Entropy question: Does a substance at 0K have no entropy?

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A discussion on the entropy of substances at absolute zero highlights Nernst's theorem, which states that as temperature approaches zero, the entropy of a non-degenerate system approaches zero. However, if a system has ground state degeneracy, its entropy can remain non-zero at absolute zero due to the various configurations of defects in the crystal lattice. The formula for entropy in such cases is k_B * log(g), where g represents the degeneracy of the ground state. Additionally, Nernst's statement clarifies that near absolute zero, reactions in a system at internal equilibrium occur without changes in entropy. Overall, the conversation emphasizes the nuances of entropy behavior as temperature approaches absolute zero.
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Is it 0 K for a substance to have no entropy?Sorry. Just had to get that off my chest.
 
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Nernst's theorem, the 3rd theorem of thermodynamics:
https://en.wikipedia.org/wiki/Third_law_of_thermodynamics
(note the graphic in the section "absolute zero")
and here's the math which directly answers your question
(Google translate isn't perfect, but it works)
https://de.wikipedia.org/wiki/Nernst-Theorem
If the basic state of the system is not degenerate, then ##g = 1## and thus
$$
\lim_{T\to 0}S(T,p,V,\ldots)=S(T=0)=S_ {0}= k_B\cdot \log g = 0
$$
Thus, the entropy of a system disappears when the temperature approaches zero.
 
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As the link given above by @fresh_42 says, if there is ground state degeneracy, then the entropy need not be zero when the temperature is at absolute zero.
 
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An imperfect crystal can have a non-zero entropy at 0 K, which I suppose is because there are many ways to distribute the same number of dislocations/defects in the lattice.
 
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The formula provided by fresh42 is correct. However, when the ground state of a system is degenerate, the entropy at 0 K is not zero, it is ##k_B \cdot log (g)## where g is the degeneracy of the ground state.
 
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fresh_42 said:
Nernst's theorem, the 3rd theorem of thermodynamics...
A slight correction.
Nernst’s statement of the third law says:
Near absolute zero, all reactions in a system in internal equilibrium take place with no change in entropy.

That means: $$\lim_{T \rightarrow 0^+} {\Delta S}=0$$
 

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