After deriving the entropy of an ideal gas we get to : $$S = Nk \left[\ln(V) + \frac{3}{2}\ln(T) + \frac{3}{2}\ln\left(\frac{2\pi mk}{h^2}\right) - \ln(N) + \frac{5}{2} \right]$$

In the zero temperature limit, we expect to have $S=0$, however, we get infinity. How can we overcome this mathematical inconsistency?

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    $\begingroup$ Have you tried a simple wikipedia search? :) $\endgroup$
    – Philip
    May 27 '17 at 21:23
  • 1
    $\begingroup$ yes, it had nothing to say germane to my question. $\endgroup$ May 27 '17 at 22:48
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    $\begingroup$ It certainly does, if you look a little closer! Anyway, I've written you an answer which is more or less an extension of what is given there. $\endgroup$
    – Philip
    May 27 '17 at 23:12

The fact that the entropy of an ideal (classical!) gas goes to infinity as the temperature $T$ approaches absolute zero is a reflection of the fact that this equation is not valid in that regime.

If I remember rightly, during the derivation of the entropy of this gas, the implicit assumption was made that the gas follows Maxwell-Boltzmann statistics. However, we know that this is not true, and that this is the 'classical' approximation.

Bosons and Fermions behave very differently near absolute zero: bosons tend to condensate into the same energy levels while fermions form a 'tower' of states because of the Pauli Exclusion principle. In order to describe such gases, one would need to use the Bose-Einstein or Fermi-Dirac distributions. Sections 5, 6 and 7 of these notes seem to address this in some detail.

An interesting side-note here is the idea of the thermal de Broglie wavelength which provides a threshold above which the classical approximation is valid. For a massive particle, this wavelength is easily calculated using the standard de Broglie wavelength:

$$\lambda_{\text{th}} = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2 \pi m k_B T}}$$

When the distance between the particles is much larger than $\lambda_\text{th}$ the gas is effectively a classical or Maxwell–Boltzmann gas. On the other hand, quantum effects dominate when the interparticle distance is of the order of or less than $\lambda_\text{th}$ and the gas must be treated as a Fermi gas or a Bose gas. If we define the average interparticle distance to be $\approx (\frac{V}{N})^{\frac{1}{3}}$, then we see that 'classical' limit is when

$$\lambda_\text{th}^3 \ll \left(\frac{V}{N}\right)$$

To connect this to your question, notice that the entropy can be rewritten as

$$\frac{S}{k_B N} = \ln\left(\frac{V}{N\lambda_\text{th}^3}\right) + \frac{5}{2}$$

Thus, since the classical approximation was used, this formula is only valid in regimes were

$$\frac{V}{N\lambda_\text{th}^3}\gg 1$$

Thus taking the limit of $T \to 0$ is outside the regime of its validity.


Entropy for an electron gas is $S={\frac {\pi }{3}}k_{B}^{2}Tn(E_{F})$. Ths goes to zero as T goes to zero.

  • $\begingroup$ we are talking about the entropy here, not the energy level. $\endgroup$ May 27 '17 at 23:13
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    $\begingroup$ I believe what @thierry is trying to say is that the dependence on temperature is obtained by using the relation for the internal energy of the gas $$U = \frac{3}{2}N k_B T$$ which goes to zero as $T\to 0$. If you rewrite the entropy in terms of the internal energy, and then impose that it have a minimal non-zero value, the infinite entropy problem does not arise. $\endgroup$
    – Philip
    May 27 '17 at 23:24
  • $\begingroup$ I must admit that while it does solve the problem of the infinity, I'm still not very convinced by this argument, since I do agree that as we approach absolute zero the number of possible microstates (for bosons atleast) does tend to 1 (all of them in the same ground state) and so it should have zero entropy as opposed to finite entropy. $\endgroup$
    – Philip
    May 27 '17 at 23:28
  • $\begingroup$ Yes, exactly, I confirm your last statement. For absolute temperature, in the limiting sense, there exists only one microstate, or one possible configuration for particles. therefore according to the famous Boltzman relation the entropy must be zero. $\endgroup$ May 27 '17 at 23:45
  • $\begingroup$ Yep, but like I explained in my answer, this is solved by actually calculating it using the right distributions (Fermi-Dirac for fermions and Bose-Einstein for bosons). $\endgroup$
    – Philip
    May 28 '17 at 0:00

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