Chapter 7

c° 2011 Ismail Tosun

Chapter 7

Fugacity of a Component in a Mixture

When two phases of a pure component are in equilibrium, fugacities in each phase must bethe same. The purpose of this chapter is to replace a pure component with a multicomponentmixture and investigate the conditions under which two phases of a multicomponent mixtureare in equilibrium with each other. For this purpose, first the fugacity definition given inChapter 5 will be extended to multicomponent mixtures; then equations to calculate fugacitiesof components making up the mixture will be developed.

7.1 FUNDAMENTAL EQUATIONS FOR A MULTICOMPONENT MIXTURE

The combined law for a single-phase, single-component open system, Eq. (4.2-9), is given as

dU = T dS − P dV + eGdn (7.1-1)

By definition, Gibbs energy is expressed as

G = U + PV − TS (7.1-2)

In differential form, Eq. (7.1-2) becomes

dG = dU + P dV + V dP − T dS − S dT (7.1-3)

Substitution of Eq. (7.1-1) into Eq. (7.1-3) leads to

dG = V dP − S dT + eGdn (7.1-4)

indicating that the Gibbs energy is dependent on pressure, temperature, and number of moles,i.e.,

G = G(P, T, n) (7.1-5)

The total differential of G is

dG =

µ∂G

∂P

¶T,n

dP +

µ∂G

∂T

¶P,n

dT +

µ∂G

∂n

¶P,T

dn (7.1-6)

Comparison of Eq. (7.1-6) with Eq. (7.1-4) yields

V =

µ∂G

∂P

¶T,n

(7.1-7)

S = −µ∂G

∂T

¶P,n

(7.1-8)

eG = µ∂G∂n

¶P,T

(7.1-9)

179

In the case of a single-phase, multicomponent open system, Gibbs energy of the phase notonly depends on pressure and temperature, but also on the number of moles of each componentexisting in the phase, i.e.,

G = G(P, T, n1, n2, ..., nk) (7.1-10)

Therefore, the total differential of G is expressed in the form

dG =

µ∂G

∂P

¶T,nj

dP +

µ∂G

∂T

¶P,nj

dT +kXi=1

µ∂G

∂ni

¶P,T,nj 6=i

dni (7.1-11)

The subscript nj in the first two terms of Eq. (7.1-11) implies that the number of moles ofeach component is held constant during differentiation. For a constant number of moles, Eqs.(7.1-7) and (7.1-8) are still valid. On the other hand, the partial molar Gibbs energy is definedby

Gi =

µ∂G

∂ni

¶P,T,nj 6=i

(7.1-12)

Thus, Eq. (7.1-11) becomes

dG = V dP − S dT +kXi=1

Gi dni (7.1-13)

Historically, for a multicomponent system, the partial molar Gibbs energy has been called thechemical potential and designated by μi, i.e., Gi = μi.

The enthalpy is related to the Gibbs energy by

H = G+ TS (7.1-14)

Differentiation of Eq. (7.1-14) gives

dH = dG+ T dS + S dT (7.1-15)

Substitution of Eq. (7.1-13) into Eq. (7.1-15) yields

dH = T dS + V dP +kXi=1

Gi dni (7.1-16)

The internal energy is related to the enthalpy by

U = H − PV (7.1-17)


dU = dH − P dV − V dP (7.1-18)

The use of Eq. (7.1-16) in Eq. (7.1-18) results in

dU = T dS − P dV +kXi=1

Gi dni (7.1-19)

180

The Helmholtz energy is related to the internal energy by

A = U − TS (7.1-20)


dA = dU − T dS − S dT (7.1-21)


dA = −P dV − S dT +kXi=1

Gi dni (7.1-22)

Equations (7.1-13), (7.1-16), (7.1-19), and (7.1-22) are the fundamental equations for a single-phase, multicomponent open system. From these equations, partial molar Gibbs energy canbe expressed in different forms as

Gi =

µ∂G

∂ni

¶T,P,nj 6=i

=

µ∂H

∂ni

¶P,S,nj 6=i

=

µ∂U

∂ni

¶V,S,nj 6=i

=

µ∂A

∂ni

¶T,V,nj 6=i

(7.1-23)

Example 7.1 Develop the Gibbs-Duhem equation for the Gibbs energy by the application ofEuler’s theorem.

Solution

Equation (7.1-19) indicates that the internal energy of a single phase, multicomponent systemdepends on S, V , and the number of moles of each component, i.e.,

U = U(S, V, n1, n2, ..., nk) (1)

Application of Euler’s theorem, Eq. (6.2-33), gives

U =

µ∂U

∂S

¶V,nj

S +

µ∂U

∂V

¶S,nj

V +kXi=1

µ∂U

∂ni

¶S,V,nj 6=i

ni (2)

From Eq. (7.1-19)µ∂U

∂S

¶V,nj

= T

µ∂U

∂V

¶S,nj

= −P

µ∂U

∂ni

¶S,V,nj 6=i

= Gi (3)

Thus, Eq. (2) takes the form

U = TS − PV +kXi=1

Gi ni (4)

Differentiation of Eq. (4) yields

dU = T dS + S dT − P dV − V dP +kXi=1

Gi dni +kXi=1

ni dGi (5)

Subtraction of Eq. (7.1-19) from Eq. (5) results in

S dT − V dP +kXi=1

ni dGi = 0 (6)

181

which is the most general form of the Gibbs-Duhem equation. It provides the relationshipbetween temperature, pressure, and the partial molar Gibbs energy.

At constant temperature and pressure, Eq. (6) reduces to

kXi=1

ni dGi = 0 constant T and P (7)

Dividing Eq. (7) by the total number of moles gives

kXi=1

xi dGi = 0 constant T and P (8)

Comment: For a single-component system, the Gibbs-Duhem equation, Eq. (6), reduces to

S dT − V dP + nd eG = 0 (9)

Dividing Eq. (9) by the number of moles, n, leads to

d eG = eV dP − eS dT (10)

which is identical with Eq. (2.2-15).

7.1.1 Criteria for Chemical Equilibrium

Consider an isolated multicomponent system containing α- and β-phases. Note that althoughthe overall system is an isolated one each phase may be considered an open system, i.e., theremay be an exchange of mass and/or energy between α- and β-phases.

In the case of no chemical reaction, the total values of internal energy, volume, and numberof moles of species i can be expressed as

U = Uα + Uβ = constant ⇒ dUα = − dUβ

V = Vα + Vβ = constant ⇒ dVα = − dVβ

ni = niα + niβ = constant ⇒ dniα = − dniβ

(7.1-24)

Suppose that an arbitrary process causes changes in the entropies of the α- and β-phases. Suchchanges can be expressed from Eq. (7.1-19) as

dSα =1

TαdUα +

PαTα

dVα −kXi=1

Gαi

Tαdniα (7.1-25)

dSβ =1

TβdUβ +

PβTβ

dVβ −kXi=1

Gβi

Tβdniβ (7.1-26)

Summation of Eqs. (7.1-25) and (7.1-26) and making use of the constraints given by Eq.(7.1-24) lead to

dS =

µ1

Tα− 1

Tβ

¶dUα +

µPαTα− Pβ

Tβ

¶dVα −

kXi=1

ÃGαi

Tα− G

βi

Tβ

!dniα (7.1-27)

182

The condition of equilibrium is satisfied, i.e., dS = 0, if the terms in parentheses in Eq. (7.1-27)are all equal to zero. Therefore, at equilibrium we have the following conditions:

Tα = Tβ Thermal equilibrium

Pα = Pβ Mechanical equilibrium

Gαi = G

βi Chemical equilibrium

(7.1-28)

Equality of partial molar Gibbs energies implies that there are no chemical driving forces tomove species from one phase to another.

7.1.2 The Gibbs Phase Rule

The Gibbs phase rule is given by Eq. (1.1-4), i.e., P + F = C + 2. We are now in a positionto derive this equation theoretically. The number of degrees of freedom, F , is defined by

F =

∙Total # of independentintensive variables

¸−∙# of independent equationsrelating these variables

¸(7.1-29)

Consider a composite system containing P phases and C components. Assume that eachcomponent exists in each phase. Hence,

For each phase : (C − 1) independent mole fractionsTotal mole fractions : P (C − 1) (7.1-30)

At equilibrium, temperature and pressure must be uniform throughout the phases. Therefore,the total number of independent intensive variables is 2+P (C− 1). The condition of chemicalequilibrium is designated by the following equations:

At equilibrium G1i = G

2i = ... = G

Pi i = 1, 2, ..., C (7.1-31)

Under these circumstances, we have

For each species : (P − 1) independent equationsTotal # of independent eqn’s : C(P − 1) (7.1-32)

Substitution of Eqs. (7.1-30) and (7.1-32) into Eq. (7.1-29) gives

F = 2 + P (C − 1)− C(P − 1) ⇒ P + F = C + 2 (7.1-33)

Note that Eq. (7.1-33) applies only to systems at equilibrium when there is no chemical reactiontaking place.

7.2 PARTIAL MOLAR GIBBS ENERGY OF AN IDEAL GAS MIXTURE

An ideal gas mixture is a gas mixture with a density so low that its molecules do not appreciablyinteract. The equation of state for an ideal gas mixture containing k species is given by

PV IGM = (n1 + n2 + ...+ nk)| {z }n

RT (7.2-1)

183

in which the superscript IGM stands for "ideal gas mixture". In an ideal gas mixture, thepartial pressure of species i, pi, is expressed as

pi = Pyi (7.2-2)

In other words, the partial pressure of species i in the ideal gas mixture is the pressure exertedby ni moles of pure species i at the same volume and temperature.

