Chapter 4

Equilibrium and Phase Stabilityin One-Component Systems

The purpose of this chapter is to provide answers to the following questions:(i) under what conditions are the two phases of a pure substance in equilibriumwith each other? and (ii) at a given temperature and pressure, in which form doesa pure substance exist–solid, liquid, or gas?

4.1 Equilibrium Criteria for Closed Systems

For a closed system, the second law of thermodynamics, Eqn (1.6-8), states that

T dS − δQ � 0. (4.1-1)

From the first law of thermodynamics

dU = δQ − P dV . (4.1-2)

Elimination of δQ between Eqns (4.1-1) and (4.1-2) leads to

T dS − dU − P dV � 0. (4.1-3)

4.1.1 Condition of Maximum Entropy

Under the conditions of constant internal energy and volume, Eqn (4.1-3) simplifiesto

dSU,V � 0, (4.1-4)

where the subscripts U and V imply that these quantities are kept constant duringa process. Under these conditions, any spontaneous process tends to increase the

The Thermodynamics of Phase and Reaction Equilibria. DOI: http://dx.doi.org/B978-0-44-4594976.00004-9© 2013 Elsevier B.V. All rights reserved. 105

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dS = 0

Equilibrium

State of System

Direction of aspontaneous process

Constant U & V

Ent

ropy

Figure 4.1: The equilibrium condition for constant internal energy and volume.

entropy of a system as shown in Fig. 4.1. When the system is at equilibrium, entropyreaches its maximum value and this condition is mathematically expressed as

dSU,V = 0 At equilibrium. (4.1-5)

4.1.2 Condition of Minimum Internal Energy

When entropy and volume are kept constant, Eqn (4.1-3) simplifies to

dUS,V � 0. (4.1-6)

Therefore, the internal energy of a system decreases as a result of a spontaneousprocess under the conditions of constant entropy and volume. This phenomenoncan be explained as follows. Spontaneous processes lead to an increase in entropy.In order to keep entropy constant, heat must be transferred to the surroundingswith a concomitant decrease in internal energy. When the system is at equilibrium,internal energy reaches its minimum value and this condition is mathematicallyexpressed as

dUS,V = 0 At equilibrium. (4.1-7)

4.1.3 Condition of Minimum Helmholtz Energy

In differential form, Helmholtz energy is given by

dA = d(U − TS) = dU − S dT − T dS. (4.1-8)

The use of Eqn (4.1-8) in Eqn (4.1-3) gives

dA + S dT + P dV � 0. (4.1-9)

Equilibrium and Phase Stability in One-Component Systems 107

When temperature and volume are kept constant, Eqn (4.1-9) simplifies to

dAT ,V � 0. (4.1-10)

Therefore, the Helmholtz energy of a system decreases as a result of a spontaneousprocess under the conditions of constant temperature and volume. When the systemis at equilibrium, Helmholtz energy reaches its minimum value and this conditionis mathematically expressed as

dAT ,V = 0 At equilibrium. (4.1-11)

4.1.4 Condition of Minimum Gibbs Energy

In differential form, Gibbs energy is given by

dG = d(H−TS) = d(U+PV −TS) = dU+P dV +V dP−T dS−S dT . (4.1-12)

The use of Eqn (4.1-12) in Eqn (4.1-3) gives

dG + S dT − V dP � 0. (4.1-13)

When temperature and pressure are kept constant, Eqn (4.1-13) simplifies to

dGT ,P � 0. (4.1-14)

Under these conditions, any spontaneous process tends to decrease the Gibbs energyof a system as shown in Fig. 4.2. When the system is at equilibrium, Gibbs energyreaches its minimum value and this condition is mathematically expressed as

dGT ,P = 0 At equilibrium. (4.1-15)

Since Gibbs energy describes equilibrium at conditions of constant temperatureand pressure, it is the most commonly used thermodynamic function in chemicalengineering applications.

