The Sackur-Tetrode equation cannot be quite right, because at low enough tempera

Question

The Sackur-Tetrode equation cannot be quite right, because at low enough temperatures, it would actually become negative! This is because its derivation assumes (see problem T8D.1) that the gas has a high enough energy that the energy levels basically seem continuous and that the gas's multiplicity is ginormous. Suppose that we start with helium gas at a fixed density of 0.165 kg/m^3 (about its density at normal atmospheric pressure and temperature) and lower the temperature until the entropy goes negative (indicating that the equation is breaking down). a) At what temperature would this happen, assuming that U = 3/2 Nk_B T (which is also not true at extremely low temperatures), and that helium remains a gas?

Explanation / Answer

The entropy for an ideal gas is : S(E, V, N) = kN 3 2 ln E N + ln V N + S0, (1) where E, V and N are the energy, volume and number of particles of the system, k is Boltzmann constant and So an undetermined constant. Sackur and Tetrode, independently, obtained an equation for the entropy of an ideal gas starting from statistical mechanics. Using this formula S = k ln W, and computing de number of microstates W

S(E, V, N) = kN 3 2 ln E N + ln V N + 3 2 ln 4m 3h 2 + 5 2

where m is the mass of a particle and h is Planck constant. Comparing this result with the previous equation

S0 = kN 3 2 ln 4m 3h 2 + 5 2

Therefore, we have found a theoretical value for the entropy constant. It is worth to say that the derivation of (2) is not trivial at all. In fact, to obtain it, two conditions must be satisfied

1. The phase space needs to be discretized in cells of volume h n, where n is the number of degrees of freedom of the system (3N).

2. The number of configurations in phase space must be divided by N! to make entropy extensive (and to take into account the indistinguishability of particles).

A modification of any of these two hypotheses leads to a different value of S0. For instance, Sackur used N N instead of N! in one of his papers and obtained 3/2 (instead of 5/2) in the last term of equation

we have a value for S0, the reader may well ask if this has any importance at all. As we always measure variations of entropy, the constant in (1) might seem irrelevant. However, if we assume the strong version of the Third Law (S(T = 0) = 0), it is indeed possible to experimentally determine this constant. In this work, I will use this version and discuss later the influence of taking the less restrictive one (S(T = 0) = const).

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