Any total property in an ideal gas mixture is the sum of individual species making up themixture, each evaluated at the mixture temperature and its own partial pressure. Therefore,we can write

U IGM(T, P ) =kXi=1

ni eU IGi (T ) (7.2-3)

HIGM(T, P ) =kXi=1

ni eHIGi (T ) (7.2-4)

SIGM(T, P ) =kXi=1

ni eSIGi (T, pi) (7.2-5)

Equations (7.2-3) and (7.2-4) imply that both the internal energy change on mixing and theenthalpy change on mixing are zero for an ideal gas mixture.

From Eq. (2.2-41), the change in the entropy of a pure ideal gas with pressure, keepingtemperature constant, is given by

deSIGi = − R

PdP at constant T (7.2-6)

Integration at constant temperature from P to pi yieldseSIGi (T, pi) = eSIG

i (T, P )−R ln³piP

´= eSIG

i (T, P )−R ln yi = eSIGi (T, P )−R ln³nin

´(7.2-7)

Substitution of Eq. (7.2-7) into Eq. (7.2-5) gives

SIGM(T, P ) =kXi=1

ni eSIGi (T, P )−R

kXi=1

ni ln³nin

´(7.2-8)

The Gibbs energy of an ideal gas mixture is

GIGM(T, P ) = HIGM(T,P )− TSIGM(T, P ) (7.2-9)


GIGM(T, P ) =kXi=1

ni eGIGi (T,P ) +RT

kXi=1

ni ln³nin

´(7.2-10)

The partial molar Gibbs energy of species i in an ideal gas mixture can be calculated as

GIGMi (T,P, yi) =

µ∂GIGM

∂ni

¶T,P,nj 6=i

= eGIGi (T, P ) +RT ln yi (7.2-11)

It is possible to rearrange Eq. (7.2-11) as

ln

µPiP

¶=

GIGMi (T, P, yi)− eGIG

i (T, P )

RT(7.2-12)

184

7.3 FUGACITY OF A COMPONENT IN A MIXTURE

Lewis extended the definition introduced by Eq. (7.2-12) to multicomponent systems, and thefugacity of a component i in a mixture, bfi, is expressed in the form

ln

" bfi(T, P, xi)Pxi

#=

Gi(T,P, xi)−GIGMi (T, P, xi)

RT(7.3-1)

subject to the following constraint

bfi/(xiP )→ 1 as P → 0 (ideal gas state) (7.3-2)

which applies to solid, liquid, and gas.The fugacity coefficient of a component i in a mixture, bφi, is defined by

bφi = bfixiP

(7.3-3)

where the symbol (b) is used to represent properties in the mixture. Although they look similarto partial properties, their definitions are not the same.

Combination of Eqs. (7.3-1), (7.2-12), and (5.1-5) leads to an alternative expression for thefugacity of component i in a mixture as

Gi = eGIGi (T, P

o)−RT lnP o +RT ln bfi (7.3-4)

orGi = λi(T ) +RT ln bfi (7.3-5)

where λi(T ) is the molar Gibbs energy of pure component i at unit fugacity, and is defined byEq. (5.2-6). Differentiation of Eq. (7.3-5) by keeping temperature and the number of moles ofeach component constant leads to

dGi = RTd ln bfi at constant T and nj (7.3-6)

When α- and β-phases of a multicomponent mixture are in equilibrium with each other,the condition of equilibrium is represented by Eq. (7.1-28), i.e.,

Tα = Tβ Pα = Pβ Gαi = G

βi (7.3-7)

Letting Tα = Tβ = T and Pα = Pβ = P , equality of partial molar Gibbs free energies leads to

λi(T ) +RT ln bfαi (T, P, xαi ) = λi(T ) +RT ln bfβi (T, P, xβi ) (7.3-8)

or bfαi (T, P, xαi ) = bfβi (T, P, xβi ) (7.3-9)

Thus, at equilibrium, fugacities of each component are equal in all phases.Differentiation of Eq. (7.3-6) with respect to pressure givesµ

∂Gi

∂P

¶T,nj

= RT

Ã∂ ln bfi∂P

!T,nj

(7.3-10)

185

The left-hand side of Eq. (7.3-10) is expressed asµ∂Gi

∂P

¶T,nj

=∂

∂P

"µ∂Gmix

∂ni

¶T,P,nj 6=i

#T,nj

(7.3-11)

Changing the order of differentiation leads toµ∂Gi

∂P

¶T,nj

=∂

∂ni

"µ∂Gmix

∂P

¶T,nj

#T,P,nj 6=i

=

µ∂Vmix

∂ni

¶T,P,nj 6=i

= V i (7.3-12)

Therefore, at constant temperature and number of moles, Eq. (7.3-10) is expressed as

RT d ln bfi = V i dP (7.3-13)

Subtraction of the term RT d ln(xiP ) from both sides of Eq. (7.3-13) and integration result in

ln

Ã bfixiP

!= ln bφi = Z P

0

µV i

RT− 1

P

¶dP at constant T and nj (7.3-14)

Noting that PV i = ZiRT , it is also possible to express Eq. (7.3-14) in the form

ln bφi = Z P

0

µZi − 1P

¶dP at constant T and nj (7.3-15)

The fugacity of a pure component can be calculated from Eq. (5.2-12) as

ln

µfiP

¶=

Z P

0

Ã eViRT− 1

P

!dP at constant T (7.3-16)

Subtraction of Eq. (7.3-16) from Eq. (7.3-14) yields

ln

Ã bfixifi

!=

1

RT

Z P

0(V i − eVi) dP at constant T and nj (7.3-17)

Equations (7.3-14), (7.3-15), and (7.3-17) can be used to calculate the fugacity of component iin a mixture, and all equations are valid for gases, liquids, and solids. Note that the evaluationof integrals in these equations requires either the equation of state or the volumetric data intabular form to be known.

7.4 IDEAL MIXTURE1

A mixture, either gaseous or liquid, is said to be an ideal mixture if it has the followingproperties:

i) It is homogeneous,ii) The sizes of the molecules are equal,iii) The forces between unlike molecules of the mixture are the same as the forces between likemolecules.

1 In the literature, ideal mixture is also referred to as ideal solution.

186

It follows from properties (ii) and (iii) that

VIMi (T, P, xi) = eVi(T,P ) (7.4-1)

On the other hand, property (iii) implies that

HIMi (T, P, xi) = eHi(T, P ) (7.4-2)

Since U i = Hi − PV i, then Eqs. (7.4-1) and (7.4-2) indicate that

UIMi (T, P, xi) = eUi(T, P ) (7.4-3)

Mixtures of two similar substances usually form an ideal mixture, i.e., benzene and toluene,o-xylene and m-xylene, butane and isobutane, n-butanol and n-pentanol, n-pentane andn-hexane.


ln

Ã bfIMixifi

!= 0 (7.4-4)

or bf IMi (T,P, xi) = xi fi(T, P ) (7.4-5)

indicating that the fugacity of component i in an ideal mixture can be calculated by multiplyingthe pure component fugacity by its mole fraction. This relationship is known as the Lewis-Randall rule. Pure component fugacities, on the other hand, can be calculated by using one ofthe methods given in Chapter 5.

The use of Eq. (7.4-5) in Eq. (7.3-3) gives

bφIMi (T, P, xi) = φi(T,P ) (7.4-6)

indicating that the fugacity coefficient of i in an ideal mixture is independent of the mixturecomposition.

Although the simplicity of the Lewis-Randall rule makes it an attractive tool in the calcula-tion of mixture fugacities, one should remember the limitations of using it. The Lewis-Randallrule is applicable under the following conditions:

• Component i is present in large excess, i.e., xi > 0.9.• Components are similar,• Gas phase is nearly ideal, i.e., pressure is low.

7.5 CALCULATION OF COMPONENT FUGACITIES IN A GAS MIXTURE

7.5.1 Fugacity of a Component in an Ideal Gas Mixture

Using Eq. (7.2-1), the partial molar volume of an ideal gas mixture is calculated as

V i =

µ∂V IGM

∂ni

¶T,P,nj 6=i

=RT

P(7.5-1)

187

Substitution of Eq. (7.5-1) into Eq. (7.3-14) gives

ln bφIGMi = 0 ⇒ bφIGMi = 1 or bfIGMi = yiP = pi (7.5-2)

Therefore, for an ideal gas mixture, the fugacity coefficient of component i is unity and thefugacity of component i is equal to its partial pressure. At high pressures and low temperatures,the fugacity coefficient of a component in a gas mixture deviates from unity, indicating nonidealbehavior.

7.5.2 Fugacity of a Component in an Ideal Mixture

In this case the Lewis-Randall rule applies and from Eqs. (7.4-5) and (7.4-6)

bfIMi (T, P, yi) = yi fi(T, P ) and bφIMi (T, P, yi) = φi(T, P ) (7.5-3)

Fugacities of pure components are determined by one of the methods described in Section 5.3.It is important not to confuse ideal gas mixture with ideal mixture2. An ideal gas

mixture always behaves as an ideal mixture. However, an ideal mixture does not necessarilybehave as an ideal gas mixture. Note that the compressibility factor for an ideal mixture isgiven by

ZIM =P eV IM

RT=

P

RT

kXi=1

yieVi = kXi=1

yi

ÃP eViRT

!=

kXi=1

yiZi (7.5-4)

On the other hand, ZIGM = 1.0.

