Chapter 6

Chapter 6

Thermodynamics of Mixtures

A mixture is a gaseous, liquid, or solid phase containing more than one component. Thechemical process industries mainly deal with the separation of mixtures, and their propertiesare needed in design calculations. Mixture properties, however, cannot be determined fromthe weighted average of the properties of the pure components comprising the mixture. Thepurpose of this chapter is to show how to calculate the mixture properties.

6.1 EQUATIONS OF STATE FOR MIXTURES

6.1.1 Virial Equation of State

At low to moderate pressures (up to ∼ 15 bar), the virial equation of state, truncated after thesecond term, for a gas mixture consisting of k species is given by

Zmix = 1 +BmixP

RT(6.1-1)

where the second virial coefficient for the mixture, Bmix, is defined as

Bmix =kXi=1

kXj=1

yiyjBij (6.1-2)

in which yi represents the mole fraction of species i. When i = j, the terms Bii and Bjj

represent second virial coefficients corresponding to the pure components i and j, respectively.When i 6= j, second virial cross coefficients are symmetric, i.e., Bij = Bji. In engineeringcalculations involving normal fluids, second virial cross coefficients, Bij , can be estimated fromEqs. (3.1-10)-(3.1-12) with the following modifications of Tc, ω, and Pc:

Tcij = (1− k∗ij)qTciTcj ωij =

ωi + ωj2

Pcij =4Tcij³eV 1/3ci + eV 1/3cj

´3ÃPcieVci

Tci+

PcjeVcj

Tcj

!(6.1-3)

where eVci and eVcj are the critical molar volumes of components i and j, respectively1. Thevalues of the binary interaction parameter2, k∗ij , have been reported by Chueh and Prausnitz(1967a), Tsonopoulos (1979), Nishiumi et al. (1988), and Meng et al. (2007). For binary

1 In the absence of data, critical molar volume can be estimated from the following formula:

eVc = RTcPc

(0.290− 0.085ω)

2Keep in mind that interaction parameters have no theoretical basis; they are simply adjustable parametersto fit experimental data.

127

mixtures of chemically similar molecules, i.e., similar size, shape, and chemical nature, k∗ij maybe considered zero. Chueh and Prausnitz (1967b) proposed the following equation to estimatek∗ij :

k∗ij = 1−8qeVci eVcj

(eV 1/3ci + eV 1/3cj )3(6.1-4)

In the absence of reliable data, the second virial cross coefficient, Bij , may be assumed tobe the arithmetic mean of Bii and Bjj , i.e., Bij ' (Bii +Bjj)/2.

Example 6.1 A rigid tank contains 3 moles of ethane (1) at 373K and 4 bar. Estimatethe pressure in the tank when 5 moles of n-butane (2) is added isothermally to the tank. Themixture is described by the virial equation of state and the following critical properties areprovided:

Component eVc ( cm3/mol)Ethane 145.5n-Butane 255.0

Solution

From Appendix A

Component Tc (K) Pc ( bar) ω

Ethane 305.3 49 0.099n-Butane 425.0 38 0.199

The final pressure in the tank can be estimated from Eq. (6.1-1), i.e.,

(Zmix)f =Pf (eVmix)f

RT= 1 +

(Bmix)fPf

RT=⇒ Pf =

RT

(eVmix)f − (Bmix)f(1)

where the subscript "f" represents the final state. The term Bmix is calculated from Eq. (6.1-2)as

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

The total tank volume, V , can be calculated from the initial conditions, i.e., when the tank isinitially filled with ethane. The use of the virial equation of state gives

Zinitial =PinitialV

ninitialRT= 1 +

B11Pinitial

RT=⇒ V = ninitial

µRT

Pinitial+B11

¶(3)

Thus, the molar volume of the mixture at the final state is

(eVmix)f =V

nf(4)

The second virial coefficients for pure components, i.e., B11 and B22, are calculated from Eqs.(3.1-10)-(3.1-12). The results are given in the table below:

Component Tr B(0) B(1) Bii ( cm3/mol)

Ethane 1.222 − 0.223 0.065 − 112.3n-Butane 0.878 − 0.437 − 0.159 − 435.7

128

To calculate the second virial cross coefficient, B12, it is first necessary to calculate the para-meters defined by Eqs. (6.1-3)-(6.1-4):

k∗12 = 1−8p(145.5) (255)¡

145.51/3 + 2551/3¢3 = 0.013

Tc12 = (1− 0.013)p(305.3)(425) = 355.5K

ω12 =0.099 + 0.199

2= 0.149

Pc12 =(4)(355.5)¡

145.51/3 + 2551/3¢3 ∙(49)(145.5)305.3

+(38)(255)

425

¸= 42.04 bar

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

(373/355.5)1.6= − 0.308

B(1)12 = 0.139−

0.172

(373/355.5)4.2= − 1.57× 10−3

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

B12 =(83.14)(355.5)

42.04

h− 0.308 + (0.149)

¡− 1.57× 10−3

¢i= − 216.7 cm3/mol

The term Bmix is calculated from Eq. (2) as

Bmix =

µ3

8

¶2(− 112.3) + 2

µ3

8

¶µ5

8

¶(− 216.7) +

µ5

8

¶2(− 435.7) = − 287.6 cm3/mol

The tank volume is calculated from Eq. (3) as

V = (3)

∙(83.14)(373)

4− 112.3

¸= 22, 922 cm3

The use of Eq. (4) gives

(eVmix)f =22, 922

8= 2865.3 cm3/mol

Substitution of the numerical values into Eq. (1) results in

Pf =(83.14)(373)

2865.3− 287.6 = 12bar

Example 6.2 A gas mixture containing 90 mol % methane (1) and 10% ethane (2) is ir-reversibly compressed from 1 bar and 310K to 30 bar and 350K in an adiabatic compressor.If the gas mixture is represented by the virial equation of state, estimate the work required permole of gas mixture. The virial coefficients are expressed as a function of temperature in theform

Bij = αij −βijT−

γij

T 2

in which the parameters are given by Estela-Uribe et al. (2003) as

129

α11 = 42.19 α12 = 100.19 α22 = 89.75

β11 = 17.13× 103 β12 = 52.82× 103 β22 = 49.80× 103

γ11 = 24.57× 105 γ12 = 12.63× 105 γ22 = 96.19× 105

with B is in cm3/mol and T is in K.

Solution

From Appendix B

eC∗P1 = 36.155− 0.511× 10−1 T + 2.215× 10−4 T 2 − 1.824× 10−7 T 3 + 4.899× 10−11 T 4eC∗P2 = 33.313− 0.111× 10−1 T + 3.566× 10−4 T 2 − 3.762× 10−7 T 3 + 11.983× 10−11 T 4System: Compressor

For a steady-state flow system, the first law gives

fWs = ∆ eH (1)

The change in enthalpy can be determined from Eq. (3.3-13), i.e.,

∆ eH =

Z 0

P1

"eVmix − T

Ã∂ eVmix

∂T

!P

#T1

dP +

Z T2

T1

eC∗PmixdT +

Z P2

0

"eVmix − T

Ã∂ eVmix

∂T

!P

#T2

dP

(2)The use of the virial equation of state, Eq. (6.1-1), leads to

Zmix =P eVmix

RT= 1 +

BmixP

RT⇒ eVmix =

RT

P+Bmix (3)

The partial derivative appearing in Eq. (2) isÃ∂ eVmix

∂T

!P

=R

P+

dBmix

dT(4)

and the integrand in Eq. (2) becomes

eVmix − T

Ã∂ eVmix

∂T

!P

= Bmix − TdBmix

dT(5)

Therefore, it is first necessary to express Bmix as a function of temperature. From Eq. (6.1-2)

Bmix = y21 B11 + 2 y1y2B12 + y22 B22 (6)

Expression of virial coefficients in terms of temperature leads to

Bmix = ϕ1 −ϕ2T− ϕ3

T 2(7)

whereϕ1 = y21 α11 + 2 y1y2 α12 + y22 α22 (8)

ϕ2 = y21 β11 + 2 y1y2 β12 + y22 β22 (9)

130

ϕ3 = y21 γ11 + 2 y1y2 γ12 + y22 γ22 (10)

The use of Eq. (7) in Eq. (5) gives

eVmix − T

Ã∂ eVmix

∂T

!P

= ϕ1 −2ϕ2T− 3ϕ3

T 2(11)

Substitution of Eq. (11) into Eq. (2) and integration give

∆ eH = ϕ1(P2 − P1)− 2ϕ2µP2T2− P1

T1

¶− 3ϕ3

ÃP2

T22

− P1

T21

!+

Z T2

T1

eC∗PmixdT (12)

The numerical values of ϕ1, ϕ2, and ϕ3 are calculated from Eqs. (8)-(10) as

ϕ1 = 53.106 cm3/mol ϕ2 = 2.388× 104 cm3.K/mol ϕ3 = 2.314× 106 cm3.K2/mol

The molar heat capacity of an ideal gas mixture is the mole fraction-weighted sum of the puremolar heat capacities, i.e.,eC∗Pmix

= y1 eC∗P1 + y2 eC∗P2= 35.871− 0.471× 10−1 T + 2.350× 10−4 T 2 − 2.018× 10−7 T 3 + 5.607× 10−11 T 4(13)

Thus, the numerical value of the integral in Eq. (12) isZ 350

310

eC∗PmixdT = 1574 J/mol

Substitution of the numerical values into Eq. (12) yields

∆ eH = 53.106× 10−6(30− 1)× 105 − 2(2.388× 10−2)µ30

350− 1

310

¶× 105

− 3(2.314)µ30

3502− 1

3102

¶× 105 + 1574 = 1171.3 J/mol

6.1.2 Cubic Equations of State

The cubic equations of state given in Table 3.1 are also valid for gas and liquid mixtures. Table6.1 summarizes these equations for mixtures. The practical question to be asked at this stage ishow to evaluate the mixture parameters, i.e., amix and bmix, once pure component parametersare calculated.

According to the van der Waals (or quadratic) mixing rules3, the mixture parameters aregiven by the following composition-dependent expressions:

amix =kXi=1

kXj=1

xixjaij (6.1-5)

bmix =kXi=1

kXj=1

xixj

µbi + bj2

¶=

kXi=1

xibi (6.1-6)

3Equations (6.1-5) and (6.1-6) are not the only equations used to calculate the parameters for the mixtures.Ghosh (1999) and Twu et al. (2002) summarized various mixing rules used in the literature. See also Orbeyand Sandler (1998).

131

Table 6.1 Cubic equations of state for mixtures4.

Eq. P aii bi

VDWRTeVmix − bmix

−amixeV 2mix

27

64

ÃR2T 2ciPci

!1

8

µRTciPci

¶

RKRTeVmix − bmix

−amixeVmix(

eVmix + bmix)√T

0.42748

ÃR2T 2.5ci

Pci

!0.08664

µRTciPci

¶

SRKRTeVmix − bmix

−amixeVmix(eVmix + bmix)

0.42748

ÃR2T 2ciPci

!αi 0.08664

µRTciPci

¶

PRRTeVmix − bmix

−amixeVmix(

eVmix + bmix) + bmix(eVmix − bmix)0.45724

ÃR2T 2ciPci

!αi 0.07780

µRTciPci

¶

where xi represents the mole fraction of species i. In Eqs. (6.1-5) and (6.1-6), the terms aii (orajj) and bi are the pure component parameters. When i 6= j, the term aij is called the unlikeinteraction parameter and is related to the pure component parameters as

aij =

⎧⎨⎩√aii ajj vDW

(1− kij)√aii ajj RK, SRK, PR

(6.1-7)

where kij is the adjustable binary interaction parameter with a symmetric property, i.e.,kij = kji.

