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For a fixed-composition simple compressible thermodynamic system with smooth equations of state, write the internal energy as $U(T,V)$ and the enthalpy as $H=U+pV$. Differentiating $H$ at constant pressure and applying the multivariable chain rule givesSubtracting the heat capacity at constant volume proves the displayed identity. For an ideal gas, $(\partial U/\partial V{)}_T=0$ and $(\partial V/\partial T{)}_p=N{k}_B/p$, so $C_p-C_V=N{k}_B$. For a nonideal material the internal energy change with volume contributes as well as expansion work.

For fixed particle number and temperature, a bubble with surface tension $\Gamma$ and area $\alpha V^{2/3}$ has effective availability $A(V)=F_{\mathrm{gas}}(T,V,N)+\Gamma \alpha V^{2/3}+p_{R}V$. Its stationary condition is $p_{\mathrm{gas}}-p_R=(2/3)\Gamma \alpha V^{-1/3}$. Its curvature is $A^{\prime \prime }=V^{-1}[{\kappa }_T^{-1}-(2/9)\Gamma \alpha V^{-1/3}]$. Positive curvature is required for stable Gaussian fluctuation approximation. The variance follows from availability curvature and equilibrium fluctuations. For an ideal gas, the restoring denominator becomes $p_R+(4/9)\Gamma \alpha V^{-1/3}>0$ for positive $p_R$. Treating the internal pressure as externally fixed instead of fixing the gas particle number changes the stability problem.

For an ideal gas mixture, the volume mixing ratio $x_i$ equals the mole fraction, number fraction, and ratio of partial to total pressure: $x_i=N_i/N=n_i/n=P_i/P$, with ${\sum }_i{x}_i=1$. This equivalence fails for general nonideal mixtures.

For positive finite opacity and nonvacuum ideal gas satisfying radiative diffusion, zero flux implies $dT/dz=0$. Write $P=\mathcal{R}\rho T$, where $\mathcal{R}=k_B/({\mu }_m{m}_p)$ for constant mean molecular weight ${\mu }_m$ in proton-mass units. If a polytropic equation of state $P=K{\rho }^{\gamma }$ with constant $K>0$ is also imposed, then $T=(K/\mathcal{R}){\rho }^{\gamma -1}$. For $\gamma \ne 1$, constant temperature would force constant density, contradicting vertical hydrostatic equilibrium $P^{\prime }=-\rho {\Omega }^{2}z$ on an open interval away from the midplane. The familiar finite polytropic slab for $\gamma >1$ is therefore a solution of the hydrostatic/adiabatic closure, not of this finite-opacity zero-radiative-flux system. For $\gamma =1$, the consistent isothermal profile is $\rho ={\rho }_m\exp [-{\Omega }^2{z}^2/(2\mathcal{R}{T}_m)]$, with no finite surface.