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Third law of thermodynamics states: "The entropy of a perfect crystal at absolute zero is exactly equal to zero". My question is why entropy at such state is $0$ ? Let say my crystal have $n$ molecules/particles, isn't third law assuming that those particels are indistinguishable? So $n$ particles at $T = 0$ form same pattern but what will happen if for me as some observer (if I could trace every single partice) it make difference whare each particular particle is positioned in pattern?

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    $\begingroup$ The 3rd law of thermodynamics states that as the temperature of a system approaches zero, then the entropy of the system approaches zero or some positive constant. If the ground state is non-degenerate, then the entropy approaches zero. If the ground state is degenerate (i.e., more than one state at the lowest energy level), then the entropy will approach a positive constant. As far as I'm aware, a perfect crystal can have a degenerate ground state, so it should be possible for the entropy of a perfect crystal to reach some non-zero value as it is cooled to zero, too. $\endgroup$ – Samuel Weir Aug 13 '17 at 3:02
  • $\begingroup$ @SamuelWeir Why wasn't this posted as an answer? $\endgroup$ – Aaron Stevens Aug 8 '18 at 3:05
  • $\begingroup$ @AaronStevens - I wasn't absolutely certain about whether a perfect crystal can have a degenerate ground state in view of things like the Jahn-Teller effect and other possible symmetry-breaking mechanisms. I'm an experimentalist, not a theorist, and I only commented because I thought that the answer below should have explicitly mentioned non-zero entropy due to degeneracy. Also thought that someone else more knowledgable about this would eventually show up to give a proper and complete answer but it looks like no one ever did. $\endgroup$ – Samuel Weir Aug 8 '18 at 4:56
  • $\begingroup$ @SamuelWeir I am pretty sure I read what you said in your first comment in my first thermo text book, so it looks correct to me :) $\endgroup$ – Aaron Stevens Aug 8 '18 at 9:17
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The term to look up is residual entropy.

If you build a solid out of a bunch of distinguishable particles, it will indeed have nonzero residual entropy because of that. Why does the third law assume that atoms of the same type are truly indistinguishable? Because that assumption is correct†!! At least in our universe!

As for the statement "The entropy of a perfect crystal at absolute zero is exactly equal to zero", that's fine, but note that there is sort of a terminology issue here. For example, consider proton disorder in (some types of) water ice. The protons (a.k.a. hydrogen nuclei) are displaced from the symmetric position in the crystal lattice, in an irregular way, and that leads to residual entropy. I guess the question is: Does this kind of ice really count as a "crystal"? If the protons are offset in an irregular way, then the configuration no longer has translational symmetry, so according to the technical definition of a crystal, this kind of ice would not qualify as a crystal. But in an everyday sense, we do say that ice is a crystal. So be aware of that.

Isotopes are an interesting case. A crystal of "pure" silicon actually consists of a mixture of two or three distinguishable nuclei (Si-28, Si-29, Si-30). Technically, this crystal has a residual entropy from this, but we usually ignore it because it has no practical relevance in everyday chemistry. (I've never seen a discussion of isotopic mixing entropy, so I'm guessing a bit here...) I think that textbook writers implicitly take the "entropy of mixing" for the natural isotopic mixture of every element, and the entropy numbers we tabulate and calculate are actually the entropy numbers beyond that baseline. Again, you can use terminology how you like. If you define the word "crystal" to imply isotopic purity by definition, then you can say that no crystal has any residual entropy of any sort. But this is clearly not how anyone uses the term "crystal" in practice.

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The entropy is zero because if you change the state of any one or more particles it will no longer have the same total energy.

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  • $\begingroup$ But is it a state change if I will switch place of two particles? If all particles are indistinguishable that should not change entropy? $\endgroup$ – Arkadiusz Wojcik Aug 12 '17 at 21:16
  • $\begingroup$ How would you even know that you switched two of the particles? Keep in mind the wave function for the system includes the indistinguishability of the particles. So you haven't actually changed anything. $\endgroup$ – Señor O Aug 12 '17 at 21:19
  • $\begingroup$ That's different from, say, if there's one particle in the first excited state, say at at $x_1$. Then there's entropy from the fact that you can switch the states at $x_3$ and $x_1$ for example, to get two different total wavefunctions that lead to the same energy. $\endgroup$ – Señor O Aug 12 '17 at 21:20
  • $\begingroup$ So does it mean that 3th law don't distinguish between particles? Ok I understand that I may not have the knowledge that I switched positions of two identical particles. But isn't that just my fault (as an observer) that I can't distingus between two different states? $\endgroup$ – Arkadiusz Wojcik Aug 12 '17 at 21:31
  • $\begingroup$ @ArkadiuszWojcik Sure, if you can distinguish between multiple distinct states at absolute zero, then entropy isn't zero. $\endgroup$ – Nat May 27 '18 at 1:36

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