1.21.6: Thermodynamic Stability: Thermal, Diffusional and Hydrostatic

On the bench in front of a chemist is a stoppered flask containing a liquid mixture, ethanol + water, at temperature \(\mathrm{T}\) and ambient pressure \(\mathrm{p}\). The chemist might wonder why the homogeneous liquid does not spontaneously separate into two liquids, say water-rich and alcohol-rich mixtures. The chemist might also wonder why the mixture does not spontaneously produce a system which comprises a warm liquid mixture and a cold liquid mixture. Yet the fact that these changes do not occur spontaneously leads to the conclusion that the conditions in operation which forbid these changes can be traced to the Second Law of Thermodynamics, prompted by the word ‘spontaneously’ used above [1].

Thermal Stability

Initially a given closed system has thermodynamic energy \(2\mathrm{U}\) and volume \(2\mathrm{V}\). We imagine that the mixture does in fact separate into two liquids, both at equilibrium, having energy \(\mathrm{U} + \delta \mathrm{U}\) with volume \(\mathrm{V}\), and energy \(\mathrm{U} - \delta \mathrm{U}\) also with volume \(\mathrm{V}\). The overall change in entropy at constant overall composition is given by equation (a).

\[\delta \mathrm{S}=\mathrm{S}(\mathrm{U}+\delta \mathrm{U}, \mathrm{V})+\mathrm{S}(\mathrm{U}-\delta \mathrm{U}, \mathrm{V})-\mathrm{S}(2 \mathrm{U}, 2 \mathrm{~V})\]

The change in entropy can be understood in terms of a Taylor expansion for a change at constant volume \(\mathrm{V}\). Thus

\[\delta S=\left(\frac{\partial^{2} S}{\partial U^{2}}\right)_{V, \xi} \,(\delta U)^{2}\]

However at constant \(\mathrm{V}\) and composition \(\xi\), the Second Law of Thermodynamics requires that \(\delta \mathrm{S}\) is positive for all spontaneous processes. The fact that such a change is not observed requires that \(\left(\frac{\partial^{2} \mathrm{~S}}{\partial \mathrm{U}^{2}}\right)\) is negative. But

\[\left(\frac{\partial \mathrm{S}}{\partial \mathrm{U}}\right)_{\mathrm{V}, \xi}=\mathrm{T}^{-1}\]

Then

\[\left(\frac{\partial^{2} \mathrm{~S}}{\partial \mathrm{U}^{2}}\right)_{\mathrm{V}, \xi}=\frac{\partial}{\partial \mathrm{U}}\left(\mathrm{T}^{-1}\right)=-\frac{1}{\mathrm{~T}^{2}} \,\left(\frac{\partial \mathrm{T}}{\partial \mathrm{U}}\right)_{\mathrm{V}, \xi}=-\frac{1}{\mathrm{~T}^{2} \, \mathrm{C}_{\mathrm{V} \xi}}\]

In order for the latter condition to hold, \(\mathrm{C}_{\mathrm{V}\xi}\) must be positive. This is therefore the condition for thermal stability. In other words we will not observe spontaneous separation into hot and cold domains in that heat capacities are positive variables.

Diffusional Stability

A given system at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) contains \(2 \, \mathrm{n}_{\mathrm{i}}\) moles of each i-chemical substance, for \(\mathrm{i} = 1, 2, 3 \ldots\). The system is divided into two parts such that each part at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) contains \(\mathrm{n}_{\mathrm{i}}\) moles of each chemical substance for \(\mathrm{i} = 2, 3, 4, \ldots\). However one part contains \(\mathrm{n}_{1} + \Delta \mathrm{n}_{1}\) moles and the other part contains \(\mathrm{n}_{1} - \Delta \mathrm{n}_{1}\) moles of chemical substance 1. Then the change in Gibbs energy \(\delta \mathrm{G}\) is given by equation (e).

\[\begin{gathered}
\delta \mathrm{G}=\mathrm{G}\left[\mathrm{T}, \mathrm{p}, \mathrm{n}_{1}+\Delta \mathrm{n}_{1}, \mathrm{n}_{2}, \mathrm{n}_{3} . .\right]+\mathrm{G}\left[\mathrm{T}, \mathrm{p}, \mathrm{n}_{1}-\Delta \mathrm{n}_{1}, \mathrm{n}_{2}, \mathrm{n}_{3} . .\right] \\
-\mathrm{G}\left[\mathrm{T}, \mathrm{p}, 2 \, \mathrm{n}_{1}, 2 \, \mathrm{n}_{2}, 2 \, \mathrm{n}_{3} . .\right]
\end{gathered}\]

In terms of a Taylor expansions,

\[\delta G=\left(\frac{\partial^{2} G}{\partial n_{1}^{2}}\right)_{T, p, n(2), n(3) . .} \,\left(\delta n_{1}\right)^{2}\]

But chemical potential,

\[\mu_{1}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(2), \mathrm{n}(3) \ldots}\]

Then

\[\left(\frac{\partial^{2} \mathrm{G}}{\partial \mathrm{n}_{1}^{2}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(2), \mathrm{n}(3) \ldots}=\left(\frac{\partial \mu_{1}}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(2), \mathrm{n}(3) \ldots .}\]

We conclude that spontaneous separation of the system into parts rich and depleted in chemical substance-1 would occur if \(\left(\frac{\partial \mu_{1}}{\partial n_{1}}\right)_{T, p, n(2), n(3)}\) is negative. But this process is never observed. Hence the condition for diffusional (or material) stability is that \(\left(\frac{\partial \mu_{1}}{\partial n_{1}}\right)_{T, p, n(2), n(3) \ldots .}>0\). Further if we add \(\delta \mathrm{n}_{1}\) moles of chemical substance to a closed system at fixed \(\mathrm{T}, \mathrm{p}, \mathrm{n}_{2}, \mathrm{n}_{3} \ldots\), chemical potential µ1 must increase.