4.2 Equilibrium Criteria for Open Systems

To develop the conditions for equilibrium, it is first necessary to develop the com-bined first and second laws of thermodynamics for an open system. The first lawof thermodynamics for an open system, Eqn (1.5-3), is given by

Hin dnin − Hout dnout + δQ + δW = d(nU)sys (4.2-1)

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State of System

dG = 0

Equilibrium

Direction of aspontaneous process

Constant T & P

Gib

bs E

nerg

yFigure 4.2: The equilibrium condition for constant temperature and pressure.

in which changes in kinetic and potential energies are considered negligible. Onthe other hand, the entropy balance, Eqn (1.6-6), is written as

δSgen = d(nS)sys + Sout dnout − Sin dnin − δQ

Tsys. (4.2-2)

To combine the first and the second law expressions, we have to restrict ourattention to processes that are reversible, i.e. δSgen = 0. Mass can be added andwithdrawn reversibly only if the properties of the entering and leaving mass areidentical to those of the system, i.e.

Tin =Tout = Tsys, Pin = Pout = Psys, Hin = Hout = Hsys, Sin = Sout = Ssys.

(4.2-3)

Under these conditions, Eqns (4.2-1) and (4.2-2) become

Hsys(dnin − dnout) + δQ + δW = dUsys, (4.2-4)

0 = dSsys + Ssys(dnout − dnin) − δQ

Tsys. (4.2-5)

For a one-component system under consideration, the mass balance is expressed inthe form

dnin − dnout = dnsys. (4.2-6)

Elimination of δQ between Eqns (4.2-4) and (4.2-5) and the use of Eqn (4.2-6) leadto

dUsys = Tsys dSsys + δW + (Hsys − TsysSsys)dnsys. (4.2-7)

If the only work is that of the expansion type of work, i.e. −Psys dVsys, Eqn (4.2-7)takes the form

dU = T dS − P dV + (H − TS)dn, (4.2-8)

Equilibrium and Phase Stability in One-Component Systems 109

where the subscript “sys” is dropped with the understanding that all propertiesbelong to the system. The coefficient of dn is nothing more than the Gibbs energyper mole, G, for a pure substance.1 Thus, Eqn (4.2-8) is expressed as

dU = T dS − P dV + G dn. (4.2-9)

Consider an isolated system containing α- and β-phases of a pure substance.Note that although the overall system is an isolated one, each phase may be consid-ered an open system, i.e. there may be an exchange of mass and/or energy betweenα- and β-phases. Since the system is isolated, then the total values of U , V , and nare all constant, i.e.

U = Uα + Uβ = constant ⇒ dUα = −dUβ,

V = Vα + Vβ = constant ⇒ dVα = −dVβ,

n = nα + nβ = constant ⇒ dnα = −dnβ.

(4.2-10)

The equilibrium criterion for an isolated system is given by Eqn (4.1-5), i.e.

dS = dSα + dSβ = 0. (4.2-11)

The total differential of entropy can be obtained from Eqn (4.2-9) as

dS = 1

TdU + P

TdV − G

Tdn, (4.2-12)

so that

dSα = 1

Tα

dUα + Pα

Tα

dVα − Gα

Tα

dnα, (4.2-13)

dSβ = 1

Tβ

dUβ + Pβ

Tβ

dVβ − Gβ

Tβ

dnβ. (4.2-14)

Substitution of Eqns (4.2-13) and (4.2-14) into Eqn (4.2-11) and making use of therelationships given by Eqn (4.2-10) lead to

dS =(

1

Tα

− 1

Tβ

)dUα +

(Pα

Tα

− Pβ

Tβ

)dVα −

(Gα

Tα

− Gβ

Tβ

)dnα = 0, (4.2-15)

which can only be satisfied if the terms in parentheses in Eqn (4.2-15) are all equalto zero. Therefore, the following conditions hold at equilibrium:

Tα = Tβ Thermal equilibriumPα = Pβ Mechanical equilibriumGα = Gβ Chemical equilibrium.

(4.2-16)

1Historically, for a single component system, Gibbs energy per mole has been called the chemical potential anddesignated by μ, i.e. G = μ.

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Note that a temperature gradient implies the performance of a thermal work, i.e.heat engine or heat pump, and a pressure gradient implies performance of mechan-ical work, i.e. expansion/contraction work. The gradient of Gibbs energy, on theother hand, implies performance of a chemical work.

