Free energy for ideal gas: constant in logarithmic expression
I'm working on an equation of state implementation based on the free energy minimization. This method had be hugely popularized in astrophysical community by <PERSON>, <PERSON> and <PERSON> (hence MHD equation of state) in a series of paper in the early 80's. However, it was originally suggested by <PERSON> and collaborators (1959JChPh..31.1211H, 1964PhRv..133..427H).
The "translational" free energy for the ideal monoatomic gas is (e.g. "Thermal Physics, 2nd ed." by <PERSON> and <PERSON> (p.163, Eq.6.24)):
$$F = NkT [\ln(n/n_Q) - 1]$$
where $k$ is the <PERSON> constant, $n$ is the particle density, $T$ is the temperature, $N$ is the number of particles and $n_Q$ is the quantum concentration. They give the same equation in the logarithmic form (Eq.6.23) where $n_Q$ is explicitly written.
$$F = NkT \left[\ln N - 1 - \ln V - \frac{3}{2}\ln(kT) + \frac{3}{2}\ln\left(\frac{2\pi \hbar^2}{M}\right)\right]$$
$$F = NkT \left[\ln N - 1 - \ln V - 1.5\ln(kT) + 1.5\ln\left(2\pi \hbar^2\right) - 1.5\ln M\right]$$
where $M$ is the mass of one particle.
<PERSON> (1964) writes the same expression for a system with different types of particles (neutral H atom, proton, electron) as a sum of free energies written for individual species:
For one particle type, obviously, $Q_1 = N\ln N$, $Q_2 = N$, $Q_3 = N \ln M$, so her equation (1) would reduce to:
$$F_1 = kT[N\ln N - (1.5 \ln(kT) + 6.974 + \ln V)N - 1.5 N\ln M]$$
$$F_1 = NkT[\ln N - \ln V - 1.5 \ln(kT) - 6.974 - 1.5\ln M]$$
When this is compared with the expression from K&K above: $$F = NkT \left[\ln N - \ln V - 1.5\ln(kT) + 1.5\ln\left(2\pi \hbar^2\right) - 1 - 1.5\ln M\right]$$
so it formally follows that:
$$C = 1.5\ln\left(2\pi \hbar^2\right) - 1 = - 6.974$$
Of course, the exact value of the constant in a logarithmic expression critically depends on the units that are used for the variables. 346
From the magnitude of $\hbar$ (in whatever common units) it's clear that the constant on R.H.S. cannot be only equal to L.H.S., but it must include some conversion factors for the units of other quantities.
<PERSON> (1980) cite the same (similar) expression and the same numerical value for the constant reappears. He specifies units in more details:
$F$ and $NkT$ must have the same units (in this case: eV/mole). If $kT$ (in $\ln kT$) is given in eV, then the reduced Planck constant must be in eV s, thus $\hbar = 6.582 \times 10^{-16}$ eV s, so that $1.5 \ln (2\pi \hbar^2) \approx - 102.1$.
If $N$ (in $\ln N$) is in moles, then (with $N_A = 6.022 \times 10^{23}$)
$$C = 1.5\ln\left(2\pi (\hbar [eV s])^2\right) + \ln N_A - 1 \approx -48.3$$
If $M$ is given in a.m.u. ($1.66 \times 10^{-24}$ g) then
$$C = 1. 374