Example 7.2 Consider an ideal binary mixture of 35 mol % n-butane (1) and 65% n-hexane(2) at 373.15K and 10 bar. Determine the fugacities of n-butane and n-hexane in the mixtureusing the Peng-Robinson equation of state.

Solution

From Appendix A

Component Tc (K) Pc ( bar) ω

n-Butane 425.0 38.0 0.199n-Hexane 507.6 30.2 0.299

The Lewis-Randall rule requires pure component fugacities. The use of the Peng-Robinsonequation of state, Eq. (5.3-12), yields the following values:

Component Tr α A B ZV φf

( bar)

n-Butane 0.878 1.086 0.170 0.023 0.837 0.8573 8.573

n-Hexane 0.735 1.245 0.349 0.035 0.047 0.2347 2.347

Thus, the fugacities of components in the mixture arebf1 = (0.35)(8.573) = 3barbf2 = (0.65)(2.347) = 1.53 bar2There is no interaction between the molecules of an ideal gas mixture. In the case of an ideal mixture,

however, the molecules interact with each other, the interaction between like molecules being almost equal tothe interaction between unlike molecules.

188

7.5.3 Fugacity from the Virial Equation of State

At low to moderate pressures, the virial equation of state of a multicomponent gas mixture,truncated after the second term, is given by

Zmix = 1 +BmixP

RT(7.5-5)

where the second virial coefficient for the mixture, Bmix, is defined by Eq. (6.1-2). The use ofEq. (7.5-5) in Eq. (7.3-15) gives the fugacity coefficient of species i in a mixture as3

ln bφVi =⎛⎝2 kX

j=1

yjBij −Bmix

⎞⎠ P

RT(7.5-6)

An alternative derivation of Eq. (7.5-6) for a binary system is given in Problem 7.4.

Example 7.3 For a mixture of n-butane (1) and n-hexane (2) at 373K, determine the fu-gacity coefficients at 1, 10, and 15 bar when y1 = 0.35. The virial coefficients at 373K aregiven as

B11 = − 422 cm3/molB22 = − 103.1 cm3/mol

B12 = B21 = − 650 cm3/molSolution

For a binary system, Eq. (6.1-2) yields

Bmix = y21 B11 + 2 y1y2B12 + y22 B22 (1)

Substitution of Eq. (1) into Eq. (7.5-6) leads to

bφ1 = exp½ P

RT

hB11 + y22 (2B12 −B11 −B22)

i¾(2)

bφ2 = exp½ P

RT

hB22 + y21 (2B12 −B11 −B22)

i¾(3)

Fugacity coefficients of n-butane (1) and n-hexane (2), calculated from Eqs. (2) and (3), aretabulated below:

P ( bar) bφ1 bφ21 0.976 0.99410 0.785 0.93815 0.696 0.909

7.5.4 Fugacity from Cubic Equations of State

Since the cubic equations of state are pressure-explicit, i.e.,

P = P (T, Vmix, n1, n2, ..., nk) (7.5-7)

then it is useful to have an equation relating the fugacity of component i in a mixture to anintegral over volume rather than pressure. To convert Eq. (7.3-14) to the form amenable tothe cubic equations of state, note that the total differential of Eq. (7.5-7) is

3See Problem 7.2

189

dP =

µ∂P

∂T

¶V,nj

dT +

µ∂P

∂Vmix

¶T,nj

dV +kXi=1

µ∂P

∂ni

¶T,V,nj 6=i

dni (7.5-8)

Differentiation of Eq. (7.5-8) with respect to ni by keeping T , P , and the number of moles ofeach component other than i constant givesµ

∂P

∂Vmix

¶T,nj

µ∂Vmix

∂ni

¶T,P,nj 6=i| {z }

V i

+

µ∂P

∂ni

¶T,V,nj 6=i

= 0 (7.5-9)

or

V i = −

µ∂P

∂ni

¶T,V,nj 6=iµ

∂P

∂Vmix

¶T,nj

= −µ∂P

∂ni

¶T,V,nj 6=i

µ∂Vmix

∂P

¶T,nj

(7.5-10)

On the other hand, integration of the equation of state, P eVmix = ZmixRT , at constant tem-perature gives

dP

P+

deVmixeVmix

=dZmix

Zmix(7.5-11)

Substitution of Eqs. (7.5-10) and (7.5-11) into Eq. (7.3-14) yields

ln bφi = 1

RT

Z ∞

eVmix

⎡⎣nµ ∂P

∂ni

¶T,V,nj 6=i

− RTeVmix

⎤⎦ deVmix − lnZmix (7.5-12)

• van der Waals equation of stateSubstitution of the van der Waals equation of state into Eq. (7.5-12) gives the fugacity coeffi-cient of species i in a mixture as

ln bφVi = Bi

ZVmix −Bmix

− ln¡ZVmix −Bmix

¢−2

kPj=1

yjAij

ZVmix

(7.5-13)

Example 7.4 Show how one can obtain Eq. (7.5-13) by the substitution of the van der Waalsequation of state into Eq. (7.5-12).

Solution

Since eV = V/n, the van der Waals equation of state takes the form

P =nRT

Vmix − nbmix− n2amix

V2mix

=nRT

Vmix − nkPi=1

yibi

−n2

kPi=1

kPj=1

yiyjaij

V2mix

=nRT

Vmix −kPi=1

nibi

−

kPi=1

kPj=1

ninjaij

V2mix

(1)

190

Differentiation of Eq. (1) with respect to ni givesµ∂P

∂ni

¶T,V,nj 6=i

=RT

Vmix −kPi=1

nibi

+nRTbiµ

Vmix −kPi=1

nibi

¶2 − 1

V2mix

∂

∂ni

⎛⎝ kXi=1

kXj=1

ninjaij

⎞⎠T,V,nj 6=i

(2)For the differentiation of the third term on the right-hand side of Eq. (2), let us carry out thedifferentiation for a binary mixture of components 1 and 2, i.e.,

∂

∂n1

⎛⎝ 2Xi=1

2Xj=1

ninjaij

⎞⎠T,V,n2

=

∙∂

∂n1

¡n21 a11 + 2n1n2 a12 + n22 a22

¢¸T,V,n2

= 2 (n1 a11 + n2 a12) = 22X

j=1

nja1j (3)

It is possible to generalize Eq. (3) as

∂

∂ni

⎛⎝ kXi=1

kXj=1

ninjaij

⎞⎠T,V,nj 6=i

= 2kX

j=1

njaij (4)

Substitution of Eq. (4) into Eq. (2) gives

µ∂P

∂ni

¶T,V,nj 6=i

=RT

Vmix − nbmix+

nRTbi(Vmix − nbmix)2

−2n

kPj=1

yjaij

V2mix

(5)

Multiplication of Eq. (5) by the total number of moles, n, results in

n

µ∂P

∂ni

¶T,V,nj 6=i

=RTeVmix − bmix

+RTbi

(eVmix − bmix)2−2

kPj=1

yjaij

eV 2mix

(6)

Substitution of Eq. (6) into Eq. (7.5-12) and integration lead to

ln bφVi = ln⎛⎝ eVmixeVmix − bmix

⎞⎠+ bieVmix − bmix

−2

kPj=1

yjaij

eVmixRT− lnZV

mix (7)

Finally, the use of dimensionless parameters reduces Eq. (7) to Eq. (7.5-13).

An alternative derivation of Eq. (7.5-13) is presented in Problem 7.6.

• Redlich-Kwong equation of state

The use of the Redlich-Kwong equation of state in Eq. (7.5-12) leads to the following expressionfor the fugacity coefficient of species i in a mixture:

ln bφVi = Bi

Bmix

¡ZVmix − 1

¢− ln

¡ZVmix −Bmix

¢− Amix

Bmix

⎛⎜⎜⎜⎝2

kPj=1

yjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ lnÃ1 +

Bmix

ZVmix

!

(7.5-14)

191

For an alternative derivation of this equation, see Problem 7.7.

Example 7.5 Calculate the fugacity coefficients of carbon dioxide (1) in a mixture containing85 mol % n-butane (2) at 444K as a function of pressure. The system is represented by theRedlich-Kwong equation of state with k12 = 0.18.