4.3 Phase Stability

Why does water exist as a liquid when 0 < T < 100◦C under atmospheric condi-tions at sea level? Why does CO2 exist as a gas under room conditions? To answersuch questions one should realize that any spontaneous process taking place at con-stant temperature and pressure leads to a decrease in the Gibbs energy of a system.The Gibbs energy reaches its minimum value when the system is at equilibrium.In other words, at a given temperature and pressure, a pure substance is stable in aphase in which it has the minimum Gibbs energy.2

In differential form, Gibbs energy of a pure substance is expressed as

dG = V dP − S dT . (4.3-1)

Therefore, variation of Gibbs energy with pressure and temperature is essential forunderstanding phase stability.

4.3.1 Change of Gibbs Energy with Pressure

Variation of Gibbs energy with pressure at constant temperature is(∂G

∂P

)T

= V . (4.3-2)

Since molar volume is always greater than zero, Eqn (4.3-2) implies that Gibbsenergy increases with increasing pressure under isothermal conditions.

For an ideal gas, integration of Eqn (4.3-2) at constant temperature gives

�GIG =∫ P2

P1

RT

PdP = RT ln

(P2

P1

). (4.3-3)

2Since G = H −TS, the value of G is dependent on the terms “H” and “TS.” At high temperatures the term “TS”dominates and, as a result, the phase with the highest value of entropy is the most stable. At low temperatures, theterm “H” dominates and the phase with the lowest value of enthalpy is the most stable.

Equilibrium and Phase Stability in One-Component Systems 111

Thus, Gibbs energy of an ideal gas increases exponentially with increasing pressure.The molar volume of liquids and solids is almost independent of pressure. In

this case, integration of Eqn (4.3-2) at constant temperature leads to

�G C = V C �P, (4.3-4)

where the superscript “C” indicates the condensed phase, i.e. either liquid or solid.Equation (4.3-4) indicates a linear variation of Gibbs energy with pressure, with aslope being equal to V C.

If m is the slope of G versus P curve at any given temperature, then the abovediscussion implies that

mV � mL > mS. (4.3-5)

While mL and mS are almost constant, mV changes exponentially with pressure.Figure 4.3a indicates the variation of G as a function of pressure for gas, liquid,and solid phases.

As shown in Fig. 4.3b, when P < Pvap, the gas phase has the lowest Gibbsenergy and the substance exists as a gas since it is most stable in this phase. AtP = Pvap, molar Gibbs energies of the gas and liquid phases are equal to each otheror, in other words, the gas and liquid phases are in equilibrium. At this point the tem-perature and pressure are called the saturation temperature (or boiling temperature)and saturation pressure (or vapor pressure), respectively. When Pvap < P < Pm,the liquid phase has the minimum Gibbs energy and the substance exists as a liquid.At P = Pm, molar Gibbs energies of the liquid and solid phases are equal to each

Gas

LiquidSolid

PressurePressure

Gas Liquid Solid

Gas

LiquidSolid

Pvap Pm

(a) (b)

Mol

ar G

ibbs

Ene

rgy

Mol

ar G

ibbs

Ene

rgy

Figure 4.3: Variation of G as a function of P at constant T and stable equilibrium states.

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Temperature

Slope = S~

Mol

ar G

ibbs

Ene

rgy

Figure 4.4: Variation of G as a function of T at constant P.

other or, in other words, the liquid and solid phases are in equilibrium. At this pointthe temperature and pressure are called the melting (or freezing) temperature andmelting pressure, respectively. When P > Pm, the solid phase is the most stablephase since it has the lowest Gibbs energy.

4.3.2 Change of Gibbs Energy with Temperature

Variation of Gibbs energy with temperature at constant pressure is given by(∂G

∂T

)P

= −S. (4.3-6)

Since the entropy is always positive, then the slope of G versus T is negative. More-over, since S increases with temperature for all substances, then G is a concave3

function of T as shown in Fig. 4.4.Variation of Gibbs energies of different phases as a function of temperature

at constant pressure is shown in Fig. 4.5a. In plotting such a graph one shouldremember that SV > SL > SS. As shown in Fig. 4.5b, solid, having the lowestGibbs energy, is the most stable phase when T < Tm. At T = Tm (melting orfreezing temperature), solid is in equilibrium with liquid and their molar Gibbsenergies are equal to each other. When Tm < T < Tb, the liquid phase is the stablephase. At T = Tb (boiling or saturation temperature), the liquid and gas phases arein equilibrium. The gas phase becomes the most stable phase for T > Tb.

