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In a hydrogen–helium gas with negligible doubly ionized helium, define the nuclear number densities $n_H=n_{H^0}+n_{H^{+}}$ and $n_{He}=n_{H{e}^0}+n_{H{e}^{+}}$. Neglecting nuclear binding energy in the abundance conversion, the baryon number density and helium mass fraction satisfyThe ionization fractions $X_{H^{+}}=n_{H^{+}}/n_H$ and $X_{H{e}^{+}}=n_{H{e}^{+}}/n_{He}$ count ionized nuclei within each species. Charge neutrality gives $n_e=n_H{X}_{H^{+}}+n_{He}{X}_{H{e}^{+}}$, henceThus complete single ionization gives $\mathcal{F}=1-3Y_p/4$, rather than one electron per baryon. This fraction uses a different denominator from $n_e/n_H$. At sufficiently high temperature to doubly ionize helium, the helium contribution must instead count two electrons per such nucleus.

Let $N=n_H+n_p$ be the conserved number density of hydrogen nuclei, whether ionized or bound, and $x=n_p/N=n_e/N$ the ionization fraction. The ideal gas pressure counts neutral atoms, free protons and free electrons:At low temperature $x\simeq 0$ and $P\simeq N{k}_{B}T$. Heating through the ionization region increases the particle count as well as their thermal energy, giving a steeper rise. At high temperature $x\simeq 1$, so $P\simeq 2N{k}_{B}T$ and the asymptotic slope is twice the low-temperature slope. There is a smooth ionization crossover, not a pressure discontinuity.

The held graph is legible. Representative readings are $P(5000\mathrm{K})\simeq 6.8\mathrm{Pa}$ and $P(15000\mathrm{K})\simeq 41.4\mathrm{Pa}$, with readout uncertainties of roughly $0.3$–$0.5\mathrm{Pa}$ and about $100\mathrm{K}$. They give $N\simeq P/(k_{B}T)$ on the neutral branch and $N\simeq P/(2k_{B}T)$ on the ionized branch, respectively. Both are consistent withThe error is a conservative graph-reading estimate, not a fitted experimental confidence interval.

For that density, the Saha equation gives $x\simeq 0.50$ at $9000\mathrm{K}$, $x\simeq 0.81$ at $10000\mathrm{K}$, and $x\simeq 0.9992$ at $15000\mathrm{K}$. Differentiating the ionization relation givesThe calculated maximum slope occurs near $9.25\times 10^3\mathrm{K}$, consistent with the source's visibly steepest region at roughly $(9.3\pm 0.4)\times 10^3\mathrm{K}$. This checks the actual slope rather than identifying it without justification with exactly half ionization.

Taking the neutral ground-state gas as the zero of internal binding energy, the matter energy density is translational kinetic energy plus ionization energy:At $15000\mathrm{K}$, with $R=13.6\mathrm{eV}$, these terms are about $62$ and $218\mathrm{J}{\mathrm{m}}^{-3}$, givingRest energies are excluded. A separate equilibrium radiation field, if the gas were in a blackbody cavity, would add its own radiation energy density; it is not part of this gas-only estimate.

Increasing the conserved nucleus density to $10N$ multiplies the low-temperature neutral-gas slope and high-temperature ionized-gas slope by ten. However, at a fixed intermediate temperature, $x^2/(1-x)=S(T)/(10N)$ is smaller, so a smaller fraction is ionized. The curved transition therefore moves to higher temperature; the new graph is not simply ten times the old graph throughout the crossover.

At a fixed chosen ionization fraction, $S(T)$ must increase tenfold. Locally this requireswhich is of order $10^3\mathrm{K}$ in this region. Directly evaluating the same Saha equation at $10^{21}{\mathrm{m}}^{-3}$ moves the maximum pressure slope from about $9250$ to $10560\mathrm{K}$. The two asymptotic slopes still have ratio two, because each hydrogen nucleus still contributes one particle when neutral and two when fully ionized.

Write $u=1/T$. Then $\log \zeta =\mathrm{constant}-(3/4)\log u-T_{H}u$, giving local slope $d{\log }_{10}\zeta /du=-(T_H+3T/4)/\log 10$. At 5000–6000 K its magnitude is about $3.6\times 10^4$ K. The low-ionization tail in the original graph drops roughly one decade per $2.8\times 10^{-5}$ K${}^{-1}$, consistent with this value.

At $u=2.0\times 10^{-4}$ K${}^{-1}$, the graph gives $T\simeq 5000$ K and $\zeta \simeq 0.004$. Invert the full equation to get $n=C{T}^{3/2}{e}^{-R/(k_{B}T)}(1-\zeta )/{\zeta }^2$, yielding $\boxed{n\sim 10^{18}{\mathrm{m}}^{-3}}$. The neighboring tail markers give the same order; plotting precision and the squared dependence on the read ionization fraction preclude claiming many significant digits.