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$C_V=\frac{1}{kT^2}[\overline{E^2} -\overline E ^2]$ where $\overline{E}^2=-\frac{1}{Z}\frac{\partial Z}{\partial \beta}$ and $\overline{E^2}=-\frac{1}{Z}\frac{\partial^2 Z}{\partial^2 \beta}$.

$Z=\sum _ie^{-\beta\cdot E_i}$ of finite sum is the partition funciton and $\beta=\frac{1}{kT}$.

The question is what is $\lim_{T\rightarrow0^+}C_V$.

According to my calculation, as $T\rightarrow 0^+$, $[\overline{E^2} -\overline E ^2]$ approach some finite number $K$. Thus $C_V$ approach $\infty$.

I didn't believe in the begining, so I tried to plot the numerical solution, I also put the simplified function into Wolfram.

However, all the result I got was $\lim_{T\rightarrow 0^+}C_V=\infty$.

And it did not make too much sense,

I found two reference

  1. http://stp.clarku.edu/notes/chap6.pdf 6.17 seem to support my argument.

  2. http://www-personal.umich.edu/~lsander/ESP/chap4.pdf page 5 otherwise.

and my professor's answer sheet says it's suppose to be $0$.

What's the correct answer?

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  • $\begingroup$ Eq. 6.17 no longer holds when temperature decreases. Gas would condense, right? $\endgroup$ – K_inverse Nov 13 '18 at 7:50
  • $\begingroup$ @K_inverse Yeah, but obviously there is a thermal stastical treatment, I couldn't figure out A. how to show $C_V=0$ with thermal statistics derivation. B. how to proof the above calculation for $C_V$ fails. $\endgroup$ – user9976437 Nov 13 '18 at 8:08
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At very low temperature, the states with the highest probability are those at very low energy, close to the ground state. In the limit $T \rightarrow 0^+$ it is only the ground state which contributes to the canonical sum over the states. If the ground state is non-degenerate, the distribution probability must have zero-variance.

Notice that such an argument fails for a classical perfect gas just because the classical description of the system assigns a macroscopic degeneracy even to the lowest energy state (zero velocity for all particles but with a huge positional degeneracy).

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