3A function f (x) is concave if the chord joining any two points lies entirely below the curve.

Equilibrium and Phase Stability in One-Component Systems 113

TemperatureTemperature

GasLiquidSolid

Tm

Gas

LiquidSolid

Gas

LiquidSolid

Tb

(a) (b)

Mol

ar G

ibbs

Ene

rgy

Mol

ar G

ibbs

Ene

rgy

Figure 4.5: Variation of G for different phases as a function of T at constant P and stable equilibriumstates.

Example 4.1 The following values are for saturated steam at 508.15 K. Showthat the liquid and vapor states of water are in equilibrium.

T (K) Pvap (MPa) HL (kJ/kg) HV (kJ/kg) SL (kJ/kg K) SV (kJ/kg K)

508.15 3.06 1013.62 2804.2 2.6558 6.1791

SolutionWhen liquid and vapor phases of water are in equilibrium, then the number ofdegrees of freedom is calculated from Eqn (1.1-4) as

F = C + 2 − P = 1 + 2 − 2 = 1.

Therefore, either temperature or pressure has to be specified to fix the state ofthe system. For saturated steam at 508.15 K, the corresponding vapor pressure(or saturation pressure) is 3.06 MPa. Under the condition of equilibrium, Gibbsenergies of liquid and vapor must be equal to each other:

GL = HL − TSL = 1013.62 − (508.15)(2.6558) = −335.9248 kJ/kg,

GV = HV − TSV = 2804.2 − (508.15)(6.1791) = −335.7097 kJ/kg.

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Problems

Problem Related to Section 4.1

4.1 When the gravitational force is appreciable, Eqn (5) of Problem 2.11 indicatesthat the change in Gibbs energy is expressed as

dG = V dP − S dT + mg dz. (1)

(a) Since Gibbs energy reaches its minimum value under the condition ofequilibrium, i.e. dG = 0, conclude that

dT = 0 ⇒ T is uniform (2)

and

V dP + mg dz = 0 ⇒ dP

dz= −ρg, (3)

which is the hydrostatic pressure distribution. Keep in mind that the dis-tance z is measured in the direction opposite to gravity.

(b) Since density, ρ, is constant for liquids, show that the integration ofEqn (3) gives

P = −ρgz + C, (4)

where C is an integration constant.(c) Density changes as a function of temperature and pressure in the case of

gases. Assuming ideal gas behavior, i.e.

ρ = PM

RT, (5)

show that the integration of Eqn (3) leads to

ln P = −Mgz

RT+ C, (6)

where M is the molecular weight and C is an integration constant.(d) Once the pressure in an oil reservoir decreases to low values, oil flow to

the well practically stops. Enhanced oil recovery (EOR) is a techniqueused to recover oil that is left in the reservoir. One of the EOR techniquesis to inject high-pressure carbon dioxide into the reservoir. When CO2 ismixed with oil, the resulting liquid has a lower viscosity, i.e. thinner, andbecomes less sticky. This process facilitates the flow of oil from the rock tothe well. Suppose that CO2 is injected into a well 4 km deep at a pressureof 140 bar. If the temperature within the well is uniform at 310 K, estimatethe pressure at the bottom of the well.

(Answer: (d) 273.4 bar)

Equilibrium and Phase Stability in One-Component Systems 115

Problems Related to Section 4.2

4.2 Complete the following table without using the steam table:

T (◦C) HL (kJ/kg) HV (kJ/kg) S L (kJ/kg K) S V (kJ/kg K)

– 567.69 2727.3 1.6870 6.9777205 875.04 – 2.3780 6.3952340 1594.2 2622.0 – 5.3357

Use T (K) = T (◦C) + 273.15.

(Answer: T = 135◦C, HV = 2795.9 kJ/kg, SL = 3.6594 kJ/kg K).

4.3 When the gravitational force is appreciable, the combined first and second lawsof thermodynamics for a closed system lead to

dU = T dS − P dV + mg dz (1)

as shown in Problem 2.11.

