1.7.13: Compressions- Isothermal- Equilibrium and Frozen

The volume of a given closed system is defined by the set of independent variables \(\mathrm{T}\), \(\mathrm{p}\) and composition \(\xi\); \(\mathrm{V}=\mathrm{V}[\mathrm{T}, \mathrm{p}, \xi]\). We assert that in this state the affinity for spontaneous chemical reaction is \(\mathrm{A}\). The system is perturbed by a change in pressure such that the system can track one of two pathways; (i) at constant \(\mathrm{A}\) or (ii) at constant \(\xi\). The differential dependences of volume on pressure are related using equation (a).

\[\left(\frac{\partial \mathrm{V}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \mathrm{A}}=\left(\frac{\partial \mathrm{V}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \xi}-\left(\frac{\partial \mathrm{A}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \xi} \,\left(\frac{\partial \xi}{\partial \mathrm{A}}\right)_{\mathrm{T}, \mathrm{p}} \,\left(\frac{\partial \mathrm{V}}{\partial \xi}\right)_{\mathrm{T}, \mathrm{p}}\]

The two differentials expressing the dependence of volume on pressure define the equilibrium isothermal compression and the frozen isothermal compression respectively [1]. For a system at equilibrium, (i.e. minimum in \(\mathrm{G}\) at fixed \(\mathrm{T}\) and \(\mathrm{p}\)) following perturbation by a change in pressure,

\[\mathrm{K}_{\mathrm{T}}(\mathrm{A}=0)=-\left(\frac{\partial \mathrm{V}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \mathrm{A}=0} \quad \mathrm{~K}_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)=-\left(\frac{\partial \mathrm{V}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \xi^{\mathrm{a}}}\]

The negative signs recognise that for all thermodynamically stable systems, the volume decreases with increase in pressure. Nevertheless there is merit in thinking of compression (and compressibility) as a positive feature of a system. Both \(\mathrm{K}_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)\) and \(\mathrm{K}_{\mathrm{T}}(\mathrm{A}=0)\) are extensive variables characterising two possible pathways. From equation (a) [2],

\[\mathrm{K}_{\mathrm{T}}(\mathrm{A}=0)=\mathrm{K}_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)-\left[\left(\frac{\partial \mathrm{V}}{\partial \xi}\right)_{\mathrm{T}, \mathrm{p}}\right]^{2} \,\left(\frac{\partial \xi}{\partial \mathrm{A}}\right)_{\mathrm{T}, \mathrm{p}}\]

But at equilibrium, \((\partial \mathrm{A} / \partial \xi)_{\mathrm{T}, \mathrm{p}}<0\). Hence, irrespective of the sign of \((\partial \mathrm{V} / \partial \xi)_{\mathrm{T}, \mathrm{p}}\), \(\mathrm{K}_{\mathrm{T}}(\mathrm{A}=0)>\mathrm{K}_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)\). Equation (c) is rewritten in terms of compressibilities [3].

\[\kappa_{\mathrm{T}}(\mathrm{A}=0)=-(1 / \mathrm{V}) \,(\partial \mathrm{V} / \partial \mathrm{p})_{\mathrm{T}, \mathrm{A}=0} \quad \kappa_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)=-(1 / \mathrm{V}) \,(\partial \mathrm{V} / \partial \mathrm{p})_{\mathrm{T}, \xi^{\text {eq }}}\]

\[\kappa_{\mathrm{T}}(\mathrm{A}=0)=\kappa_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)-(1 / \mathrm{V}) \,\left[(\partial \mathrm{V} / \partial \xi)_{\mathrm{T}, \mathrm{p}}\right]^{2} \,(\partial \xi / \partial \mathrm{A})_{\mathrm{T}, \mathrm{p}}\]

Because \((\partial \mathrm{A} / \partial \xi)_{\mathrm{T}, \mathrm{p}}<0\), for all stable systems, \(\kappa_{\mathrm{T}}(\mathrm{A}=0)>\kappa_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)\). According therefore to equation (e) the volume decrease accompanying a given change in pressure is more dramatic under condition that \(\mathrm{A} = 0\) than under the condition where \(\xi\) remains constant at \(\xi^{\mathrm{eq}}\) [4]. Both \(\kappa_{\mathrm{T}}(\mathrm{A}=0)\) and \(\kappa_{\mathrm{T}}\left(\xi^{\mathrm{eq}}\right)\) are volume intensive properties of a solution [5].

Footnotes

[1] The contrast between the two conditions is familiar to anyone who has dived into a swimming pool and “got it wrong”. Hitting the wall of water is similar to the conditions for\(\mathrm{K}_{\mathrm{T}}(\xi)\) whereas for a good dive the conditions resemble \(\mathrm{K}_{\mathrm{T}}(\mathrm{A}=0)\); the water molecules move apart to allow a smooth entry into the water.

[2] Consider \(\left(\frac{\partial^{2} \mathrm{G}}{\partial \mathrm{p} \, \mathrm{d} \xi}\right)=\left(\frac{\partial^{2} \mathrm{G}}{\partial \xi \, \mathrm{dp}}\right)\). Then, \(-\left(\frac{\partial \mathrm{A}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \xi}=\left(\frac{\partial \mathrm{V}}{\partial \xi}\right)_{\mathrm{T}, \mathrm{p}}\)

[3] \(\mathrm{K}_{\mathrm{T}}=\left[\mathrm{m}^{3} \mathrm{~Pa}^{-1}\right] ; \kappa_{\mathrm{T}}=\left[\mathrm{Pa}^{-1}\right]\)

[4] Equation (e) forms the basis of the pressure-jump fast reaction technique. A rapid change in pressure produces a “frozen” system which relaxes to the equilibrium state at a rate characteristic of the system.

[5] For information concerning \(\mathrm{D}_{2}\mathrm{O}(\ell)\), see R. A. Fine and F. J. Millero, J.Chem.Phys.,1975,63,89.