The parameter αi in the Soave-Redlich-Kwong (SRK) and Peng-Robinson (PR) equationsof state is defined by

αi =

⎧⎪⎨⎪⎩£1 +

¡0.480 + 1.574ωi − 0.176ω2i

¢ ¡1−

pTri¢¤2

SRK

£1 +

¡0.37464 + 1.54226ωi − 0.26992ω2i

¢ ¡1−

pTri¢¤2

PR

(6.1-8)

On the other hand, the compressibility factor for the mixture, Zmix, is obtained from

Z3mix + pZ2mix + q Zmix + r = 0 (6.1-9)

where the coefficients p, q, and r are given in Table 6.2. The dimensionless parameters for themixture are defined by

Amix =kXi=1

kXj=1

xixjAij (6.1-10)

Bmix =kXi=1

xiBi (6.1-11)

in which the terms Aij and Bi are defined in Table 6.3.

4VDW - van der Waals, RK - Redlich-Kwong, SRK - Soave-Redlich-Kwong, PR - Peng-Robinson

132

Table 6.2 Parameters in Eq. (6.1-9).

Equation p q r

van der Waals − 1−Bmix Amix −AmixBmix

Redlich-Kwong − 1 Amix −Bmix −B2mix −AmixBmix

Soave-Redlich-Kwong − 1 Amix −Bmix −B2mix −AmixBmix

Peng-Robinson − 1 +Bmix Amix − 2Bmix − 3B2mix −AmixBmix +B2mix +B3mix

Table 6.3 Dimensionless parameters Aij and Bi.

Equation Aii Aij Bi

van der Waals27

64

ÃPr

T2r

!i

pAiiAjj

1

8

µPr

Tr

¶i

Redlich-Kwong 0.42748

ÃPr

T2.5r

!i

(1− kij)pAiiAjj 0.08664

µPr

Tr

¶i

Soave-Redlich-Kwong 0.42748

ÃPr

T2r

!i

αi (1− kij)pAiiAjj 0.08664

µPr

Tr

¶i

Peng-Robinson 0.45724

ÃPr

T2r

!i

αi (1− kij)pAiiAjj 0.07780

µPr

Tr

¶i

Example 6.3 Calculate the molar volume of a gas mixture consisting of 50 mol % propane(1), 20% n-butane (2), and 30% n-pentane (3) at 393K and 40 bar using

a) van der Waals equation of state,b) Peng-Robinson equation of state (k12 = 0.003, k13 = 0.027, k23 = 0.017).

Solution

From Appendix A


Propane 369.9 42.5 0.153n-Butane 425.0 38.0 0.199n-Pentane 469.8 33.6 0.251

The molar volume of the mixture is calculated from

eVmix =ZVmixRT

P

Therefore, it is first necessary to estimate the compressibility factor for the mixture, ZVmix. The

reduced temperature and pressure values are

133

Component Tr Pr

Propane 1.062 0.941n-Butane 0.925 1.053n-Pentane 0.837 1.190

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

A11 = 0.352 A12 = A21 = 0.427 B1 = 0.111A22 = 0.519 A13 = A31 = 0.502 B2 = 0.142A33 = 0.717 A23 = A32 = 0.610 B3 = 0.178

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

Amix = 0.482 and Bmix = 0.137

Therefore, Eq. (6.1-9) becomes

Z3mix − 1.137Z2mix + 0.482Zmix − 0.066 = 0 ⇒ ZVmix = 0.26

The molar volume of the mixture is

eVmix =ZVmixRT

P=(0.26)(83.14)(393)

40= 212.4 cm3/mol

b) The use of Eq. (6.1-8) gives the parameter α as

Component α

Propane 0.963n-Butane 1.052n-Pentane 1.131

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

A11 = 0.367 A12 = A21 = 0.465 B1 = 0.069A22 = 0.592 A13 = A31 = 0.553 B2 = 0.089A33 = 0.878 A23 = A32 = 0.709 B3 = 0.111


Amix = 0.538 and Bmix = 0.086


Z3mix − 0.914Z2mix + 0.344Zmix − 0.038 = 0 ⇒ ZVmix = 0.179

The molar volume of the mixture is

eVmix =ZVmixRT

P=(0.179)(83.14)(393)

40= 146.2 cm3/mol

134

6.1.2.1 Calculation of the change in internal energy

The equations developed in Section 3.2 can be extended to mixtures with the following modi-fications.

• van der Waals equation of stateFor mixtures, Eq. (3.2-16) becomes

∆eU = R

µAmix1T1

Zmix1

−Amix2T2

Zmix2

¶+

Z T2

T1

eC∗VmixdT (6.1-12)

where eC∗Vmix=

kXi=1

xi eC∗Vi (6.1-13)

• Redlich-Kwong equation of state

For mixtures, Eq. (3.2-17) becomes

∆eU = 3R

2

∙Amix1T1

Bmix1

ln

µ1 +

Bmix1

Zmix1

¶−

Amix2T2

Bmix2

ln

µ1 +

Bmix2

Zmix2

¶¸+

Z T2

T1


• Peng-Robinson equation of state


∆eU = T2 (damix/dT )T2 − amix2√8 bmix

ln

"Zmix2 +

¡1 +√2¢Bmix2

Zmix2 +¡1−√2¢Bmix2

#

−T1 (damix/dT )T1 − amix2√

8 bmix

ln

"Zmix1 +

¡1 +√2¢Bmix1


#+

T2ZT1


wheredamix

dT= − 1

2T

kXi=1

kXj=1

xixjaij(Γi + Γj) (6.1-16)

in which

Γi =¡0.37464 + 1.54226ωi − 0.26992ω2i

¢rTriαi

(6.1-17)

Substitution of Eq. (6.1-16) into Eq. (6.1-15) gives

∆eU = RT2Ωmix2√8Bmix2

ln

"Zmix2 +

¡1 +√2¢Bmix2


#−

RT1Ωmix1√8Bmix1

ln

"Zmix1 +

¡1 +√2¢Bmix1


#

+

Z T2

T1

eC∗VmixdT

(6.1-18)where

Ωmix = −1

2

kXi=1

kXj=1

xixjAij(Γi + Γj)−Amix (6.1-19)

135

6.1.2.2 Calculation of the change in enthalpy


• van der Waals equation of state


∆ eH = RT2

µZmix2 − 1−

Amix2

Zmix2

¶−RT1

µZmix1 − 1−

Amix1

Zmix1

¶+

Z T2

T1

eC∗PmixdT (6.1-20)

where eC∗Pmix=

kXi=1

xi eC∗Pi (6.1-21)



∆ eH = RT2

∙Zmix2 − 1−

3

2

Amix2

Bmix2

ln

µ1 +

Bmix2

Zmix2

¶¸

−RT1

∙Zmix1 − 1−

3

2

Amix1

Bmix1

ln

µ1 +

Bmix1

Zmix1

¶¸+

Z T2

T1

eC∗PmixdT

(6.1-22)

• Peng-Robinson equation of stateFor mixtures, Eq. (3.3-23) becomes

∆ eH = RT2

(Zmix2 − 1 +

Ωmix2√8Bmix2

ln

"Zmix2 +

¡1 +√2¢Bmix2


#)

−RT1

(Zmix1 − 1 +

Ωmix1√8Bmix1

ln

"Zmix1 +

¡1 +√2¢Bmix1


#)+

Z T2

T1

eC∗PmixdT

(6.1-23)where Ωmix is defined by Eq. (6.1-19).

6.1.2.3 Calculation of the change in entropy


• van der Waals equation of stateFor mixtures, Eq. (3.4-13) becomes

∆eS = R ln

µZmix2 −Bmix2

Zmix1 −Bmix1

¶+

Z T2

T1

eC∗Pmix

TdT −R ln

µP2P1

¶(6.1-24)

136

• Redlich-Kwong equation of stateFor mixtures, Eq. (3.4-14) becomes

∆eS = R ln

µZmix2 −Bmix2

Zmix1 −Bmix1

¶− R

2

∙Amix2

Bmix2

ln

µ1 +

Bmix2

Zmix2

¶−

Amix1

Bmix1

ln

µ1 +

Bmix1

Zmix1

¶¸

+

Z T2

T1

eC∗Pmix

TdT −R ln

µP2P1

¶(6.1-25)

• Peng-Robinson equation of stateFor mixtures, Eq. (3.4-15) becomes

∆eS = R ln

µZmix2 −Bmix2

Zmix1 −Bmix1

¶+

RΘmix2√8Bmix2

ln

"Zmix2 +

¡1 +√2¢Bmix2


#

−RΘmix1√8Bmix1

ln

"Zmix1 +

¡1 +√2¢Bmix1


#+

Z T2

T1

eC∗Pmix

TdT −R ln

µP2P1

¶(6.1-26)

where Θmix is defined byΘmix = Ωmix +Amix (6.1-27)

Example 6.4 A gas mixture consisting of 50 mol % propane (1), 20% n-butane (2), and30% n-pentane (3) is compressed irreversibly from 300K and 1 bar to 500K and 40 bar byan adiabatic compressor. Estimate the work input per mole of gas mixture passing throughthe compressor if the mixture obeys the Peng-Robinson equation of state with k12 = 0.003,k13 = 0.027, and k23 = 0.017.

Solution

From Appendix A



From Appendix BeC∗P1 = 29.595 + 0.838× 10−1 T + 3.256× 10−4 T 2 − 3.958× 10−7 T 3 + 13.129× 10−11 T 4eC∗P2 = 24.258 + 2.335× 10−1 T + 1.279× 10−4 T 2 − 2.437× 10−7 T 3 + 8.552× 10−11 T 4eC∗P3 = 33.780 + 2.485× 10−1 T + 2.535× 10−4 T 2 − 3.838× 10−7 T 3 + 12.977× 10−11 T 4System: Compressor

This is a steady-state flow process and from the first law

∆ eH = eQ|z0

+ fWs (1)

Therefore, the work input to the compressor is equal to the change in enthalpy, which can bedetermined from Eq. (6.1-23).

137

• State 1 (T1 = 300K, P1 = 1bar)The values of reduced temperature, reduced pressure, and α are

Component Tr Pr α



A11 = 0.019 A12 = A21 = 0.023 B1 = 2.302× 10−3A22 = 0.029 A13 = A31 = 0.028 B2 = 2.865× 10−3A33 = 0.044 A23 = A32 = 0.035 B3 = 3.653× 10−3


Amix1 = 0.027 Bmix1 = 2.82× 10−3


Z3mix1 − 0.997Z2mix1 + 0.021Zmix1 − 6.817× 10−5 = 0 ⇒ Zmix1 = 0.976

The parameter Γ is calculated from Eq. (6.1-17) as

Γ1 = 0.513 Γ2 = 0.509 Γ3 = 0.518

The use of Eq. (6.1-19) givesΩmix1 = − 0.041

• State 2 (T2 = 500K, P2 = 40bar)The values of reduced temperature, reduced pressure, and α are

Component Tr Pr α



A11 = 0.191 A12 = A21 = 0.243 B1 = 0.054A22 = 0.310 A13 = A31 = 0.288 B2 = 0.070A33 = 0.459 A23 = A32 = 0.371 B3 = 0.087


Amix2 = 0.281 Bmix2 = 0.067


Z3mix2 − 0.933Z2mix2 + 0.134Zmix2 − 0.014 = 0 ⇒ Zmix2 = 0.785

138

The parameter Γ is calculated from Eq. (6.1-17) as

Γ1 = 0.779 Γ2 = 0.771 Γ3 = 0.787

The use of Eq. (6.1-19) givesΩmix2 = − 0.5

The heat capacity of the mixture can be calculated aseC∗Pmix= y1 eC∗P1 + y2 eC∗P2 + y3 eC∗P3= 29.783 + 1.631× 10−1 T + 2.644× 10−4 T 2 − 3.618× 10−7 T 3 + 12.168× 10−11 T 4

so that Z 500

300

eC∗PmixdT = 23, 428 J/mol

Substitution of the numerical values into Eq. (6.1-23) gives

∆ eH = − 3343 + 164.3 + 23, 428 = 20, 249 J/mol

6.1.3 Departure Functions

The departure functions developed for pure components in Section 3.6 can be extended tomixtures. For mixtures, Eqs. (3.6-3), (3.6-4), and (3.6-5) take the form

∆ eH = − ( eHmix − eHIGM)T1,P1 +

Z T2

T1

eC∗PmixdT + ( eHmix − eHIGM)T2,P2 (6.1-28)

∆eS = − (eSmix − eSIGM)T1,P1 + Z T2

T1

eC∗Pmix

TdT −R ln

µP2P1

¶+ (eSmix − eSIGM)T2,P2 (6.1-29)

∆eU = − (eUmix − eU IGM)T1,P1 +

Z T2

T1

eC∗VmixdT + (eUmix − eU IGM)T2,P2 (6.1-30)

where the departure functions are given by the following equations depending on the type ofthe equation of state.

