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I just have a question concerning the third law of thermodynamics.

The third law describes that the entropy should be a well defined constant if the system reaches the ground state which depends only on the temperature. Beside this fact we now that the temperature is independent on the measurement system we can assume that $S(T\rightarrow 0) = 0$.
This is not difficult to understand. If you take a look on the definition of the entropy $S = k_B \log{\Omega}$ then this means that the number of microstates is equal to a constant or in other words, the system is in a well defined state. If you would measure $x$-times the system it will not change and you get $x$-times the same results. I'm right so far?

Okay. Now we take a look on one example - the ideal gas. If we calculate the partition function we will get something like: $$ Z(T,V,N) \propto T^{3N/2} $$ And for the entropy we will get: $$ S(T,V,N) \propto \ln{T^{3/2}} $$

Both doesn't really fullfil the third law. Or is my assumption wrong? I'm mean that the entropy goes to zero? Maybe the ideal gas doesn't fulfill the third law, but my concern is that the calculation for the partition function would be almost the same ($\propto T^\alpha$) if we calculated it for an other system, based on the definition.

Has anybody maybe something like a thumb rule for checking if the system fulfill the third law after calculating the entropy?

Thank you for your help.

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Related: and links therein. – Qmechanic Oct 21 '13 at 8:36
up vote 1 down vote accepted

Well yes, for the ideal gas model, $pV=Nk_BT$, you find

$$U=\frac{3}{2}Nk_BT\propto T,$$



This itself is a violation of the third law.

What does it say to us? The lengthy discussion in the comments of this question might help you understand the problem.

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This is great. Thank you. – PateToni Feb 11 '12 at 19:08

You derivation is correct.

It proves that no substance at T=0K can be an ideal gas, nor any of the other systems you may have in mind whose partition function doesn't have the right limiting behavior.

Indeed, all known materials are solid or fluid at sufficiently low temperature.

But it hasn't have to be a perfect crystal either: Really perfect crystals of macroscopic size are very rare, and no matter how much you cool an imperfect crystal, you can never turn it that way into a perfect crystal.

Because of the formula $S=kB\log \Omega$, the third law says in fact that $\Omega=1$ at zero temperature, i.e., a system at zero temperature is necessarily in a pure quantum state, while systems at positive temperature are always in a mixed state. This is the reason why absolute zero is not reachable, as the quantum dynamics implies that one can never can clean a macroscopic system from all admixtures.

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The mathematical formulation of the third law is as follows: if the equations of state are $$T=f(p,V), \quad S=g(p,V)$$ for two functions $f$ and $g$ of two variables and we solve these to get $S$ as a function of $p$ and $T$, then the resulting function $S=h(p,T)$ has the property that for each $p$, $\lim_{T\to 0}h(p,T) = 0$. For simple models (e.g., the ideal gas and the van der Waals gas), these computations can be carried out by hand and the result is, as suggested in the query, that the law fails. Curiously, we have not seen these calculations carried out in the literature and, further, we have not encountered any model there which does satisfy the third law. There is, however, a slight perturbation of the ideal gas law which does, namely if we suppose that the isotherms have the form $p^aV^b=const$ and the adiabats $p^cV^d=const$ for suitable indices---the ideal gas is, of course, the case where $a=b=c=1$, then an elementary, but rather tedious calculation, shows that $$ S=\frac 1 {\sqrt J}\left ( \frac 1 {a-b-J+1}\right )p^{c(a-b-J+1)}\left (\sqrt J T(d-c-J+1) p^{-a(d-c-J+1}\right )^\frac 1{b(d-c-J+1)} $$ where $J=ad-bc $. We have displayed this formula in all its gory details to show that the underlying method is not entirely superficial but the important point is that the dependence of $S$ on $T$ is a power law and so, for suitable choices of the indices, will satisfy Nernst's law. If we choose $b$ almost, but not quite, equal to $a$ we will obtain a small perturbation of the ideal gas which satisfies the third law. We are not suggesting that this is a realistic model for any real thermodynamical substance, although the fact that at very high pressures the factor $p$ in Boyle's law might be better modelled by a non-trivial power does not seem to be intrinsically improbable.

The model used above was introduced in the arXiv paper 1102.1520 where the methods used for the above computation are explicated.

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Im not too great at statistical thermodynamics, but I think your flaw is the assumption that entropy of an IG is zero at zero K. The third law states that S of a perfectly crystalline solid is zero at zero K. To get entropy of IG at zero K, you will have to convert solid to liquid to gas, and those processes will have entropies. An ideal gas at 0K cannot be viewed as a perfect crystalline solid (even thougb molecules are statiomary), as the molecules are not evenly distributed. Ironically, this comes from the definition of IG itself. Since there are no intramolecular forces, the position of one particle has no bearing on the positions of others. So we still have a degree of randomness, which is the distribution of particles.

Wikipedia says that the concept of an IG breaks down at low value of V/N (last para of entropy section). Guess that works, too.

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Thank you. Okay, the point of the assumption was that the zero point and the sign of the temperature just have only a absolute meaning. I think about the story that $\hat{S} = \alpha S + \alpha_0$ where you transform the one entropy $S$ to a new one $\hat{S}$ (and vice versa) with $\alpha = T/\hat{T}$. – PateToni Feb 11 '12 at 19:17

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