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I assume the following statement for the "third law" of thermodynamics:

$$\lim_{N \to \infty} \lim_{T \to 0} \frac{S}{N} = 0 \tag{1}\label{1}$$

That is to say, I am considering those systems with a sub-exponential ground state degeneracy to satisfy the third law (see this answer of mine for more details about what I mean).

My question is: is there any real-life system that does not satisfy \ref{1}, while also remaining at equilibrium as $T \to 0$?

Of course, there are examples of systems which don't satisfy \ref{1}, but all the examples I know concern non-equilibrium systems, like glasses (see residual entropy). For these system, however, no definition of entropy is in principle available, since $S$ can only be defined at equilibrium$^*$. Therefore, we cannot really say that they don't satisfy \ref{1}.

Note: I am not interested in model systems, but in real life systems, that can be studied experimentally.

$^*$ Of course people have proposed possible out-of-equilibrium definitions of $S$, but as far as I know there is no consensus, so let's not talk about those. For more details about the problems related to the definition of entropy for out-of-equilibrium states, see for example here.

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    $\begingroup$ It's nonsense to say that entropy can only be defined in equilibrium, because then the second law would not have any content (after all, it is about the universe being not in equilibrium but having a small entropy which then raises over time). Also, Boltzmann's $S=k_B \log(W)$ to define entropy clearly works for non-equilibrium systems. Real life systems do not fit very well to your equation (1) which contains two mathematical limits! $\endgroup$
    – Luke
    Commented Feb 14, 2018 at 11:32
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    $\begingroup$ @Luke 1) There is no problem with the second law: you perform a process between two eq. states A and B, where you can calculate $S$. Between A and B the system is not at eq., but you can still calculate $\Delta S = S_B - S_A$ because initial and final states are at eq. 2) $S=k \log W$ is only valid at eq. For the problems related to the definition of entropy in out of eq. states, see for ex. here 3) Such limits are routinely performed in thermodynamics and statistical mechanics, and I don't really see what's wrong with them. $\endgroup$
    – valerio
    Commented Feb 14, 2018 at 11:40
  • $\begingroup$ Let me try again to get the two points across, but then I stop since this is not a discussion page: 1. In a typical process described by the second law in real life, the system is not in equilibrium. So bad for the approach you cite if it doesn't work out of equilibrium. 2. How do you intend to check if (1) is correct by a real experiment? You can't, because you only have finitely many particles. That's what I'm saying: it's too mathematical. One has to rephrase the third law in order to empirically test it, e.g. by "Absolute 0K cannot be reached." $\endgroup$
    – Luke
    Commented Feb 14, 2018 at 11:52
  • $\begingroup$ @Luke Your arguments are quite...bizarre. Of course limits such as those can never be exactly realized in practice, but they can still tell us a great deal about real physical systems. After all, almost all of the calculations of statistical mechanics are based on the thermodynamic limit ($N,V\to\infty$), and it doesn't seem to me that this undermines in any way its predictive power. What we mean by $N\to \infty$ is usually that the system is macroscopic; from the point of view of a physicist, there is not much difference $N\to\infty$ and $N=10^{23}$... $\endgroup$
    – valerio
    Commented Feb 14, 2018 at 23:17
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    $\begingroup$ @Luke Also, it seems that you have some serious misunderstanding about entropy. We can for sure talk about entropy for an out of equilibrium system, but we have no mathematical expression of it available. However, I am asking about systems which are at equilibrium, so there is no problem here: I am assuming that entropy can be defined. $\endgroup$
    – valerio
    Commented Feb 14, 2018 at 23:21

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