• van der Waals equation of state

For mixtures, Eqs. (3.6-12), (3.6-13), and (3.6-14) become³ eHmix − eHIGM´T,P

RT= Zmix − 1−

Amix

Zmix(6.1-31)

³eSmix − eSIGM´T,P

R= ln (Zmix −Bmix) (6.1-32)

³eUmix − eU IGM´T,P

RT= − Amix

Zmix(6.1-33)

139



RT= Zmix − 1−

3

2

Amix

Bmixln

µ1 +

Bmix

Zmix

¶(6.1-34)


R= ln (Zmix −Bmix)−

1

2

Amix

Bmixln

µ1 +

Bmix

Zmix

¶(6.1-35)


RT= − 3

2

Amix

Bmixln

µ1 +

Bmix

Zmix

¶(6.1-36)

• Peng-Robinson equation of state


RT= Zmix − 1 +

Ωmix√8Bmix

ln

"Zmix +

¡1 +√2¢Bmix

Zmix +¡1−√2¢Bmix

#(6.1-37)


R= ln(Zmix −Bmix) +

Θmix√8Bmix

ln

"Zmix +

¡1 +√2¢Bmix


#(6.1-38)


RT=

Ωmix√8Bmix

ln

"Zmix +

¡1 +√2¢Bmix


#(6.1-39)

where Ωmix and Θmix are given by Eqs. (6.1-19) and (6.1-27), respectively.

Example 6.5 Solve Example 6.2 using departure functions if the gas mixture obeys theRedlich-Kwong equation of state.

Solution

From Appendix A

Component Tc (K) Pc ( bar)

Methane 190.6 46.1Ethane 305.3 49.0

For a steady-state flow system, the first law givesfWs = ∆ eH (1)

The change in enthalpy can be calculated from Eq. (6.1-28).

• State 1 (T1 = 310K, P1 = 1bar)The values of reduced temperature and reduced pressure are

140

Component Tr Pr

Methane 1.626 0.022Ethane 1.015 0.020

Using the equations given in Table 6.3 and taking kij to be equal to zero, the dimensionlessparameters are calculated as

A11 = 2.790× 10−3 A12 = A21 = 4.794× 10−3 B1 = 1.172× 10−3A22 = 8.237× 10−3 B2 = 1.707× 10−3


Amix1 = 3.205× 10−3 Bmix1 = 1.225× 10−3


Z3mix1 − Z2mix1 + 1.978× 10−3 Zmix1 − 3.926× 10−6 = 0 ⇒ Zmix1 = 0.998

The enthalpy departure function is calculated from Eq. (6.1-34) as

( eHmix − eHIGM)T1,P1 = (8.314)(310)

∙0.998− 1− 3

2

µ3.205× 10−31.225× 10−3

¶ln

µ1 +

1.225× 10−30.998

¶¸= − 17.6 J/mol

• State 2 (T2 = 350K, P2 = 30bar)The values of reduced temperature and reduced pressure are

Component Tr Pr

Methane 1.836 0.651Ethane 1.146 0.612


A11 = 0.061 A12 = A21 = 0.106 B1 = 0.031A22 = 0.186 B2 = 0.046


Amix2 = 0.07 Bmix2 = 0.033


Z3mix2 − Z2mix2 + 0.036Zmix2 − 2.31× 10−3 = 0 ⇒ Zmix2 = 0.965

The enthalpy departure function is calculated from Eq. (6.1-34) as

( eHmix − eHIGM)T2,P2 = (8.314)(350)

∙0.964− 1− 3

2

µ0.07

0.033

¶ln

µ1 +

0.033

0.965

¶¸= − 416.1 J/mol

141

From Example 6.2 Z 350

310

eC∗PmixdT = 1574 J/mol

Substitution of the numerical values into Eq. (6.1-28) leads to

∆ eH = 17.6 + 1574− 416.1 = 1175.5 J/mol

Example 6.6 A carbon dioxide stream at 10 bar and 350K and a methane stream at 5 barand 270K are continuously fed to an adiabatic mixer in the mole ratio 1:3. If the exit gasmixture is at 3 bar, estimate its temperature. Assume that the pure gases as well as the mixtureobey the Redlich-Kwong equation of state.

Solution

From Appendix A, the critical constants are given below:

Component Tc (K) Pc ( bar)

Carbon dioxide 304.2 73.8Methane 190.6 46.1

From Appendix BeC∗PCO2 = 29.268− 0.224× 10−1 T + 2.653× 10−4 T 2 − 4.153× 10−7 T 3 + 20.057× 10−11 T 4eC∗PCH4 = 36.155− 0.511× 10−1 T + 2.215× 10−4 T 2 − 1.824× 10−7 T 3 + 4.899× 10−11 T 4For a steady-state flow system, the first law gives

∆H = Q|z0

+ Ws|z0

= 0 (1)

The process path for calculating enthalpy change is shown in the figure below:

F

E

D C

B A

Pure CO2 at 10 bar & 350 K

Pure ideal CO2 at 10 bar & 350 K

Pure CH4 at 5 bar & 270 K

Pure ideal CH4 at 5 bar & 270 K

Pure ideal CO2 at 3 bar & T3

Pure ideal CH4 at 3 bar & T3

Ideal gas mixture at 3 bar & T3

Gas mixture at 3 bar & T3

Taking 1mol of CO2 as a basis, the enthalpy change is expressed as

∆H = ∆ eHA + 3∆ eHB +∆ eHC + 3∆ eHD + 4∆ eHE + 4∆ eHF (2)

Considering the following states

142

State T (K) P ( bar)

1 350 102 270 53 T3 3

Eq. (2) is expressed in the form

∆H = − ( eHCO2 − eHIGCO2)T1,P1 − 3 ( eHCH4 − eHIG

CH4)T2,P2 +

Z T3

350

eC∗PCO2dT + 3Z T3

270

eC∗PCH4dT+ 4 ( eHmix − eHIGM)T3,P3 (3)

Since there is no interaction between ideal gas molecules, mixing of ideal gases produces nochange in enthalpy. Thus, ∆ eHE = 0. Substitution of Eq. (3) into Eq. (1) gives

− ( eHCO2 − eHIGCO2)T1,P1 − 3 ( eHCH4 − eHIG

CH4)T2,P2 +

Z T3

350

eC∗PCO2dT + 3Z T3

270

eC∗PCH4dT+ 4 ( eHmix − eHIGM)T3,P3 = 0 (4)

Departure function for pure CO2

At 350K and 10 barA = 0.041 B = 0.010 Z = 0.969

The use of Eq. (3.6-15) gives

( eHCO2 − eHIGCO2)T1,P1 = − 274 J/mol (5)

Departure function for pure CH4

At 270K and 5 bar

A = 0.019 B = 6.634× 10−3 Z = 0.987


( eHCH4 − eHIGCH4)T2,P2 = − 121.6 J/mol (6)

The remaining terms in Eq. (4) cannot be calculated since T3 is not known. Therefore, atrial-and-error procedure will be used to estimate T3.

Assuming T3 = 288.3K, the values of the integrals in Eq. (4) areZ 288.3

350

eC∗PCO2dT = − 2327 J/mol (7)

Z 288.3

270

eC∗PCH4dT = 649.3 J/mol (8)

Departure function for the mixture

Taking k12 to be equal to zero, the dimensionless parameters and the compressibility factorbecome

Amix = 0.012 Bmix = 3.725× 10−3 Zmix = 0.992

143


( eHmix − eHIGM)T3,P3 = − 63.4 J/mol (9)

If the assumed temperature is correct, then the summation of the terms on the left-hand side ofEq. (4) should be zero. Substitution of Eqs. (5), (6), (7), (8), and (9) into the left-hand sideof Eq. (4) gives

274 + 3 (121.6)− 2327 + 3 (649.3)− 4 (63.4) = 6.1which is close enough to zero.

6.1.4 Which Equation of State to Use?

The two central questions in solving problems encountered in real life are: (i) Which equationof state should be used for a given mixture?, (ii) Which mixing rule should be employedin the calculations? Unfortunately, there is no clearcut recipe for answering such questions.Simulation packages given in Appendix H provide decision trees for making the proper selectionby taking into account various factors, some of which are given below:

• Nature of the chemical species (Polar/Nonpolar, Electrolyte/Nonelectrolyte)• Real or pseudocomponents5,• Value of pressure (less than or greater than 10 bar)• Availability of interaction parameters.The Peng-Robinson and the Soave-Redlich Kwong equations of state are mostly preferred inthe natural-gas, petroleum, and petrochemical industries (Prausnitz and Tavares, 2004). Formore details on the subject, the reader should refer to Chen and Mathias (2002), Poling et al.,(2004), de Hemptinne and Behar (2006), and Edwards (2008).

6.2 PARTIAL MOLAR QUANTITY

Consider two separate beakers containing water of volumes 20 cm3 and 50 cm3. If the contentsof these two beakers are mixed, the total volume of water will be 70 cm3. Since volumes of apure component are additive, this is an expected result.

When 20 cm3 of water is mixed with 50 cm3 of ethanol, the total volume of the solutionturns out to be 67 cm3, less than the sum of pure liquid volumes. The reason for this is thefact that water and ethanol molecules interact with each other and pack differently than withlike molecules6. If 1 and 2 represent water and ethanol, respectively, then

Vmix 6= n1 eV1 + n2 eV2 (6.2-1)

in which eV1 and eV2 represent the molar volumes of pure 1 and 2, respectively.In engineering calculations, however, we want to have an equation similar to Eq. (6.2-1) to

calculate the total volume of a mixture. In other words, we have to come up with a quantityV i so that the additive property holds, i.e.,

Vmix = n1V 1 + n2V 2 (6.2-2)

5 In complex mixtures, such as crude oil, it is not possible to identify each component making up the mixture.In that case, some components are lumped into a so-called pseudo-component. Reduction in the number ofcomponents makes the mixture amenable to modeling. The criteria used in lumping and the estimation ofpseudo-component properties are beyond the scope of this text.

6Consider a box of volume V1 filled with small metal balls and another box of volume V2 filled with rice. Ifthe contents of these two boxes are mixed, the final volume will be less than V1 + V2.

