1.7.17: Compressions- Isothermal- Solutes- Partial Molar Compressions

A given aqueous solution at temperature \(\mathrm{T}\) and near ambient pressure \(\mathrm{p}\) contains a solute \(j\) at molality \(\mathrm{m}_{j}\). The chemical potential \(\mu_{j}(\mathrm{aq})\) is related to the molality \(\mathrm{m}_{j}\) using equation (a).

\[\mu_{\mathrm{j}}(\mathrm{aq})=\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)+\mathrm{R} \, \mathrm{T} \, \ln \left(\gamma_{\mathrm{j}}\right) \nonumber \]

Then

\[\mathrm{V}_{\mathrm{j}}(\mathrm{aq})=\mathrm{V}_{\mathrm{j}}^{\infty}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \,\left[\partial \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{p}\right]_{\mathrm{T}} \nonumber \]

By definition, the partial molar isothermal compression of solute \(j\),[1]

\[K_{T_{j}}(a q)=-\left(\frac{\partial V_{j}(a q)}{\partial p}\right)_{T} \nonumber \]

Then

\[\mathrm{K}_{\mathrm{Tj}}(\mathrm{aq})=\mathrm{K}_{\mathrm{TJ}}^{\infty}(\mathrm{aq})-\mathrm{R} \, \mathrm{T} \,\left[\partial^{2} \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{p}^{2}\right]_{\mathrm{T}} \nonumber \]

Thus by definition,

\[\operatorname{limit}\left(\mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \mathrm{K}_{\mathrm{T}_{\mathrm{j}}}(\mathrm{aq})=\mathrm{K}_{\mathrm{T}_{\mathrm{j}}}^{\infty}(\mathrm{aq}) \nonumber \]

Hence the difference between \(\mathrm{K}_{\mathrm{Tj}_{\mathrm{j}}}(\mathrm{aq})\) and \(\mathrm{K}_{\mathrm{T}_{j}}^{\infty}(\mathrm{aq})\) depends on the second differential of \(\ln \left(\gamma_{j}\right)\) with respect to pressure.

---