Solution

From Appendix A

Component Tc (K) Pc ( bar)

Carbon dioxide 304.2 73.8n-Butane 425.0 38.0

Once the dimensionless parameters, Amix and Bmix, and the compressibility factor for thevapor mixture, ZV

mix, are calculated using the equations given in Section 6.1.2, the fugacity

coefficient of carbon dioxide in the mixture, bφVCO2,can be determined from Eq. (7.5-14). On theother hand, if the mixture behaves ideally, then the Lewis-Randall rule applies and the fugacitycoefficient of CO2 in the mixture is equal to the fugacity coefficient of pure CO2, φVCO2 . Thefugacity coefficient of pure CO2 is calculated by using Eq. (5.3-10). The results are presentedin tabular form and also plotted as shown below:

P ( bar) Amix Bmix ZVmix

bφVCO2 φVCO2

1 8.333× 10−3 1.976× 10−3 0.994 1.002 0.99910 0.083 0.020 0.934 1.021 0.98650 0.417 0.099 0.602 1.250 0.93160 0.500 0.119 0.505 1.392 0.91870 0.583 0.138 0.453 1.498 0.90580 0.667 0.158 0.445 1.519 0.89290 0.750 0.178 0.456 1.498 0.880100 0.833 0.198 0.474 1.462 0.868150 1.250 0.296 0.593 1.277 0.813200 1.667 0.395 0.721 1.151 0.767250 2.083 0.494 0.847 1.069 0.730300 2.500 0.593 0.970 1.015 0.701

0 100 200 3000.6

0.8

1

1.2

1.4

1.61.519

0.701

x

y

3001 P

192

As can be seen from the figure, the fugacity coefficient of CO2 is strongly dependent on pressurewith a maximum point. An ideal mixture assumption fails at all pressures, since the molefraction of carbon dioxide is small. For more details on this problem see Chueh and Prausnitz(1967).

• Soave-Redlich-Kwong equation of stateThe use of the Soave-Redlich-Kwong equation of state in Eq. (7.5-12) leads to the followingexpression for the fugacity coefficient of species i in a mixture:

ln bφVi = Bi

Bmix

¡ZVmix − 1

¢− ln

¡ZVmix −Bmix

¢− Amix

Bmix

⎛⎜⎜⎜⎝2

kPj=1

yjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ lnÃ1 +

Bmix

ZVmix

!

(7.5-15)In terms of dimensionless parameters, Eq. (7.5-15) is identical with the one given for theRedlich-Kwong equation of state, Eq. (7.5-14).

• Peng-Robinson equation of stateThe use of the Peng-Robinson equation of state in Eq. (7.5-12) leads to the following expressionfor the fugacity coefficient of species i in a mixture:

ln bφVi = Bi

Bmix

¡ZVmix − 1

¢− ln

¡ZVmix −Bmix

¢

− Amix√8Bmix

⎛⎜⎜⎜⎝2

kPj=1

yjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ ln"ZVmix + (1 +

√2)Bmix

ZVmix + (1−

√2)Bmix

# (7.5-16)

Example 7.6 Determine the fugacity coefficients of n-butane (1) and n-hexane (2) in a35 mol % n-butane, 65% n-hexane mixture at 373.15K and 10 bar using the Peng-Robinsonequation of state. Take k12 = − 0.006.

Solution

The properties at the critical point as well as the acentric factors were given in Example 7.2.The values of the reduced temperature and pressure are

Component Tr Pr

n-Butane 0.878 0.263n-Hexane 0.735 0.033

a) Using the equations given in Table 6.3, the dimensionless parameters are calculated as

A11 = 0.170 A12 = A21 = 0.245 B1 = 0.023A22 = 0.349 B2 = 0.035

193

The dimensionless parameters for the mixture are calculated from Eqs. (6.1-10) and (6.1-11)as

Amix = 0.279 and Bmix = 0.031

Therefore, Eq. (6.1-9) becomes

Z3mix − 0.969Z2mix + 0.215Zmix − 0.0077 = 0 ⇒ ZVmix = 0.662

Therefore, the fugacity coefficients of components are calculated from Eq. (7.5-16) asbφ1 = 0.886 bφ2 = 0.692Comment: The fugacities of components in the mixture arebf1 = (0.886)(10)(0.35) = 3.1 barbf2 = (0.692)(10)(0.65) = 4.5 barComparison of these values with the ones calculated in Example 7.2 indicates that the use ofthe Lewis-Randall rule introduces a serious error, especially for n-hexane.

7.6 CALCULATION OF COMPONENT FUGACITIES IN A LIQUID MIXTURE

Cubic equations of state can also describe P -eV -T relations of liquid mixtures. However, es-timation of mixture parameters is not a straightforward task. Depending on the componentsmaking up the mixture, different mixing rules may be used. When van der Waals mixing rules,i.e., Eqs. (6.1-6) and (6.1-10), are used, the equations given in Section 7.5.4 are also applicablefor liquid mixtures.

•van der Waals equation of state

ln bφLi = Bi

ZLmix −Bmix

− ln¡ZLmix −Bmix

¢−2

kPj=1

xjAij

ZLmix

(7.6-1)

•Redlich-Kwong equation of state

ln bφLi = Bi

Bmix

¡ZLmix − 1

¢− ln

¡ZLmix −Bmix

¢− Amix

Bmix

⎛⎜⎜⎜⎝2

kPj=1

xjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ lnÃ1 +

Bmix

ZLmix

!

(7.6-2)

• Soave-Redlich-Kwong equation of state

ln bφLi = Bi

Bmix

¡ZLmix − 1

¢− ln

¡ZLmix −Bmix

¢− Amix

Bmix

⎛⎜⎜⎜⎝2

kPj=1

xjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ lnÃ1 +

Bmix

ZLmix

!

(7.6-3)

194

• Peng-Robinson equation of state

ln bφLi = Bi

Bmix

¡ZLmix − 1

¢− ln

¡ZLmix −Bmix

¢

− Amix√8Bmix

⎛⎜⎜⎜⎝2

kPj=1

xjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ ln"ZLmix + (1 +

√2)Bmix

ZLmix + (1−

√2)Bmix

# (7.6-4)

Example 7.7 Shibata and Sandler (1989) reported the following data for the coexisting vaporand liquid phases of nitrogen (1) and cyclohexane (2) mixture:

T = 366.4K P = 138.76 bar x1 = 0.1286 y1 = 0.9721

The system is represented by the Peng-Robinson equation of state with k12 = 0.109.

a) Show that the vapor and liquid phases are indeed in equilibrium with each other.b) Estimate the densities of vapor and liquid phases.

Solution

From Appendix A

Component M ( g/mol) Tc (K) Pc ( bar) ω

Nitrogen 28.01 126.2 34.0 0.039Cyclohexane 84.16 554.0 40.7 0.212

a) The condition of equilibrium is given by

bfVi = bfLi ⇒ yi bφVi = xi bφLi (1)

Using the equations given in Table 6.3, the dimensionless parameters are calculated as

A11 = 0.107 A12 = A21 = 0.620 B1 = 0.109A22 = 4.541 B2 = 0.401

• Vapor PhaseUsing y1 = 0.9721 and y2 = 0.0279, the dimensionless parameters for the mixture are calculatedfrom Eqs. (6.1-10) and (6.1-11) as

Amix =2X

i=1

2Xj=1

yiyjAij = 0.138 and Bmix =2X

i=1

yiBi = 0.118

The compressibility factor for the vapor phase is calculated from Eq. (6.1-9) as

ZVmix = 1.019

Substitution of the numerical values into Eq. (7.5-16) leads to

bφV1 = 1.021 bφV2 = 0.493195

• Liquid PhaseUsing x1 = 0.1286 and x2 = 0.8714, the dimensionless parameters for the mixture are calculatedfrom Eqs. (6.1-10) and (6.1-11) as

Amix =2X

i=1

2Xj=1

xixjAij = 3.589 and Bmix =2X

i=1

xiBi = 0.364

The compressibility factor for the liquid phase is calculated from Eq. (6.1-9) as

ZLmix = 0.471

Substitution of the numerical values into Eq. (7.6-4) leads to

bφL1 = 7.623 bφL2 = 0.017The use of Eq. (1) gives

y1 bφV1 = (0.9721)(1.021) = 0.993 x1 bφL1 = (0.1286)(7.623) = 0.980y2 bφV2 = (0.0279)(0.493) = 0.014 x2 bφL2 = (0.8714)(0.017) = 0.015

Since yi bφVi ' xi bφLi , vapor and liquid phases are in equilibrium with each other.

b) The molar volume of the vapor phase is

eV Vmix =

ZVmixRT

P=(1.019)(83.14)(366.4)

138.76= 223.7 cm3/mol

The molecular weight of the vapor mixture is

MV = (0.9721)(28.01) + (0.0279)(84.16) = 29.577 g/mol

The density of the vapor is

ρV =29.577

223.7= 0.132 g/ cm3

The molar volume of the liquid phase is

eV Lmix =

ZLmixRT

P=(0.471)(83.14)(366.4)

138.76= 103.4 cm3/mol

The molecular weight of the liquid mixture is

ML = (0.1286)(28.01) + (0.8714)(84.16) = 76.939 g/mol

The density of the liquid is

ρL =76.939

103.4= 0.744 g/ cm3

In most applications, however, liquid mixtures cannot be expressed by equations of state. Inthat case, component fugacities in liquid mixtures are determined by using activity coefficientsas will be explained in the next chapter.

