# Stability criteria

(Redirected from Stability Criteria)

Equation dSE,V = 0 from Equilibrium Criteria for Pure Substances demonstrates that for a constant-volume isolated system, the equilibrium condition requires that the entropy of the system must be stationary, i.e., dS = 0. Equation dSE,V = 0 from Equilibrium Criteria for Pure Substances is also valid if the entropy of the system is at either maximum or minimum. To ensure that the system is at stable equilibrium, i.e., equilibrium in the system can be maintained after a small perturbation, it is necessary for the system to satisfy eq. $\Delta {S_{E,V}} \ge 0$ from Equilibrium Criteria for Pure Substances as well. All other equilibrium criteria listed in eqs. dFT,V = 0dHS,p = 0 from Equilibrium Criteria for Pure Substances are similarly inadequate to ensure stable equilibrium, and the inequalities stated in eqs. $\Delta {F_{T,V}} \le 0$ $\Delta {H_{S,p}} \le 0$ from Equilibrium Criteria for Pure Substances are necessary in addition. We will now focus on the use of the energy minimum principle (in E, F, G, and H representation) to address the stability of a simple system in equilibrium. For a simple system to be in equilibrium, as analyzed in detail above, the system must be stable. If the system is not stable, it will spontaneously change state to become stable. Stability for a single-phase system can be broken down into three distinct types: (1) thermal, (2) mechanical, and (3) chemical.

## Thermal Stability

A simple system (homogeneous, single-phase) confined by an adiabatic, rigid, and impermeable boundary is shown in Fig. 1. Under these constraints, the energy and volume of the system remain constant during any process. In its initial state, the system is divided by an adiabatic partition into two halves of equal volume, with the left side at a slightly higher temperature than the right. This temperature difference is accounted for in the representation of the internal energy by (E + ΔE) for the left half and (E − ΔE) for the right half. At some arbitrary time, heat is allowed to be exchanged between the two halves and sufficient time elapses for the system to reach equilibrium. Since no energy has been added to the system, the total energy in the initial state matches the total energy in the final state, i.e., ${E_i} = \left( {E + \Delta E} \right) + \left( {E - \Delta E} \right) = 2E = {E_f}\qquad \qquad(1)$

If the entropies of the left and right halves in the final state are SL and SR, respectively, they must be identical, because the two halves possess the same internal energy and volume at the final state, i.e., ${S_L} = {S_R} = S(E,V)\qquad \qquad(2)$

Since the volume of the left half is not changed between the initial and final states, the entropy of the left half in the initial state can be found by Taylor series expansion as follows: ${S_{L,i}} = S(E + \Delta E,V) = S + {\left( {\frac{{\partial S}}{{\partial E}}} \right)_V}\left( {\Delta E} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}S}}{{\partial {E^2}}}} \right)_V}{\left( {\Delta E} \right)^2} + ...\qquad \qquad(3)$

Similarly, the entropy in the right half at the initial state can be represented by ${S_{R,i}} = S(E - \Delta E,V) = S - {\left( {\frac{{\partial S}}{{\partial E}}} \right)_V}\left( {\Delta E} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}S}}{{\partial {E^2}}}} \right)_V}{\left( {\Delta E} \right)^2} - ...\qquad \qquad(4)$

According to eq. (68), the final system entropy must be equal to or greater than the initial total system entropy. In other words, the entropy generated by this process must not be negative. ${S_{gen}} = \left( {{S_L} + {S_R}} \right) - \left( {{S_{L,i}} + {S_{R,i}}} \right) \ge 0\qquad \qquad(5)$

Combining eqs. (2) – (5) results in the following expression: ${S_{gen}} = - {\left( {\frac{{{\partial ^2}S}}{{\partial {E^2}}}} \right)_V}{\left( {\Delta E} \right)^2} \ge 0\qquad \qquad(6)$

Since (ΔE)2 will always be finite positive, approaching zero as the two halves steadily approach uniform total internal energy, the second order partial derivative in eq. (6) must be negative and finite. ${\left( {\frac{{{\partial ^2}S}}{{\partial {E^2}}}} \right)_V} < 0\qquad \qquad(7)$

