Chemical Engineering Thermodynamics II ( ) 02 - The Molar Gibbs Free Energy & Fugacity of a Pure Component

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1 Chemial Engineering Thermodynamis II (090533) 0 - The Molar Gibbs Free Energy & Fugaity of a ure Component Dr. Ali Khalaf Al-matar Chemial Engineering Department University of Jordan banihaniali@yahoo.om Fugaity & Fugaity Coeffiient The produt of pressure times exponential of the Gibbs free energy differene divided by RT is termed the FUGACITY IG gt (, ) g ( T, ) 1 RT f = exp = exp v d RT RT 0 The fugaity divided by pressure is the FUGACITY COEFFICIENT IG f g( T, ) g ( T, ) 1 RT φ = = exp = exp v d RT RT 0 Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 1

2 roperties of Fugaity and its Coeffiient Fugaity has units of pressure. lim f 0 = Fugaity oeffiient is dimensionless f limφ = lim = Fugaity is very useful in phase equilibria alulations. Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 3 Equilibrium Criterion Using Fugaity: Derivation Derive an expression for the equilibrium riterion based on fugaity. Start with the three equilibrium onditions. Note that T and are fixed and the Gibbs free energy must be equal for the oexisting phases I II g ( T, ) = g ( T, ) I II IG f ( T, ) IG f ( T, ) g ( T, ) + RTln = g ( T, ) + RTln Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 4

3 Equilibrium Criterion Using Fugaity: Final Form The ideal gas molar Gibbs free energy has the same value at the same T and I II f ( T, ) f ( T, ) ln = ln I II f ( T, ) = f ( T, ) I II φ ( T, ) = φ ( T, ) At equilibrium: the fugaity must be equal in the oexisting phases. also, fugaity oeffiients must be equal. Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 5 Fugaity and hase Equilibria Sine fugaity is related to EOS. It provides an approah to solving phase equilibria problems The problem with using the definition of fugaity diretly is that the integration requires the EOS to be in the form v = f(t,). The EOS so far use = f(t, v). With the nonlinear forms of EOS it is diffiult to transform them to the form required by fugaity Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 6 3

4 Transformation of Integration ariable Convert the integration from volume expliit to pressure expliit using the following transformation 1 d = d ( v ) dv = dz dv v v Z v Using this transformation the expression for the fugaity oeffiient beomes v f( T, ) 1 RT lnφ = ln = dv ln Z ( Z 1) RT + v= v Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 7 Effet of T and on the Fugaity The effet of temperature and pressure upon the fugaity an be derived and are given by: ln f ( T, ) g RT = v = T T [ ] ln f( T, ) / ht (, ) h IG ( T, ) = T RT Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 8 4

5 Fugaity of a ure gaseous Speies For gases, use an EOS ombined with the definition of the fugaity. The EOS is used to provide the dependene of pressure on volume. Frequently, you may need to numerially integrate ertain numerial data instead of using an EOS Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 9 Fugaity of a Gas Using the R EOS v Z RT / f ( T, ) 1 = RT lnφ = ln = dv ln Z ( Z 1) RT + v v= f A Z + (1+ ) B ln = ( Z 1) ln ( Z B) ln B Z + (1 ) B a A = ( RT ) b B = RT Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 10 5

6 Representative EOS Soave-Redlih-Kwong (SRK) RT a( T ) = v b v ( v + b ) ( RT ) at ( ) = α ( T) T α( T ) = 1+ κ 1 T κ = ω 0.176ω RT b = eng-robinson (R) RT a( T ) = v b v ( v + b ) + b ( v b ) ( RT ) at ( ) = α( T) T α( T ) = 1+ κ 1 T κ = ω 0.699ω RT b = Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 11 Cubi EOS Coeffiients α β γ Z vdw 1 B A AB Z Z Z + α + β + γ = 0 SRK R B A B B A B 3B AB AB + B + B a A = ( RT ) B = b RT Initial Guess for solution apor (apor like): ideal gas (Z = 1). iquid: Redued ovolume (Z = B). Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 1 6

7 Solution Methodology for R EOS 1. Obtain (T ; ;ω).. Find b in R EOS. 3. Find a in R EOS 1. Determine κ.. Determine α. 3. Determine a(t). 4. Determine redued parameters A and B. 5. Evaluate the ubi onstants in the Z expression. 6. Solve the ubi for the roots and determine if they fall in the subooled liquid, superheated vapor or the two phase oexistene region. RT b = ( RT ) at ( ) = α( T) a A = ( RT ) κ = ω 0.699ω T α( T ) = 1+ κ 1 T b B = RT 3 Z + ( 1 + BZ ) + ( A B 3 B) Z = 3 ( AB B B ) 0 Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 13 Fugaity Using Corresponding States riniple (CS) The priniple of orresponding states may be used to obtain the fugaity utilizing the redued temperature and pressure as parameters. A third fator may be used to enhane the auray. This third fator may be itzer s aentri fator, or the ritial ompressibility. Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 14 7

8 Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 15 Fugaity of a ure iquid You an use an EOS to get the fugaity of a liquid in a manner similar to that for a gas e.g., eng- Robinson EOS yields f A Z + (1+ ) B ln = ( Z 1) ln ( Z B) ln B Z + (1 ) B Alternatively, one an derive an expression for fugaity from the equilibrium ondition of the equality of fugaity to obtain 1 vap f ( T, ) = ( f / ) exp vd sat, T RT vap Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 16 8

9 f ( T, ) = exp oynting Corretion vap is the vapor pressure at the speified T, and the (f/) sat is the fugaity oeffiient of the fluid (liquid = vapor) at saturation. The exponential term is referred to as the oynting pressure orretion. It is important At high pressures At low temperatures (ryogeni systems) Assuming that liquids are inompressible vap ( ) sat, vap f v sat, T RT Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 17 Fugaity of a ure Solid (hase) Solids may undergo several phase transitions. Usually dealing with solids that undergo sublimation; whih is what the supersript sat refers to in this ase. Similar to the derivation of liquids assuming that J + 1 S sat f 1 J f ( T, ) = ( T) exp v d sat, T RT J = 1 J Solids sublimation pressures are very small (φ=1). Solids are inompressible sat ( ( )) S v T S sat f ( T, ) = ( T)exp RT Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 18 9

10 apor ressure from EOS We know that f and f an be obtained via an EOS using the liquid and vapor ompressibility respetively. For the R EOS f A Z + (1+ ) B ln = ( Z 1) ln ( Z B) ln B Z + (1 ) B f A Z + (1+ ) B ln = ( Z 1) ln ( Z B) ln B Z + (1 ) B At equilibrium f ( T, vap ) = f ( T, vap ) Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 19 Algorithm for vapor ressure Using EOS Consequently, solve for the vapor pressure using any EOS. This solution is iterative in nature: Assume a pressure Solve the ubi for Z and Z Evaluate f and f If the two fugaities are equal to within a ertain onvergene riteria then the pressure assumed is the vapor pressure Always hek for trivial solutions (too low or too high pressures leading to single phase). Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 0 10

11 Thermo II: 0-The Molar Gibbs Free Energy & Fugaity of a ure Component 1 11