(a) In the case of an open system, show that the combined first and secondlaws of thermodynamics give

dU = T dS − P dV + mg dz + (G + Mgz)dn, (2)

where m and M represent the mass and the molecular weight, respectively.(b) Show that the change in Gibbs energy is given by

dG = V dP − S dT + mg dz + (G + Mgz)dn. (3)

(c) Under equilibrium conditions show that

dT = 0, (4)

V dP + mg dz = 0, (5)

G + Mgz = 0. (6)

(d) Differentiate Eqn (6) under constant temperature and conclude thatEqns (5) and (6) both lead to hydrostatic pressure distribution as given byEqn (3) in Problem 4.1.

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Problems Related to Section 4.3

4.4 A function f (x) is convex if the chord joining any two points lies entirelyabove the curve. Explain why Gibbs energy cannot be a convex function oftemperature at constant pressure. In other words, show that the followingplot of G versus T at constant pressure is impossible.

Temperature

4.5 An element X exists in nature in two different solid forms, Xα and Xβ . Thestandard Gibbs energy of formations of Xα and Xβ at 298 K and atmosphericpressure are 510 J/mol and 485 J/mol, respectively.

(a) Which form of X is the more stable at 298 K? Explain.(b) The molar volumes of Xα and Xβ are 25 cm3/mol and 30 cm3/mol, re-

spectively. Keeping the temperature constant at 298 K, does the increasein pressure make Xα more stable than Xβ? If so, at what pressure will thetransition occur?

(c) The standard molar entropies of Xα and Xβ are 23.6 J/mol K and34.8 J/mol K, respectively. Under atmospheric pressure, does the increasein temperature make Xα more stable than Xβ? If so, at what temperaturewill the transition occur?

(Answer: (a) Xβ , (b) 51 bar and (c) No)

4.6 Tin exists in two solid forms, white and gray tin. While white tin has a body-centered tetragonal structure, the structure of gray tin is diamond cubic. Theenthalpy and entropy changes associated with the transformation of gray tin towhite tin at 298 K and atmospheric pressure are 2090 J/mol and 7.3 J/mol K,respectively.

Equilibrium and Phase Stability in One-Component Systems 117

(a) Which form of tin is the more stable at 298 K?(b) Under atmospheric pressure, estimate the temperature at which white and

gray tin coexist in equilibrium. Indicate your assumptions.

(Answer: (b) 286.3 K)

4.7 Phase transitions of SiO2 under high pressure are of interest to earth scientists.Atake et al. (2000) reported that the enthalpy and entropy changes during α-quartz to coesite (polymorph of quartz) transition at 298 K and atmosphericpressure are 3400 J/mol and −2.16 J/mol K, respectively. The molar volumeof α-quartz is 22.688 × 10−6 m3/mol and the molar volume of coesite is20.6 × 10−6 m3/mol. At 298 K, what is the pressure at which these two formsof SiO2 will be in equilibrium?

(Answer: 1.94 GPa)

4.8 The vapor pressure of water as a function of temperature is given by

ln PvapH2O = 13.8365 − 6.6075 × 103

T+ 10.1855 × 105

T 2− 17.7039 × 107

T 3,

300 � T � 600,

where PvapH2O is in bar and T is in K.

(a) Estimate the minimum pressure at which water exists as a liquid at 350 K.(b) Estimate the maximum pressure at which water exists as a vapor at 500 K.

(Answer: (a) 0.425 bar and (b) 26.56 bar)

4.9 A pure substance X flows through a heat exchanger under steady conditions asshown in the figure below. The inlet and outlet streams are single-phase andthe ideal gas heat capacity is C∗

P = 11.5R.

T = 350 K T = 250 K

P = 1 barP = 8 bar

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Estimate the amount of heat added to or removed from substance X usingthe following information provided from the solutions of the Peng-Robinsonequation of state at the inlet and outlet conditions:

Z(

H − HIG)

T ,P/RT

(S − S IG

)T ,P

/R

0.0334 −5.868 −6.088Inlet 0.0895 −3.107 −3.465

0.8573 −0.410 −0.275

0.0043 −10.451 −9.988Outlet 0.0334 −2.673 −4.033

0.9588 −0.106 −0.066

(Answer: −30,090 J/mol)

ReferenceAtake, T., I. Noriko, H. Kawaji, K. Matsuzaka and M. Akaogi, 2000, J. Chem. Thermodyn., 32,

217-227.