 Solution

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.