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Phase Equilibrium: Pure Substances D.F. Mendoza Universidad de Antioquia

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Equilibrium and Spontaneity Thermodynamic laws dU = δQ − δW δQ dS = + S gen T from the first law we obtain: δQ = dU + pdV Replacing δQ in the second law dS =

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

dU p + dV + S gen T T

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Introduction

Maximum entropy principle

The combined first and second law equation dS =

dU p + dV + S gen T T

for an isolated system where V and U are constant becomes dS = S gen since S gen ≥ 0, an spontaneous processes increases the entropy of the isolated system, and reaches a maximum (dS = 0) where it attains the thermodynamic equilibrium.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Minimum internal energy principle

The combined first and second law equation using U -explicit formulation is: dU = T dS − pdV − T S gen for a system at constant S and V , the internal energy change is dU = −T S gen since S gen ≥ 0, an spontaneous processes at constant S and V reduces the internal energy of the system and this system attains the thermodynamic equilibrium when U is minimum (dU = 0).

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Minimum enthalpy principle

The combined first and second law equation using H-explicit formulation is: dH = T dS + V dp − T S gen for a system at constant S and p, the enthalpy change is dH = −T S gen since S gen ≥ 0, an spontaneous processes at constant S and p reduces the enthalpy of the system and this system attains the thermodynamic equilibrium when H is minimum (dH = 0).

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Minimum Helmholtz principle

The combined first and second law equation using A-explicit formulation is: dA = −SdT − pdV − T S gen for a system at constant T and V , the Helmholtz energy change is dA = −T S gen since S gen ≥ 0, an spontaneous processes at constant T and V reduces the Helmholtz energy of the system and this system attains the thermodynamic equilibrium when A is minimum (dA = 0).

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Minimum Gibbs energy principle

The combined first and second law equation using G-explicit formulation is: dG = −SdT + V dp − T S gen for an isolated system where T and p are constant becomes dG = −T S gen since S gen ≥ 0, an spontaneous processes at constant T and p reduces the Gibbs energy of the system and this system attains the thermodynamic equilibrium when G is minimum (dG = 0).

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Summary of equilibrium and spontaneity principles

Variable U S H A G

Constant S, V U, V S, p T, V T, p

D.F. Mendoza (Universidad de Antioquia)

Spontaneous dU < 0 dS > 0 dH < 0 dA < 0 dG < 0

Equilibrium dU = 0; minimum dS = 0; maximum dH = 0; minimum dA = 0; minimum dG = 0; minimum

Phase Equilibrium: Pure Substances

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Introduction

Thermodynamic relations for open systems The fundamental equation for an open-single phase system is: dS =

p g 1 dU + dV − dn T T T

we note that the thermodynamic description of an open system requires the number of moles as new variable, also we note that the molar Gibbs energy (also known as chemical potential, µ) is a new ”intensive” potential in open systems. The fundamental equation can be arranged in different forms: Forms of the fundamental equation for open systems

dU = T dS − pdV + gdn dA = −SdT − pdV + gdn dH = T dS + V dp + gdn dG = −SdT + V dp + gdn

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Phase equilibrium and the phase rule For an isolated, heterogeneous system (involving π phases), the thermodynamic equilibrium implies Thermal, mechanical, and chemical equilibrium T1 = T2 = ⋯ = Tπ p1 = p2 = ⋯ = pπ g1 = g2 = ⋯ = gπ The thermodynamic degrees of freedom F to solve the intensive variables of a phase equilibrium problem, involving one component, is given by the phase rule: F =3−π where π is the number of phases.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Gibbs-Duhem equation The internal energy form of the fundamental equation, for single-phase, open system is, dU = T dS − pdV + gdn integrating through an isothermal, isobaric path from n1 = 0 to n2 = n moles, we obtain: U = T S − pV + gn deriving U : dU = T dS + SdT − pdV − V dp + gdn + ndg comparing with the fundamental equation we get: SdT − V dp + ndg = 0

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Clapeyron equation The equilibrium state between phases 1 and 2 requires: g (1) (T sat , psat ) = g (2) (T sat , psat ) If we move along the equilibrium line we have dg (1) = dg (2) using Gibbs-Duhem (dg = −sdT + vdp) for each phase, −s(1) dT sat + v (1) dpsat = −s(2) dT sat + v (2) dpsat grouping terms (

dp sat s(2) − s(1) ∆sphase h(2) − h(1) ∆hphase ) = (2) = = sat (2) = sat phase phase (1) (1) dT ∆v v −v T (v − v ) T ∆v

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Clapeyron equation can be used in any two-phase equilibrium • Liquid-vapor equilibrium (liquid-vapor line) dp sat s(v) − s(l) h(v) − h(l) ) = (v) = sat (v) (l) dT v −v T (v − v (l) )

(

• Solid-gas equilibrium1 (sublimation equilibrium line) (

dp sat s(v) − s(c) h(v) − h(c) ) = (v) = sat (v) (c) dT v −v T (v − v (c) )

