density $\rho$ and a gas particle mass $Am_{\rm H}$, the Coulomb energy per particle is $e_{\rm C}=\tilde{Z}^2\mathrm{e}^2/(4\pi\epsilon_0\mathcal{D})$, where $\tilde{Z}\mathrm{e}$ is the particle charge, $\epsilon_0$ is the permittivity, $\mathcal{D}=(Am_{\rm H}/\rho)^{1/3}$ is the mean inter-particle distance, $m_{\rm H}$ is the atomic mass unit. At temperature $T$, the kinetic energy per particle has the form $e_{\rm k}=3kT/2$, where $k$ is the Boltzmann constant. The ratio of Coulomb energy to kinetic energy can be estimated as $$\frac{e_{\rm C}}{e_{\rm
k}}=\left[\frac{\tilde{Z}^2\mathrm{e}^2}{6\pi\epsilon_0 k(Am_{\rm
H})^{1/3}}\right] \frac{\rho^{1/3}}{T},$$ where $A$ and $\tilde{Z}$ are of the order of unit, the term in square brackets is approximately $10^5$. For our modeled cases, the ratio of $\rho^{1/3}/T$ is less than $10^{-6.5}$, which means $e_{\rm C}/e_{\rm k}\approx 0.01$. This indicates that the particle interactions are dominated by collisions and the nonideal effects caused by the Coulomb force can be safely neglected. Hence, for the present calculation, the gas is treated as a fully ionized ideal gas. The internal energy $e$ and total pressure $p$ can be expressed as: $$\label{eos}
e=aT^4+\frac{3}{2}p_{\rm g}+\rho e_{\rm i},\\
p=p_{\rm g}+p_{\rm r},$$ respectively, where $p_{\rm g}=\beta p$ is the gas pressure, $p_{\rm
r}$ is the radiation pressure, and $e_{\rm i}$ is the ionization energy per mass unit. Using Eqs. (\[eos\]) 