Equation (7) can be rewritten as follows with the use of eq. ${\left( {\frac{{\partial E}}{{\partial S}}} \right)_{V,{n_i}}} = T$ from Closed Systems with Compositional Change: $\frac{\partial }{{\partial E}}{\left( {\frac{1}{T}} \right)_V} = - \frac{1}{{{T^2}}}{\left( {\frac{{\partial T}}{{\partial E}}} \right)_V} < 0\qquad \qquad(8)$

i.e., $\frac{1}{{{T^2}}}{\left( {\frac{{\partial T}}{{\partial E}}} \right)_V} > 0\qquad \qquad(9)$

so thermal stability requires that internal energy of the system must be an increasing function of temperature. Considering the definition of specific heat at constant volume, ${c_v} = \frac{1}{m}{\left( {\frac{{\partial E}}{{\partial T}}} \right)_V}\qquad \qquad(10)$

eq. (9) can be written as $\frac{1}{{{T^2}m{c_v}}} > 0\qquad \qquad(11)$

Since both T2 and m in eq. (11) are positive, eq. (11) implies that the specific heat of the system at constant volume must always be positive in order for the system to move from its initial state – as defined above – to its final state, rather than in the opposite direction. ${c_v} > 0\qquad \qquad(12)$

In other words, positive cv ensures that the system cannot spontaneously segregate itself into two thermally-dissimilar regions. A simple system at equilibrium is thermally stable if, at constant volume, the temperature of the system increases with increasing internal energy.

## Mechanical Stability

The mechanical stability criteria can be determined by considering a simple system that initially has an internal pressure discontinuity. Fig. 2 shows a simple system with constant total volume and temperature. The constant temperature is maintained by contact with a thermal reservoir at a constant temperature, T. Initially, the system is divided into two parts of slightly-different volume by an off-center partition held in place by a locking mechanism. The left side is at a slightly higher pressure and lower volume. When the locking mechanism is disengaged, the partition will gradually float to the midpoint of the system, at which point the pressures and volumes of the two halves are equalized. In a manner similar to the analysis of thermal stability, the total volume of the system in the initial and final states can be represented as ${V_i} = \left( {V - \Delta V} \right) + \left( {V + \Delta V} \right) = V + V = {V_f}\qquad \qquad(13)$

Since the temperature and total volume of this system remain constant, the minimum Helmholtz free energy principle, eq. $\Delta {F_{T,V}} \le 0$ from Equilibrium Criteria for Pure Substances, can be applied to this problem. In the final state, the volumes and temperatures of the two halves are the same; therefore, the Helmholtz free energies of the left and right halves in the final state must be the same, i.e., ${F_L} = {F_R} = F(V,T)\qquad \qquad(14)$

where F is the final Helmholtz free energy of either half of the system. As was the case with entropy in the preceding thermal stability analysis, the Helmholtz free energies of the initial two parts can be related to the final values by noting the constant-temperature volume changes experienced by each of the two parts: ${F_{L,i}} = F(V - \Delta V,T) = F - {\left( {\frac{{\partial F}}{{\partial V}}} \right)_T}\left( {\Delta V} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}F}}{{\partial {V^2}}}} \right)_T}{\left( {\Delta V} \right)^2} - ...\qquad \qquad(15)$ ${F_{R,i}} = F(V + \Delta V,T) = F + {\left( {\frac{{\partial F}}{{\partial V}}} \right)_T}\left( {\Delta V} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}F}}{{\partial {V^2}}}} \right)_T}{\left( {\Delta V} \right)^2} + ...\qquad \qquad(16)$

From the equilibrium analysis above, it can be concluded that the Helmholtz free energy of the final system must be less than or equal to the initial total system Helmholtz free energy: $\left( {{F_L} + {F_R}} \right) - \left( {{F_{L,i}} + {F_{R,i}}} \right) \le 0\qquad \qquad(17)$

Combining eqs. (14) – (17) results in the following expression: ${\left( {\frac{{{\partial ^2}F}}{{\partial {V^2}}}} \right)_T}{\left( {\Delta V} \right)^2} \ge 0\qquad \qquad(18)$

Since (ΔV)2 will always be finite positive, approaching zero as the Helmholtz free energies of the two portions steadily approach uniformity, the second order partial derivative of eq. (18) must be positive and finite: ${\left( {\frac{{{\partial ^2}F}}{{\partial {V^2}}}} \right)_T} > 0\qquad \qquad(19)$

Equation (19) can be rewritten in the following form by considering eq. $- p = {\left( {\frac{{\partial E}}{{\partial V}}} \right)_{S,{n_i}}} = {\left( {\frac{{\partial F}}{{\partial V}}} \right)_{T,{n_i}}}$ from Closed Systems with Compositional Change: $- {\left( {\frac{{\partial p}}{{\partial V}}} \right)_T} = \frac{1}{{{\kappa _T}V}} > 0\qquad \qquad(20)$

where κT is isothermal compressibility: ${\kappa _T} = - \frac{1}{V}{\left( {\frac{{\partial V}}{{\partial p}}} \right)_T}\qquad \qquad(21)$

From eq. (20), the criterion for mechanical stability of the system is found to be ${\kappa _T} > 0\qquad \qquad(22)$

Therefore, a simple system at equilibrium is mechanically stable if the isothermal compressibility factor is positive, i.e., the volume of the system shrinks with increasing pressure.

