Entropy of an ideal gas in $T\to 0$ limit 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?
 A: 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.
A: 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.