144

or, in general,

Vmix =kXi=1

niV i (6.2-3)

The term V i is called the partial molar volume of species i and is defined by

V i =

µ∂Vmix

∂ni

¶T,P,nj 6=i

(6.2-4)

The subscripts on the partial derivative mean that pressure, temperature, and number of molesof all components other than i are kept constant.

In general, a partial molar property, ϕi, is the rate at which an extensive property of theentire mixture, ϕmix, changes with the number of moles of component i in the mixture whentemperature, pressure, and number of moles of all components other than i are kept constant,i.e.,

ϕi =

µ∂ϕmix

∂ni

¶T,P,nj 6=i

(6.2-5)

It should be kept in mind that ϕi is a property of the mixture and not simply a property ofcomponent i. The partial molar quantity is an intensive property that is generally dependenton the composition of the mixture as well as temperature and pressure.

The total property of the mixture is expressed in terms of the partial molar properties as

ϕmix =kXi=1

ni ϕi (6.2-6)

or dividing each term by the total number of moles gives

eϕmix =kXi=1

xi ϕi (6.2-7)

Example 6.7 Pecar and Dolecek (2005) experimentally determined the densities and thepartial molar volumes of a binary mixture of ethanol (1) and water (2) as a function of tem-perature and pressure. In a mixture with an ethanol mole fraction of 0.35 at 298K and 1 bar,the mixture density and the partial molar volume of ethanol are determined as 891.9 kg/m3

and 57.31 cm3/mol, respectively. Calculate the partial molar volume of water.

Solution

From Eq. (6.2-7) we can write eVmix = x1V 1 + x2V 2 (1)

Note that the molar volume of the mixture is given by

eVmix =Mmix

ρmix

(2)

in which Mmix and ρmix represent the molecular weight and the density of the mixture, respec-tively. The molecular weight of the mixture is given by

Mmix = x1M1 + x2M2 = (0.35)(46.07) + (0.65)(18.02) = 27.84 g/mol

145

Therefore, the molar volume of the mixture is calculated from Eq. (2) as

eVmix =27.84

0.8919= 31.21 cm3/mol

The partial molar volume of water is then determined from Eq. (1) as

31.21 = (0.35)(57.31) + 0.65V 2 ⇒ V 2 = 17.16 cm3/mol

Consider the definition of enthalpy, i.e.,

H = U + PV (6.2-8)

Differentiation of Eq. (6.2-8) with respect to ni by keeping temperature, pressure, and numberof moles of all components other than i constant givesµ

∂H

∂ni

¶T,P,nj 6=i

=

µ∂U

∂ni

¶T,P,nj 6=i

+ P

µ∂V

∂ni

¶T,P,nj 6=i

(6.2-9)

Application of the definition of a partial molar quantity leads to

Hi = U i + PV i (6.2-10)

Similarly, we can write

Ai = U i − TSi (6.2-11)

Gi = Hi − TSi (6.2-12)

6.2.1 Determination of Partial Molar Quantities

For simplicity let us consider the determination of partial molar volumes in a binary mixture.In general, partial molar volumes can be determined by two different approaches.

• Method of tangent slopeIn this method, a known amount of a compound, say one mole of component 1, is placed in agraduated cylinder. By keeping temperature and pressure constant throughout the experiment,the volume of the mixture is recorded by adding increasing amounts of component 2. Theneither the total volume of the mixture, Vmix, is plotted as a function of n2 or an equation isfitted to the volume data in the form Vmix = eV1 +αn2 + β n1.52 + γ n22 + ... , in which eV1 is themolar volume of pure 1.

At any given value of n2, the slope of the tangent to the curve gives the partial molarvolume of component 2, i.e.,

Slope =µ∂Vmix

∂n2

¶T,P,n1

= V 2 (6.2-13)

If an equation is fitted, then differentiation of the equation with respect to n2 gives the partialmolar volume of component 2 as

V 2 =

µ∂Vmix

∂n2

¶T,P,n1

= α+ 1.5β n0.52 + 2 γ n2 + ... (6.2-14)

146

Once V 2 is determined, the partial molar volume of component 1 can be obtained from Eq.(6.2-6) as

V 1 =Vmix − n2 V 2

n1(6.2-15)

Example 6.8 When NaCl is dissolved in water at 298K, the volume of the resulting mixtureis represented as a function of molality7, m, in the form

Vmix = 1003 + 16.62m+ 1.77m1.5 + 0.12m2 (1)

where Vmix is in cm3. For a 0.01 m solution of NaCl, calculate:

a) The total solution volume,b) The partial molar volumes of NaCl and water.

Solution

a) From Eq. (1)

Vmix = 1003 + 16.62 (0.01) + 1.77 (0.01)1.5 + 0.12 (0.01)2 = 1003.2 cm3

b) Let the subscripts 1 and 2 designate H2O and NaCl, respectively. Choosing 1 kg of wateras a basis, m = n2. Thus, differentiation of Eq. (1) with respect to m gives the partial molarvolume of NaCl as

V 2 = 16.62 + 2.655m0.5 + 0.24m

= 16.62 + 2.655(0.01)0.5 + 0.24(0.01) = 16.89 cm3/mol

From Eq. (6.2-15)

V 1 =Vmix − n2 V 2

n1=1003.2− (0.01)(16.89)

1000/18= 18.05 cm3/mol

• Method of tangent interceptsThe total volume of a binary mixture is given by

Vmix = (n1 + n2) eVmix (6.2-16)

The partial molar volume of component 1 is

V 1 =

µ∂Vmix

∂n1

¶T,P,n2

=

½∂

∂n1

h(n1 + n2) eVmix

i¾T,P,n2

(6.2-17)

Carrying out the differentiation gives

V 1 = eVmix + n

Ã∂ eVmix

∂n1

!T,P,n2

(6.2-18)

By the application of the chain rule, Eq. (6.2-18) becomes

V 1 = eVmix + n

Ã∂ eVmix

∂x2

!T,P,n2

µ∂x2∂n1

¶T,P,n2

(6.2-19)

7Molality is the number of moles of solute dissolved in 1 kg of solvent.

147

Since x2 = n2/(n1 + n2), then µ∂x2∂n1

¶T,P,n2

= − x2n

(6.2-20)

so that Eq. (6.2-19) takes the form

V 1 = eVmix − x2

Ã∂ eVmix

∂x2

!T,P,n2

(6.2-21)

Suppose that eVmix varies as a function of x2 as shown in Figure 6.1. At any given value of x2,let us draw a tangent to the curve. Note that the slope of the tangent line is given by

Slope =∂ eVmix

∂x2= − tanα = − BC

AC(6.2-22)

C

B

1~V

mixV~

2~V

α A

x2 0 1 D

Figure 6.1 Graphical determination of partial molar volumes.

Substitution of Eq. (6.2-22) into Eq. (6.2-21) leads to

V 1 = eVmix − x2

Ã∂ eVmix

∂x2

!T,P,n2

= CD −AC

µ− BC

AC

¶= CD +BC = BD (6.2-23)

Thus, the intercept of the tangent line at x2 = 0 is the partial molar volume of component1. Similarly, the intercept of the tangent line at x2 = 1 gives the partial molar volume ofcomponent 2.

For a binary system8, Eq. (6.2-21) can be generalized as

ϕ1 = eϕmix − x2

µdeϕmix

dx2

¶constant T and P (6.2-24)

and

ϕ2 = eϕmix − x1

µdeϕmix

dx1

¶constant T and P (6.2-25)

Since x1 + x2 = 1 for a binary system, then

deϕmix

dx1= − deϕmix

dx2(6.2-26)

8See Problem 6.10 for the generalization of Eq. (6.2-21) for a k-component system.

148

Thus, depending on the functional form of eϕmix, the derivatives in Eqs. (6.2-24) and (6.2-25)can be interchanged according to Eq. (6.2-26).

In the limiting cases of the mole fractions, i.e., xi → 0 or xi → 1, the partial molar propertiesbecome

limxi→0

ϕi = ϕ∞i and limxi→1

ϕi = eϕi (6.2-27)

where ϕ∞i is called the partial molar quantity of component i at infinite dilution.

Example 6.9 It is required to prepare 1 L of a solution containing 10 weight % of component1 and 90% of component 2. The stock solution available in the laboratory has the compositionof 60 mol % component 1 and 40% of component 2. Determine the volumes of the stocksolution and pure component 2 that must be mixed.

Data: The molar volume of the mixture (in cm3/mol) is experimentally determined aseVmix = 60x1 + 110x2 + 5x1x2

The molecular weights of components 1 and 2 are 40 and 60 g/mol, respectively.

Solution

Let X and Y be the volumes (in cm3) of the stock solution and pure component 2, respectively,that must be mixed to produce 1 L of desired mixture. The conservation statements for thenumber of moles of components 1 and 2 are expressed asÃ

XeV stockmix

!(0.6) =

⎛⎝ 1000eV finalmix

⎞⎠xfinal1 (1)

ÃXeV stockmix

!(1− 0.6) + YeV2 =

⎛⎝ 1000eV finalmix

⎞⎠³1− xfinal1

´(2)

Determination of the molar volumes of the stock and final solutions requires the mole fractionsto be known. In a binary mixture, the mole fraction, x, is related to the weight fraction, ω, as

x1 =

ω1M1

ω1M1

+ω2M2

where M1 and M2 are the molecular weights of components 1 and 2, respectively. Therefore,the mole fraction of component 1 in the final solution is

xfinal1 =

0.1

40

0.1

40+0.9

60

= 0.143

The molar volume of the final solution iseV finalmix = (60)(0.143) + (110)(1− 0.143) + (5)(0.143)(1− 0.143) = 103.46 cm3/mol

The molar volume of the stock solution iseV stockmix = (60)(0.6) + (110)(1− 0.6) + (5)(0.6)(1− 0.6) = 81.2 cm3/mol

149

The molar volume of pure component 2 is

eV2 = limx2→1

eVmix = 110 cm3/mol

Substitution of the numerical values into Eq. (1) gives

X = 187.1 cm3

Then the use of Eq. (2) results inY = 809.8 cm3

Note that X + Y = 187 + 809.8 = 996.8 cm3 < 1000 cm3, because of volume increase uponmixing.

Example 6.10 The molar enthalpy of a binary liquid mixture of 1 and 2 at constant temper-ature and pressure is given by

eHmix = 400x1 + 250x2 + x1x2 (15x1 + 45x2)

where eHmix is in J/mol. Calculate

a) The molar enthalpies of pure components 1 and 2,

b) The partial molar enthalpies of components 1 and 2 at x1 = 0.35,

c) The partial molar enthalpies of components 1 and 2 at infinite dilution.

Solution

a) Molar enthalpies of pure components are

eH1 = limx1→1

eHmix = 400 J/mol eH2 = limx2→1

eHmix = 250 J/mol

b) From Eq. (6.2-24)

H1 = eHmix − x2

Ãd eHmix

dx2

!(1)

The derivative of eHmix with respect to x2 is

d eHmix

dx2= − 400 + 250 + (1− 2x2)(15x1 + 45x2) + x1x2(− 15 + 45)

= − 150 + 15x1 + 45x2 − 90x22 (2)

Thus, Eq. (1) becomes

H1 = 400x1 + 250x2 + x1x2 (15x1 + 45x2)− x2(− 150 + 15x1 + 45x2 − 90x22)= 400− 15x22 + 60x32 (3)

The partial molar enthalpy of component 2 can be calculated from Eq. (6.2-25)

H2 = eHmix − x1

Ãd eHmix

dx1

!(4)

150

Sinced eHmix

dx1= − d eHmix

dx2(5)

Eq. (4) becomes

H2 = eHmix + x1

Ãd eHmix

dx2

!(6)

or

H2 = 400x1 + 250x2 + x1x2 (15x1 + 45x2) + x1(− 150 + 15x1 + 45x2 − 90x22)= 250 + 75x21 − 60x31 (7)

When x1 = 0.35H1 = 410.14 J/mol and H2 = 256.62 J/mol

c) The partial molar enthalpies of components 1 and 2 at infinite dilution are

H∞1 = lim

x1→0H1 = 445 J/mol

H∞2 = lim

x2→0H2 = 265 J/mol

6.2.2 Homogeneous Functions and Partial Molar Quantities

A function f(x, y) is said to be homogeneous of degree n if

f(λx, λy) = λn f(x, y) (6.2-28)

for all λ. For a homogeneous function of degree n, Euler’s theorem states that

n f(x1, x2, ..., xk) =kXi=1

µ∂f

∂xi

¶xi (6.2-29)

The extensive properties in thermodynamics can be regarded as homogeneous functions ofdegree "1". Therefore, for every extensive property we can write

f(x1, x2,..., xk) =kXi=1

µ∂f

∂xi

¶xi When f is an extensive property (6.2-30)

On the other hand, the intensive properties are homogeneous functions of degree "0" and canbe expressed as

0 =kXi=1

µ∂f

∂xi

¶xi When f is an intensive property (6.2-31)

Any extensive thermodynamic property of a mixture, ϕmix, is dependent on temperature,pressure, and the number of moles of each component present in the mixture, i.e.,

ϕmix = ϕmix(T, P, n1, n2, ..., nk) (6.2-32)

151

At constant temperature and pressure, application of Eq. (6.2-30) gives

ϕmix =kXi=1

µ∂ϕmix

∂ni

¶T,P,nj 6=i

ni (6.2-33)

or

ϕmix =kXi=1

ϕi ni (6.2-34)

which is identical with Eq. (6.2-6).

6.3 PROPERTY CHANGES ON MIXING

When two non-identical liquids are mixed, the total property of the mixture, ϕmix, is differentfrom the sum of the properties of the pure liquids, i.e.,

ϕmix 6=kXi=1

ni eϕi (6.3-1)

The inequality sign in Eq. (6.3-1) can be removed by the introduction of a "correction factor",∆ϕmix, so that

ϕmix =kXi=1

ni eϕi +∆ϕmix (6.3-2)

The term ∆ϕmix is called property change on mixing. In other words, property change onmixing represents the difference between the actual property of the mixture and the totalproperty of the unmixed pure components at the same temperature and pressure.

Dividing each term in Eq. (6.3-2) by the total number of moles gives

eϕmix =kXi=1

xi eϕi +∆eϕmix (6.3-3)

Rearrangement of Eq. (6.3-3) gives

∆eϕmix = eϕmix −kXi=1

xi eϕi = kXi=1

xi (ϕi − eϕi) (6.3-4)

As shown in Figure 6.2, property change on mixing may take positive or negative values de-pending on the interactions between molecules. Note that the dotted line represents the casewhen ∆eϕmix is zero.

mix~ϕ

2ϕ~

1ϕ~ 0~ <Δ mixϕ

0>Δ mix~ϕ

x2 0 1

mix 1 1 2 2 1 2 2 1φ x φ x φ φ x (φ φ )= + = + −% % % % % %

Figure 6.2 Property change on mixing for a binary system.

152

Among the property changes on mixing, chemical engineers are particularly interested inthe isothermal volume change on mixing, ∆eVmix, and the isothermal enthalpy change on mixing(or simply the heat of mixing), ∆ eHmix, both of which have to be determined experimentally.

When ∆Vmix>0, the interactions between unlike molecules are weaker than those betweenlike molecules. On the other hand, if the interactions between unlike molecules are strongerthan those between like molecules, then molecules pack themselves more tightly so as∆Vmix<0.

In the case of heat of mixing, note that the enthalpy change is associated with the changein bond energies. If pure components A and B are mixed to form a mixture, it is necessary tobreak A-A and B-B bonds to create A-B bonds as shown in Figure 6.3. Bond breaking requiresor absorbs energy, i.e., an endothermic process. On the other hand, bond formation releasesenergy, i.e., an exothermic process.

+AA − BB −

BA −

Figure 6.3 Formation of a binary mixture of species A and B.

When A-B bonds are stronger than A-A and B-B bonds, components in the mixture aremore stable than pure components and ∆Hmix < 0. On the other hand, when A-A and B-Bbonds are stronger than A-B bonds, pure components are more stable than the components inthe mixture and ∆Hmix > 0.

Example 6.11 When 50 cm3 of liquid A is mixed with 60 cm3 of liquid B, the resultingsolution has a volume of 113 cm3.What can be said about the relative magnitudes of V A andeVA, as well as V B and eVB?Solution

Since ∆Vmix = 113 − (50 + 60) = 3 cm3 > 0, a eVmix versus composition diagram can bequalitatively drawn as shown in the figure below. From the graph one can easily conclude that

V A > eVA and V B > eVBmixV~

BV

AV

BV~

AV~

0 1 Ax

*Ax

6.3.1 Determination of the Volume Change on Mixing

Once the mixture is prepared by specifying the mole fractions of each species, its density aswell as the densities of the pure components are measured by a densitometer. Then the volumechange on mixing is determined from the following equation:

153

∆eVmix =

kXi=1

xiMi

ρmix

−kXi=1

xiMi

ρi=

kXi=1

xiMi

µ1

ρmix

− 1

ρi

¶(6.3-5)

where Mi is the molecular weight of species i and ρi is the density of pure i.

Example 6.12 The following data are reported by Grenner et al. (2006) for a binary mixtureof cyclohexanamine (1) and n-heptane (2) at 303.15K under atmospheric pressure:

x1 0.0000 0.3006 0.5000 0.7996 1.0000

ρmix ( g/ cm3 ) 0.67532 0.71971 0.75383 0.81268 0.85820

The molecular weights of cyclohexanamine and n-heptane are 99.17 and 100.20 g/mol, respec-tively. Determine the volume change on mixing as a function of composition.

Solution

From the given data, pure component densities for cyclohexanamine and n-heptane are 0.85820and 0.67532 g/ cm3, respectively. The volume change on mixing can be calculated from Eq.(6.3-5), i.e.,

∆eVmix = x1M1

µ1

ρmix

− 1

ρ1

¶+ x2M2

µ1

ρmix

− 1

ρ2

¶The calculated values are given below:

x1 0.3006 0.5000 0.7996

∆eVmix ( cm3/mol) 0.284 0.273 0.150

6.3.2 Determination of the Heat of Mixing

The heat of mixing data can be obtained by the use of an isothermal flow calorimeter as shownin Figure 6.4. Note that the flow calorimeter is simply a steady-state mixing device.

Pure 2 at T

Pure 1 at T

Mixture at T

Heating or cooling

Figure 6.4 An isothermal flow calorimeter.

154

Pure fluid 1 and pure fluid 2, both at a temperature T , enter the calorimeter at molar flowrates of n1 and n2, respectively. They are mixed thoroughly by a stirrer to maintain uniformtemperature within the calorimeter. Heat is added or removed through the coils placed in thecalorimeter so as to keep the temperature of the mixture equal to that of the entering streams.Taking the contents of the calorimeter to be the system, the mass and energy balances are

n1 + n2 = nmix (6.3-6)

andnmix

eHmix − n1 eH1 − n2 eH2 = Q+ Ws (6.3-7)

Assuming Ws to be negligible and dividing Eq. (6.3-7) by nmix give

eHmix −³x1 eH1 + x2 eH2

´| z

∆ eHmix

=Q

n1 + n2(6.3-8)

Therefore, once the molar flow rates of the pure streams are fixed, one can determine the heatof mixing from Eq. (6.3-8) by measuring the heat transfer rate, Q. Carrying out experimentsat different flow ratios of the two streams yields the heat of mixing as a function of composition.

Experimental determination of the heat of mixing enables one to calculate the molar en-thalpy of the mixture from the following equation:

eHmix =kXi=1

xi eHi +∆ eHmix (6.3-9)

Thus, it is possible to plot the enthalpy of the mixture as a function of composition. Suchplots, also known as enthalpy-concentration diagrams, are very useful in determining energybalances.

Example 6.13 An isothermal mixer at 294K is fed with a pure sulfuric acid stream of flowrate 160 kg/min and an aqueous solution stream containing 38 weight % sulfuric acid of flowrate 150 kg/min. Estimate the rate of heat that must be added or removed so as to keep thetemperature of the exit stream at 294K under steady conditions.

The heat of mixing data for the sulfuric acid (1) and water (2) mixture at 294K is representedby Ross (1952) as

∆ eHmix ( kJ/mol) = x1x2

h− 53.531 + 20.869 (x1 − x2)

iThe pure component enthalpies for sulfuric acid and water are 1.596 and 1.591 kJ/mol, re-spectively.

Solution

The flow diagram of the mixing process is given below:

ω1 = 0.38 150 kg/min

Pure H2SO4

160 kg/min

MIXER

T = 294 K

CA

B

155

The heat that must be removed or added to the mixer can be determined from the steady-stateenergy balance as

∆H = Q+ Ws|z∼ 0

(1)

orQ = nC eHC − nA eHA − nB eHB (2)

The molar enthalpies from streams B and C can be calculated from

eHmix = x1 eH1 + x2 eH2 +∆ eHmix

= 1.596x1 + 1.591x2 + x1x2

h− 53.531 + 20.869 (x1 − x2)

i(3)

To calculate the weight fraction of sulfuric acid in stream C, first it is necessary to write theoverall material balance as

mA + mB = mC ⇒ mC = 160 + 150 = 310 kg/min (4)

On the other hand, the material balance for sulfuric acid is

mA + mB(ω1)B = mC(ω1)C ⇒ (ω1)C =160 + (150)(0.38)

310= 0.7 (5)

The use of Eq. (3) requires mole fractions of components to be known. The relationship betweenthe weight and mole fractions is given by

x1 =

ω1M1

ω1M1

+ω2M2

(6)

Thus, the mole fractions of sulfuric acid in streams B and C are

(x1)B =

0.38

98

0.38

98+1− 0.3818

= 0.1 (x1)C =

0.7

98

0.7

98+1− 0.718

= 0.3

The use of Eq. (3) yields

Stream An eHA = 1.596 kJ/mol

Stream B

½eHB = (1.596)(0.1) + (1.591)(0.9) + (0.1)(0.9)£− 53.531 + (20.869)(0.1− 0.9)

¤= − 4.729 kJ/mol

Stream C

½eHC = (1.596)(0.3) + (1.591)(0.7) + (0.3)(0.7)£− 53.531 + (20.869)(0.3− 0.7)

¤= − 11.402 kJ/mol

To determine the molar flow rates of streams B and C, we have to calculate their molecularweights:

MB = (0.1)(98) + (0.9)(18) = 26 g/mol = 26× 10−3 kg/mol

MC = (0.3)(98) + (0.7)(18) = 42 g/mol = 42× 10−3 kg/mol

156

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

Q =

µ310

42× 10−3¶(− 11.402)−

µ160

98× 10−3¶(1.596)−

µ150

26× 10−3¶(− 4.729)

= − 59, 512 kJ/min

indicating that heat must be removed from the system.

Example 6.14 Using the data given in Example 6.13 first plot the enthalpy-concentrationdiagram for sulfuric acid (1) and water (2) mixtures at 294K. Then prove that it is alwayssafer to add acid to water rather than water to acid.

Solution

The molar enthalpy of the mixture is given by

eHmix( kJ/mol) = 1.596x1 + 1.591x2 + x1x2

h− 53.531 + 20.869 (x1 − x2)

i(1)

The plot of the molar enthalpy of the mixture, eHmix, versus mole fraction of water, x2, is shownbelow.

0 0.2 0.4 0.6 0.8 115−

10−

5−

0

51.596

12.265−

Hmix x2( )

10 x2

Using Eqs. (6.2-24) and (6.2-25), the partial molar enthalpies are obtained as a function ofcomposition in the form

H1 = 1.596− 53.531x22 + 62.607x1x22 − 20.869x32 (2)

H2 = 1.591− 53.531x21 − 62.607x21x2 + 20.869x31 (3)

The partial molar enthalpies at infinite dilution are

H∞1 = lim

x1→0H1 = 1.596− 53.531− 20.869 = − 72.804 kJ/mol

H∞2 = lim

x2→0H2 = 1.591− 53.531 + 20.869 = − 31.071 kJ/mol

The partial molar enthalpies at infinite dilution can also be determined from the intercepts ofthe tangents drawn to the curve at x2 = 0 and x2 = 1 as shown in the figure below.

157

∞2H

0

mixH~

∞1H

1H~ 2H~

1

0

x2

The heat of mixing is given by

∆ eHmix = x1(H1 − eH1) + x2(H2 − eH2) (4)

Now, let us consider two cases:

Process A (Adding a small amount of acid to a very large amount of water)

If 1 g of H2SO4 is added to 1000 g of H2O, then

x1 =

1

98

1

98+1000

18

= 1.84× 10−4

Since x1 ¿ x2, then H1 = H∞1 and H2 ' eH2. Thus, Eq. (4) simplifies to

∆ eHmix

¯Process A

= x1(H∞1 − eH1) =

¡1.84× 10−4

¢(− 72, 804− 1596) = − 13.7 J/mol (3)

Process B (Adding a small amount of water to a very large amount of acid)

If 1 g of H2O is added to 1000 g of H2SO4, then

x2 =

1

18

1000

98+1

18

= 5.42× 10−3

Since x2 ¿ x1, then H1 ' eH1 and H2 = H∞2 . Thus, Eq. (4) simplifies to

∆ eHmix

¯Process B

= x2(H∞2 − eH2) =

¡5.42× 10−3

¢(− 31, 071− 1591) = − 177 J/mol

Note that mixing is exothermic for both processes. However, more heat is released when wateris added to acid. This may lead to violent boiling of the solution and splashing of the acid.Therefore, acid should be slowly added to water to prevent boiling.

Once property change on mixing, ∆eϕmix, is determined experimentally as a function ofmole fraction, the next step is to fit the data and to express property change on mixing as a

158

function of composition. For a binary system, for example, such a relationship is expressed bythe following equation:

∆eϕmix = x1x2 f(x1, x2) (6.3-10)

The term x1x2 originates from the fact that ∆eϕmix becomes zero for a pure component, i.e.,x1 = 1 (or x2 = 0) and x2 = 1 (or x1 = 0). In the literature, it is customary to expressf(x1, x2) in polynomial form as

∆eϕmix

x1x2=

NXi=1

Ai(x1 − x2)i−1 = A1 +A2(x1 − x2) +A3(x1 − x2)

2 + ... (6.3-11)

which is known as the Redlich-Kister type expansion.

Example 6.15 Conti et al. (1997) reported the following experimental values of ∆ eHmix fora binary system of chloroform (1) and cyclohexane (2) at 298K:

x1 ∆ eHmix ( J/mol) x1 ∆ eHmix ( J/mol) x1 ∆ eHmix ( J/mol)

0.0563 150.8 0.2666 520.2 0.5740 618.10.0563 147.4 0.3661 609.6 0.6690 567.70.1643 375.2 0.4732 630.9 0.7587 478.30.2520 500.1 0.4732 644.3 0.8435 351.50.2666 519.0 0.5740 618.7 0.9238 199.7

Fit the data to the equation of the form

∆ eHmix = x1x2

4Xi=1

Ai(x1 − x2)i−1 (1)

and evaluate the coefficients A1, A2, A3, and A4.

Solution

The coefficients can be estimated by the method of least squares. Note that Eq. (1) can berearranged in the form of

∆ eHmix

x1x2= A1 +A2(x1 − x2) +A3(x1 − x2)

2 +A4(x1 − x2)3 (2)

ory = A1 +A2 z +A3 z

2 +A4 z3 (3)

where

y =∆ eHmix

x1x2and z = x1 − x2 (4)

The coefficients Ai must be chosen such that the sum of the squares of the deviations, S, givenby

S =NXi=1

hyi − (A1 +A2 zi +A3 z

2i +A4 z

3i )i2

(5)

159

is minimum. Note that N represents the number of data points. Minimization of S is accom-plished by differentiating Eq. (5) with respect to Ai (i = 1, 2, 3, 4), and setting the derivativesequal to zero. The result is

A1N +A2NPi=1

zi +A3NPi=1

z2i +A4NPi=1

z3i =NPi=1

yi (6)

A1NPi=1

zi +A2NPi=1

z2i +A3NPi=1

z3i +A4NPi=1

z4i =NPi=1

zi yi (7)

A1NPi=1

z2i +A2NPi=1

z3i +A3NPi=1

z4i +A4NPi=1

z5i =NPi=1

z2i yi (8)

A1NPi=1

z3i +A2NPi=1

z4i +A3NPi=1

z5i +A4NPi=1

z6i =NPi=1

z3i yi (9)

Equations (6)-(9) can be solved for Ai using the matrix algebra, i.e.,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

A1

A2

A3

A4

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

NNPi=1

ziNPi=1

z2iNPi=1

z3i

NPi=1

ziNPi=1

z2iNPi=1

z3iNPi=1

z4i

NPi=1

z2iNPi=1

z3iNPi=1

z4iNPi=1

z5i

NPi=1

z3iNPi=1

z4iNPi=1

z5iNPi=1

z6i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

−1 ⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

NPi=1

yi

NPi=1

zi yi

NPi=1

z2i yi

NPi=1

z3i yi

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(10)

or ⎡⎢⎢⎣A1A2A3A4

⎤⎥⎥⎦ =⎡⎢⎢⎣

15 − 1.565 4.401 − 0.928− 1.565 4.401 − 0.928 2.4294.401 − 0.928 2.429 − 0.681− 0.928 2.429 − 0.681 1.601

⎤⎥⎥⎦−1 ⎡⎢⎢⎣

3.979× 104− 4.389× 1031.206× 104− 2.597× 103

⎤⎥⎥⎦

=

⎡⎢⎢⎣2.546× 103− 125.091361.22197.836

⎤⎥⎥⎦Comment: The correlation coefficient for the third-order polynomial fit is 0.979.

6.3.3 Determination of Partial Molar Quantities from ∆eϕmix

One of the practical questions to ask is how to determine partial molar quantities once aproperty change on mixing is determined as a function of composition. For a binary mixture,the molar property of a mixture is given by

eϕmix = (x1eϕ1 + x2eϕ2) +∆eϕmix (6.3-12)

160

On the other hand, Eq. (6.2-24) gives the partial molar property of component 1 as

ϕ1 = eϕmix − x2

µ∂eϕmix

∂x2

¶T,P,n2

(6.3-13)

Substitution of Eq. (6.3-12) into Eq. (6.3-13) gives

ϕ1 = x1eϕ1 + x2eϕ2 +∆eϕmix − x2

"− eϕ1 + eϕ2 +µ∂∆eϕmix

∂x2

¶T,P,n2

#(6.3-14)

Simplification of Eq. (6.3-14) leads to

ϕ1 − eϕ1 = ∆eϕmix − x2

µ∂∆eϕmix

∂x2

¶T,P,n2

(6.3-15)

Since Eqs. (6.2-24) and (6.3-15) have similar forms, graphical interpretation of Eq. (6.3-15) isthe same as that of Eq. (6.2-24) as shown in Figure 6.5.

1 1φ φ− %

2 2φ φ− %

2x

0

mixΔφ%

0 1

Figure 6.5 Determination of partial molar properties from a property change on mixing.

6.4 THE GIBBS-DUHEM EQUATION

The Gibbs-Duhem equation gives a relationship between the partial molar properties of differentcomponents in a mixture. For any thermodynamic property, the total property of the mixtureis expressed as

ϕmix =kXi=1

ni ϕi (6.4-1)

Differentiation of Eq. (6.4-1) gives

dϕmix =kXi=1

ni dϕi +kXi=1

ϕi dni (6.4-2)

On the other hand, the total property of the mixture, ϕmix, is a function of temperature,pressure, and the number of moles of each component present in the mixture, i.e.,

ϕmix = ϕmix(T, P, n1, n2, ..., nk) (6.4-3)

161

The total differential of ϕmix is

dϕmix =

µ∂ϕmix

∂T

¶P,nj

dT +

µ∂ϕmix

∂P

¶T,nj

dP +kXi=1

µ∂ϕmix

∂ni

¶T,P,nj 6=i| z

ϕi

dni (6.4-4)

At constant temperature and pressure Eq. (6.4-4) simplifies to

dϕmix =kXi=1

ϕi dni (6.4-5)

The use of Eq. (6.4-5) in Eq. (6.4-2) leads to

kXi=1

ni dϕi = 0 (6.4-6)

or dividing each term by the total number of moles gives

kXi=1

xi dϕi = 0 constant T and P (6.4-7)

which is known as the Gibbs-Duhem equation under the conditions of constant temperatureand pressure.

For simplicity let us consider a binary mixture and exploit the practical uses of the Gibbs-Duhem equation. For a binary mixture, Eq. (6.4-7) simplifies to

x1 dϕ1 + x2 dϕ2 = 0 (6.4-8)

Differentiation of Eq. (6.4-8) with respect to x1 results in

x1dϕ1dx1

+ x2dϕ2dx1

= 0 constant T and P (6.4-9)

The Gibbs-Duhem equation, Eq. (6.4-9), can be used for the following purposes:

• If the partial molar quantity of one of the species is known, the partial molar quantity of theother species can be calculated from Eq. (6.4-9).

• If the partial molar quantities of both species are experimentally determined, the expressionsfor ϕ1 and ϕ2 should satisfy the Gibbs-Duhem equation. Therefore, the Gibbs-Duhem equationis quite useful for checking whether the experimental data are thermodynamically consistentor not.

Example 6.16 If the partial molar volume of component 1 in a binary mixture at constantT and P is given by

V 1 = eV1 + β x22

where eV1 is the molar volume of pure 1, find the corresponding equation for V 2.

Solution

For ϕ = V , the Gibbs-Duhem equation, Eq. (6.4-9), is written as

x1dV 1dx1

+ x2dV 2dx1

= 0 constant T & P (1)

162

The term dV 1/dx1 becomes

dV 1dx1

=d

dx1

h eV1 + β (1− x1)2i= − 2β(1− x1) = − 2β x2 (2)

It is also possible to evaluate the same term as

dV 1dx1

= − dV 1dx2

= − d

dx2

³eV1 + β x22

´= − 2β x2

Substitution of Eq. (2) into Eq. (1) gives

dV 2dx1

= 2β x1 (3)

Integration of Eq. (3) leads toV 2 = β x21 + C (4)

The constant C is evaluated by using the fact that as x1 → 0, V 2 → eV2. This gives C = eV2and the partial molar volume of species 2 is expressed in the form

V 2 = eV2 + β x21 (5)

The generalized form of Eq. (6.4-9) for a k-component system is given in Problem 6.29.

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Conti, G., P. Gianni, M. Tramati, L. Lepori and E. Matteoli, 1997, J. Chem. Thermodynamics,29, 865-877.

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PROBLEMS

Problems related to Section 6.1

6.1 Estimate the molar volume of a binary mixture containing 30 mol % propane (1) and70% n-butane (2) at 350K and 3 bar. The mixture is represented by the virial equationstate, and the critical molar volumes of propane and n-butane are 2.00 × 10−4m3/mol and2.55× 10−4m3/mol, respectively.(Answer: 9.27× 10−3m3/mol)

6.2 A binary gaseous mixture at T1 and P1 enters a throttling valve and leaves it at P2. Themixture is represented by the virial equation of state with the virial coefficients depending ontemperature in the form

Bij = αij + βij T

164

Show that the exit temperature, T2, is given by

T2 = T1 +αmix(P1 − P2)eC∗Pmix

whereαmix = y21 α11 + 2 y1y2 α12 + y22 α22

eC∗Pmix= y1 eC∗P1 + y2 eC∗P2

in which y1 and y2 represent mole fractions of components 1 and 2, respectively.

6.3 A rigid cylinder holds a gas mixture containing 75 mol % propane (1) and 25% n-butane(2) at 300K and 50 bar. Calculate the molar volume of the gas mixture (k12 = 0.003) using

a) Redlich-Kwong equation of state,b) Peng-Robinson equation of state.(Answer: a) 99.53 cm3/mol b) 86.02 cm3/mol)

6.4 It is required to store 10 kg of ethane (1), 25 kg of ethylene (2), and 40 kg of propane at400K and 35 bar. Estimate the volume of the tank if the mixture obeys the Peng-Robinsonequation of state. Take k12 = 0.010, k13 = 0.001, and k23 = 0.

(Answer: 1.712m3)

6.5 A rigid tank is divided into two parts by a rigid partition. One side contains 2 kmol ofmethane at 300K and 100 bar, and the other side contains 3 kmol of nitrogen at 300K and100 bar. The partition is removed and the gases mix with each other. If the temperature iskept constant at 300K, estimate the final pressure. The system is represented by the van derWaals equation of state.

(Answer: 101.1 bar)

6.6 The term (∂ eV /∂T )P is needed in the calculation of the Joule-Thomson coefficient (seeProblem 2.1). If a gas mixture is represented by the Peng-Robinson equation of state, thenthis term is calculated indirectly by the use of Eq. (2.2-50), i.e.,Ã

∂ eV∂T

!P

= −(∂P/∂T )eV(∂P/∂ eV )T (1)

a) Show thatµ∂P

∂T

¶eV =

P

T

∙1

Zmix −Bmix−

Θmix

Zmix(Zmix +Bmix) +Bmix(Zmix −Bmix)

¸(2)

µ∂P

∂ eV¶T

=P 2

RT

(− 1

(Zmix −Bmix)2+

2Amix(Zmix +Bmix)£Zmix(Zmix +Bmix) +Bmix(Zmix −Bmix)

¤2)

(3)

where Θmix is defined by Eq. (6.1-27).

b) For a binary gas mixture consisting of 75 mol % propane (1) and 25% n-butane (2) at 310Kand 10 bar, calculate the value of (∂ eV /∂T )P . Take k12 = 0.003.(Answer: b) 12.383 cm3/mol.K)

165

6.7 A gas mixture consisting of 65 mol % methane (1) and 35% isobutane (2) is initially at400K and 60 bar. Estimate the temperature of the gas mixture if it is throttled to 5 bar. Usethe Peng-Robinson equation of state with k12 = 0.026.

(Answer: 372.6K)

6.8 This problem is the extension of Problem 3.33 to mixtures. For mixtures, the departurefunction for molar heat capacity at constant volume is defined by³ eCVmix − eCIGM

V

´T,P

=∂

∂T

∙³eUmix − eU IGM´T,P

¸eV (1)

a) For a gas mixture represented by the Peng-Robinson equation of state, show that the useof Eq. (6.1-39) in Eq. (1) leads to³ eCVmix − eCIGM

V

´T,P

R=

ψmix√8Bmix

ln

"Zmix +

¡1 +√2¢Bmix


#(2)

where

ψmix =1

4

kXi=1

kXj=1

xixjAij(Γi + 2ΓiΓj + Γj) (3)

in which the term Γi is defined by Eq. (6.1-17).

b) Combine Eq. (2) with Eqs. (2.2-51) and (3.2-12) to obtain³ eCPmix − eCIGMP

´T,P

R=

ψmix√8Bmix

ln

"Zmix +

¡1 +√2¢Bmix


#

+

∙1

Zmix −Bmix−

Θmix


¸21

(Zmix −Bmix)2−

2Amix(Zmix +Bmix)£Zmix(Zmix +Bmix) +Bmix(Zmix −Bmix)

¤2− 1 (4)

where Θmix is defined by Eq. (6.1-27). Note that the molar heat capacity of an ideal gasmixture is the mole fraction weighted average of the pure component values, i.e.,

eCIGMP =

kXi=1

yi eC∗pi (5)

c) For a binary gas mixture consisting of 75 mol % propane (1) and 25% n-butane (2) at 310Kand 10 bar, calculate the departure functions for molar heat capacities at constant volume andpressure. The mixture obeys the Peng-Robinson equation of state with k12 = 0.003.

(Answer: c) 0.775 J/mol.K and 10.173 J/mol.K)


6.9 It is required to prepare 35 L of a mixture containing 60 mol % component 1 and 40%component 2 at 298K. The molar volume of the mixture as a function of composition is shownbelow.

166

Molar volume (cm3/mol)

0.9 0.7 0.5 0.3 0.1

90

80

70

60

50

40

30

x1

T = 298 K

a) Estimate the volumes of pure components that must be mixed for this purpose.b) Are the like interactions stronger or weaker than the unlike interactions?(Answer: a) V1 = 12 L, V2 = 16 L)

6.10 Consider a k-component mixture in which n represents the total number of moles ofmaterial.

a) Show that the substitution ofϕmix = n eϕmix (1)

into Eq. (6.2-5) gives

ϕi = eϕmix + n

µ∂eϕmix

∂ni

¶T,P,nj 6=i

(2)

b) Use the chain rule of differentiation and express Eq. (2) as

ϕi = eϕmix +k−1Xj=1j 6=i

µ∂eϕmix

∂xj

¶T,P

µ∂xj

∂ni

¶+ n

µ∂eϕmix

∂xi

¶T,P

µ∂xi

∂ni

¶(3)

c) Noting that∂xj

∂ni= −

nj

n2and

∂xi

∂ni=1

n−

ni

n2(4)

show that Eq. (3) takes the form

ϕi = eϕmix −k−1Xj=1j 6=i

xj

µ∂eϕmix

∂xj

¶T,P

+ (1− xi)

µ∂eϕmix

∂xi

¶T,P

(5)

6.11 For a ternary system of components 1, 2, and 3, the molar volume of the mixture isgiven by eVmix = eV1 x1 + eV2 x2 + eV3 x3 +Ax1 x2 x3

where A is a function of temperature, and the terms eV1, eV2, and eV3 represent the molarvolumes of pure components 1, 2, and 3, respectively. Show that the partial molar volumes ofcomponents 1, 2, and 3 are expressed as a function of composition by the following equations:

V 1 = eV1 +Ax2 x3 (1− 2x1)

167

V 2 = eV2 +Ax1 x3 (1− 2x2)V 3 = eV3 +Ax1 x2 (1− 2x3)

6.12 The molar enthalpy of a binary liquid mixture of A and B is represented byeHmix = (α1 + β1xA)xA + (α2 + β2xB)xB

where α1, α2, β1, and β2 are dependent on temperature. Show that

HA = (α1 − β2) + 2(β1 + β2)xA − (β1 + β2)x2A

HB = (α2 + β2)− (β1 + β2)x2A

6.13 To prevent freezing of water in a car radiator, ethylene glycol (MW = 62.07 g/mol) isadded as an antifreeze. Ray and Nemethy (1973) provided the following data for the mixtureof water (1) and ethylene glycol (2) at 298K:

ρmix = 0.99722 + 0.42114x2 − 0.72022x22 + 0.62036x32 − 0.20911x42where ρmix is in g/ cm

3.

a) Calculate the molar volumes of pure water and ethylene glycol.b) If 1 L of ethylene glycol is added to 5 L of water, what will be the total volume of theresulting mixture?

c) Determine the partial molar volumes of water and ethylene glycol at x2 = 0.2.(Answer: a) 18.07 cm3/mol and 55.95 cm3/mol b) 5980 cm3 c) 17.37 cm3/mol and57.45 cm3/mol)

6.14 Partial molar volume of species i can be calculated from Eq. (6.2-4) once the equationof state is specified.

a) Since cubic equations of state are pressure-explicit, show that it is more convenient to expressEq. (6.2-4) in the form

V i = −

µ∂P

∂ni

¶T,V,nj 6=iµ

∂P

∂V

¶T,nj

(1)

b) Use the Redlich-Kwong equation of state to obtain

V i =RT

P

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

Zmix −Bmix

µ1 +

Bi

Zmix −Bmix

¶−

2kX

j=1

xjAij −AmixBi

Zmix +Bmix

Zmix (Zmix +Bmix)

1

(Zmix −Bmix)2− Amix (2Zmix +Bmix)

Z2mix (Zmix +Bmix)2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦(2)

c) A ternary gas mixture containing 45 mol % methane (1), 35% ethane (2), and 20% carbondioxide is at 350K and 25 bar. Estimate the partial molar volumes of each component if themixture obeys the Redlich-Kwong equation of state. Take kij = 0.

(Answer: 1146 cm3/mol, 1035 cm3/mol, 1072 cm3/mol)

168

6.15 For a mixture obeying the Peng-Robinson equation of state, use Eq. (1) of Problem6.14 and show that the partial molar volume of species i is given by

V i =RT

P

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

Zmix −Bmix

µ1 +

Bi

Zmix −Bmix

¶

−

2kX

j=1

xjAij −2AmixBi(Zmix −Bmix)



1

(Zmix −Bmix)2− 2Amix (Zmix +Bmix)£

Zmix(Zmix +Bmix) +Bmix(Zmix −Bmix)¤2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Consider a binary mixture consisting of 75 mol % ethane (1) and 25% n-decane (2) at 444Kand 138 bar. Estimate the partial molar volume of ethane using the Peng-Robinson equationof state. Take kij = 0.

(Answer: 153.4 cm3/mol)


6.16 For a binary mixture of components 1 and 2, the molar enthalpy of the mixture (inJ/mol) is given by eHmix = 750x1 + 380x2 + 10x1(5x1 + 17x2)

a) Show that the partial molar enthalpies of components 1 and 2 are expressed as

H1 = 920− 240x1 + 120x21 and H2 = 380 + 120x21

b) Show that∆ eHmix = 120x1x2

c) Calculate H1, H2, and ∆ eHmix at x1 = 0.4.

(Answer: c) 843.2 J/mol, 399.2 J/mol, 28.8 J/mol)

6.17 Using the following data reported by Moravkova and Linek (2005) for a binary mixtureof benzene (1) and acetophenone (2) at 25 C, determine the volume change on mixing (incm3/mol) at the given mole fractions of benzene. The molecular weights of benzene andacetophenone are 78.1 and 120.15 g/mol, respectively.

x1 ρmix ( g/ cm3 ) x1 ρmix( g/ cm

3) x1 ρmix( g/ cm3)

0.00000 1.02311 0.37918 0.97658 0.72894 0.923840.10571 1.01102 0.46261 0.96498 0.80033 0.911560.18602 1.00134 0.54838 0.95249 0.89527 0.894230.28547 0.98894 0.62477 0.94071 1.00000 0.87364

169

6.18 An isothermal mixer at 300K is fed with a stream containing 35 mol % A of flow rate2mol/ s and a stream containing 80 mol % A of flow rate 1mol/ s. Calculate the heat thatmust be removed or added per mole of solution leaving the mixer under steady conditions soas to keep the outlet stream temperature constant at 300K. The heat of mixing data for abinary liquid mixture of A and B at 300K are represented by the following equation:

∆ eHmix = xA xB(10xA + 30xB)

where ∆ eHmix is in J/mol.

(Answer: 0.765 J/mol of solution leaving the mixer)

6.19 Suppose that the pure component enthalpies of A and B in Problem 6.18 are 180 and250 J/mol, respectively.

a) Calculate the partial molar enthalpies of components A and B at infinite dilution.

b) Plot eHmix versus xA and evaluate H∞A and H

∞B graphically.

(Answer: a) 210 and 260 J/mol)

6.20 The enthalpy of solution (or heat of solution) is defined as the change in enthalpy thatresults when one mole of solute (component 1) is dissolved in a solvent (component 2). Ogawaet al. (1997) reported the following data for the enthalpy of solution, ∆Hsol, for benzene (1)in cyclohexane (2) at 298.15K:

m1 (mol/ kg) ∆Hsol ( kJ/mol of benzene)

0.0912 3.150.1529 3.160.1544 3.160.1569 3.13

where m1 is the molality of benzene. Determine the heat of mixing as a function of the molefraction of benzene.

(Answer: 24 J/mol, 40.1 J/mol, 40.5 J/mol, 40.8 J/mol)

6.21 Constantinescu and Wichterle (2002) reported the following experimental values of∆eVmix for a binary system of ethanol (1) and methyl propanoate (2) at 298.15K

x1∆eVmix

( cm3/mol)x1

∆eVmix

( cm3/mol)x1

∆eVmix

( cm3/mol)

0.0718 0.04714 0.3976 0.12064 0.7065 0.089950.1378 0.07003 0.4838 0.12086 0.7531 0.080790.2216 0.10068 0.5411 0.11743 0.8285 0.062070.2947 0.10946 0.5873 0.10664 0.9115 0.034380.3441 0.11500 0.6496 0.09741

Fit the data to the following equation

∆eVmix = x1x2

3Xi=1

Ai(x1 − x2)i−1 (1)

170

and evaluate the coefficients A1, A2, and A3.

(Answer: A1 = 0.4660, A2 = − 0.1352, A3 = 0.1226)

6.22 Serbanovic et al. (2006) reported the following data for the densities of a binary mixtureof methanol (1) and benzene (2) at 298.15K:

x1 ρmix ( g/ cm3 ) x1 ρmix( g/ cm

3) x1 ρmix( g/ cm3)

0.0000 0.873582 0.2030 0.864285 0.5503 0.8426000.0516 0.871283 0.2607 0.861329 0.5810 0.8400820.0768 0.870172 0.2978 0.859308 0.6499 0.8339590.1011 0.869074 0.3312 0.857433 0.6993 0.8290970.1241 0.868029 0.3497 0.856357 0.8000 0.8177090.1501 0.866825 0.4019 0.853164 0.8992 0.8040200.1989 0.864485 0.4493 0.850021 1.0000 0.7866940.2004 0.864414 0.4990 0.846501

a) Fit the data to the following equation:

∆eVmix = x1x2

4Xi=1

Ai(x1 − x2)i−1 (1)

and show that the coefficients are given as

A1 = − 0.0199 A2 = − 0.1342 A3 = 0.1740 A4 = − 0.1906

b) Use Eq. (6.3-15) and show that the partial molar volumes of components 1 and 2 areexpressed as a function of composition in the form

V 1 = eV1 + x22

4Xi=1

Ai(x1 − x2)i−1 + 2x1x

22

3Xi=1

iAi+1(x1 − x2)i−1 (2)

V 2 = eV2 + x21

4Xi=1

Ai(x1 − x2)i−1 − 2x21x2

3Xi=1

iAi+1(x1 − x2)i−1 (3)

and conclude that

V∞1 = eV1 + 4X

i=1

(− 1)i−1Ai and V∞2 = eV2 + 4X

i=1

Ai (4)

Also calculate the numerical values of V∞1 and V

∞2 .

(Answer: b) 41.206 cm3/mol and 89.243 cm3/mol)

6.23 The molar volume of a mixture containing methanol (1) and n-hexane (2) at 298.15Kis given by Orge et al. (1997) as

eVmix ( cm3/mol) = 40.74x1 + 131.55x2 + x1x2

3Xi=1

Ai(x1 − x2)i−1

withA1 = 2.0741 A2 = 0.3195 A3 = 1.7733

171

a) Determine the volumes of pure methanol and n-hexane required to form 600 cm3 of a solutioncontaining 30 mol % methanol.

b) Determine the total volume of the mixture when 5 moles of methanol are mixed with 3moles of n-hexane. What would the volume be if the mixture behaved ideally?

(Answer: a) 70 cm3 and 527.3 cm3 b) 602.6 cm3, 598.4 cm3)

6.24 Lepori et al. (2002) determined volume change on mixing for a binary mixture ofperfluorohexane (C6F14) and n-octane at 298.15K using a vibrating-tube densimeter. Theexperimental results are fitted to an equation of the form

∆eVmix ( cm3/mol) = x1x2

h16.00 + 3.67 (x1 − x2) + 11.70 (x1 − x2)

2i

where subscripts 1 and 2 represent perfluorohexane and n-octane, respectively. The molarvolumes of pure components are reported as

eV1 = 201.64 cm3/mol eV2 = 163.50 cm3/mola) Estimate the total volume of the mixture when 500 cm3 of perfluorohexane is mixed with800 cm3 of n-octane.

b) Calculate the partial molar volume of each component at the resulting mixture composition.c) Calculate the partial molar volume of each component at infinite dilution.(Answer: a) 1326 cm3 b) 207.53 cm3/mol and 165.92 cm3/mol c) 225.67 cm3/mol and194.87 cm3/mol)

6.25 The heat of mixing data for a binary mixture of tetramethyl-1,3-butanediamine (TMBD)and n-heptane at 298.15K were correlated by Dahmani et al. (2002) in the form

∆ eHmix ( J/mol) = x1x2

h1186.86− 338.28 (x1 − x2)

iwhere subscripts 1 and 2 represent TMBD and n-heptane, respectively.

a) Plot ∆ eHmix versus x1 and estimate the difference between the partial molar and purecomponent enthalpies of TMBD and n-heptane at x1 = 0.4 graphically.

b) Repeat the calculations by using Eq. (6.3-15).(Answer: 354.2 J/mol and 265.7 J/mol)

6.26 The heat of mixing data for water (1) and tetrahydrofuran (2) mixtures at 298.15K arereported by Kiyohara and Benson (1977) as follows:

x1 ∆ eHmix ( J/mol ) x1 ∆ eHmix( J/mol) x1 ∆ eHmix( J/mol)

0.04 151.83 0.38 23.22 0.70 − 578.390.06 208.51 0.40 − 14.21 0.80 − 706.590.10 283.04 0.42 − 52.31 0.90 − 708.480.20 289.35 0.50 − 210.49 0.94 − 581.550.30 160.86 0.58 − 367.94 0.96 − 455.060.34 94.26 0.62 − 443.05 0.98 − 266.58

172

a) Plot ∆ eHmix versus x1 and determine H1 − eH1 and H2 − eH2 graphically at x1 = 0.5.

b) How much heat would be released (or absorbed) upon mixing 6 moles of water with 14moles of tetrahydofuran at a constant temperature of 298.15K.

(Answer: a) H1 − eH1 = − 1110 J/mol H2 − eH2 = 705 J/mol b) 3217 J)


6.27 After carrying out experiments on volume change on mixing of binary systems, yourfriend proposes the following expression to express the molar volume of a mixture as a functionof composition: eVmix = (α1 + β1x

21)x1 + (α2 + β2x

22)x2 (1)

where αi and βi are constants. Since Eq. (1) has the formeVmix = V 1 x1 + V 2 x2 (2)

your friend compares Eqs. (1) and (2) and suggests that

V 1 = α1 + β1x21 and V 2 = α2 + β2x

22

Do you agree? Why?

6.28 Pure components 1 and 2 have the molar enthalpies of 200 J/mol and 350 J/mol, re-spectively. When these two components are mixed at constant temperature and pressure, thepartial molar enthalpy of component 1 is reported as

H1 ( J/mol) = 200 + 40x22 (1)

Your boss wants you to express the molar enthalpy of the mixture as a function of composition.

a) First use the Gibbs-Duhem equation, i.e.,

x1dH1

dx1+ x2

dH2

dx1= 0 (2)

to obtain the partial molar enthalpy of component 2 as

H2 = 350 + 40x21 (3)

Then use eHmix = x1H1 + x2H2 (4)

to show that eHmix = 350− 150x1 + 40x1x2 (5)

b) Rearrange Eq. (6.2-24) for enthalpy as

d eHmix

dx2− 1

x2eHmix = −

H1

x2(6)

which is a linear equation with an integrating factor of 1/x2. Multiply Eq. (6) by an integratingfactor to transform it into

d

dx2

Ã eHmix

x2

!= − H1

x22

(7)

173

Integrate Eq. (7) to obtain the solution as

eHmix = −x2

ZH1

x22

dx2 +C x2 (8)

where C is an integration constant. Show that the substitution of Eq. (1) into Eq. (8) andintegration also lead to Eq. (5).

6.29 The generalized form of the Gibbs-Duhem equation can be derived as follows:

a) Use ϕmix = n eϕmix in Eq. (6.4-4) to obtain

dϕmix = n

µ∂eϕmix

∂T

¶P,nj

dT + n

µ∂eϕmix

∂P

¶T,nj

dP +kXi=1

ϕi dni (1)

b) Show that the combination of Eq. (1) with Eq. (6.4-2) gives

−µ∂eϕmix

∂T

¶P,xj

dT −µ∂eϕmix

∂P

¶T,xj

dP +kXi=1

xi dϕi = 0 (2)

which is known as the generalized Gibbs-Duhem equation.

c) Useϕi = ϕi(T, P, x1, x2, .., xk−1) (3)

to obtain

dϕi =

µ∂ϕi∂T

¶P,xj

dT +

µ∂ϕi∂P

¶T,xj

dP +k−1Xm=1

µ∂ϕi∂xm

¶T,P

dxm (4)

d) Show that the substitution of Eq. (4) into Eq. (2) leads to

−µ∂eϕmix

∂T

¶P,xj

dT −µ∂eϕmix

∂P

¶T,xj

dP +kXi=1

xi

µ∂ϕi∂T

¶P,xj

dT +kXi=1

xi

µ∂ϕi∂P

¶T,xj

dP

+kXi=1

xi

k−1Xm=1

µ∂ϕi∂xm

¶T,P

dxm = 0 (5)

e) Note that

kXi=1

xi

µ∂ϕi∂T

¶P,xj

dT =kXi=1

∙∂(xiϕi)

∂T

¸P,xj

dT =

Ã∂

∂T

kXi=1

xiϕi

!P,xj

dT =

µ∂eϕmix

∂T

¶P,xj

dT

(6)Similarly,

kXi=1

xi

µ∂ϕi∂P

¶T,xj

dP =

µ∂eϕmix

∂P

¶T,xj

dP (7)

f) Use Eqs. (6) and (7) in Eq. (5) and conclude that

kXi=1

xi

k−1Xm=1

µ∂ϕi∂xm

¶T,P

dxm = 0 (8)

which is the generalized form of Eq. (6.4-9) for a k-component system.

174

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