## Chemical Stability

For a simple system in equilibrium to be stable, the system must also have chemical stability. In other words, certain conditions will prevent the system from spontaneously separating into two or more subsystems of varying chemical composition. As an aid to analyzing the criteria for chemical stability, the system shown in Fig. 3 is presented. The simple system depicted is in contact with a constant temperature reservoir and its boundary is impermeable to all species present. Two frictionless pistons ensure that the pressure of the system is always a constant value. Internally, the system in its initial state consists of a semipermeable membrane partition that prevents the movements of only species i between the two portions of the system. Initially, the left portion of the system

contains more moles of species i than the right portion. At some arbitrary time, the membrane is made permeable to permit the flow of species i between the two compartments, and the system reaches an equilibrium condition with respect to matter flow. Conservation of the number of moles of species i during the process shown in Fig. 3 can be written as ${n_{i,i}} = \left( {{n_i} + \Delta {n_i}} \right) + \left( {{n_i} - \Delta {n_i}} \right) = 2{n_i} = {n_{i,f}}\qquad \qquad(23)$

Since this system maintains constant temperature and pressure, the energy minimum principle that governs this system is the Gibbs free energy principle shown in eq. $\Delta {G_{T,p}} \le 0$ from Equilibrium Criteria for Pure Substances. Therefore, denoting the Gibbs free energy of the left and right halves at the equilibrium final state as GL and GR respectively, the following can be stated: ${G_L} = {G_R} = G({n_i},T,p)\qquad \qquad(24)$

where G is the final Gibbs free energy of either half of the system. As demonstrated above with entropy in the thermal stability analysis, the Gibbs free energies of the initial two parts can be related to the final values by noting the change of the Gibbs free energy with the change in the mole number i of the left- and right-hand sides when temperature, pressure, and all other mole numbers are held constant: ${G_{L,i}} = G({n_i} + \Delta {n_i},T,p) = G + {\left( {\frac{{\partial G}}{{\partial {n_i}}}} \right)_{T,p,{n_j}}}\left( {\Delta {n_i}} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}G}}{{\partial n_i^2}}} \right)_{T,p,{n_j}}}{\left( {\Delta {n_i}} \right)^2} + ...\qquad \qquad(25)$ ${G_{R,i}} = G({n_i} + \Delta {n_i},T,p) = G - {\left( {\frac{{\partial G}}{{\partial {n_i}}}} \right)_{T,p,{n_j}}}\left( {\Delta {n_i}} \right) + \frac{1}{2}{\left( {\frac{{{\partial ^2}G}}{{\partial n_i^2}}} \right)_{T,p,{n_j}}}{\left( {\Delta {n_i}} \right)^2} - ...\qquad \qquad(26)$

In the equilibrium analysis followed above, it was established that the Gibbs free energy of the final system must be less than or equal to the initial total system Gibbs free energy, i.e., $\left( {{G_L} + {G_R}} \right) - \left( {{G_{L,i}} + {G_{R,i}}} \right) \le 0\qquad \qquad(27)$

Combining eqs. (23) – (27) results in the following expression: ${\left( {\frac{{{\partial ^2}G}}{{\partial n_i^2}}} \right)_{T,p,{n_j}}}{\left( {\Delta {n_i}} \right)^2} \ge 0\qquad \qquad(28)$

Since (Δni)2 will always be finite positive, approaching zero as the Gibbs free energy of the two portions steadily approaches uniformity, the second-order partial derivative in eq. (28) must be positive and finite: ${\left( {\frac{{{\partial ^2}G}}{{\partial n_i^2}}} \right)_{T,p,{n_j}}} > 0\qquad \qquad(29)$

Substituting eq. ${\mu _i} = {\left( {\frac{{\partial H}}{{\partial {n_i}}}} \right)_{p,S,{n_{j \ne i}}}} = {\left( {\frac{{\partial F}}{{\partial {n_i}}}} \right)_{T,V,{n_{j \ne i}}}} = {\left( {\frac{{\partial G}}{{\partial {n_i}}}} \right)_{T,p,{n_{j \ne i}}}}$ from Closed Systems with Compositional Change into eq. (29) results in the chemical stability criterion for the simple system: ${\left( {\frac{{\partial {\mu _i}}}{{\partial {n_i}}}} \right)_{T,p,{n_j}}} > 0\qquad \qquad(30)$

Therefore, a simple system in equilibrium, such as the final state of the system, is chemically stable if the chemical potential of the ith species increases with an increase in mole number of the ith species.

## References

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA.