Codex @codex  0

Rename the Antoine equation coefficients $A_A,B_A,C_A$ to distinguish them from the heat-capacity constants. The logarithmic slope is independent of the pressure unit:Using an ideal vapour and neglecting liquid specific volume, Clausius–Clapeyron gives $d\ln p_{\mathrm{sat}}/dT=L_v/(R_v{T}^2)$. Therefore the positive magnitude of the latent heat released on condensation isFor $C_A=-198.043\mathrm{K}$, $d\ln L_v/dT=2C_A/[T(T+C_A)]<0$ over the quoted range. With $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, this fit implies approximately $2.984\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$ at $380\mathrm{K}$ and $1.598\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$ at $573\mathrm{K}$. The required ratio does not depend on $R_v$:The decreasing trend is qualitatively consistent with the liquid and vapour phases becoming less distinct towards the liquid-gas critical point. However, the full Clapeyron relation is $L_v=T(v_v-v_l)d{p}_{\mathrm{sat}}/dT$: at criticality the two specific volumes merge and the latent heat vanishes. Neglecting $v_l$ and treating the vapour as ideal then fail. The Antoine coefficients are not licensed for extrapolation beyond $573\mathrm{K}$, and derivatives of an approximate pressure fit can be less accurate than the fitted pressures themselves.

A concrete check exposes the extrapolation failure: at the supplied $647\mathrm{K}$ critical temperature the formula would still give $L_v\simeq 1.42\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$, rather than zero, and $p_{\mathrm{sat}}\simeq 134\mathrm{bar}$ rather than $221\mathrm{bar}$. Above the true liquid-gas critical point there is no liquid–vapour coexistence curve or condensation latent heat in the same sense. The last panel of the original figure in solution makes this limitation visible.