196

7.7 VARIATION OF COMPONENT FUGACITY WITH PRESSURE ANDTEMPERATURE

The variation of component fugacity with pressure by keeping temperature and compositionconstant can be easily obtained from Eq. (7.3-13) asÃ

∂ ln bfi∂P

!T,xj

=V i

RT(7.7-1)

On the other hand, differentiation of Eq. (7.3-1) with respect to temperature by keepingpressure and composition constant givesÃ

∂ ln bfi∂T

!P,xj

=1

R

∂

∂T

ÃGi

T

!P,xj

− 1

R

∂

∂T

ÃGIGMi

T

!P,xj

(7.7-2)

The Gibbs-Helmholtz equation developed for pure components, Eq. (5.7-8), is also valid forpartial molar properties, i.e.,

∂

∂T

ÃGi

T

!P,xj

= − Hi

T 2(7.7-3)

Therefore, Eq. (7.7-2) reduces toÃ∂ ln bfi∂T

!P,xj

= −Hi −H

IGMi

RT 2(7.7-4)

or Ã∂ ln bfi∂T

!P,xj

= −Hi − eHIG

i

RT 2(7.7-5)

Variation in ln bfi is expressed in the form of an exact differential as

d ln bfi = Ã∂ ln bfi∂P

!T

dP +

Ã∂ ln bfi∂T

!P

dT (7.7-6)


d ln bfi = Ã V i

RT

!dP −

ÃHi − eHIG

i

RT 2

!dT (7.7-7)

Note that Eq. (7.7-7) is analogous to Eq. (5.8-5) in which molar quantities are replaced bypartial molar quantities.

7.8 THE USE OF FUGACITY IN PHASE EQUILIBRIUM CALCULATIONS

When two phases of a pure component are in equilibrium with each other, compositions ofthe phases are the same. For example, when water vapor is in equilibrium with liquid waterat 373K under atmospheric pressure, there is no difference in compositions of phases. In thecase of multicomponent mixtures, on the other hand, compositions of the phases that are inequilibrium with each other are not the same. Estimation of the compositions of componentsin each phase is of practical interest to chemical engineers.

197

7.8.1 Vapor-Liquid Equilibrium

When pure liquid A evaporates into vapor B, evaporation takes place as long as

fLA(T,P ) >bfVA (T, P, yA) Mass transfer from liquid to vapor (7.8-1)

In writing Eq. (7.8-1) it is implicitly assumed that the solubility of B in the liquid is negligible,i.e., liquid exists as pure A. The mass transfer of A to the vapor phase stops when the systemreaches equilibrium, i.e.,

fLA(T,P ) =bfVA (T, P, y∗A) Condition of equilibrium (7.8-2)

Under these conditions vapor is saturated with A, y∗A being the corresponding mole fraction. Inother words, when vapor is saturated with A, it contains the maximum amount of A at a giventemperature and pressure. The use of Eqs. (5.4-8) and (7.3-4) for the left- and right-hand sidesof Eq. (7.8-2), respectively, results in

P vapA φVA(T, P

vapA ) exp

⎡⎣ eV LA (P − P vap

A )

RT

⎤⎦ = y∗A P bφVA(T, P, y∗A) (7.8-3)

If pressure is not very high, the following simplifying assumptions are plausible:

• Poynting correction factor is unity,• The vapor of A at T and P vap

A is ideal, i.e., φVA(T, PvapA ) = 1,

• Vapor is an ideal gas mixture, i.e., bφVA(T, P, y∗A) = 1.Under these circumstances, Eq. (7.8-3) simplifies to

P vapA = y∗A P (7.8-4)

indicating that vapor is saturated with A when its partial pressure equals vapor pressure atthe given temperature.

For example, vapor pressure of liquid water at 298K is 3.169 kPa. Air at 298K and 101.3 kPais saturated with water vapor (or relative humidity is 100%) when the partial pressure of watervapor equals 3.169 kPa. Under these conditions, air holds the maximum water with a molefraction of

y∗H2O =P vapH2O

P=3.169

101.3= 0.0313

Example 7.8 Three liters of toluene accidentally spills on the laboratory floor. Determinethe fraction of toluene vaporized over a long period of time, say 24 h, if the volume of thelaboratory is 180m3. Assume that the laboratory temperature remains constant at 298K.

Solution

From Appendix ATc = 593K Pc = 41bar ω = 0.263

From Appendix C

lnP vap = 9.3935− 3096.52

T − 53.67

198

The toluene spilled on the floor will vaporize into air as long as

fLtoluene(T, P ) >bfVtoluene(T, P, ytoluene) (1)

In writing Eq. (1) it is implicitly assumed that air has a negligible solubility in liquid toluene sothat the liquid phase is pure toluene. Under equilibrium conditions, i.e., when air is saturatedwith toluene, vaporization stops and we have

fLtoluene(298K, 1.013 bar) =bfVtoluene(298K, 1.013 bar, y∗toluene) (2)

where y∗toluene represents the mole fraction of toluene in air saturated with toluene at 298K andatmospheric pressure, i.e., 1.013 bar. Using the same assumptions in simplifying Eq. (7.8-3),Eq. (2) reduces to

P vaptoluene = P y∗toluene ⇒ y∗toluene =

P vaptoluene

P(3)

At 298K, the vapor pressure of toluene is

P vaptoluene = exp

µ9.3935− 3096.52

298− 53.67

¶= 0.038 bar

Substitution of numerical values into Eq. (3) gives

y∗toluene =0.038

1.013= 3.75× 10−2

In other words, liquid toluene vaporizes until its mole fraction in the vapor phase reaches3.75× 10−2.

Now it is necessary to determine whether the spilled toluene causes air saturated with tolueneor not. From Table 5.2, the molar volume of liquid toluene is approximately 107 cm3/mol.Therefore, the number of moles of toluene is

ntoluene =VbottleeV L

=3000

107= 28.04mol

The number of moles of air is

nair =PVlabRT

=(1.013)(180)

(8.314× 10−5)(298) = 7359.6mol

If all toluene were evaporated, then its mole would be

ytoluene =28.04

28.04 + 7359.6= 3.8× 10−3

which is less than the saturation value of 3.75× 10−2. Therefore, all toluene is evaporated.

It should be kept in mind that Eq. (7.8-3) simplifies to Eq. (7.8-4) as long as pressureis low. When pressure is moderate or high, then it is necessary to use Eq. (7.8-3) as will beshown in the next example.

Example 7.9 A cylinder contains propylene (1) and nitrogen (2) at 298K and 18 bar. Esti-mate the vapor phase composition if the solubility of nitrogen in liquid propylene is negligible.Assume that the vapor phase is represented by the virial equation of state. Critical molar vol-umes of propylene and nitrogen are 1.81×10−4m3/mol and 0.895×10−4m3/mol, respectively.The vapor pressure of propylene at 298K is 11.53 bar.

199

Solution

From Appendix A


Propylene 365.2 46 0.144Nitrogen 126.2 34 0.039

Since the solubility of nitrogen in liquid propylene is negligible, the liquid phase consists of purepropylene and the condition of equilibrium is given by Eq. (7.8-2) as

fL1 (T,P ) =bfV1 (T, P, y∗1) (1)

or, from Eq. (7.8-3),

P vap1 φV1 (T, P

vap1 ) exp

⎡⎣ eV L1 (P − P vap

1 )

RT

⎤⎦ = y∗1 P bφV1 (T, P, y∗1) (2)

Calculation of the fugacity coefficients requires virial coefficients, B11, B22, and B12, to becalculated.

• Calculation of second virial coefficients for pure components

From Eqs. (3.1-10)-(3.1-13)

B11 =(83.14)(365.2)

46

½∙0.083− 0.422

(298/365.2)1.6

¸+ (0.144)

∙0.139− 0.172

(298/365.2)4.2

¸¾= − 356 cm3/mol

B22 =(83.14)(126.2)

34

½∙0.083− 0.422

(298/126.2)1.6

¸+ (0.039)

∙0.139− 0.172

(298/126.2)4.2

¸¾= − 5.7 cm3/mol

• Calculation of second virial cross coefficient

The parameters defined by Eqs. (6.1-3) and (6.1-4) are calculated as

k∗12 = 1−8

qeVc1 eVc2(eV 1/3c1 + eV 1/3c2 )3

= 1− 8p(1.81× 10−4) (0.895× 10−4)h

(1.81× 10−4)1/3 + (0.895× 10−4)1/3i3 = 0.020

Tc12 = (1− k∗12)pTc1Tc2 = (1− 0.020)

p(365.2)(126.2) = 210.4K

ω12 =ω1 + ω22

=0.144 + 0.039

2= 0.092

Pc12 =(4)(210.4)h

(1.81× 10−4)1/3 + (0.895× 10−4)1/3i3 ∙(46)(1.81× 10−4)365.2

+(34)(0.895× 10−4)

126.2

¸= 37.98 bar

200

The parameters B(0)12 and B(1)12 are calculated from Eqs. (3.1-11) and (3.1-12), respectively, as

B(0)12 = 0.083−

0.422

(298/210.4)1.6= − 0.159

B(1)12 = 0.139−

0.172

(298/210.4)4.2= 0.099

Finally, the second virial cross coefficient B12 is calculated from Eq. (3.1-10) as

B12 =(83.14)(210.4)

37.98

h− 0.159 + (0.092) (0.099)

i= − 69.04 cm3/mol

• Calculation of fugacity coefficientsThe fugacity coefficient of pure propylene vapor is calculated from Eq. (5.3-5) as

φV1 (T, Pvap1 ) = exp

ÃB11P

vap1

RT

!= exp

∙(− 356)(11.53)(83.14)(298)

¸= 0.847

The fugacity coefficient of propylene in the vapor phase is calculated from Eq. (7.5-6) as

bφV1 (T, P, y∗1) = exp½ P

RT

hB11 + (1− y∗1)

2 (2B12 −B11 −B22)i¾

= exp

µ(18)

(83.14)(298)

n− 356 + (1− y∗1)

2h(2)(− 69.04) + 356 + 5.7

io¶= exp

h0.259 + 0.162 (1− y∗1)

2i

(3)

The molar volume of liquid propylene is calculated from the Rackett equation, Eq. (5.4-9), as

ZRA = 0.29056− (0.08775)(0.144) = 0.278

eV L1 =

(83.14)(365.2)

46(0.278)1+[1−(298/365.2)]

2/7

= 83.3 cm3/mol

Substitution of the numerical values into Eq. (2) gives

(11.53)(0.847) exp

∙(83.3)(18− 11.53)(83.14)(298)

¸= y∗1 (18) exp

h0.259 + 0.162 (1− y∗1)

2i

(4)

The solution of Eq. (4) by MATHCADR°gives y∗1 = 0.404.

