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Chapter 6 Thermodynamics of Mixtures A mixture is a gaseous, liquid, or solid phase containing more than one component. The chemical process industries mainly deal with the separation of mixtures, and their properties are needed in design calculations. Mixture properties, however, cannot be determined from the weighted average of the properties of the pure components comprising the mixture. The purpose 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 the second term, for a gas mixture consisting of k species is given by Zmix = 1 +

Bmix P RT

(6.1-1)

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

k k X X

yi yj Bij

(6.1-2)

i=1 j=1

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 engineering calculations involving normal fluids, second virial cross coefficients, Bij , can be estimated from Eqs. (3.1-10)-(3.1-12) with the following modifications of Tc , ω, and Pc : Ã ! q 4 Tcij Pcj Vecj Pci Veci ωi + ωj ∗ Pcij = ³ ω ij = + Tcij = (1 − kij ) Tci Tcj ´ 2 Tci Tcj 1/3 1/3 3 e e Vci + Vcj (6.1-3) where Veci and Vecj are the critical molar volumes of components i and j, respectively1 . The ∗ , have been reported by Chueh and Prausnitz values of the binary interaction parameter2 , kij (1967a), Tsonopoulos (1979), Nishiumi et al. (1988), and Meng et al. (2007). For binary 1

2

In the absence of data, critical molar volume can be estimated from the following formula: RTc (0.290 − 0.085 ω) Vec = Pc

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

127

∗ may mixtures of chemically similar molecules, i.e., similar size, shape, and chemical nature, kij be considered zero. Chueh and Prausnitz (1967b) proposed the following equation to estimate ∗: kij q 8 Veci Vecj ∗ =1− (6.1-4) kij 1/3 1/3 (Veci + Vecj )3

In the absence of reliable data, the second virial cross coefficient, Bij , may be assumed to be 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 373 K and 4 bar. Estimate the pressure in the tank when 5 moles of n-butane (2) is added isothermally to the tank. The mixture is described by the virial equation of state and the following critical properties are provided: Component Ethane n-Butane

Vec ( cm3 / mol) 145.5 255.0

Solution From Appendix A Component Ethane n-Butane

Tc ( K)

Pc ( bar)

ω

305.3 425.0

49 38

0.099 0.199

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

(Bmix )f Pf Pf (Vemix )f =1+ RT RT

=⇒

Pf =

RT (Vemix )f − (Bmix )f

(1)

where the subscript "f" represents the final state. The term Bmix is calculated from Eq. (6.1-2) as (2) Bmix = y12 B11 + 2 y1 y2 B12 + y22 B22 The total tank volume, V , can be calculated from the initial conditions, i.e., when the tank is initially filled with ethane. The use of the virial equation of state gives µ ¶ B11 Pinitial Pinitial V RT =1+ =⇒ V = ninitial (3) Zinitial = + B11 ninitial RT RT Pinitial Thus, the molar volume of the mixture at the final state is (Vemix )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 Ethane n-Butane

Tr

B (0)

B (1)

1.222 0.878

− 0.223 − 0.437

0.065 − 0.159

128

Bii ( cm3 / mol) − 112.3 − 435.7

To calculate the second virial cross coefficient, B12 , it is first necessary to calculate the parameters defined by Eqs. (6.1-3)-(6.1-4): p 8 (145.5) (255) ∗ k12 = 1 − ¡ ¢3 = 0.013 145.51/3 + 2551/3 p Tc12 = (1 − 0.013) (305.3)(425) = 355.5 K

0.099 + 0.199 = 0.149 2 ¸ ∙ (49)(145.5) (38)(255) (4)(355.5) + = 42.04 bar =¡ ¢3 305.3 425 145.51/3 + 2551/3

ω 12 = Pc12

(0)

(1)

The parameters B12 and B12 are calculated from Eqs. (3.1-11) and (3.1-12), respectively, as (0)

B12 = 0.083 − (1)

B12 = 0.139 −

0.422 = − 0.308 (373/355.5)1.6

0.172 = − 1.57 × 10−3 (373/355.5)4.2

Finally, the second virial cross coefficient B12 is calculated from Eq. (3.1-10) as ¡ ¢i (83.14)(355.5) h − 0.308 + (0.149) − 1.57 × 10−3 = − 216.7 cm3 / mol B12 = 42.04

The term Bmix is calculated from Eq. (2) as µ ¶2 µ ¶2 µ ¶µ ¶ 5 5 3 3 (− 216.7) + Bmix = (− 112.3) + 2 (− 435.7) = − 287.6 cm3 / mol 8 8 8 8

The tank volume is calculated from Eq. (3) as ¸ ∙ (83.14)(373) − 112.3 = 22, 922 cm3 V = (3) 4 The use of Eq. (4) gives (Vemix )f =

22, 922 = 2865.3 cm3 / mol 8

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

(83.14)(373) = 12 bar 2865.3 − 287.6

Example 6.2 A gas mixture containing 90 mol % methane (1) and 10% ethane (2) is irreversibly compressed from 1 bar and 310 K to 30 bar and 350 K in an adiabatic compressor. If the gas mixture is represented by the virial equation of state, estimate the work required per mole of gas mixture. The virial coefficients are expressed as a function of temperature in the form γ ij β ij − 2 Bij = αij − T T in which the parameters are given by Estela-Uribe et al. (2003) as 129

α11 = 42.19 β 11 =

α12 = 100.19

17.13 × 103

β 12 = 52.82

γ 11 = 24.57 × 105

α22 = 89.75

× 103

β 22 = 49.80 × 103

γ 12 = 12.63 × 105

γ 22 = 96.19 × 105

with B is in cm3 / mol and T is in K. Solution From Appendix B eP∗ = 36.155 − 0.511 × 10−1 T + 2.215 × 10−4 T 2 − 1.824 × 10−7 T 3 + 4.899 × 10−11 T 4 C 1

eP∗ = 33.313 − 0.111 × 10−1 T + 3.566 × 10−4 T 2 − 3.762 × 10−7 T 3 + 11.983 × 10−11 T 4 C 2

System: Compressor

For a steady-state flow system, the first law gives fs = ∆H e W

(1)

The change in enthalpy can be determined from Eq. (3.3-13), i.e., Ã Ã ! # ! # Z T2 Z P2 " Z 0" emix emix ∂ V ∂ V ∗ eP dT + e= dP + C Vemix − T Vemix − T ∆H mix ∂T ∂T T1 0 P1 P T1

dP

P T2

(2)

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

Bmix P P Vemix =1+ RT RT

⇒

RT Vemix = + Bmix P

The partial derivative appearing in Eq. (2) is ! Ã ∂ Vemix R dBmix = + ∂T P dT

(3)

(4)

P

and the integrand in Eq. (2) becomes à ! emix ∂ V dBmix Vemix − T = Bmix − T ∂T dT

(5)

P

Therefore, it is first necessary to express Bmix as a function of temperature. From Eq. (6.1-2) Bmix = y12 B11 + 2 y1 y2 B12 + y22 B22

(6)

Expression of virial coefficients in terms of temperature leads to Bmix = ϕ1 −

ϕ ϕ2 − 32 T T

(7)

where ϕ1 = y12 α11 + 2 y1 y2 α12 + y22 α22

(8)

ϕ2 = y12 β 11 + 2 y1 y2 β 12 + y22 β 22

(9)

130

ϕ3 = y12 γ 11 + 2 y1 y2 γ 12 + y22 γ 22

(10)

The use of Eq. (7) in Eq. (5) gives Vemix − T

Ã

∂ Vemix ∂T

!

P

= ϕ1 −

2ϕ2 3ϕ3 − 2 T T

Substitution of Eq. (11) into Eq. (2) and integration give à ! Z µ ¶ T2 P P P P 2 1 2 1 e = ϕ1 (P2 − P1 ) − 2ϕ2 eP∗ ∆H − 3ϕ3 C − − dT + mix 2 2 T2 T1 T1 T2 T1

(11)

(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 pure molar heat capacities, i.e., eP∗ = y1 C eP∗ + y2 C eP∗ C 1 2 mix

4 = 35.871 − 0.471 × 10−1 T + 2.350 × 10−4 T 2 − 2.018 × 10−7 T 3 + 5.607 × 10−11 T(13)

Thus, the numerical value of the integral in Eq. (12) is Z 350 eP∗ dT = 1574 J/ mol C mix 310

Substitution of the numerical values into Eq. (12) yields ¶ µ 1 30 −6 5 −2 e − × 105 ∆H = 53.106 × 10 (30 − 1) × 10 − 2(2.388 × 10 ) 350 310 µ ¶ 30 1 − 3(2.314) − × 105 + 1574 = 1171.3 J/ mol 3502 3102

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. Table 6.1 summarizes these equations for mixtures. The practical question to be asked at this stage is how to evaluate the mixture parameters, i.e., amix and bmix , once pure component parameters are calculated. According to the van der Waals (or quadratic) mixing rules3 , the mixture parameters are given by the following composition-dependent expressions: amix =

k k X X

xi xj aij

(6.1-5)

i=1 j=1

bmix =

k k X X i=1 j=1

xi xj

µ

bi + bj 2

3

¶

=

k X

xi bi

(6.1-6)

i=1

Equations (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 Orbey and Sandler (1998).

131

Table 6.1 Cubic equations of state for mixtures4 . P

Eq. RT

VDW

RT

RK

Vemix − bmix

−

RT

SRK

PR

Vemix − bmix

RT

Vemix − bmix

Vemix − bmix

−

−

amix

27 64

2 Vemix

amix

√ Vemix (Vemix + bmix ) T

−

amix

Vemix (Vemix + bmix ) amix

Vemix (Vemix + bmix ) + bmix (Vemix − bmix )

Ã

0.42748

aii

bi ! 2

R2 Tci Pci Ã

0.42748

Ã

0.45724

Ã

R2 Tc2.5 i Pci

R2 Tc2i Pci R2 Tc2i Pci

! !