Hydrostatic Stability

A given system at temperature \(\mathrm{T}\) and chemical composition \(\xi\) has volume \(2\mathrm{V}\). We imagine that a infinitely thin partition exists separating the system into two parts having equal volumes \(\mathrm{V}\). The Helmholtz energy of the system volume \(2\mathrm{V}\) is given by equation (i).

\[\mathrm{F}=\mathrm{F}[\mathrm{T}, 2 \mathrm{~V}, \xi]\]

The Helmholtz energy of the two parts, volume \(\mathrm{V}\) is given by equation (j).

\[\mathrm{F}=\mathrm{F}[\mathrm{T}, \mathrm{V}, \xi]\]

The partition is envisaged as moving to produce two parts having volumes \((\mathrm{V}+\delta \mathrm{V})\) and \((\mathrm{V}-\delta \mathrm{V})\). Then at constant composition the change in Helmholtz energy is given by equation (k).

\[\delta \mathrm{F}=\mathrm{F}[\mathrm{T}, \mathrm{V}+\delta \mathrm{V}]+\mathrm{F}[\mathrm{T}, \mathrm{V}-\delta \mathrm{V}]-\mathrm{F}[\mathrm{T}, 2 \mathrm{~V}]\]

Then using Taylor’s theorem,

\[\delta \mathrm{F}=\left(\partial^{2} \mathrm{~F} / \partial \mathrm{V}^{2}\right)_{\mathrm{T}} \,(\partial \mathrm{V})^{2}\]

Hence [2]

\[\left(\partial^{2} \mathrm{~F} / \partial \mathrm{V}^{2}\right)_{\mathrm{T}}=-(\partial \mathrm{p} / \partial \mathrm{V})_{\mathrm{T}, \xi}\]

Irrespective of the sign of (\(\delta \mathrm{V}\)), we conclude that \(\delta \mathrm{F}\) would be negative in the event that \((\partial \mathrm{p} / \partial \mathrm{V})_{\mathrm{T}, \xi}\) is positive. But we never witness such a spontaneous separation of a system into two parts. In other words \((\partial \mathrm{p} / \partial \mathrm{V})_{\mathrm{T}, \xi}<0\). Hence,

\[\kappa_{\mathrm{T}}=-\left(\frac{1}{\mathrm{~V}}\right) \,\left(\frac{\partial \mathrm{V}}{\partial \mathrm{p}}\right)_{\mathrm{T}, \xi}>0\]

Therefore for a system at fixed composition and temperature, if we as observers of this system increase the pressure \(\mathrm{p}\), the volume of the system decreases. This is the condition of hydrostatic (or, mechanical) stability [1].

Physical Consequences of Stability

Taken together, conditions for thermal stability \(\left(\mathrm{C}_{\mathrm{V} \xi}>0\right)\) and for mechanical stability \(\left(\kappa_{\mathrm{T}}>0\right)\) have further consequences [3]. Since,

\[\sigma=\mathrm{C}_{\mathrm{V} \xi} / \mathrm{V}+\frac{\mathrm{T} \,\left(\alpha_{\mathrm{p}}\right)^{2}}{\kappa_{\mathrm{T}}}\]

Then isobaric heat capacities and heat capacitances , \(\sigma\left(=\mathrm{C}_{\mathrm{pg}} / \mathrm{V}\right)\) are positive for stable phases.

Furthermore, heat capacities and compressibilities are related by equation (p).

\[\frac{\mathrm{K}_{\mathrm{S}}}{\mathrm{K}_{\mathrm{T}}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{p}}}\]

Hence the isentropic compressibility \(\kappa_{\mathrm{S}}\) of a stable phase must also be positive. In summary,

\[\mathrm{C}_{\mathrm{pm}} \geq \mathrm{C}_{\mathrm{Vm}_{\mathrm{m}}}>0\]

\[\kappa_{\mathrm{T}} \geq \kappa_{\mathrm{S}}>0\]

Footnotes

[1] M. L. McGlashan, Chemical Thermodynamics, Academic Press, London, 1979, chapter 7.

[2] By definition, \(\mathrm{F}=\mathrm{F}[\mathrm{T}, \mathrm{V}, \xi]\). The total differential of the latter equation is as follows.

\[\mathrm{dF}=\left(\frac{\partial \mathrm{F}}{\partial \mathrm{T}}\right)_{\mathrm{V}, \xi} \, \mathrm{dT}+\left(\frac{\partial \mathrm{F}}{\partial \mathrm{V}}\right)_{\mathrm{T}, \xi} \, \mathrm{dV}+\left(\frac{\partial \mathrm{F}}{\partial \xi}\right)_{\mathrm{T}, \mathrm{V}} \, \mathrm{d} \xi\]

But (the ‘all-minus’ equation)

\[\mathrm{dF}=-\mathrm{S} \, \mathrm{dT}-\mathrm{p} \, \mathrm{dV}-\mathrm{A} \, \mathrm{d} \xi\]

Then,

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

Or,

\[\left(\frac{\partial^{2} \mathrm{~F}}{\partial \mathrm{V}^{2}}\right)_{\mathrm{T}, \xi}=-\left(\frac{\partial \mathrm{p}}{\partial \mathrm{V}}\right)_{\mathrm{T}, \xi}\]

[3] H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, Wiley, New York, 2nd edn., 1985, pp.209-210.