I don't completely understand the proof that is given for the claim that the heat capacity goes to zero, if the temperature approaches $0K$.
They do it as follows, if $C_x$ is the heat capacity where the parameter $x$ is taken constant, and we heat the system from $0K$ to $T_0K$, then :
$$S(T_0) = \int_0^{T_0}{dS} = \int_0^{T_0}{\frac{dQ}{T}} = \int_0^{T_0}{\frac{C_xdT}{T}} $$
(where the differential in the second step is not a real differential)
The only way for this integral to converge, is when $C_x$ is zero in the limit of $T\to 0K$.
But I have 2 problems with this. First they claim also that it is impossible to approach zero kelvin, so in fact there doesn't exist a path from $0K$ to $T_0 K$, so it should be impossible to heat the system from $0K$ to $T_0K$. This is worrying me, because in the proof of the claim that it is impossible to reach $0K$, they use a very similar reasoning showing that the integral can not exist.
Secondly the path must be reversible, so even if there exists a path from $0K$ to $T_0K$, how do we know there also exists a reversible one?
Best Answer
The best way to see this is to realize that the zero heat capacity is a quantum effect. Classically, the heat capacity does not go to zero.
Quantally what happens is that at low enough temperatures all the particles are in their lowest possible energy states. To get even one particle into a higher energy state requires a small but finite energy $\epsilon$. If the gas is at a finite temperature such that the total energy of the gas is less than $\epsilon$, there can be no movement to a higher energy state, the energy is then constant, and the heat capacity is then zero.
In the equation for entropy in terms of heat capacity, note that the heat capacity has to go to zero faster than $T$. For example $C_V = a\ln T$ won't work.
But in general your technique is not the way to prove that the heat capacity is zero. The actual calculation is statistical in nature and has two cases, one for bosons such as $^4He$ and another for fermions such as electrons.