1 8

0.08664

αi

0.08664

µ

0.07780

µ

αi

On the other hand, the compressibility factor for the mixture, Zmix , is obtained from 3 2 + p Zmix + q Zmix + r = 0 Zmix

(6.1-9)

where the coefficients p, q, and r are given in Table 6.2. The dimensionless parameters for the mixture are defined by k k X X Amix = xi xj Aij (6.1-10) i=1 j=1

xi Bi

i=1

in which the terms Aij and Bi are defined in Table 6.3. 4

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

132

Pci µ

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) equations of state is defined by ⎧£ p ¢¤2 ¢¡ ¡ ⎪ SRK ⎨ 1 + 0.480 + 1.574 ω i − 0.176 ω 2i 1 − Tri (6.1-8) αi = p ¢¤2 ¡ ¢¡ ⎪ ⎩£ 2 1 + 0.37464 + 1.54226 ωi − 0.26992 ω i 1 − Tri PR

k X

RTci

!

where xi represents the mole fraction of species i. In Eqs. (6.1-5) and (6.1-6), the terms aii (or ajj ) and bi are the pure component parameters. When i 6= j, the term aij is called the unlike interaction parameter and is related to the pure component parameters as ⎧√ vDW ⎨ aii ajj aij = (6.1-7) ⎩ √ RK, SRK, PR (1 − kij ) aii ajj

Bmix =

µ

(6.1-11)

¶

RTci Pci RTci Pci RTci Pci

¶ ¶ ¶

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

q

r

van der Waals

− 1 − Bmix

Amix

− Amix Bmix

Redlich-Kwong

−1

2 Amix − Bmix − Bmix

− Amix Bmix

Soave-Redlich-Kwong

−1

2 Amix − Bmix − Bmix

− Amix Bmix

− 1 + Bmix

2 Amix − 2Bmix − 3Bmix

2 3 − Amix Bmix + Bmix + Bmix

Equation

Peng-Robinson

Table 6.3 Dimensionless parameters Aij and Bi . Aii à ! 27 Pr 64 Tr2

Equation van der Waals

Aij p Aii Ajj

i

Redlich-Kwong

0.42748

Soave-Redlich-Kwong

Peng-Robinson

Ã

Ã

0.42748

Ã

0.45724

!

Pr 2.5

Tr

Pr 2

Tr

Pr 2

Tr

p (1 − kij ) Aii Ajj

i

!

p (1 − kij ) Aii Ajj

αi

i

!

p (1 − kij ) Aii Ajj

αi

i

Bi µ ¶ 1 Pr 8 Tr i 0.08664

0.08664

0.07780

µ

Pr Tr

¶

µ

Pr Tr

¶

µ

Pr Tr

¶

i

i

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 393 K 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 Component

Tc ( K)

Pc ( bar)

ω

Propane n-Butane n-Pentane

369.9 425.0 469.8

42.5 38.0 33.6

0.153 0.199 0.251

The molar volume of the mixture is calculated from Vemix =

V RT Zmix

P

V . The Therefore, it is first necessary to estimate the compressibility factor for the mixture, Zmix reduced temperature and pressure values are

133

Component

Tr

Pr

Propane n-Butane n-Pentane

1.062 0.925 0.837

0.941 1.053 1.190

a) Using the equations given in Table 6.3, the dimensionless parameters are calculated as A11 = 0.352 A22 = 0.519 A33 = 0.717

A12 = A21 = 0.427 A13 = A31 = 0.502 A23 = A32 = 0.610

B1 = 0.111 B2 = 0.142 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 3 2 Zmix − 1.137 Zmix + 0.482 Zmix − 0.066 = 0

⇒

V Zmix = 0.26

The molar volume of the mixture is Vemix =

V RT Zmix

P

=

(0.26)(83.14)(393) = 212.4 cm3 / mol 40

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

α

Propane n-Butane n-Pentane

0.963 1.052 1.131

Using the equations given in Table 6.3, the dimensionless parameters are calculated as A11 = 0.367 A22 = 0.592 A33 = 0.878

A12 = A21 = 0.465 A13 = A31 = 0.553 A23 = A32 = 0.709

B1 = 0.069 B2 = 0.089 B3 = 0.111

The dimensionless parameters for the mixture are calculated from Eqs. (6.1-10) and (6.1-11) as and Bmix = 0.086 Amix = 0.538 Therefore, Eq. (6.1-9) becomes 3 2 Zmix − 0.914 Zmix + 0.344 Zmix − 0.038 = 0

⇒

V Zmix = 0.179

The molar volume of the mixture is Vemix =

V RT Zmix

P

=

(0.179)(83.14)(393) = 146.2 cm3 / mol 40

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 modifications. • van der Waals equation of state

For mixtures, Eq. (3.2-16) becomes ¶ Z µ Amix1 T1 Amix2 T2 e + ∆U = R − Zmix1 Zmix2 where

eV∗ C = mix

• Redlich-Kwong equation of state

k X i=1

T2 T1

e∗ dT C Vmix

(6.1-12)

eV∗ xi C i

(6.1-13)

For mixtures, Eq. (3.2-17) becomes ∙ ¶ ¶¸ Z µ µ Amix2 T2 Bmix1 Bmix2 3R Amix1 T1 e − + ∆U = ln 1 + ln 1 + 2 Bmix1 Zmix1 Bmix2 Zmix2

T2 T1

e∗ dT (6.1-14) C Vmix

• Peng-Robinson equation of state For mixtures, Eq. (3.2-18) becomes e= ∆U

" # √ ¢ ¡ T2 (damix /dT )T2 − amix2 Zmix2 + 1 + 2 Bmix2 √ ¢ √ ¡ ln 8 bmix Zmix2 + 1 − 2 Bmix2 " # ZT2 √ ¢ ¡ T1 (damix /dT )T1 − amix2 Zmix1 + 1 + 2 Bmix1 eV∗ dT √ √ ¢ ¡ ln − + C mix 8 bmix Zmix1 + 1 − 2 Bmix1

(6.1-15)

T1

where

k k 1 XX damix =− xi xj aij (Γi + Γj ) dT 2T

(6.1-16)

i=1 j=1

in which

Γi =

¡

0.37464 + 1.54226 ω i − 0.26992 ω2i

¢

r

Tri αi

(6.1-17)

Substitution of Eq. (6.1-16) into Eq. (6.1-15) gives " " # # √ ¢ √ ¢ ¡ ¡ RT2 Ωmix2 RT1 Ωmix1 Z Z + 1 + 2 B + 1 + 2 B mix mix mix mix 2 2 1 1 e= √ √ ¢ √ ¢ ¡ ¡ ∆U ln ln − √ 8 Bmix2 Zmix2 + 1 − 2 Bmix2 8 Bmix1 Zmix1 + 1 − 2 Bmix1 + where

Z

T2 T1

e∗ dT C Vmix

(6.1-18) k

Ωmix = −

k

1 XX xi xj Aij (Γi + Γj ) − Amix 2 i=1 j=1

135

(6.1-19)

6.1.2.2 Calculation of the change in enthalpy The equations developed in Section 3.3 can be extended to mixtures with the following modifications. • van der Waals equation of state For mixtures, Eq. (3.3-21) becomes µ ¶ µ ¶ Z Amix2 Amix1 e − RT1 Zmix1 − 1 − + ∆H = RT2 Zmix2 − 1 − Zmix2 Zmix1

where

eP∗ = C mix

k X i=1

T2 T1

e∗ C Pmix dT

eP∗ xi C i

(6.1-20)

(6.1-21)

• Redlich-Kwong equation of state For mixtures, Eq. (3.3-22) becomes ∙ ¶¸ µ Bmix2 3 Amix2 e ∆H = RT2 Zmix2 − 1 − ln 1 + 2 Bmix2 Zmix2

∙ ¶¸ Z µ Bmix1 3 Amix1 − RT1 Zmix1 − 1 − + ln 1 + 2 Bmix1 Zmix1

(6.1-22) T2 T1

e∗ C Pmix dT

• Peng-Robinson equation of state For mixtures, Eq. (3.3-23) becomes e = RT2 ∆H

(

Zmix2

− RT1

(

#) " √ ¢ ¡ Ωmix2 Zmix2 + 1 + 2 Bmix2 √ ¢ ¡ −1+ √ ln 8 Bmix2 Zmix2 + 1 − 2 Bmix2

Zmix1

" #) Z √ ¢ ¡ Ωmix1 Zmix1 + 1 + 2 Bmix1 √ ¢ ¡ −1+ √ ln + 8 Bmix1 Zmix1 + 1 − 2 Bmix1

T2 T1

e∗ C Pmix dT

(6.1-23)

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

6.1.2.3 Calculation of the change in entropy The equations developed in Section 3.4 can be extended to mixtures with the following modifications. • van der Waals equation of state For mixtures, Eq. (3.4-13) becomes ∆Se = R ln

µ

Zmix2 − Bmix2 Zmix1 − Bmix1

¶

+

136

Z

T2 T1

e∗ C Pmix T

dT − R ln

µ

P2 P1

¶

(6.1-24)

• Redlich-Kwong equation of state

For mixtures, Eq. (3.4-14) becomes ∙ ¶ ¶ ¶¸ µ µ µ Amix1 Zmix2 − Bmix2 Bmix2 Bmix1 R Amix2 e − − ∆S = R ln ln 1 + ln 1 + Zmix1 − Bmix1 2 Bmix2 Zmix2 Bmix1 Zmix1 +

Z

T2 T1

e∗ C Pmix T

dT − R ln

µ

P2 P1

¶

(6.1-25)

• Peng-Robinson equation of state

For mixtures, Eq. (3.4-15) becomes " # √ ¢ ¡ ¶ µ R Θmix2 Zmix2 − Bmix2 Zmix2 + 1 + 2 Bmix2 √ ¢ ¡ +√ ∆Se = R ln ln Zmix1 − Bmix1 8 Bmix2 Zmix2 + 1 − 2 Bmix2 " # Z √ ¢ ¡ Zmix1 + 1 + 2 Bmix1 R Θmix1 √ ¢ ¡ −√ ln + 8 Bmix1 Zmix1 + 1 − 2 Bmix1

T2 T1

e∗ C Pmix T

dT − R ln

µ

P2 P1

¶ (6.1-26)

where Θmix is defined by

Θmix = Ωmix + Amix

(6.1-27)

Example 6.4 A gas mixture consisting of 50 mol % propane (1), 20% n-butane (2), and 30% n-pentane (3) is compressed irreversibly from 300 K and 1 bar to 500 K and 40 bar by an adiabatic compressor. Estimate the work input per mole of gas mixture passing through the 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 Component

Tc ( K)

Pc ( bar)

ω

Propane n-Butane n-Pentane

369.9 425.0 469.8

42.5 38.0 33.6

0.153 0.199 0.251

From Appendix B eP∗ = 29.595 + 0.838 × 10−1 T + 3.256 × 10−4 T 2 − 3.958 × 10−7 T 3 + 13.129 × 10−11 T 4 C 1 eP∗ = 24.258 + 2.335 × 10−1 T + 1.279 × 10−4 T 2 − 2.437 × 10−7 T 3 + 8.552 × 10−11 T 4 C 2 eP∗ = 33.780 + 2.485 × 10−1 T + 2.535 × 10−4 T 2 − 3.838 × 10−7 T 3 + 12.977 × 10−11 T 4 C 3

System: Compressor

This is a steady-state flow process and from the first law e= Q e +W fs ∆H |{z}

(1)

0

Therefore, the work input to the compressor is equal to the change in enthalpy, which can be determined from Eq. (6.1-23). 137

• State 1 (T1 = 300 K, P1 = 1 bar) The values of reduced temperature, reduced pressure, and α are Component Propane n-Butane n-Pentane

Tr

Pr

α

0.811 0.706 0.639

0.024 0.026 0.030

1.124 1.226 1.322

Using the equations given in Table 6.3, the dimensionless parameters are calculated as A11 = 0.019 A22 = 0.029 A33 = 0.044

A12 = A21 = 0.023 A13 = A31 = 0.028 A23 = A32 = 0.035

B1 = 2.302 × 10−3 B2 = 2.865 × 10−3 B3 = 3.653 × 10−3

The dimensionless parameters for the mixture are calculated from Eqs. (6.1-10) and (6.1-11) as Bmix1 = 2.82 × 10−3 Amix1 = 0.027 Therefore, Eq. (6.1-9) becomes 3 2 Zmix − 0.997 Zmix + 0.021 Zmix1 − 6.817 × 10−5 = 0 1 1

⇒

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) gives Ωmix1 = − 0.041 • State 2 (T2 = 500 K, P2 = 40 bar) The values of reduced temperature, reduced pressure, and α are Component Propane n-Butane n-Pentane

Tr

Pr

α

1.352 1.176 1.064

0.941 1.053 1.190

0.813 0.890 0.953

Using the equations given in Table 6.3, the dimensionless parameters are calculated as A11 = 0.191 A22 = 0.310 A33 = 0.459

A12 = A21 = 0.243 A13 = A31 = 0.288 A23 = A32 = 0.371

B1 = 0.054 B2 = 0.070 B3 = 0.087

The dimensionless parameters for the mixture are calculated from Eqs. (6.1-10) and (6.1-11) as Bmix2 = 0.067 Amix2 = 0.281 Therefore, Eq. (6.1-9) becomes 3 2 Zmix − 0.933 Zmix + 0.134 Zmix2 − 0.014 = 0 2 2

138

⇒

Zmix2 = 0.785

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) gives Ωmix2 = − 0.5 The heat capacity of the mixture can be calculated as eP∗ + y2 C eP∗ + y3 C eP∗ eP∗ = y1 C C 1 2 3 mix

= 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

eP∗ dT = 23, 428 J/ mol C mix

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

e = − 3343 + 164.3 + 23, 428 = 20, 249 J/ mol ∆H 6.1.3 Departure Functions The departure functions developed for pure components in Section 3.6 can be extended to mixtures. For mixtures, Eqs. (3.6-3), (3.6-4), and (3.6-5) take the form e = − (H e mix − H e IGM )T ,P + ∆H 1 1 ∆Se = − (Semix − SeIGM )T1 ,P1 +

Z

T2 T1

Z

T2 T1

e∗ C Pmix T

e = − (U emix − U e IGM )T ,P + ∆U 1 1

Z

T2 T1

e e IGM )T ,P e∗ C 2 2 Pmix dT + (Hmix − H

dT − R ln

µ

P2 P1

¶

+ (Semix − SeIGM )T2 ,P2

emix − U e IGM )T ,P e∗ dT + (U C 2 2 Vmix

(6.1-28)

(6.1-29)

(6.1-30)

where the departure functions are given by the following equations depending on the type of the equation of state. • van der Waals equation of state For mixtures, Eqs. (3.6-12), (3.6-13), and (3.6-14) become ³ ´ e IGM e mix − H H

T,P

= Zmix − 1 −

Amix Zmix

(6.1-31)

T,P

= ln (Zmix − Bmix )

(6.1-32)

RT

³ ´ Semix − SeIGM R

³ ´ e IGM emix − U U

T,P

RT

139

=−

Amix Zmix

(6.1-33)

• Redlich-Kwong equation of state For mixtures, Eqs. (3.6-15), (3.6-16), and (3.6-17) become ³ ´ e IGM e mix − H ¶ µ H 3 Amix Bmix T,P = Zmix − 1 − ln 1 + RT 2 Bmix Zmix ³ ´ IGM e e Smix − S

¶ µ 1 Amix Bmix = ln (Zmix − Bmix ) − ln 1 + R 2 Bmix Zmix ³ ´ e IGM emix − U ¶ µ U 3 Amix Bmix T,P =− ln 1 + RT 2 Bmix Zmix T,P

(6.1-34)

(6.1-35)

(6.1-36)

• Peng-Robinson equation of state For mixtures, Eqs. (3.6-18), (3.6-19), and (3.6-20) become ³ ´ " # √ ¢ ¡ e IGM e mix − H H Ωmix Zmix + 1 + 2 Bmix T,P √ ¢ ¡ = Zmix − 1 + √ ln RT 8 Bmix Zmix + 1 − 2 Bmix ³ ´ Semix − SeIGM

T,P

R

" # √ ¢ ¡ Zmix + 1 + 2 Bmix Θmix √ ¢ ¡ = ln(Zmix − Bmix ) + √ ln 8 Bmix Zmix + 1 − 2 Bmix

³ ´ e IGM emix − U U

T,P

RT

" # √ ¢ ¡ Ωmix Zmix + 1 + 2 Bmix √ ¢ ¡ =√ ln 8 Bmix Zmix + 1 − 2 Bmix

(6.1-37)

(6.1-38)

(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 the Redlich-Kwong equation of state. Solution From Appendix A Component

Tc ( K)

Pc ( bar)

Methane Ethane

190.6 305.3

46.1 49.0

For a steady-state flow system, the first law gives fs = ∆H e W

The change in enthalpy can be calculated from Eq. (6.1-28). • State 1 (T1 = 310 K, P1 = 1 bar)

The values of reduced temperature and reduced pressure are 140

(1)

Component Methane Ethane

Tr

Pr

1.626 1.015

0.022 0.020

Using the equations given in Table 6.3 and taking kij to be equal to zero, the dimensionless parameters are calculated as A11 = 2.790 × 10−3 A22 = 8.237 × 10−3

A12 = A21 = 4.794 × 10−3

B1 = 1.172 × 10−3 B2 = 1.707 × 10−3

The dimensionless parameters for the mixture are calculated from Eqs. (6.1-10) and (6.1-11) as Bmix1 = 1.225 × 10−3 Amix1 = 3.205 × 10−3 Therefore, Eq. (6.1-9) becomes 3 2 − Zmix + 1.978 × 10−3 Zmix1 − 3.926 × 10−6 = 0 Zmix 1 1

⇒

Zmix1 = 0.998

The enthalpy departure function is calculated from Eq. (6.1-34) as µ ¶¸ ¶ µ ∙ 1.225 × 10−3 3 3.205 × 10−3 IGM e e ln 1 + )T1 ,P1 = (8.314)(310) 0.998 − 1 − (Hmix − H 2 1.225 × 10−3 0.998 = − 17.6 J/ mol • State 2 (T2 = 350 K, P2 = 30 bar) The values of reduced temperature and reduced pressure are Component Methane Ethane

Tr

Pr

1.836 1.146

0.651 0.612

Using the equations given in Table 6.3, the dimensionless parameters are calculated as A11 = 0.061 A22 = 0.186

A12 = A21 = 0.106

B1 = 0.031 B2 = 0.046

The dimensionless parameters for the mixture are calculated from Eqs. (6.1-10) and (6.1-11) as Bmix2 = 0.033 Amix2 = 0.07 Therefore, Eq. (6.1-9) becomes 3 2 − Zmix + 0.036 Zmix2 − 2.31 × 10−3 = 0 Zmix 2 2

⇒

Zmix2 = 0.965

The enthalpy departure function is calculated from Eq. (6.1-34) as µ ¶ µ ¶¸ ∙ 0.07 0.033 3 IGM e e mix − H ln 1 + )T2 ,P2 = (8.314)(350) 0.964 − 1 − (H 2 0.033 0.965 = − 416.1 J/ mol 141

From Example 6.2

Z

350

310

eP∗ dT = 1574 J/ mol C mix

Substitution of the numerical values into Eq. (6.1-28) leads to e = 17.6 + 1574 − 416.1 = 1175.5 J/ mol ∆H Example 6.6 A carbon dioxide stream at 10 bar and 350 K and a methane stream at 5 bar and 270 K are continuously fed to an adiabatic mixer in the mole ratio 1:3. If the exit gas mixture is at 3 bar, estimate its temperature. Assume that the pure gases as well as the mixture obey the Redlich-Kwong equation of state. Solution From Appendix A, the critical constants are given below: Component

Tc ( K)

Pc ( bar)

Carbon dioxide Methane

304.2 190.6

73.8 46.1

From Appendix B eP∗ = 29.268 − 0.224 × 10−1 T + 2.653 × 10−4 T 2 − 4.153 × 10−7 T 3 + 20.057 × 10−11 T 4 C CO 2

eP∗ C CH4

= 36.155 − 0.511 × 10−1 T + 2.215 × 10−4 T 2 − 1.824 × 10−7 T 3 + 4.899 × 10−11 T 4

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

∆H = Q + Ws = 0 |{z} |{z}

(1)

0

0

The process path for calculating enthalpy change is shown in the figure below: Pure CO2 at 10 bar & 350 K

Pure CH4 at 5 bar & 270 K

A

B

Pure ideal CO2 at 10 bar & 350 K

Pure ideal CH4 at 5 bar & 270 K

C

D

Pure ideal CO2 at 3 bar & T3

Pure ideal CH4 at 3 bar & T3

E Ideal gas mixture at 3 bar & T3

F Gas mixture at 3 bar & T3

Taking 1 mol of CO2 as a basis, the enthalpy change is expressed as e B + ∆H e C + 3 ∆H e D + 4 ∆H e E + 4 ∆H eF e A + 3 ∆H ∆H = ∆H

Considering the following states

142

(2)

State

T ( K)

P ( bar)

1 2 3

350 270 T3

10 5 3

Eq. (2) is expressed in the form IG IG e CO − H e CO e CH − H e CH ∆H = − (H ) − 3 (H ) + 2 4 2 T1 ,P1 4 T2 ,P2

Z

T3

350

eP∗ dT + 3 C CO 2

Z

T3

eP∗ dT C CH

270 e IGM

e mix − H + 4 (H

4

)T3 ,P3 (3)

Since there is no interaction between ideal gas molecules, mixing of ideal gases produces no e E = 0. Substitution of Eq. (3) into Eq. (1) gives change in enthalpy. Thus, ∆H IG IG e CO − H e CO e CH − H e CH − (H ) − 3 (H ) + 2 4 2 T1 ,P1 4 T2 ,P2

Z

T3

350

Departure function for pure CO2

eP∗ dT + 3 C CO 2

Z

T3

270

eP∗ dT C CH

e IGM

e mix − H + 4 (H

4

)T3 ,P3 = 0 (4)

At 350 K and 10 bar A = 0.041

B = 0.010

Z = 0.969

The use of Eq. (3.6-15) gives IG e CO e CO − H ) = − 274 J/ mol (H 2 2 T1 ,P1

(5)

Departure function for pure CH4 At 270 K and 5 bar

B = 6.634 × 10−3

A = 0.019

Z = 0.987

The use of Eq. (3.6-15) gives IG e CH e CH4 − H ) = − 121.6 J/ mol (H 4 T2 ,P2

(6)

The remaining terms in Eq. (4) cannot be calculated since T3 is not known. Therefore, a trial-and-error procedure will be used to estimate T3 . Assuming T3 = 288.3 K, the values of the integrals in Eq. (4) are Z 288.3 eP∗ dT = − 2327 J/ mol (7) C CO 2

350

Z

288.3

270

eP∗ dT = 649.3 J/ mol C CH 4

(8)

Departure function for the mixture Taking k12 to be equal to zero, the dimensionless parameters and the compressibility factor become Bmix = 3.725 × 10−3 Zmix = 0.992 Amix = 0.012 143

The use of Eq. (6.1-33) gives e IGM )T ,P = − 63.4 J/ mol e mix − H (H 3 3

(9)

If the assumed temperature is correct, then the summation of the terms on the left-hand side of Eq. (4) should be zero. Substitution of Eqs. (5), (6), (7), (8), and (9) into the left-hand side of Eq. (4) gives 274 + 3 (121.6) − 2327 + 3 (649.3) − 4 (63.4) = 6.1 which 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 equation of state should be used for a given mixture?, (ii) Which mixing rule should be employed in 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 selection by 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 in the natural-gas, petroleum, and petrochemical industries (Prausnitz and Tavares, 2004). For more 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 contents of these two beakers are mixed, the total volume of water will be 70 cm3 . Since volumes of a pure 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 solution turns out to be 67 cm3 , less than the sum of pure liquid volumes. The reason for this is the fact that water and ethanol molecules interact with each other and pack differently than with like molecules6 . If 1 and 2 represent water and ethanol, respectively, then Vmix 6= n1 Ve1 + n2 Ve2

(6.2-1)

Vmix = n1 V 1 + n2 V 2

(6.2-2)

in which Ve1 and Ve2 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 quantity V i so that the additive property holds, i.e.,

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 of components makes the mixture amenable to modeling. The criteria used in lumping and the estimation of pseudo-component properties are beyond the scope of this text. 6 Consider a box of volume V1 filled with small metal balls and another box of volume V2 filled with rice. If the contents of these two boxes are mixed, the final volume will be less than V1 + V2 .

144

or, in general, Vmix =

k X

ni V i

(6.2-3)

i=1

The term V i is called the partial molar volume of species i and is defined by ¶ µ ∂Vmix Vi = ∂ni T,P,nj6=i

(6.2-4)

The subscripts on the partial derivative mean that pressure, temperature, and number of moles of all components other than i are kept constant. In general, a partial molar property, ϕi , is the rate at which an extensive property of the entire mixture, ϕmix , changes with the number of moles of component i in the mixture when temperature, pressure, and number of moles of all components other than i are kept constant, i.e., ¶ µ ∂ϕmix ϕi = (6.2-5) ∂ni T,P,nj6=i It should be kept in mind that ϕi is a property of the mixture and not simply a property of component i. The partial molar quantity is an intensive property that is generally dependent on 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 =

k X

ni ϕi

(6.2-6)

i=1

or dividing each term by the total number of moles gives ϕ e mix =

k X

xi ϕi

(6.2-7)

i=1

Example 6.7 Pecar and Dolecek (2005) experimentally determined the densities and the partial molar volumes of a binary mixture of ethanol (1) and water (2) as a function of temperature and pressure. In a mixture with an ethanol mole fraction of 0.35 at 298 K 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

Vemix = x1 V 1 + x2 V 2

(1)

Note that the molar volume of the mixture is given by Mmix Vemix = ρmix

(2)

in which Mmix and ρmix represent the molecular weight and the density of the mixture, respectively. The molecular weight of the mixture is given by Mmix = x1 M1 + x2 M2 = (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 27.84 = 31.21 cm3 / mol Vemix = 0.8919

The partial molar volume of water is then determined from Eq. (1) as 31.21 = (0.35)(57.31) + 0.65 V 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 number of moles of all components other than i constant gives µ µ ¶ ¶ ¶ µ ∂U ∂V ∂H = +P (6.2-9) ∂ni T,P,nj6=i ∂ni T,P,nj6=i ∂ni T,P,nj6=i Application of the definition of a partial molar quantity leads to Hi = U i + P V i

(6.2-10)

Ai = U i − T S i

(6.2-11)

Gi = H i − T S i

(6.2-12)

Similarly, we can write

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 slope In this method, a known amount of a compound, say one mole of component 1, is placed in a graduated cylinder. By keeping temperature and pressure constant throughout the experiment, the volume of the mixture is recorded by adding increasing amounts of component 2. Then either the total volume of the mixture, Vmix , is plotted as a function of n2 or an equation is 2 e fitted to the volume data in the form Vmix = Ve1 + α n2 + β n1.5 2 + γ n2 + ... , in which V1 is the molar volume of pure 1. At any given value of n2 , the slope of the tangent to the curve gives the partial molar volume of component 2, i.e., ¶ µ ∂Vmix =V2 (6.2-13) Slope = ∂n2 T,P,n1 If an equation is fitted, then differentiation of the equation with respect to n2 gives the partial molar volume of component 2 as ¶ µ ∂Vmix V2 = = α + 1.5 β n0.5 (6.2-14) 2 + 2 γ n2 + ... ∂n2 T,P,n1 146

Once V 2 is determined, the partial molar volume of component 1 can be obtained from Eq. (6.2-6) as Vmix − n2 V 2 V1 = (6.2-15) n1 Example 6.8 When NaCl is dissolved in water at 298 K, the volume of the resulting mixture is represented as a function of molality7 , m, in the form Vmix = 1003 + 16.62 m + 1.77 m1.5 + 0.12 m2

(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 H2 O and NaCl, respectively. Choosing 1 kg of water as a basis, m = n2 . Thus, differentiation of Eq. (1) with respect to m gives the partial molar volume of NaCl as V 2 = 16.62 + 2.655 m0.5 + 0.24 m = 16.62 + 2.655(0.01)0.5 + 0.24(0.01) = 16.89 cm3 / mol From Eq. (6.2-15) V1 =

Vmix − n2 V 2 1003.2 − (0.01)(16.89) = 18.05 cm3 / mol = n1 1000/18

• Method of tangent intercepts The total volume of a binary mixture is given by Vmix = (n1 + n2 ) Vemix

The partial molar volume of component 1 is ¶ µ ½ i¾ ∂Vmix ∂ h e (n1 + n2 ) Vmix V1 = = ∂n1 T,P,n2 ∂n1 T,P,n2

(6.2-16)

(6.2-17)

Carrying out the differentiation gives

V 1 = Vemix + n

Ã

∂ Vemix ∂n1

!

T,P,n2

By the application of the chain rule, Eq. (6.2-18) becomes à ! µ ¶ emix ∂ V ∂x2 e V 1 = Vmix + n ∂x2 ∂n1 T,P,n2 T,P,n2

7

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

147

(6.2-18)

(6.2-19)

Since x2 = n2 /(n1 + n2 ), then

µ

∂x2 ∂n1

¶

T,P,n2

=−

x2 n

(6.2-20)

!

(6.2-21)

so that Eq. (6.2-19) takes the form V 1 = Vemix − x2

Ã

∂ Vemix ∂x2

T,P,n2

Suppose that Vemix 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 = ~ Vmix

BC ∂ Vemix = − tan α = − ∂x2 AC

~ V1

~ V2

B

Figure 6.1

A

α

C

D

(6.2-22)

0

x2

1

Graphical determination of partial molar volumes.

Substitution of Eq. (6.2-22) into Eq. (6.2-21) leads to à ! emix ∂ V V 1 = Vemix − x2 ∂x2 T,P,n2 ¶ µ BC = CD + BC = BD = CD − AC − AC

(6.2-23)

Thus, the intercept of the tangent line at x2 = 0 is the partial molar volume of component 1. Similarly, the intercept of the tangent line at x2 = 1 gives the partial molar volume of component 2. For a binary system8 , Eq. (6.2-21) can be generalized as µ ¶ de ϕmix ϕ1 = ϕ e mix − x2 constant T and P (6.2-24) dx2 and

ϕ2 = ϕ e mix − x1

µ

de ϕmix dx1

¶

constant T and P

(6.2-25)

Since x1 + x2 = 1 for a binary system, then de ϕ de ϕmix = − mix dx1 dx2 8

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

148

(6.2-26)

Thus, depending on the functional form of ϕ e mix , 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 properties become and lim ϕi = ϕ ei (6.2-27) lim ϕi = ϕ∞ i where

ϕ∞ i

xi →0

xi →1

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 component 1 and 90% of component 2. The stock solution available in the laboratory has the composition of 60 mol % component 1 and 40% of component 2. Determine the volumes of the stock solution and pure component 2 that must be mixed. Data: The molar volume of the mixture (in cm3 / mol) is experimentally determined as Vemix = 60 x1 + 110 x2 + 5 x1 x2

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 the number of moles of components 1 and 2 are expressed as ⎞ ⎛ ! Ã 1000 X ⎠ xf1 inal (1) (0.6) = ⎝ stock f inal e V Ve mix

Ã

X

stock Vemix

!

mix

(1 − 0.6) +

⎛

⎞

´ 1000 ⎠ ³ Y =⎝ 1 − xf1 inal Ve Ve f inal 2

(2)

mix

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

x1 =

ω1 M1 ω1 ω2 + M1 M2

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

xf1 inal =

0.1 40 0.1 0.9 + 40 60

= 0.143

The molar volume of the final solution is f inal = (60)(0.143) + (110)(1 − 0.143) + (5)(0.143)(1 − 0.143) = 103.46 cm3 / mol Vemix

The molar volume of the stock solution is

stock Vemix = (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 Ve2 = lim Vemix = 110 cm3 / mol x2 →1

Substitution of the numerical values into Eq. (1) gives X = 187.1 cm3 Then the use of Eq. (2) results in Y = 809.8 cm3

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

Example 6.10 The molar enthalpy of a binary liquid mixture of 1 and 2 at constant temperature and pressure is given by e mix = 400 x1 + 250 x2 + x1 x2 (15 x1 + 45 x2 ) H

e mix is in J/ mol. Calculate where H

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 e 2 = lim H e mix = 250 J/ mol H

e mix = 400 J/ mol e 1 = lim H H x1 →1

x2 →1

b) From Eq. (6.2-24)

e mix − x2 H1 = H

e mix with respect to x2 is The derivative of H

Ã

e mix dH dx2

!

e mix dH = − 400 + 250 + (1 − 2 x2 )(15 x1 + 45 x2 ) + x1 x2 (− 15 + 45) dx2 = − 150 + 15 x1 + 45 x2 − 90 x22

(1)

(2)

Thus, Eq. (1) becomes H 1 = 400 x1 + 250 x2 + x1 x2 (15 x1 + 45 x2 ) − x2 (− 150 + 15 x1 + 45 x2 − 90 x22 ) = 400 − 15 x22 + 60 x32

The partial molar enthalpy of component 2 can be calculated from Eq. (6.2-25) Ã ! e mix d H e mix − x1 H2 = H dx1 150

(3)

(4)

Since

e mix e mix dH dH =− dx1 dx2

Eq. (4) becomes

e mix + x1 H2 = H

or

Ã

e mix dH dx2

(5) !

(6)

H 2 = 400 x1 + 250 x2 + x1 x2 (15 x1 + 45 x2 ) + x1 (− 150 + 15 x1 + 45 x2 − 90 x22 ) = 250 + 75 x21 − 60 x31

(7)

When x1 = 0.35 H 1 = 410.14 J/ mol

and

H 2 = 256.62 J/ mol

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

H 1 = lim H 1 = 445 J/ mol x1 →0

∞

H 2 = lim H 2 = 265 J/ mol x2 →0

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 ) =

¶ k µ X ∂f i=1

∂xi

xi

(6.2-29)

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

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

¶ k µ X ∂f i=1

∂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 can be expressed as ¶ k µ X ∂f 0= (6.2-31) xi When f is an intensive property ∂xi i=1

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 )

151

(6.2-32)

At constant temperature and pressure, application of Eq. (6.2-30) gives ¶ k µ X ∂ϕmix ϕmix = ni ∂ni T,P,nj6=i

(6.2-33)

i=1

or

ϕmix =

k X

ϕi ni

(6.2-34)

i=1

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 different from the sum of the properties of the pure liquids, i.e., ϕmix 6=

k X i=1

ni ϕ ei

(6.3-1)

The inequality sign in Eq. (6.3-1) can be removed by the introduction of a "correction factor", ∆ϕmix , so that k X ni ϕ e i + ∆ϕmix (6.3-2) ϕmix = i=1

The term ∆ϕmix is called property change on mixing. In other words, property change on mixing represents the difference between the actual property of the mixture and the total property 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 ϕ e mix =

k X i=1

Rearrangement of Eq. (6.3-3) gives

e mix − ∆e ϕmix = ϕ

xi ϕ e i + ∆e ϕmix

k X i=1

xi ϕ ei =

k X i=1

(6.3-3)

xi (ϕi − ϕ ei)

(6.3-4)

As shown in Figure 6.2, property change on mixing may take positive or negative values depending on the interactions between molecules. Note that the dotted line represents the case when ∆e ϕmix is zero. % mix = x1φ % 1 + x 2φ %2 = φ % 1 + x2 (φ %2 − φ % 1) φ

ϕ~mix ϕ~2

Δϕ~mix > 0

Δϕ~mix < 0

ϕ~1

0

Figure 6.2

x2

1

Property change on mixing for a binary system.

152

Among the property changes on mixing, chemical engineers are particularly interested in the isothermal volume change on mixing, ∆Vemix , and the isothermal enthalpy change on mixing e mix , both of which have to be determined experimentally. (or simply the heat of mixing), ∆H When ∆Vmix > 0, the interactions between unlike molecules are weaker than those between like molecules. On the other hand, if the interactions between unlike molecules are stronger than 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 change in bond energies. If pure components A and B are mixed to form a mixture, it is necessary to break A-A and B-B bonds to create A-B bonds as shown in Figure 6.3. Bond breaking requires or absorbs energy, i.e., an endothermic process. On the other hand, bond formation releases energy, i.e., an exothermic process.

+ A−A

B −B A −B

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 are more stable than pure components and ∆Hmix < 0. On the other hand, when A-A and B-B bonds are stronger than A-B bonds, pure components are more stable than the components in the mixture and ∆Hmix > 0. Example 6.11 When 50 cm3 of liquid A is mixed with 60 cm3 of liquid B, the resulting solution has a volume of 113 cm3 .What can be said about the relative magnitudes of V A and VeA , as well as V B and VeB ? Solution

Since ∆Vmix = 113 − (50 + 60) = 3 cm3 > 0, a Vemix versus composition diagram can be qualitatively drawn as shown in the figure below. From the graph one can easily conclude that V A > VeA and V B > VeB ~ Vmix

VA

~ VA

VB

~ VB 0

x*A

xA

1

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 as well as the densities of the pure components are measured by a densitometer. Then the volume change on mixing is determined from the following equation: 153

∆Vemix =

k X

xi Mi

i=1

ρmix

−

k X xi Mi i=1

ρi

=

k X

xi Mi

i=1

µ

1 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 mixture of cyclohexanamine (1) and n-heptane (2) at 303.15 K 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, respectively. 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 0.85820 and 0.67532 g/ cm3 , respectively. The volume change on mixing can be calculated from Eq. (6.3-5), i.e., µ ¶ µ ¶ 1 1 1 1 e − − ∆Vmix = x1 M1 + x2 M2 ρmix ρ1 ρmix ρ2 The calculated values are given below:

x1

0.3006

0.5000

0.7996

∆Vemix ( 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 shown in Figure 6.4. Note that the flow calorimeter is simply a steady-state mixing device.

Pure 1 at T Mixture at T Pure 2 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 flow rates of n˙ 1 and n˙ 2 , respectively. They are mixed thoroughly by a stirrer to maintain uniform temperature within the calorimeter. Heat is added or removed through the coils placed in the calorimeter 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

and

n˙ 1 + n˙ 2 = n˙ mix

(6.3-6)

˙s e mix − n˙ 1 H e 1 − n˙ 2 H e 2 = Q˙ + W n˙ mix H

(6.3-7)

˙ s to be negligible and dividing Eq. (6.3-7) by n˙ mix give Assuming W ´ ³ Q˙ e mix − x1 H e 1 + x2 H e2 = H {z } n˙ 1 + n˙ 2 |

(6.3-8)

e mix ∆H

Therefore, once the molar flow rates of the pure streams are fixed, one can determine the heat ˙ Carrying out experiments of mixing from Eq. (6.3-8) by measuring the heat transfer rate, Q. at 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 enthalpy of the mixture from the following equation: e mix = H

k X i=1

e mix e i + ∆H xi H

(6.3-9)

Thus, it is possible to plot the enthalpy of the mixture as a function of composition. Such plots, also known as enthalpy-concentration diagrams, are very useful in determining energy balances. Example 6.13 An isothermal mixer at 294 K is fed with a pure sulfuric acid stream of flow rate 160 kg/ min and an aqueous solution stream containing 38 weight % sulfuric acid of flow rate 150 kg/ min. Estimate the rate of heat that must be added or removed so as to keep the temperature of the exit stream at 294 K under steady conditions. The heat of mixing data for the sulfuric acid (1) and water (2) mixture at 294 K is represented by Ross (1952) as h i e mix ( kJ/ mol) = x1 x2 − 53.531 + 20.869 (x1 − x2 ) ∆H

The pure component enthalpies for sulfuric acid and water are 1.596 and 1.591 kJ/ mol, respectively. Solution The flow diagram of the mixing process is given below:

A

Pure H2SO4 160 kg/min MIXER T = 294 K

B ω1 = 0.38 150 kg/min

155

C

The heat that must be removed or added to the mixer can be determined from the steady-state energy balance as ˙s (1) ∆H˙ = Q˙ + W |{z} ∼0

or

e C − n˙ A H e A − n˙ B H eB Q˙ = n˙ C H

(2)

The molar enthalpies from streams B and C can be calculated from e mix e 1 + x2 H e 2 + ∆H e mix = x1 H H

h i = 1.596 x1 + 1.591 x2 + x1 x2 − 53.531 + 20.869 (x1 − x2 )

(3)

To calculate the weight fraction of sulfuric acid in stream C, first it is necessary to write the overall material balance as ˙B =m ˙C m ˙ A+m

⇒

m ˙ C = 160 + 150 = 310 kg/ min

(4)

On the other hand, the material balance for sulfuric acid is m ˙ A+m ˙ B (ω 1 )B = m ˙ C (ω 1 )C

⇒

(ω 1 )C =

160 + (150)(0.38) = 0.7 310

(5)

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

x1 =

ω1 M1

(6)

ω1 ω2 + M1 M2

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

(x1 )B =

0.38 98 0.38 1 − 0.38 + 98 18

= 0.1

(x1 )C =

0.7 98 0.7 1 − 0.7 + 98 18

= 0.3

The use of Eq. (3) yields n e A = 1.596 kJ/ mol Stream A H

½ £ ¤ e B = (1.596)(0.1) + (1.591)(0.9) + (0.1)(0.9) − 53.531 + (20.869)(0.1 − 0.9) H Stream B = − 4.729 kJ/ mol ½ £ ¤ e C = (1.596)(0.3) + (1.591)(0.7) + (0.3)(0.7) − 53.531 + (20.869)(0.3 − 0.7) H Stream C = − 11.402 kJ/ mol

To determine the molar flow rates of streams B and C, we have to calculate their molecular weights: 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 ¶ µ ¶ µ ¶ µ 160 150 310 ˙ (− 11.402) − (1.596) − (− 4.729) Q= 42 × 10−3 98 × 10−3 26 × 10−3 = − 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-concentration diagram for sulfuric acid (1) and water (2) mixtures at 294 K. Then prove that it is always safer to add acid to water rather than water to acid. Solution The molar enthalpy of the mixture is given by i h e mix ( kJ/ mol) = 1.596 x1 + 1.591 x2 + x1 x2 − 53.531 + 20.869 (x1 − x2 ) H

(1)

e mix , versus mole fraction of water, x2 , is shown The plot of the molar enthalpy of the mixture, H below. 5 1.596

0

Hmix( x2) − 5

− 10

− 12.265 − 15

0

0.2

0.4

0

0.6 x2

0.8

1 1

Using Eqs. (6.2-24) and (6.2-25), the partial molar enthalpies are obtained as a function of composition in the form H 1 = 1.596 − 53.531 x22 + 62.607 x1 x22 − 20.869 x32

(2)

H 2 = 1.591 − 53.531 x21 − 62.607 x21 x2 + 20.869 x31

(3)

The partial molar enthalpies at infinite dilution are ∞

H 1 = lim H 1 = 1.596 − 53.531 − 20.869 = − 72.804 kJ/ mol x1 →0

∞

H 2 = lim H 2 = 1.591 − 53.531 + 20.869 = − 31.071 kJ/ mol x2 →0

The partial molar enthalpies at infinite dilution can also be determined from the intercepts of the tangents drawn to the curve at x2 = 0 and x2 = 1 as shown in the figure below. 157

~ H1

~ H2

0

~ H mix

H2∞

H1∞

0

x2

1

The heat of mixing is given by e mix = x1 (H 1 − H e 1 ) + x2 (H 2 − H e2) ∆H

(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 H2 SO4 is added to 1000 g of H2 O, then 1 98 = 1.84 × 10−4 x1 = 1000 1 + 98 18 ∞ e 2 . Thus, Eq. (4) simplifies to Since x1 ¿ x2 , then H 1 = H 1 and H 2 ' H ¯ ¢ ¡ ∞ e mix ¯¯ e 1 ) = 1.84 × 10−4 (− 72, 804 − 1596) = − 13.7 J/ mol ∆H = x1 (H 1 − H Process A

(3)

Process B (Adding a small amount of water to a very large amount of acid) If 1 g of H2 O is added to 1000 g of H2 SO4 , then 1 18 = 5.42 × 10−3 x2 = 1000 1 + 98 18

e 1 and H 2 = H ∞ Since x2 ¿ x1 , then H 1 ' H 2 . Thus, Eq. (4) simplifies to ¯ ¡ ¢ ∞ e mix ¯¯ e 2 ) = 5.42 × 10−3 (− 31, 071 − 1591) = − 177 J/ mol ∆H = x2 (H 2 − H Process B

Note that mixing is exothermic for both processes. However, more heat is released when water is 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, ∆e ϕmix , is determined experimentally as a function of mole 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 by the following equation: (6.3-10) ∆e ϕmix = x1 x2 f (x1 , x2 ) The term x1 x2 originates from the fact that ∆e ϕmix 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 express f (x1 , x2 ) in polynomial form as N ∆e ϕmix X = Ai (x1 − x2 )i−1 = A1 + A2 (x1 − x2 ) + A3 (x1 − x2 )2 + ... x1 x2

(6.3-11)

i=1

which is known as the Redlich-Kister type expansion. e mix for Example 6.15 Conti et al. (1997) reported the following experimental values of ∆H a binary system of chloroform (1) and cyclohexane (2) at 298 K: x1 0.0563 0.0563 0.1643 0.2520 0.2666

e mix ( J/ mol) ∆H

x1

150.8 147.4 375.2 500.1 519.0

0.2666 0.3661 0.4732 0.4732 0.5740

e mix ( J/ mol) ∆H 520.2 609.6 630.9 644.3 618.7

x1 0.5740 0.6690 0.7587 0.8435 0.9238

e ∆H mix ( J/ mol) 618.1 567.7 478.3 351.5 199.7

Fit the data to the equation of the form e mix = x1 x2 ∆H

4 X i=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 be rearranged in the form of

or

e mix ∆H = A1 + A2 (x1 − x2 ) + A3 (x1 − x2 )2 + A4 (x1 − x2 )3 x1 x2

(2)

y = A1 + A2 z + A3 z 2 + A4 z 3

(3)

where y=

e mix ∆H x1 x2

and

z = x1 − x2

(4)

The coefficients Ai must be chosen such that the sum of the squares of the deviations, S, given by N h i2 X yi − (A1 + A2 zi + A3 zi2 + A4 zi3 ) (5) S= i=1

159

is minimum. Note that N represents the number of data points. Minimization of S is accomplished by differentiating Eq. (5) with respect to Ai (i = 1, 2, 3, 4), and setting the derivatives equal to zero. The result is A1 N + A2

N P

i=1

A1

N P

zi + A2

i=1

A1

N P

i=1

A1

N P

i=1

N P

i=1

zi2 + A2 zi3 + A2

N P

i=1 N P

N P

zi + A3

i=1

i=1

zi2 + A3 zi3 + A3 zi4 + A3

zi2 + A4

N P

i=1

zi3 + A4

i=1 N P

i=1 N P

N P

i=1

zi4 + A4 zi5 + A4

N P

zi3 =

N P

i=1 N P

i=1 N P

i=1

yi

(6)

zi yi

(7)

zi2 yi

(8)

zi3 yi

(9)

i=1

zi4 = zi5 = zi6 =

N P

i=1 N P

i=1 N P

i=1

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

⎢ N ⎡ ⎤ ⎢ ⎢ A1 ⎢ N ⎢ ⎥ ⎢ P ⎢ ⎥ ⎢ ⎢ ⎢ A2 ⎥ ⎢ i=1 zi ⎢ ⎥ ⎢ ⎢ ⎥=⎢ ⎢ ⎥ ⎢ ⎢ A3 ⎥ ⎢ P N ⎢ ⎥ ⎢ zi2 ⎣ ⎦ ⎢ i=1 ⎢ ⎢ A4 ⎢ N ⎣P zi3 i=1

or

N P

zi

i=1 N P

i=1 N P

i=1 N P

i=1

N P

i=1

zi2 zi3 zi4

N P

i=1 N P

i=1 N P

i=1

zi2 zi3 zi4 zi5

N P

⎤−1 ⎡ 3 zi ⎥ ⎢

⎥ ⎥ ⎥ ⎥ N P zi4 ⎥ ⎥ ⎥ i=1 ⎥ ⎥ ⎥ N P ⎥ 5 zi ⎥ ⎥ i=1 ⎥ ⎥ ⎥ N ⎦ P 6 zi i=1

i=1

N P

⎤

yi ⎥ ⎢ i=1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢P ⎥ ⎢N z y ⎥ ⎢ i i⎥ ⎢ i=1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢N ⎥ ⎢P 2 ⎥ ⎢ z y ⎥ ⎢ i=1 i i ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ N ⎣P ⎦ 3 zi yi

(10)

i=1

⎡

⎤−1 ⎡ ⎤ ⎡ ⎤ A1 15 − 1.565 4.401 − 0.928 3.979 × 104 3⎥ ⎢ A2 ⎥ ⎢ − 1.565 ⎢ 4.401 − 0.928 2.429 ⎥ ⎢ ⎥=⎢ ⎥ ⎢ − 4.389 × 10 ⎥ ⎣ A3 ⎦ ⎣ 4.401 − 0.928 2.429 − 0.681 ⎦ ⎣ 1.206 × 104 ⎦ − 0.928 2.429 − 0.681 1.601 A4 − 2.597 × 103 ⎡ ⎤ 2.546 × 103 ⎢ − 125.091 ⎥ ⎥ =⎢ ⎣ 361.22 ⎦ 197.836

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

6.3.3 Determination of Partial Molar Quantities from ∆e ϕmix One of the practical questions to ask is how to determine partial molar quantities once a property change on mixing is determined as a function of composition. For a binary mixture, the molar property of a mixture is given by e 1 + x2 ϕ e 2 ) + ∆e ϕmix ϕ e mix = (x1 ϕ 160

(6.3-12)

On the other hand, Eq. (6.2-24) gives the partial molar property of component 1 as µ ¶ ∂e ϕmix ϕ1 = ϕ e mix − x2 (6.3-13) ∂x2 T,P,n2 Substitution of Eq. (6.3-12) into Eq. (6.3-13) gives "

ϕ1 = x1 ϕ e 1 + x2 ϕ e 2 + ∆e ϕmix − x2 − ϕ e2 + e1 + ϕ

µ

∂∆e ϕmix ∂x2

¶

T,P,n2

#

(6.3-14)

Simplification of Eq. (6.3-14) leads to

ϕ1 − ϕ e 1 = ∆e ϕmix − x2

µ

∂∆e ϕmix ∂x2

¶

(6.3-15)

T,P,n2

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

0

%2 φ2 − φ φ1 − φ% 1

0

Figure 6.5

1

x2

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 different components in a mixture. For any thermodynamic property, the total property of the mixture is expressed as k X ϕmix = ni ϕi (6.4-1) i=1

Differentiation of Eq. (6.4-1) gives

dϕmix =

k X

ni dϕi +

i=1

k X

ϕi dni

(6.4-2)

i=1

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 ) 161

(6.4-3)

The total differential of ϕmix is ¶ ¶ µ µ ¶ k µ X ∂ϕmix ∂ϕmix ∂ϕmix dϕmix = dT + dP + dni ∂T ∂P ∂ni T,P,nj6=i P,nj T,nj i=1 | {z }

(6.4-4)

ϕi

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

k X

ϕi dni

(6.4-5)

i=1

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

ni dϕi = 0

(6.4-6)

i=1

or dividing each term by the total number of moles gives k X

xi dϕi = 0

constant T and P

(6.4-7)

i=1

which is known as the Gibbs-Duhem equation under the conditions of constant temperature and pressure. For simplicity let us consider a binary mixture and exploit the practical uses of the GibbsDuhem 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 x1

dϕ1 dϕ + x2 2 = 0 dx1 dx1

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 the other species can be calculated from Eq. (6.4-9). • If the partial molar quantities of both species are experimentally determined, the expressions for ϕ1 and ϕ2 should satisfy the Gibbs-Duhem equation. Therefore, the Gibbs-Duhem equation is quite useful for checking whether the experimental data are thermodynamically consistent or not. Example 6.16 If the partial molar volume of component 1 in a binary mixture at constant T and P is given by V 1 = Ve1 + β x22 where Ve1 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 x1

dV 1 dV 2 + x2 =0 dx1 dx1

constant T & P

162

(1)

The term dV 1 /dx1 becomes i dV 1 d he = V1 + β (1 − x1 )2 = − 2 β(1 − x1 ) = − 2 β x2 dx1 dx1

(2)

It is also possible to evaluate the same term as

´ dV 1 dV 1 d ³e =− =− V1 + β x22 = − 2 β x2 dx1 dx2 dx2

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

dV 2 = 2 β x1 dx1

(3)

V 2 = β x21 + C

(4)

Integration of Eq. (3) leads to The constant C is evaluated by using the fact that as x1 → 0, V 2 → Ve2 . This gives C = Ve2 and the partial molar volume of species 2 is expressed in the form V 2 = Ve2 + β x21

(5)

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

REFERENCES Chen, C.C. and P.M. Mathias, 2002, AIChE Journal, 48 (2), 194-200. Chueh, P.L. and J.M. Prausnitz, 1967a, Ind. Eng. Chem. Fundam., 6, 492-498. Chueh, P.L. and J.M. Prausnitz, 1967b, AIChE Journal, 13, 1099-1107. Constantinescu, D. and I. Wichterle, 2002, Fluid Phase Equilibria, 203, 71-82. Conti, G., P. Gianni, M. Tramati, L. Lepori and E. Matteoli, 1997, J. Chem. Thermodynamics, 29, 865-877. Dahmani, A., I. Mokbel and J. Jose, 2002, Fluid Phase Equilibria, 203, 193-204. de Hemptinne, J.C. and E. Behar, 2006, Oil & Gas Science and Technology, 61 (3), 303-317. Edwards, J.E., 2008, Process Modelling Selection of Thermodynamic Methods, P&I Design Ltd. (http://www.chemstations.com/content/documents/Technical_Articles/thermo.pdf ) Estela-Uribe, J.F., J. Jaramillo, M.A. Salazar and J.P.M. Trusler, 2003, Fluid Phase Equilibria, 204, 169-182. Ghosh, P., 1999, Chem. Eng. Technol., 22 (5), 379-399. Grenner, A., M. Klauck, M. Kramer and J. Schmelzer, 2006, J. Chem. Eng. Data 51, 176-180. 163

Kiyohara, O. and G.C. Benson, 1977, Can. J. Chem., 55, 1354-1359. Lepori, L., E. Matteoli, A. Spanedda, C. Duce and M.R. Tine, 2002, Fluid Phase Equilibria, 201, 119-134. Meng, L., Y.Y. Duan and X.D. Wang, 2007, Fluid Phase Equilibria, 260, 354-358. Moravkova, L. and J. Linek, 2005, J. Chem. Thermodynamics, 37, 814-819. Nishiumi, H., T. Arai and K. Takeuchi, 1988, Fluid Phase Equilibria, 42, 43-62. Ogawa, H., S. Murakami, T. Takigawa and M. Ohba, 1997, Fluid Phase Equilibria, 136, 279287. Orbey, H. and S.I. Sandler, 1998, Modeling Vapor-Liquid Equilibria: Cubic Equations of State and Their Mixing Rules, Cambridge University Press, New York. Orge, B., M. Iglesias, A. Rodriguez, J.M. Canosa and J. Tojo, 1997, Fluid Phase Equilibria, 133, 213-227. Pecar, D. and V. Dolecek, 2005, Fluid Phase Equilibria, 230, 36-44. Poling, B.E., J.M. Prausnitz and J.P. O’Connell, 2004, The Properties of Gases and Liquids, 5th Ed., Chapter 5, McGraw-Hill, New York. Ray, A. and G. Nemethy, 1973, J. Chem. Eng. Data, 18 (3), 309-311. Ross, W.D., 1952, Chem. Eng. Progr., 43, 314. Serbanovic, S.P., M.L. Kijevcanin, I.R. Radoviv and B.D. Djordjevic, 2006, Fluid Phase Equilibria, 239, 69-82. Tsonopoulos, C., 1979, Equations of state in engineering and research, Adv. Chem. Ser., 182, 143-162. Twu, C.H., W.D. Sim and V. Tassone, 2002, Chemical Engineering Progress, 98 (11), 58-65.