• Solid-liquid equilibrium (melting line) (

dp sat s(l) − s(c) h(l) − h(c) ) = (l) = dT v − v (c) T sat (v (l) − v (c) )

• Solid-solid equilibrium ( 1

h(c2 ) − h(c1 ) dp sat s(c2 ) − s(c1 ) ) = (c ) = sat (c ) (c ) dT v 2 −v 1 T (v 2 − v (c1 ) )

c means solid phase

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

psat empirical equations • General approach

( ln (

l psat T αi ) = 1 + ∑ ai [1 − ( ) ] pn Tn i=1 l psat T αi ) = ∑ ai [1 − ( ) ] pn Tn i=1

where: Tn , pn are normalizing values for temperature and pressure; ai and αi are calculated fitting experimental data. Wagner, Saul, Pruβ. International equation for the pressure along the melting and along the sublimation curve for ordinary water substance. J. Phys. Chem. Ref. Data, 23:515-527, 1994.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

• Sublimation line2 (water: 190 ≤ T ≤ 273.16 K)

ln (

psub ) = a1 (1 − θ−1.5 ) + a2 (1 − θ−1.25 ) pt

where: θ = T /Tt , a1 = −13.928169, a2 = 34.7078238, Tt = 273.16 K, pt = 611.657 Pa. • Melting line (ice I: 251.165 ≤ T ≤ 256.164 K)

pm = 1 + a1 (1 − θ−3 ) + a2 (1 − θ21.2 ) pt where: θ = T /Tt , a1 = −0.626 × 106 , a2 = 0.197135 × 106 , Tt = 273.16 K, pt = 0.000611657 MPa. 2

Wagner, Saul, Pruβ. International equation for the pressure along the melting and along the sublimation curve for ordinary water substance. J. Phys. Chem. Ref. Data, 23:515-527, 1994. D.F. Mendoza (Universidad de Antioquia)

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Introduction

• Vapor-liquid line • Antoine equation B C +T where: A, B and C are constants obtained fitting experimental data. • Wagner aτ + bτ 1.5 + cτ 2.5 + dτ 5 ln pvpr = Tr ln pvap = A −

where: Tr = 1/Tc , pvpr = pvap /pc , τ = 1 − Tr , a, b and c are constants obtained fitting experimental data.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Example Obtain A, B, and C coefficients of the Antoine equation that represent the following experimental data for n-decane. T sat (°C) 1 5 10 20 40 60 100 200 400 760

psat (mmHg) 16.5 42.3 55.7 69.8 85.5 95.5 108.6 128.4 150.6 174.1

Hint: linearize the Antoine equation y = a0 + a1 x1 + a2 x2 ln p = A + D.F. Mendoza (Universidad de Antioquia)

AC − B ln p −C T T

Phase Equilibrium: Pure Substances

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Introduction

Clausius-Clapeyron equation This is a simplified equation derived from the Clapeyron equation for (vapor-liquid and gas-solid phase equilibrium), where we assume: • v (v) >> v (l) (vapor-liquid equilibrium case). • v (v) >> v (c) (gas-solid equilibrium case). • gas-phase follows the ideal gas behavior (i.e., v (v) = RT sat /psat ).

Vapor-liquid equilibrium (

d ln p sat s(v) − s(l) h(v) − h(l) ) = = dT RT RT 2

Gas-solid equilibrium (

d ln p sat s(v) − s(c) h(v) − h(c) ) = = dT RT RT 2

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Example Trimethyl gallium, Ga(CH3 )3 , can be used as a feed gas to grow films of Gas. Use Clausius-Clapeyron equation to estimate the enthalpy of vaporization of Ga(CH3)3 .3 ln psat =

−4222.1 + 17.556 T

where psat in kPa, T in K.

3

Koretsky Engineering and chemical thermodynamics

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Molar Gibbs energy (chemical potential) of a pure substance The chemical potential (molar or specific Gibbs energy) of a pure substance can be expressed as function of p and T , using the Gibbs-Duhem expression dg = dµ = −sdT + vdp where s is the molar (or specific) entropy and v is the molar (or specific) volume. µ (or g) is a state function, therefore: (

∂g ∂µ ) =( ) = −s ∂T p ∂T p

D.F. Mendoza (Universidad de Antioquia)

and

(

∂µ ∂g ) =( ) =v ∂p T ∂p T

Phase Equilibrium: Pure Substances

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Introduction

integrating this equation we get µ(T, p) = µ(T ref , pref ) − ∫

T T ref

or g(T, p) = g(T ref , pref ) − ∫

p

sdT + ∫

pref

T T

vdp

p

sdT + ∫ ref

pref

vdp

where superscript ref denotes an arbitrary reference state. Grouping the two terms at the right-hand, µ(T, p) = µ(T, pref ) + ∫ or g(T, p) = g(T, pref ) + ