, known as Stirling’s approximation, not for N! but for ln N!, which is exactly what is needed in Eq. 20-19. Stirling’s approximation is
The Stirling of this approximation was an English mathematician and not the Robert Stirling of engine fame.
CHECK POINT 5 A box contains 1 mol of a gas. Consider two configurations: (a) each half of the box contains half the molecules and (b) each third of the box contains one-third of the molecules. Which configuration has more microstates?
In Sample Problem 20-1 we showed that when n moles of an ideal gas doubles its volume in a free expansion, the entropy increase from the initial state i to the final state f is Sf − Si = nR ln 2. Derive this result with statistical mechanics.
Solution: One Key Idea here is that we can relate the entropy S of any given configuration of the molecules in the gas to the multiplicity W of microstates for that configuration, using Eq. 20-19 (S = k ln W). We are interested in two configurations: the final configuration f (with the molecules occupying the full volume of their container in Fig. 20-1b) and the initial configuration i (with the molecules occupying the left half of the container).
A second Key Idea is that, because the molecules are in a closed container, we can calculate the multiplicity W of their microstates with Eq. 20-18. Here we have N molecules in the n moles of the gas. Initially, with the molecules all in the left half of the container, their (n1, n2) configuration is (N, 0). Then, Eq. 20-18 gives their multiplicity as
Finally, with the molecules spread through the full volume, their (n1, n2) configuration is (N/2, N/2). Then, Eq. 20-18 gives their multiplicity as
From Eq. 20-19, the initial and final entropies are
Si = k ln Wi = k ln 1 = 0
and
In writing Eq. 20-21, we have used the relation
Now, applying Eq. 20-20 to evaluate Eq. 20-21, we find that
From Eq. 19-8 we can substitute nR for Nk, where R is the universal gas constant. Equation 20-22 then becomes
Sf = nR ln 2.
The change in entropy from the initial state to the final is thus
which is what we set out to show. In Sample Problem 20-1 we calculated this entropy increase for a free expansion with thermodynamics by finding an equivalent reversible process and calculating the entropy change for that process in terms of temperature and heat transfer. In this sample problem, we calculate the same increase in entropy with statistical mechanics using the fact that the system consists of molecules.
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