2
$\begingroup$

Thermal expansion coefficient is defined as: \begin{equation} \alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p \end{equation} We can prove through the third law of thermodynamics that: \begin{equation} \lim_{T\to0}\alpha=0 \end{equation} Now, consider ideal gases equation: \begin{equation} pV=nRT\implies \alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p=\frac{1}{V}\left(\frac{\partial}{\partial T}\frac{nRT}{p}\right)_p=\frac{nR}{pV}=\frac{1}{T} \end{equation} So we have: \begin{equation} \lim_{T\to0}\alpha=\lim_{T\to0}\frac{1}{T}=+\infty \end{equation}

How do we justify this contradiction? As far as I know the Third Law is strongly based on experimental evidences and experiment-based assumptions (well, this is what I know from an introdoctury physics course). Also since ideal gases are a good approximation for real gases only at high temperatures, can we conclude this is the reason why $\alpha$ does not go to $0$? On the other hand, if the third law had been based on ideal models first, then what I found above would be contradictory. Is this right?

$\endgroup$

1 Answer 1

3
$\begingroup$

A real gas will condense if you get close enough to $T=0$, no matter how low the pressure, so in practice there won't be a conflict. But this is an unsatisfying answer.

The basis of the third law is that a system at absolute zero has zero entropy, because there is only one possible lowest-energy state. In the case of an ideal gas at absolute zero, all its point-like, non-interacting particles will be collapsed at a single point with zero volume. This is an unphysical state.

The specific heat of a gas is another example where you would expect $C_p\rightarrow 0$ whereas it is constant for a perfect gas (a perfect gas is an ideal gas with a constant specific heat, so it's even more restricted than an ideal gas).

The escape from this contradiction according to The Wiki (no references provided unfortunately) is that Bose-Einstein or Fermi-Dirac statistics start becoming relevant for ideal/perfect gases near absolute zero.

The surprise is that statistical mechanics and thermodynamics are not complete without quantum mechanics.

$\endgroup$
2
  • $\begingroup$ Nice answer! As the Wiki has no reference, I'll provide one from Reif, Fundamentals of Statistical and Thermal Physics, pg 170. When discussing the third law he notes that the difference between Cp and Cv approaches 0, rather than R as expected for an ideal gas. He writes there is no contradiction in this: "...when T --> 0 and the system approaches its ground state, quantum mechanical effects become very important. Hence the classical equation of state is no longer valid, even if the interactions between the particles in a gas are so small that the gas can be treated as ideal". $\endgroup$
    – CGS
    Jul 17, 2020 at 19:37
  • $\begingroup$ Interesting fact. I had found that $c_p-c_v$ approaches zero on Pauli's book but I hadn't noticed that "contradiction" as well. $\endgroup$ Jul 18, 2020 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.