Suppose we want to compute the exact standard molar electronic entropy of helium at a finite but not very high temperature, say $T = 298.15 \textrm{ K}$. By "standard" I mean the standard state, i.e. the helium is an ideal gas at 1 bar, where all interactions between different helium atoms are neglected, or equivalently speaking, the helium is at an infinitesimal pressure and the resulting entropy is converted to 1 bar by applying the ideal gas formula:

$$\tag{1}S_1 - S_0 = R \ln \frac{p_0}{p_1}$$.

The molar electronic entropy is given as follows: \begin{align}\tag{2} S_e = -R \sum_i p_i \ln p_i \\ = -R \sum_i \frac{1}{Q} e^{-\frac{E_i-E_0}{kT}} \ln \frac{1}{Q} e^{-\frac{E_i-E_0}{kT}} \tag{3}\\ = R \sum_i \frac{1}{Q} e^{-\frac{E_i-E_0}{kT}} (\frac{E_i-E_0}{kT} + \ln Q),\tag{4} \end{align} where $E_i$ is energy of the electronic state $i$, and $Q$ is the partition function: $$\tag{5} Q = \sum_i e^{-\frac{E_i-E_0}{kT}} $$ Now, for the helium atom, there are an infinite number of electronic states below the first ionization potential of helium, $I_1$. Thus, $$ Q > \sum_i^{+\infty} e^{-\frac{I_1-E_0}{kT}} = +\infty\tag{6} $$ Consequently, we have \begin{align} S_e > R \sum_i \frac{1}{Q} e^{-\frac{E_i-E_0}{kT}} (\frac{E_0-E_0}{kT} + \ln Q) \tag{7}\\ = R \sum_i \frac{1}{Q} e^{-\frac{E_i-E_0}{kT}} \ln Q \tag{8}\\ = R \ln Q = +\infty \end{align} However, this would mean that the total standard molar entropy of helium is also infinite. Actually we may substitute helium by practically any gaseous molecule, and conclude that the standard molar entropies of all molecular gasses are infinite! Of course the entropy diverges very slowly, because $e^{-\frac{I_1-E_0}{kT}}$ is extremely small at room temperature and the entropy is the logarithm of the sum, but since I apparently didn't make any approximation, this infinity does pose a theoretic problem.

My intuition is that as long as the interactions between the helium atoms are taken into account, the summation becomes finite, however small the interactions are (but I cannot prove this at the moment). But an ideal gas reference state is, by definition, a hypothetical state where the atoms cannot see each other at all, as if the gas is infinitely thin.

My questions are thus:

  1. Are the above derivations correct?
  2. If yes, does this imply that it is inherently flawed to define the standard molar entropy of a gas using an ideal gas reference state?
  3. Are there any alternative definitions of the standard molar entropy of a gas, that can get rid of this divergence? One may think of choosing the reference state as a state with very small but non-zero pressure, small enough so that the effect of intermolecular interaction on translational entropy can be neglected but large enough so that the divergence of the electronic partition function is effectively suppressed. But then, how should one choose the value of the pressure, without introducing arbitrariness into the definition of the standard molar entropy?
  • 1
    $\begingroup$ Related: phys.libretexts.org/Bookshelves/Astronomy__Cosmology/…. Also: scienceblogs.com/builtonfacts/2011/01/25/…. For the second link, there is some good discussion in the comments (mixed in with some spam accounts). "Schroeder's text 'An Introduction to Thermal Physics' discusses this exact issue in Problem 6.9 (page 227)". They suggest the form of the partition function is approximate and corrections fix the divergence. $\endgroup$
    – Tyberius
    Commented Jul 10, 2021 at 15:29
  • $\begingroup$ @Tyberius Thanks! This perfectly answers my question. In particular I'm satisfied with the approach where the helium atom is taken to be in a box that is as large as our universe. This is a completely unambiguous solution as long as we are only studying physics and chemistry that happen in the Anthropocene, rather than those that happen right before the heat death of the universe. By extension we can redefine an ideal gas as a hypothetical gas so thin that there is only one molecule per universe, rather than a gas that is mathematically infinitely thin. $\endgroup$
    – wzkchem5
    Commented Jul 10, 2021 at 16:23
  • $\begingroup$ +1. I just added in the equation labels because even if you don't need to refer to them in the question, someone else 5 years from now might want to. Notice we have the "cite" button so that all questions and answers can be cited and picked up by Google Scholar, so these are like "mini-publications". $\endgroup$ Commented Jul 12, 2021 at 6:24

1 Answer 1


I'll try to summarize the argument mentioned in the comments from Daniel Schroeder's Introduction to Thermal Physics.

Your derivation is correct within a certain approximation commonly made when dealing with atomic sized systems. Consider an isolated system of an hydrogen atom in thermal equilibrium with a reservoir (which could in principle be the rest of the universe). The Boltzmann factor for an electronic state $i$ of the atom arises from the Boltzmann entropy equation for the reservoir: $$S_R(i)=k\ln(\Omega_R(i))\to\Omega_R(i)=\exp(kS_R(i))\tag{1}$$ Where $S_R(i)$ is the entropy of the reservoir when the atom is in the state $i$ and $\Omega_R(i)$ is the number of microstates. The partition function is just the sum of the microstates for each macrostate, $Q=\sum_i\Omega_R(i)=\exp(kS_R(i))$. Going forward, I will actually use entropy differences relative to some reference state in these expressions. This simply corresponds to multiplying the Boltzmann factors (and thus the partition function) by a fixed constant, so it won't affect any of the derived results.

To get the expression for the partition function in terms of the electronic energy of the atom, we can use the general relation: $$\mathrm{d} S_R=T\mathrm{d} U_R+P\mathrm{d} V_R-\mu\mathrm{d} N_R=-T\mathrm{d} E-P\mathrm{d} V_r+\mu\mathrm{d} N_r\tag{2}$$ The second equality arises from the fact that any of the changes to the reservoir $R$ must be coming at the expense of the single atom subsystem $r$. In this case, $\mu\mathrm{d}N_r=0$ as we are not exchanging this single atom.

Here is where we get to the approximation at the root of the divergence you found. Typically, the $P\mathrm{d} V$ is also neglected because the volume change of an atom becoming excited, and thus the PV work, is typically miniscule in comparison to the electronic energy change. However, this is not always true. Schroeder gives $a_0n^2$ as a rough approximation of hydrogens radius for an excited state $n$, which will lead to significant volume change for very large $n$. By including this term, the contribution of the higher lying Rydberg states to the partition function will rapidly go to zero, allowing the sum to converge.


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