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Say that a closed system has $n$ dimensions and is in the shape of a $n$-ball with a radius of 1, it's volume will be

$$\frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2}+1)}$$

which tends to 0 yet is not empty as n tends to infinity.

My question might not make any sense, and my understanding of entropy might make things even worse, but if it's sensible, what will become of the entropy of such a system?

If my first exposé really makes no sense, I would reformulate as follows: let's say that an unfathomable force compresses a closed system such that it's volume decreases towards zero, what will become of the entropy of such a system? (I realise it might not be the same problem/question)

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2 Answers 2

up vote 5 down vote accepted

Let me begin with the second question where you don't change the dimensionality, just the volume.

The entropy never decreases when you actually compress gas. The compression means that the walls are mostly moving against the colliding molecules which means that they're recoiled backwards at higher velocities. The molecules' kinetic energy increases so they occupy a larger volume in the momentum space (in macroscopic language, a gas heats up while being compressed) which at least compensates the decrease of the volume in the position space.

The other answer is incorrect. The second laws says not only that systems exhibit some activity indicating that they don't like a decreasing entropy; instead, it says that whatever activity physical systems display, they will never achieve a macroscopic decrease of the entropy. It's just impossible. To compress gas by 70% is possible, to decrease the entropy by a macroscopic amount is not.

Now, the interesting first question. If you could change the effective dimensionality, it would still be true in any consistent theory that the entropy can't decrease. So if your theory were just able to add dimensions like that while keeping a molecule in a sphere of the increasing dimension, the second law of thermodynamics would imply that such an addition of dimensions isn't physically possible – it would be another, more sophisticated example of the perpetual motion machine of the second kind.

In some sense, it is true that the second law encourages physical systems to lose the dimensions (a way to increase the entropy, given your formula for the higher-dimensional spherical volumes). When the energy dissipates, the energy per degree of freedom effectively goes down which allows us to use a lower-dimensional "effective" description. For example, a gas full of Kaluza-Klein particles probing (moving in) extra dimensions will tend dissipate its energy and decay to many lower-energy quanta which are effectively living just in 3+1 dimensions.

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The entropy decreases to zero eventually. It is harder and harder to do as the number of particles is kept fixed. Said differently, the system does not like having its entropy decreased and as a consequence, the pressure increases.

To answer to the edited version, if the temperature is kept fixed, a real system will crystalize and the pressure will diverge at close packing. In this case, the entropy is just the number of ways to arrange your particles in the crystal but that is volume independent. If you want to be consistent with the third law of thermodynamics at the same time (this closed packed structure is the same that you would get when T is zero) then it means that the entropy is the reference state and the actual thermodynamic entropy is zero.

If you were only considering a classical ideal gaz then, the partition function is multiplied by the accessible volume. This implies that the entropy goes as the log of the volume that becomes -infinity when V tends to zero. This is more an artifact of the model that anything else though.

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The entropy not only refuses to decrease to zero, it never decreases (by a macroscopic amount) at all! Not for a picosecond. The second law says not that the entropy doesn't want to drop to zero; it says that it never drops. Compression of a gas is a mundane, allowed thing; a decrease of the entropy is not possible. They're completely different things. –  Luboš Motl May 8 '13 at 8:54
    
@LubošMotl: well it wasn't clear what was being fixed during the process. I understodod the question as: "let us consider a system at equilibrium, I can compute its entropy as a function of its state variables, what is the value of this entropy when the volume tends to zero?". I think this is a fair question and that obviously the answer will depend on the volume. –  gatsu May 8 '13 at 9:02

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