Condensation is the reverse of vaporization. Let us calculate the temperature at whichcondensation starts when the vapor composition and pressure are known. Suppose that airat 298K and 101.3 kPa has a relative humidity of 80%. The partial pressure of water vapor,PH2O, in air is

pH2O = yH2OPvapH2O = (0.80)(3.169) = 2.535 kPa

Condensation, or mass transfer from the vapor to the liquid, takes place if

bfVH2O(T, P, yH2O) > fLH2O(T, P ) ⇒ pH2O > P vapH2O

In writing the above equation, it is implicitly assumed that the solubility of air in water isnegligible. In our case since pH2O < P vap

H2O, no condensation occurs. As the temperature is

201

decreased, the amount of water vapor in air does not change. The first liquid droplet appearswhen the temperature drops to T ∗, at which point4

pH2O = P vapH2O = 2.535 kPa

In other words, air becomes saturated with water vapor. The use of the Antoine equation givesthe dew point temperature as

T ∗ = 46.13 +3816.44

11.6834− ln(2.535/100) = 294.6K (21.6◦C)

Figure 7.1 illustrates the dew point temperature of an air-water system.

Pressure

Saturation Line

Tdew

2H OP

2

vapH OP

Temperature

T

Figure 7.1 Dew point temperature of an air-water system.

Example 7.10 A piston-cylinder arrangement holds a gas mixture of 20 mol % chloroform(1) and 80% nitrogen (2) at 340K and 1.013 bar. The system is cooled under constant pressure.Estimate the temperature at which condensation of chloroform takes place.

Solution

From Appendix C

lnP vap1 = 8.3530− 2696.79

T − 46.16Let T ∗ be the temperature at which the first liquid droplet of chloroform appears. If the solubilityof nitrogen in liquid chloroform is assumed negligible, i.e., the liquid phase is pure chloroform,then the condition of equilibrium is expressed as

fL1 (T∗, 1.013 bar) = bfV1 (T ∗, 1.013 bar, y1) (1)

or,

P vap1 (T ∗)φV1

hT ∗, P vap

1 (T ∗)iexp

( eV L1

£1.013− P vap

1 (T ∗)¤

RT ∗

)= (1.013) y1 bφV1 (T ∗, 1.013 bar, y1)

(2)Since the pressure is atmospheric, the following simplifying assumptions are plausible:

• Poynting correction factor is unity,• Vapor phase is an ideal gas mixture, i.e., bφV1 (T ∗, 1.013 bar, y1) ' 1• φV1

£T ∗, P vap

1 (T ∗)¤' 1

4This is known as the dew point temperature. Such calculations will be covered in detail in Chapter 9.

202

Therefore, Eq. (2) simplifies to

P vap1 (T ∗) = 1.013 y1 = (1.013)(0.2) = 0.203 bar

The temperature, T ∗, is calculated from the Antoine equation as

T ∗ = 46.16 +2696.79

8.3530− ln(0.203) = 317.3K

Comment: Since the pressure is atmospheric, simplifying assumptions lead to a straightfor-ward calculation of T ∗. In the case of high pressures, one should use a trial-and-error procedureto estimate T ∗, i.e., assume T ∗ and check whether Eq. (2) is satisfied.

7.8.2 Solid-Liquid Equilibrium

Let us add a small amount of sugar to a cup filled with tea. If we stir the tea in the cup,the sugar will disappear. This means the tea is unsaturated with sugar or, in other words, thefugacity of the solid sugar is greater than the fugacity of the sugar in the tea, i.e.,

fSsugar(T, P ) >bfLsugar(T,P, xsugar) Mass transfer from solid to liquid (7.8-5)

If we keep continue adding sugar to the tea at constant temperature, a point comes where theliquid is saturated with sugar

fSsugar(T, P ) =bfLsugar(T, P, x∗sugar) Condition of equilibrium (7.8-6)

The saturated condition, i.e., xsugar = x∗sugar, represents the maximum amount of sugar thatcan be present in the tea at the given temperature and pressure. After this point, any additionalsugar will not dissolve and will appear as a solid phase within the cup. Solid-liquid equilibriumwill be discussed in detail in Chapter 12.

7.8.3 Solid-Vapor Equilibrium

Consider sublimation of species A into vapor B. Sublimation takes place as long as

fSA(T,P ) >bfVA (T, P, yA) Mass transfer from solid to vapor (7.8-7)

In writing Eq. (7.8-7) it is implicitly assumed that the solid phase is pure A, i.e., solubility ofB in the solid phase is negligible. Sublimation stops when vapor B is saturated with A, i.e.,yA = y∗A, at the given temperature and pressure. Under these conditions

fSA(T, P ) =bfVA (T, P, y∗A) Condition of equilibrium (7.8-8)

The use of Eqs. (5.5-8) and (7.3-4) for the left- and right-hand sides, respectively, of Eq. (7.8-8)leads to

P subA exp

" eV SA (P − P sub

A )

RT

#= y∗A P bφVA(T, P, y∗A) (7.8-9)

or

y∗A =

ÃP subA

P

!E (7.8-10)

203

where the enhancement factor, E, is defined by

E =

exp

" eV SA (P − P sub

A )

RT

#bφVA (7.8-11)

As stated in Section 5.6, supercritical fluids are at a temperature and pressure greater thanor equal to the critical temperature and pressure of fluids. At high pressures, supercritical fluidspossess solvent properties that are superior to those of liquid solvents. As a result, they maybe an alternative to replace toxic and environmentally hazardous solvents used in separationand/or purification processes. Being nontoxic, inexpensive, and nonflammable, and having alow critical temperature, carbon dioxide is one of the most widely used supercritical fluids forextraction purposes. The design of extraction processes involving supercritical solvents requiresthe understanding of solid-vapor equilibrium.

Example 7.11 It is required to estimate the solubility of naphthalene (2) in carbon dioxide(1) at 118.6 bar and 328K for a supercritical extraction. Assume that the vapor phase isrepresented by the Peng-Robinson equation of state with k12 = 0.08246. For naphthalene, thesublimation pressure at 328K and the molar volume are 15.9 × 10−4 bar and 111.9 cm3/mol,respectively.

Solution

From Appendix A


Carbon dioxide 304.2 73.8 0.239Naphthalene 748.0 41.0 0.302

Note that the given temperature is less than the triple point temperature of naphthalene, whichis 353.42K from Table 5.3. Therefore, when the naphthalene content of vapor exceeds itssolubility limit, solid naphthalene is formed by desublimation. If the solid phase consists of purenaphthalene, the condition of phase equilibrium states that the fugacity of pure solid naphthaleneis equal to the fugacity of naphthalene in the gas phase. Thus, Eq. (7.8-9) is expressed as

P sub2 exp

" eV S2 (P − P sub

2 )

RT

#= y∗2 P bφV2 (T, P, y∗2) (1)

or

15.9× 10−4 exp∙(111.9)(118.6− 15.9× 10−4)

(83.14)(328)

¸| {z }

2.587×10−3

= y∗2 P bφV2 (T, P, y∗2) (2)

Since the fugacity coefficient of component 2 is dependent on the mole fraction, determination

of y∗2 from Eq. (2) requires a trial-and-error solution. Once the value of y∗2 is assumed, bφV2is determined from Eq. (7.5-16). If the assumed y∗2 value is correct, then the right-hand side(RHS) of Eq. (2) should be equal to 2.587×10−3. The calculation procedure is tabulated below :

204

y∗2 × 103 Amix Bmix ZVmix

bφV2 RHS of Eq. (3)

7.00 0.622 0.119 0.383 2.877× 10−3 2.388× 10−39.30 0.631 0.120 0.373 2.541× 10−3 2.803× 10−38.01 0.626 0.119 0.379 2.723 × 10−3 2.587× 10−3

Therefore, the mole fraction of naphthalene in CO2 at 118.6 bar and 328K is 8.01×10−3. Theexperimental value reported by McHugh and Paulaitis (1980) is 12.29× 10−3.Comment: When the mole fraction of one of the components is very small, i.e., y1 ' 1 (ory2 ' 0), the following simplifying assumptions are plausible:

Amix = A11 Bmix = B1 ZVmix = ZV

1 (3)

and the fugacity coefficient of component 2 takes the form

ln³bφV2 ´

approx=

B2B1

¡ZV1 − 1

¢− ln

¡ZV1 −B1

¢− A11√

8B1

µ2A12A11

− B2B1

¶ln

"ZV1 + (1 +

√2)B1

ZV1 + (1−

√2)B1

#(4)

which is independent of composition. Thus, Eq. (2) becomes

2.587× 10−3 = y∗2 P³bφV2 ´

approx(5)

The numerical values of the parameters in the Peng-Robinson equation of state are

A11 = 0.597 B1 = 0.116A12 = 2.372 B2 = 0.513

For pure CO2, Eq. (6.1-9) becomes

Z31 − 0.884Z21 + 0.325Z1 − 0.054 = 0 ⇒ ZV1 = 0.414

The approximate value of the fugacity coefficient of component 2 is calculated from Eq. (4) as³bφV2 ´approx

= 4.195× 10−3

Substitution of the numerical values into Eq. (5) gives

2.587× 10−3 = y∗2(118.6)(4.195× 10−3) ⇒ y∗2 = 5.2× 10−3

As far as the order of magnitude is concerned, this value is not very far from the previouslycalculated one.

