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 D"appen, Hummer and Mihalas (hence MHD equation of state) in a series of paper in the early 80's. However, it was originally suggested by Harris 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 Kittel and Kruger (p.163, Eq.6.24)):
$$F = NkT [\ln(n/n_Q) - 1]$$
where $k$ is the Boltzmann 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.
Harris (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. 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.
Dappen (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
$$C = 1.5\ln\left(2\pi (\hbar [eV s])^2\right) + \ln N_A - 1.5 \ln a.m.u. - 1 = -33.7$$
(it's $\approx 44$ if a.m.u. is in kg).
I need to get what this number is to be able to compare my results to those of Harris and others. As cannot recover their constant, I'm getting worried that there is something elementary in the definition of the quantities that I am missing, so I beg for another pair of eyes here?
Was this equation (constant) a common knowledge in the 60's? Does it come from some of the famous old books on SE (I've checked Fowler's textbooks, nothing there). Or they used some other assumption that was common and now I'm not seeing it? Or I'm just making some basic mistake in my calculation?
Any help would be greatly appreciated. Thanks!!
 A: The translational Helmholtz free energy of a monatomic gas is sometimes written succinctly as
$$F=NkT\ln\left[\left(\frac{2\pi\hbar^2}{mkT}\right)^{3/2}\frac{N}{eV}\right]$$
for molecule number $N$, Boltzmann constant $k$, temperature $T$, reduced Planck constant $\hbar$, molecule mass $m$, exponential constant $e$, and volume $V$. This can be expanded as
$$F=NkT\ln\left[\frac{3}{2}\ln\left(\frac{2\pi\hbar^2}{k}\right)-1-\frac{3}{2}\ln T-\frac{3}{2}\ln m+\ln N-\ln V\right].$$
Some (mid-20th-century) authors replaced the first two terms in brackets ($\frac{3}{2}\ln\left(\frac{2\pi\hbar^2}{k}\right)-1$) with $-6.974$ while replacing $m$ with the molecular weight $M$. How so?
As you note, the assumed units are electron volts per molecule. The constant therefore appears to be calculated as follows:

*

*Planck's reduced constant is $\hbar=6.582\times 10^{-16}\,\text{eV}\cdot\text{s}$.

*Boltzmann's constant is $k = 8.617\times 10^{-5}\,\text{eV}\cdot\text{K}$.

*Avagadro's constant is $N_A=6.022\times 10^{23}\,\text{mol}^{-1}$.

Then, stripping units,
$$\frac{3}{2}\ln\frac{2\pi\hbar^2}{k}-1+\frac{3}{2}\ln N_A=-6.943.$$
I attribute the difference of less than half a percent to the difference between ca. 1960 and current estimates of the physical constants.