PROBLEMS Problems related to Section 6.1 6.1 Estimate the molar volume of a binary mixture containing 30 mol % propane (1) and 70% n-butane (2) at 350 K and 3 bar. The mixture is represented by the virial equation state, and the critical molar volumes of propane and n-butane are 2.00 × 10−4 m3 / mol and 2.55 × 10−4 m3 / mol, respectively. (Answer: 9.27 × 10−3 m3 / mol)

6.2 A binary gaseous mixture at T1 and P1 enters a throttling valve and leaves it at P2 . The mixture is represented by the virial equation of state with the virial coefficients depending on temperature in the form Bij = αij + β ij T 164

Show that the exit temperature, T2 , is given by T2 = T1 +

αmix (P1 − P2 ) e∗ C Pmix

where

αmix = y12 α11 + 2 y1 y2 α12 + y22 α22 e∗ = y1 C e∗ + y2 C e∗ C Pmix P1 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 300 K 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 at 400 K and 35 bar. Estimate the volume of the tank if the mixture obeys the Peng-Robinson equation of state. Take k12 = 0.010, k13 = 0.001, and k23 = 0. (Answer: 1.712 m3 ) 6.5 A rigid tank is divided into two parts by a rigid partition. One side contains 2 kmol of methane at 300 K and 100 bar, and the other side contains 3 kmol of nitrogen at 300 K and 100 bar. The partition is removed and the gases mix with each other. If the temperature is kept constant at 300 K, estimate the final pressure. The system is represented by the van der Waals equation of state. (Answer: 101.1 bar) 6.6 The term (∂ Ve /∂T )P is needed in the calculation of the Joule-Thomson coefficient (see Problem 2.1). If a gas mixture is represented by the Peng-Robinson equation of state, then this term is calculated indirectly by the use of Eq. (2.2-50), i.e., ! Ã (∂P/∂T )Ve ∂ Ve =− (1) ∂T (∂P/∂ Ve ) P T

a) Show that ¶ ∙ ¸ µ Θmix 1 ∂P P = − ∂T Ve T Zmix − Bmix Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix ) ) ( µ ¶ 2 Amix (Zmix + Bmix ) ∂P P2 1 = +£ − ¤2 RT (Zmix − Bmix )2 Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix ) ∂ Ve T

(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 310 K and 10 bar, calculate the value of (∂ Ve /∂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 at 400 K and 60 bar. Estimate the temperature of the gas mixture if it is throttled to 5 bar. Use the Peng-Robinson equation of state with k12 = 0.026. (Answer: 372.6 K) 6.8 This problem is the extension of Problem 3.33 to mixtures. For mixtures, the departure function for molar heat capacity at constant volume is defined by ∙³ ´ ´ ¸ ³ ∂ IGM IGM e e emix − U eV = (1) U C mix − CV ∂T T,P T,P Ve a) For a gas mixture represented by the Peng-Robinson equation of state, show that the use of Eq. (6.1-39) in Eq. (1) leads to ´ ³ " # √ ¢ ¡ eIGM eV − C C mix V ψ mix Zmix + 1 + 2 Bmix T,P √ ¢ ¡ =√ ln (2) R 8 Bmix Zmix + 1 − 2 Bmix

where

k

ψ mix =

k

1 XX xi xj Aij (Γi + 2 Γi Γj + Γj ) 4

(3)

i=1 j=1

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 obtain ³ ´ eIGM eP − C C mix P

T,P

R

+

∙

" # √ ¢ ¡ ψ mix Zmix + 1 + 2 Bmix √ ¢ ¡ =√ ln 8 Bmix Zmix + 1 − 2 Bmix

Θmix 1 − Zmix − Bmix Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix )

¸2

2 Amix (Zmix + Bmix ) 1 −£ ¤2 2 (Zmix − Bmix ) Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix )

− 1 (4)

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

k X i=1

ep∗ yi C i

(5)

c) For a binary gas mixture consisting of 75 mol % propane (1) and 25% n-butane (2) at 310 K and 10 bar, calculate the departure functions for molar heat capacities at constant volume and pressure. 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) Problems related to Section 6.2 6.9 It is required to prepare 35 L of a mixture containing 60 mol % component 1 and 40% component 2 at 298 K. The molar volume of the mixture as a function of composition is shown below. 166

T = 298 K 90

80

70 Molar volume 60 (cm3/mol)

50

40

30

0.1

0.3

0.5

0.7

0.9

x1

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 of material. a) Show that the substitution of e mix ϕmix = n ϕ

into Eq. (6.2-5) gives

ϕi = ϕ e mix + n

µ

∂e ϕmix ∂ni

(1)

¶

(2)

T,P,nj6=i

b) Use the chain rule of differentiation and express Eq. (2) as ¶ µ ¶ µ ¶ µ ¶ k−1 µ X ∂e ϕmix ∂ϕ e mix ∂xj ∂xi ϕi = ϕ e mix + +n ∂xj T,P ∂ni ∂xi T,P ∂ni

(3)

j=1 j6=i

c) Noting that

ni nj ∂xi ∂xj 1 =− 2 and = − 2 ∂ni n ∂ni n n show that Eq. (3) takes the form µ ¶ ¶ µ k−1 X ∂ϕ e mix ∂ϕ e mix ϕi = ϕ e mix − xj + (1 − xi ) ∂xj T,P ∂xi T,P

(4)

(5)

j=1 j6=i

6.11 For a ternary system of components 1, 2, and 3, the molar volume of the mixture is given by Vemix = Ve1 x1 + Ve2 x2 + Ve3 x3 + A x1 x2 x3 where A is a function of temperature, and the terms Ve1 , Ve2 , and Ve3 represent the molar volumes of pure components 1, 2, and 3, respectively. Show that the partial molar volumes of components 1, 2, and 3 are expressed as a function of composition by the following equations: V 1 = Ve1 + A x2 x3 (1 − 2 x1 ) 167

V 2 = Ve2 + A x1 x3 (1 − 2 x2 ) V 3 = Ve3 + A x1 x2 (1 − 2 x3 )

6.12 The molar enthalpy of a binary liquid mixture of A and B is represented by e mix = (α1 + β 1 xA ) xA + (α2 + β 2 xB ) xB H

where α1 , α2 , β 1 , and β 2 are dependent on temperature. Show that H A = (α1 − β 2 ) + 2(β 1 + β 2 )xA − (β 1 + β 2 )x2A H B = (α2 + β 2 ) − (β 1 + β 2 )x2A 6.13 To prevent freezing of water in a car radiator, ethylene glycol (MW = 62.07 g/ mol) is added as an antifreeze. Ray and Nemethy (1973) provided the following data for the mixture of water (1) and ethylene glycol (2) at 298 K: ρmix = 0.99722 + 0.42114 x2 − 0.72022 x22 + 0.62036 x32 − 0.20911 x42 where ρmix is in g/ cm3 . 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 the resulting 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 and 57.45 cm3 / mol) 6.14 Partial molar volume of species i can be calculated from Eq. (6.2-4) once the equation of state is specified. a) Since cubic equations of state are pressure-explicit, show that it is more convenient to express Eq. (6.2-4) in the form ¶ µ ∂P ∂ni T,V,nj6=i ¶ Vi =− µ (1) ∂P ∂V T,nj b) Use the Redlich-Kwong equation of state to obtain ⎡

⎤ Amix Bi xj Aij − ⎢ ⎥ µ ¶ 2 ⎢ Zmix + Bmix ⎥ 1 Bi j=1 ⎢ ⎥ 1+ − ⎢ ⎥ ⎥ Zmix − Bmix Zmix − Bmix Zmix (Zmix + Bmix ) RT ⎢ ⎢ ⎥ Vi = ⎢ ⎥ P ⎢ 1 Amix (2 Zmix + Bmix ) ⎥ − ⎢ ⎥ 2 ⎢ ⎥ Zmix (Zmix + Bmix )2 (Zmix − Bmix )2 ⎣ ⎦ k X

(2)

c) A ternary gas mixture containing 45 mol % methane (1), 35% ethane (2), and 20% carbon dioxide is at 350 K and 25 bar. Estimate the partial molar volumes of each component if the mixture 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 Problem 6.14 and show that the partial molar volume of species i is given by µ ¶ ⎤ ⎡ 1 Bi 1+ ⎥ ⎢ Zmix − Bmix Zmix − Bmix ⎥ ⎢ k ⎥ ⎢ X 2 Amix Bi (Zmix − Bmix ) ⎥ ⎢ 2 xj Aij − ⎥ ⎢ ⎢ Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix ) ⎥ j=1 ⎥ ⎢ − ⎥ ⎢ ⎥ Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix ) RT ⎢ ⎥ ⎢ Vi = ⎥ ⎢ P ⎢ 1 2 Amix (Zmix + Bmix ) ⎥ − ⎥ ⎢ ¤ £ 2 ⎥ ⎢ (Zmix − Bmix )2 Zmix (Zmix + Bmix ) + Bmix (Zmix − Bmix ) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎦ ⎣

Consider a binary mixture consisting of 75 mol % ethane (1) and 25% n-decane (2) at 444 K and 138 bar. Estimate the partial molar volume of ethane using the Peng-Robinson equation of state. Take kij = 0. (Answer: 153.4 cm3 / mol) Problems related to Section 6.3 6.16 For a binary mixture of components 1 and 2, the molar enthalpy of the mixture (in J/ mol) is given by e mix = 750 x1 + 380 x2 + 10 x1 (5 x1 + 17 x2 ) H

a) Show that the partial molar enthalpies of components 1 and 2 are expressed as H 1 = 920 − 240 x1 + 120 x21 b) Show that

and

H 2 = 380 + 120 x21

e mix = 120 x1 x2 ∆H

e mix at x1 = 0.4. c) Calculate H 1 , H 2 , and ∆H

(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 mixture of benzene (1) and acetophenone (2) at 25 ◦ C, determine the volume change on mixing (in cm3 / mol) at the given mole fractions of benzene. The molecular weights of benzene and acetophenone are 78.1 and 120.15 g/ mol, respectively.

x1

ρmix ( g/ cm3 )

x1

ρmix ( g/ cm3 )

x1

ρmix ( g/ cm3 )

0.00000 0.10571 0.18602 0.28547

1.02311 1.01102 1.00134 0.98894

0.37918 0.46261 0.54838 0.62477

0.97658 0.96498 0.95249 0.94071

0.72894 0.80033 0.89527 1.00000

0.92384 0.91156 0.89423 0.87364

169

6.18 An isothermal mixer at 300 K is fed with a stream containing 35 mol % A of flow rate 2 mol/ s and a stream containing 80 mol % A of flow rate 1 mol/ s. Calculate the heat that must be removed or added per mole of solution leaving the mixer under steady conditions so as to keep the outlet stream temperature constant at 300 K. The heat of mixing data for a binary liquid mixture of A and B at 300 K are represented by the following equation: e mix = xA xB (10 xA + 30 xB ) ∆H

e mix is in J/ mol. where ∆H (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 and 250 J/ mol, respectively. a) Calculate the partial molar enthalpies of components A and B at infinite dilution. ∞ e mix versus xA and evaluate H ∞ b) Plot 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 that results when one mole of solute (component 1) is dissolved in a solvent (component 2). Ogawa et al. (1997) reported the following data for the enthalpy of solution, ∆Hsol , for benzene (1) in cyclohexane (2) at 298.15 K: m1 ( mol/ kg)

∆Hsol ( kJ/ mol of benzene)

0.0912 0.1529 0.1544 0.1569

3.15 3.16 3.16 3.13

where m1 is the molality of benzene. Determine the heat of mixing as a function of the mole fraction 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 ∆Vemix for a binary system of ethanol (1) and methyl propanoate (2) at 298.15 K x1

0.0718 0.1378 0.2216 0.2947 0.3441

∆Vemix ( cm3 / mol) 0.04714 0.07003 0.10068 0.10946 0.11500

x1

0.3976 0.4838 0.5411 0.5873 0.6496

∆Vemix ( cm3 / mol) 0.12064 0.12086 0.11743 0.10664 0.09741

x1

0.7065 0.7531 0.8285 0.9115

∆Vemix ( cm3 / mol) 0.08995 0.08079 0.06207 0.03438

Fit the data to the following equation ∆Vemix = x1 x2

3 X i=1

170

Ai (x1 − x2 )i−1

(1)

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 mixture of methanol (1) and benzene (2) at 298.15 K: x1

ρmix ( g/ cm3 )

x1

ρmix ( g/ cm3 )

x1

ρmix ( g/ cm3 )

0.0000 0.0516 0.0768 0.1011 0.1241 0.1501 0.1989 0.2004

0.873582 0.871283 0.870172 0.869074 0.868029 0.866825 0.864485 0.864414

0.2030 0.2607 0.2978 0.3312 0.3497 0.4019 0.4493 0.4990

0.864285 0.861329 0.859308 0.857433 0.856357 0.853164 0.850021 0.846501

0.5503 0.5810 0.6499 0.6993 0.8000 0.8992 1.0000

0.842600 0.840082 0.833959 0.829097 0.817709 0.804020 0.786694

a) Fit the data to the following equation: ∆Vemix = x1 x2

4 X i=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 are expressed as a function of composition in the form V 1 = Ve1 + x22 V 2 = Ve2 + x21

and conclude that

V

∞ 1

4 X i=1

4 X i=1

= Ve1 +

i−1

Ai (x1 − x2 )

3 X

i Ai+1 (x1 − x2 )i−1

(2)

3 X

i Ai+1 (x1 − x2 )i−1

(3)

i=1

i−1

Ai (x1 − x2 )

+ 2 x1 x22

− 2 x21 x2

4 X (− 1)i−1 Ai

and

i=1

∞ V2

i=1

∞

∞

Also calculate the numerical values of V 1 and V 2 . (Answer: b) 41.206 cm3 / mol and 89.243 cm3 / mol)

= Ve2 +

4 X

Ai

(4)

i=1

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

with

Vemix ( cm3 / mol) = 40.74 x1 + 131.55 x2 + x1 x2 A1 = 2.0741

A2 = 0.3195 171

3 X i=1

Ai (x1 − x2 )i−1

A3 = 1.7733

a) Determine the volumes of pure methanol and n-hexane required to form 600 cm3 of a solution containing 30 mol % methanol. b) Determine the total volume of the mixture when 5 moles of methanol are mixed with 3 moles 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 of perfluorohexane (C6 F14 ) and n-octane at 298.15 K using a vibrating-tube densimeter. The experimental results are fitted to an equation of the form h i ∆Vemix ( cm3 / mol) = x1 x2 16.00 + 3.67 (x1 − x2 ) + 11.70 (x1 − x2 )2

where subscripts 1 and 2 represent perfluorohexane and n-octane, respectively. The molar volumes of pure components are reported as Ve2 = 163.50 cm3 / mol

Ve1 = 201.64 cm3 / mol

a) Estimate the total volume of the mixture when 500 cm3 of perfluorohexane is mixed with 800 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 194.87 cm3 / mol)

b) 207.53 cm3 / mol and 165.92 cm3 / mol c) 225.67 cm3 / mol and

6.25 The heat of mixing data for a binary mixture of tetramethyl-1,3-butanediamine (TMBD) and n-heptane at 298.15 K were correlated by Dahmani et al. (2002) in the form h i e mix ( J/ mol) = x1 x2 1186.86 − 338.28 (x1 − x2 ) ∆H where subscripts 1 and 2 represent TMBD and n-heptane, respectively.

e mix versus x1 and estimate the difference between the partial molar and pure a) Plot ∆H component 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 298.15 K are reported by Kiyohara and Benson (1977) as follows:

x1 0.04 0.06 0.10 0.20 0.30 0.34

e mix ( J/ mol ) ∆H 151.83 208.51 283.04 289.35 160.86 94.26

x1 0.38 0.40 0.42 0.50 0.58 0.62

e mix ( J/ mol) ∆H 23.22 − 14.21 − 52.31 − 210.49 − 367.94 − 443.05

172

x1 0.70 0.80 0.90 0.94 0.96 0.98

e mix ( J/ mol) ∆H − 578.39 − 706.59 − 708.48 − 581.55 − 455.06 − 266.58

e mix versus x1 and determine H 1 − H e 1 and H 2 − H e 2 graphically at x1 = 0.5. a) Plot ∆H

b) How much heat would be released (or absorbed) upon mixing 6 moles of water with 14 moles of tetrahydofuran at a constant temperature of 298.15 K. e 1 = − 1110 J/ mol H 2 − H e 2 = 705 J/ mol b) 3217 J) (Answer: a) H 1 − H Problems related to Section 6.4

6.27 After carrying out experiments on volume change on mixing of binary systems, your friend proposes the following expression to express the molar volume of a mixture as a function of composition: (1) Vemix = (α1 + β 1 x21 ) x1 + (α2 + β 2 x22 ) x2

where αi and β i are constants. Since Eq. (1) has the form Vemix = V 1 x1 + V 2 x2

(2)

your friend compares Eqs. (1) and (2) and suggests that V 1 = α1 + β 1 x21

and

V 2 = α2 + β 2 x22

Do you agree? Why? 6.28 Pure components 1 and 2 have the molar enthalpies of 200 J/ mol and 350 J/ mol, respectively. When these two components are mixed at constant temperature and pressure, the partial molar enthalpy of component 1 is reported as H 1 ( J/ mol) = 200 + 40 x22

(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., x1

dH 1 dH 2 + x2 =0 dx1 dx1

(2)

to obtain the partial molar enthalpy of component 2 as

Then use to show that

H 2 = 350 + 40 x21

(3)

e mix = x1 H 1 + x2 H 2 H

(4)

e mix = 350 − 150 x1 + 40 x1 x2 H

(5)

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

e mix dH H1 1 e − Hmix = − dx2 x2 x2

(6)

which is a linear equation with an integrating factor of 1/x2 . Multiply Eq. (6) by an integrating factor to transform it into à ! e mix H1 d H (7) =− 2 dx2 x2 x 2

173

Integrate Eq. (7) to obtain the solution as e mix = − x2 H

Z

H1

dx2 + C x2

2

x2

(8)

where C is an integration constant. Show that the substitution of Eq. (1) into Eq. (8) and integration also lead to Eq. (5). 6.29 The generalized form of the Gibbs-Duhem equation can be derived as follows: e mix in Eq. (6.4-4) to obtain a) Use ϕmix = n ϕ dϕmix = n

µ

∂e ϕmix ∂T

¶

dT + n

P,nj

µ

∂ϕ e mix ∂P

¶

dP +

T,nj

k X

ϕi dni

(1)

i=1

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

µ

∂e ϕmix ∂T

¶

P,xj

dT −

µ

∂ϕ e mix ∂P

¶

dP +

T,xj

k X

xi dϕi = 0

(2)

i=1

which is known as the generalized Gibbs-Duhem equation. c) Use ϕi = ϕi (T, P, x1 , x2 , .., xk−1 ) to obtain dϕi =

µ

∂ϕi ∂T

¶

dT +

P,xj

µ

∂ϕi ∂P

¶

dP +

T,xj

(3)

¶ k−1 µ X ∂ϕ i

∂xm

m=1

dxm

(4)

T,P

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

µ

∂ϕ e mix ∂T

¶

P,xj

dT −

µ

∂ϕ e mix ∂P

¶

dP +

T,xj

k X

µ

xi

i=1

∂ϕi ∂T

¶

dT +

P,xj

+

k X

k X

xi

i=1

xi

i=1

µ

∂ϕi ∂P

¶ k−1 µ X ∂ϕ

¶

i

m=1

∂xm

dP

T,xj

dxm = 0 (5)

T,P

e) Note that k X i=1

xi

µ

∂ϕi ∂T

à ! ¸ k ∙ k X ∂ X ∂(xi ϕi ) dT = dT = xi ϕi ∂T ∂T P,xj P,xj

¶

i=1

i=1

Similarly, k X i=1

xi

µ

∂ϕi ∂P

¶

dP =

T,xj

µ

∂ϕ e mix ∂P

dT =

P,xj

¶

dP

µ

∂ϕ e mix ∂T

¶

dT

P,xj

(6) (7)

T,xj

f ) Use Eqs. (6) and (7) in Eq. (5) and conclude that k X i=1

xi

¶ k−1 µ X ∂ϕ i

m=1

∂xm

dxm = 0

T,P

which is the generalized form of Eq. (6.4-9) for a k-component system. 174

(8)