In the above examples, our analysis is restricted to cases in which equilibrium takes placebetween a pure component in one phase and a multicomponent mixture in another phase.Obviously, this is not necessarily the case in general.

205

REFERENCES

Chueh, P.L. and J.M. Prausnitz, 1967, Ind. Eng. Chem. Fundam., 6 (4), 492-498.

Eggeman, T. and S. Chafin, 2005, Chem. Eng. Prog., 101 (3), 39-44.

Firoozabadi, A., 1999, Thermodynamics of Hydrocarbon Reservoirs, McGraw-Hill, New York.

Giauque, W.F. and C.J. Egan, 1937, J. Chem. Phys., 5 (1), 45-54.

Gurdial, G.S. and N.R. Foster, 1991, Ind. Eng. Chem. Res., 30 (3), 575-580.

McHugh, M. and M.E. Paulaitis, 1980, J. Chem. Eng. Data, 25, 326-329.

Mendes, R.L., B.P. Nobre, J.P. Coelho and A.F. Palavra, 1999, J. Supercritical Fluids, 16,99-106.

Shibata, S.K. and S.I. Sandler, 1989, J. Chem. Eng. Data, 34, 419-424.

ZareNezhad, B. and T. Eggeman, 2006, Cryogenics, 46, 840-845.

PROBLEMS

Problem related to Section 7.2

7.1 Derive Eq. (7.2-11).

Problems related to Section 7.5

7.2 Derive Eq. (7.5-6) as follows:

a) By definition, the partial molar compressibility factor of component i is given by

Zi =

∙∂(nZmix)

∂ni

¸T,P,nj 6=i

(1)

Show that the substitution of Eq. (7.5-5) into Eq. (1) and rearrangement give

Zi =

∙∂

∂ni

µn+

n2BmixP

nRT

¶¸T,P,nj 6=i

(2)

Carry out the differentiation to obtain

Zi = 1−BmixP

RT+

P

nRT

⎛⎝ ∂

∂ni

kXi=1

kXj=1

ninjBij

⎞⎠T,P,nj 6=i

(3)

b) To differentiate the third term on the right-hand side of Eq. (3), first consider a binarymixture of components 1 and 2, and show that⎛⎝ ∂

∂n1

2Xi=1

2Xj=1

ninjBij

⎞⎠T,P,n2

= 2 (n1B11 + n2B12) = 22X

j=1

njB1j (4)

206

Generalize Eq. (4) as ⎛⎝ ∂

∂ni

kXi=1

kXj=1

ninjBij

⎞⎠T,P,nj 6=i

= 2kX

j=1

njBij (5)

Therefore, Eq. (3) takes the form

Zi = 1 +P

RT

⎛⎝2 kXj=1

yjBij −Bmix

⎞⎠ (6)

c) Show that the substitution of Eq. (6) into Eq. (7.3-15) leads to Eq. (7.5-6).

7.3 Derive an expression to represent the overall fugacity coefficient of the mixture, φmix, asfollows:

a) Start with

eGmix =kXi=1

xiGi (1)

and show that eGmix =kXi=1

xiλi +RTkXi=1

xi ln bfi (2)

b) The fugacity of the mixture, fmix, is defined byeGmix = λmix +RT ln fmix (3)

with the following constraint

φmix = fmix/P → 1 as P → 0 (4)

Equate Eqs. (2) and (3) to get

λmix +RT ln fmix =kXi=1

xiλi +RTkXi=1

xi ln bfi (5)

Since Eq. (5) is valid at all pressures, evaluate Eq. (5) as P → 0 to conclude that

λmix =kXi=1

xiλi +kXi=1

xi lnxi (6)

c) Show that the use of Eq. (6) in Eq. (5) leads to

lnφmix =kXi=1

xi ln bφi (7)

or

φmix =kYi=1

bφxii (8)

207

Note that the form of Eq. (3) is similar to the one given for a pure component, Eq. (5.2-6).Therefore, the pure component fugacity coefficient equations given in Chapter 5 can also beused to express the overall fugacity coefficient of mixtures.

7.4 Equation (7) of Problem 7.3 suggests that the term ln bφi can be interpreted as the partialmolar property of lnφmix, i.e.,

ln bφi = ∙ ∂

∂ni(n lnφmix)

¸T,P,nj 6=i

(1)

Therefore, for a binary mixture, show that the application of Eqs. (6.1-24) and (6.1-25) gives

ln bφ1 = lnφmix − x2d lnφmix

dx2(2)

ln bφ2 = lnφmix − x1d lnφmix

dx1(3)

a) Note that Eq. (5.3-5) can be extended to mixtures as

lnφmix =BmixP

RT(4)

b) Use Eq. (4) in Eqs. (2) and (3) to obtain

ln bφ1 = P

RT

hB11 + y22 (2B12 −B11 −B22)

i(5)

ln bφ2 = P

RT

hB22 + y21 (2B12 −B11 −B22)

i(6)

7.5 In the case of a pressure-explicit equation of state, the use of Eq. (1) in Problem 7.4, i.e.,

ln bφi = ∙ ∂

∂ni(n lnφmix)

¸T,P,nj 6=i

(1)

is not practical in the calculation of component fugacities. An alternative expression can bedeveloped as follows:

a) In generaln lnφmix = f(T, P, Vmix, n1, n2, ..., nk) (2)

In Eq. (1), temperature, pressure, and the number of moles of each component other than iare kept constant during differentiation. Therefore, Eq. (2) can be expressed as

n lnφmix = g(Vmix, ni) (3)

b) First express the total differential of Eq. (3) as

d(n lnφmix) =

∙∂(n lnφmix)

∂Vmix

¸T,P,nj

dVmix +

∙∂(n lnφmix)

∂ni

¸T,P,Vmix,nj 6=i

dni (4)

and then differentiate Eq. (4) with respect to ni, by keeping T , P , and nj 6=i constant, to obtain∙∂(n lnφmix)

∂ni

¸T,P,nj 6=i

=

∙∂(n lnφmix)

∂Vmix

¸T,P,nj

µ∂Vmix

∂ni

¶T,P,nj 6=i| {z }

V i

+

∙∂(n lnφmix)

∂ni

¸T,P,Vmix,nj 6=i

(5)

208

c) Note that Eq. (5.3-1) can be extended to mixtures as

lnφmix =

Z P

0

Ã eVmix

RT− 1

P

!dP at constant T and nj (6)

Hence, conclude that ∙∂(n lnφmix)

∂Vmix

¸T,P,nj

= 0 (7)

d) Combine Eqs. (1), (5), and (7) to obtain

ln bφi = ∙∂(n lnφmix)

∂ni

¸T,P,Vmix,nj 6=i

(8)

7.6 From Eq. (5.3-9), the overall fugacity coefficient of the mixture obeying the van derWaals equation of state is given by

lnφmix = Zmix − 1− ln(Zmix −Bmix)−Amix

Zmix(1)

a) Show that the substitution of Eq. (1) into Eq. (8) of Problem 7.5 and rearrangement give

ln bφi = ∙∂(nZmix)

∂ni

¸T,P,Vmix,nj 6=i

− 1−∙

∂

∂ni

µn2Amix

nZmix

¶¸T,P,Vmix,nj 6=i

− ln(Zmix −Bmix)

− n

Zmix −Bmix

(∙∂

∂ni

µnZmix

n

¶¸T,P,Vmix,nj 6=i

−∙

∂

∂ni

µnBmix

n

¶¸T,P,Vmix,nj 6=i

)(2)

b) Show that ∙∂(nZmix)

∂ni

¸T,P,Vmix,nj 6=i

= 0 (3)

∙∂

∂ni

µn2Amix

nZmix

¶¸T,P,Vmix,nj 6=i

=

2kP

j=1xjAij

Zmix(4)

∙∂

∂ni

µnZmix

n

¶¸T,P,Vmix,nj 6=i

= − Zmix

n(5)

∙∂

∂ni

µnBmix

n

¶¸T,P,Vmix,nj 6=i

=Bi −Bmix

n(6)

Substitute Eqs. (3) - (6) into Eq. (2) to obtain Eq. (7.5-13).

