Today, while doing some statistical mechanics, I came across a problem regarding the entropy of an ideal gas. The entropy of an ideal gas is determined by the Sackur-Tetrode equation $$\frac{S}{k_b N} = \ln \left[\frac{V}{N}\left(\frac{4\pi m E}{3 h^2 N}\right)^{3/2}\right] + \frac{5}{2},$$ which appears to be violating the third law of thermodynamics. If I plug in the formula for the energy $E = \frac{3}{2}N k_b T$ and take the limit $T \rightarrow 0$, the entropy diverges to $-\infty$, although -- according to Nernst's theorem -- it should tend to a constant value $S_0$. So my question is why does the Sackur-Tetrode equation apparently fails to abide by the third law.


2 Answers 2


Let $\bar \lambda = \frac{h}{p}$ denote the rms de Broglie wavelength of the atoms of the mono-atomic ideal gas whose momentum is $p$. As long as $\lambda \ll \bar \ell$, where $\bar \ell$ is the mean free path (see wiki) between the atoms, you may assume that they are independent and the gas is ideal. The mean free path is a monotonic function of the temperature and as the gas cools it gets smaller.

Since $\frac{p^2}{2m} \approx k_B T$ the wavelegth $\bar \lambda$ gets longer (on average slower atoms), hence the atoms will likely interact more and the gas cannot be considered "ideal", consequently the Sackur-Tetrode formula becomes invalid at low temperatures.


The foundational assumptions underlying the Sackur-Tetrode equation are that the total energy of the system is given by

$$H(\mathbf p_i,\mathbf x_i) = \sum_{i=1}^N \frac{p_i^2}{2m}$$

where $\mathbf p_i$ and $\mathbf x_i$ are the momentum and position of the $i^{th}$ constituent, and that the number of accessible states $\Omega(E,V,N)$ is proportional to the volume of the phase space region for which $H(\mathbf p_i,\mathbf x_i)\leq E$.

In this case, since the energy is position-independent, the volume in question is equal to $V^N$ times the volume of the $3N$-dimensional hypersphere of radius $\sqrt{2mE}$. If we let $E\rightarrow 0$, the volume of the momentum hypersphere also goes to zero, which means that there are no zero-energy states available.

This doesn't seem to match our intent. One can easily imagine (classically) a system of particles which are all exactly stationary, so there should be at least one zero-energy state (in fact, infinitely many of them, corresponding to the different positions the stationary particles could have). The reason this fails is that we are treating momentum as a continuous variable and integrating, but the set of zero-energy phase space points has measure zero.

One solution to this would be to discretize the phase space. Let the momenta take the form $$\mathbf p = \frac{\pi\hbar}{L}\left(n,m,l\right)$$ where $n,m,l$ are integers. The spatial coordinates can be discretized e.g. in cubes of side length $\delta$.

The Hamiltonian becomes

$$\mathcal H(\mathbf p_i,\mathbf x_i) = \sum_{i=1}^N \frac{\pi^2\hbar^2}{2mL^2}(n_i^2+m_i^2+l_i^2)$$

Rather than considering an integral over phase space, the number of states is given by $(V/\delta^3)^N$ times the number of integer solutions to the inequality

$$\sum_i n_i^2+m_i^2+l_i^2 \leq \frac{2mL^2}{\pi^2\hbar^2} E$$

If the right hand side is extremely large then this is, to extremely good approximation, just the total volume of the $3N$-dimensional hypersphere of radius $E/\left(\frac{\pi^2\hbar^2}{2mL^2}\right)$. Making this approximation will bring us back to Sackur-Tetrode. For small energies, however, there is a clear departure - in particular, if $E<\frac{\pi^2\hbar^2}{2mL^2}$, then there is exactly one solution, namely $(n_i,m_i,l_i)=(0,0,0)$.

In the limit as $L\rightarrow \infty$, the spacing between allowed values of the momentum goes to zero. However, we are still performing a finite sum rather than an integral, and we always retain the fact that the limit of the number of available states as $E\rightarrow 0$ is

$$\lim_{E\rightarrow 0} \Omega(E,V,N) = \left(\frac{V}{\delta^3}\right) ^N \cdot 1$$

and the entropy becomes $$\lim_{E\rightarrow 0} \frac{S(E,V,N)}{k_b} = \lim_{E\rightarrow 0}\log\left(\frac{\Omega(E,V,N)}{N!}\right) \approx N\log\left(\frac{V}{N \delta^3}\right)$$

in accordance with the third law.

The standard explanation for the fact that Sackur-Tetrode fails at low temperature is that the ideal gas model fails because interactions and quantum effects become too important to ignore. Certainly the latter part of the expression is true - real gases must depart from the ideal gas assumptions at low temperature - but the real reason that Sackur-Tetrode fails is that our method for counting the energetically accessible states of the system falls apart at low energies. The phase space integral approach tells us that there is no way to arrange our particles to have zero energy, but this is obviously not true - just set every momentum to zero and place each particle wherever you'd like.

I must reiterate that this is not a physically useful model to describe real gases. Quantum effects and interactions become too important to ignore in real systems. However, I think it's important to understand our models even outside their regimes of applicability.


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.