1.14.74: Vaporization
A given chemical substance \(j\) can exist in phases I and II. For phase I,
\[\mathrm{U}_{\mathrm{j}}^{*}(\mathrm{I})=\mathrm{H}_{\mathrm{j}}^{*}(\mathrm{I})-\mathrm{p} \, \mathrm{V}_{\mathrm{j}}^{*}(\mathrm{I}) \label{a} \]
\(\mathrm{U}_{\mathrm{j}}^{*}(\mathrm{I}), \mathrm{~H}_{\mathrm{j}}^{*}(\mathrm{I}) \text { and } \mathrm{V}_{\mathrm{j}}^{*}(\mathrm{I})\) are the molar thermodynamic energy, enthalpy and volume respectively of chemical substance \(j\) in phase I at pressure \(\mathrm{p}\). Chemical substance \(j\) can also exist in phase II at the same pressure \(\mathrm{p}\).
\[\mathrm{U}_{\mathrm{j}}^{*}(\mathrm{II})=\mathrm{H}_{\mathrm{j}}^{*}(\mathrm{II})-\mathrm{p} \, \mathrm{V}_{\mathrm{j}}^{*}(\mathrm{II}) \label{b} \]
Equations \ref{a} and \ref{b} are quite general. In an important application we identify phase II as the vapor phase which we assume to have the properties of a perfect gas. Phase I is the liquid state. For the process `liquid → vapor' ( i.e. vaporization) at temperature \(\mathrm{T}\),
\[\Delta_{\text {vap }} \mathrm{U}_{\mathrm{j}}^{*}(\mathrm{~T})=\Delta_{\mathrm{vap}} \mathrm{H}_{\mathrm{j}}^{*}(\mathrm{~T})-\mathrm{p} \,\left[\mathrm{V}_{\mathrm{j}}^{*}(\mathrm{~g})-\mathrm{V}_{\mathrm{j}}^{*}(\ell)\right] \nonumber \]
But at temperature \(\mathrm{T}\), \(\mathrm{V}_{\mathrm{j}}^{*}(\mathrm{~g})-\mathrm{V}_{\mathrm{j}}^{*}(\ell) \gg 0\) Also for one mole of a perfect gas, \(\mathrm{p} \, \mathrm{V}_{\mathrm{j}}^{*}(\mathrm{~g})=\mathrm{R} \, \mathrm{T}\).
Hence,
\[\Delta_{\text {vap }} \mathrm{U}_{\mathrm{j}}^{*}(\mathrm{~T})=\Delta_{\text {vap }} \mathrm{H}_{\mathrm{j}}^{*}(\mathrm{~T})-\mathrm{R} \, \mathrm{T} \nonumber \]
\(\Delta_{\text {vap }} \mathrm{H}_{\mathrm{j}}^{\mathrm{N}}(\mathrm{T})\) is obtained from the dependence of vapour pressure on temperature; see Clausius - Clapeyron Equation. Hence we obtain the molar thermodynamic energy of vaporisation.