7.7 From Eq. (5.3-9), the overall fugacity coefficient of the mixture obeying the Redlich-Kwong equation of state is given by

lnφmix = Zmix − 1− ln(Zmix −Bmix)−Amix

Bmixln

µ1 +

Bmix

Zmix

¶(1)

209

a) Show that the substitution of Eq. (1) into Eq. (8) of Problem 7.5 and rearrangement give

ln bφi = ∙∂(nZmix)

∂ni

¸T,P,Vmix,nj 6=i

− 1− ln(Zmix −Bmix)

− n

Zmix −Bmix

(∙∂

∂ni

µnZmix

n

¶¸T,P,Vmix,nj 6=i

−∙

∂

∂ni

µnBmix

n

¶¸T,P,Vmix,nj 6=i

)

−lnµ1 +

Bmix

Zmix

¶ ∙∂

∂ni

µn2Amix

nBmix

¶¸T,P,Vmix,nj 6=i

− nAmixZmix

Bmix(Zmix +Bmix)

∙∂

∂ni

µnBmix

nZmix

¶¸T,P,Vmix,nj 6=i

(2)

b) Show that

∙∂

∂ni

µn2Amix

nBmix

¶¸T,P,Vmix,nj 6=i

=

2kP

j=1xjAij

Bmix− AmixBi

B2mix

(3)

∙∂

∂ni

µnBmix

nZmix

¶¸T,P,Vmix,nj 6=i

=Bi

nZmix(4)

Substitute Eqs. (3) and (4) together with Eqs. (3), (5), and (6) of Problem 7.6 into Eq. (2) toobtain

ln bφi = Bi

Zmix −Bmix− Amix

Bmix

µBi

Zmix +Bmix

¶− ln (Zmix −Bmix)

− Amix

Bmix

⎛⎜⎜⎜⎝2

kPj=1

xjAij

Amix− Bi

Bmix

⎞⎟⎟⎟⎠ lnµ1 +

Bmix

Zmix

¶(5)

c) Combining the first two terms on the right-hand side of Eq. (5) and making use of theRedlich-Kwong equation of state, i.e.,

Z3mix − Z2mix + (Amix −Bmix −B2mix)Zmix −AmixBmix = 0 (6)

show that Eq. (5) reduces to Eq. (7.5-14).

7.8 Twenty-eight moles of methane (1), 16moles of ethane (2), and 36moles of carbon dioxide(3) are mixed in a rigid tank at 320K and 10 bar. Calculate the fugacities of each componentin the gas mixture using

a) The Lewis-Randall rule if pure gases obey the Peng-Robinson equation of state,b) The virial equation of state with

B11 = − 35.17 cm3/mol B22 = − 159.42 cm3/mol B33 = − 104.54 cm3/molB12 = − 76.71 cm3/mol B13 = − 54.02 cm3/mol B23 = − 105.65 cm3/mol

(Answer: a) bf1 = 3.440 bar, bf2 = 1.867 bar, bf3 = 4.305 bar b) bf1 = 3.467 bar, bf2 = 1.902 bar,bf3 = 4.341 bar)7.9 A ternary mixture consisting of 50 mol % methane (1), 30% ethane (2), and 20% propane(3) is at 380K and 200 bar. The mixture is represented by the van der Waals equation of state.

210

a) Estimate the fugacities of each component in the mixture.b) Repeat the calculations for an ideal mixture.(Answer: a) bf1 = 92.132 bar, bf2 = 31.860 bar, bf3 = 14.875 bar b) bf1 = 86.362 bar,bf2 = 31.209 bar, bf3 = 13.572 bar)7.10 A gaseous mixture consisting of 15 mol % methane (1), 25% ethylene (2), 20% ethane(3), and 40% propane (4) is at 400K and 150 bar. Calculate the component fugacities in themixture using the Peng-Robinson equation of state with

k12 = 0.022 k13 = − 0.003 k14 = 0.016 k23 = 0.010 k24 = 0 k34 = 0.001

(Answer: bf1 = 25.935 bar, bf2 = 27.508 bar, bf3 = 19.058 bar, bf4 = 24.077 bar)7.11 A rigid tank is divided into two parts by a rigid semi-permeable membrane throughwhich carbon dioxide can pass. One side contains a gaseous mixture consisting of 70 mol %carbon dioxide (1) and 30% methane (2) at 323K and 60 bar. The other side contains purecarbon dioxide at 323K. Estimate the pressure of pure carbon dioxide so as to stop diffusionof carbon dioxide through the membrane if the gas phase obeys the Peng-Robinson equationof state with k12 = 0.09.

(Answer: 38.3 bar)

7.12 This is the continuation of Problem 4.3 for a multicomponent system.

a) For a multicomponent system, show that Eq. (3) of Problem 4.3 takes the form

dG = V dP − S dT +mg dz +kXi=1

(Gi +Migz) dni (1)

b) Under equilibrium conditions, i.e., dG = 0, conclude that

dT = 0 ⇒ T is uniform (2)

V dP +mg dz = 0 ⇒ dP

dz= − ρg (3)

Gi +Migz = 0 (4)

Keep in mind that the distance z is measured in the direction opposite to gravity.

c) Differentiate Eq. (4) by keeping temperature constant to obtain

dGi = RT d ln bfi = −Mig dz (5)

which is known as the Gibbs sedimentation expression (Firoozabadi, 1999).

d) Show that the integration of Eq. (5) gives

ln bfi = −Migz

RT+ C (6)

where C is an integration constant. If the temperature is uniform in a well of depth H, showthat Eq. (6) takes the form bfi,bottombfi,top = exp

µMigH

RT

¶(7)

211

e) The composition at the top of a natural gas well of depth 3 km is 90 mol % methane (1)and 10% ethane (2). The temperature within the well is uniform at 310K and the pressure atthe top of the well is 120 bar. It is required to estimate the composition and the pressure atthe bottom of the well.

i) Assuming natural gas to be an ideal gas mixture, show that Eq. (7) reduces toµy1

y2

¶bottom

=

µy1

y2

¶top

exp

∙(M1 −M2)gH

RT

¸(8)

First use Eq. (8) with y1 + y2 = 1 and estimate the composition at the bottom of the well.Then calculate the pressure at the bottom of the well.

ii) Assuming natural gas to be an ideal mixture, show that Eq. (7) takes the form

y1,bottom = y1,top(f1P )top

(f1P )bottomexp

µM1gH

RT

¶(9)

y2,bottom = y2,top(f2P )top

(f2P )bottomexp

µM2gH

RT

¶(10)

First assume the pressure at the bottom of the well and calculate pure component fugacitiesusing the Peng-Robinson equation of state. Then determine the mole fractions using Eqs. (9)and (10). Repeat this procedure until y1 + y2 = 1.

(Answer: e) y1 = 0.8847 P = 146.5 bar; y1 = 0.85 P = 139.7 bar)

Problems related to Section 7.8

7.13 Since the sublimation pressure of solids is very small, its determination by experimentaltechniques is often difficult. One way of determining the sublimation pressure is to use thesolubility data. Mendes et al. (1999) reported that the solubility of β-carotene (2) in carbondioxide (1) at 313.15K and 201 bar is y2 = 13× 10−8. Using the following data for β-carotene

Tc = 1177K Pc = 12.4 bar ω = 1.04 eV S = 536.5 cm3/mol

and the Peng-Robinson equation of state with k12 = 0.1013, estimate the sublimation pressureof β-carotene at 313.15K. Mendes et al. reported the sublimation pressure as 1.72×10−19 bar.(Answer: 1.68× 10−19 bar)

7.14 Determine the solubility of salicylic acid (2) in supercritical carbon dioxide (1) at 313Kand 192.5 bar. Assume that the vapor phase is represented by the Peng-Robinson equation ofstate with k12 = 0.233. For salicyclic acid

Tc = 864.79K Pc = 50.24 bar ω = 0.785

At 313K, the sublimation pressure and the molar volume of solid salicyclic acid are 2.86 ×10−6 bar and 95.70 cm3/mol, respectively. The experimental value reported by Gurdial andFoster (1991) is y2 = 3.64× 10−4.(Answer: 3.50× 10−4)

7.15 A natural gas mainly consists of methane. Cryogenic processes are used in natural gasplants for the separation of methane from heavier alkanes. The presence of carbon dioxide, dueto its relatively high triple point, in a methane stream causes freezing, leading to clogging of

212

pipes and other operational problems. Consider a mixture of methane (1) and carbon dioxide(2) in which the mole fraction of CO2 is 0.1. If the pressure is 50 bar, estimate the temperatureat which carbon dioxide freezes out using the Peng-Robinson equation of state. The followingdata are provided by ZareNezhad and Eggeman (2006):

ρS2 = 1562 kg/m3

P sub2 = 9.44× 106 exp

µ− 3108.2

T

¶where P sub

2 is in bar and T is in K. The interaction parameter is given as a function oftemperature in the form

k12 = 0.0998 +5.4835

T− 36.134

T 2

Repeat the calculations when y2 = 0.01. For more details on the problem, see Eggeman andChafin (2005).

(Answer: 189.8K, 152K)

7.16 Consider vapor-liquid equilibrium between nitrogen (1) and water (2) in a rigid containerat low temperature. The vapor phase nonideality is represented by the virial equation of state.Note that the following assumptions are plausible for this problem:

• Solubility of nitrogen in liquid water is negligible, i.e., x2 ' 1.• Since the temperature is low, then the vapor pressure of water is also low, i.e., y2 ¿ y1.

Under these circumstances, show that the equality of water fugacities in the liquid and vaporphases leads to

y2 =P vap2

Pexp

⎧⎪⎨⎪⎩h(eV L2 +B11)− 2B12

iP

RT

⎫⎪⎬⎪⎭

213

Chapter 7

Documents

Transcript of Chapter 7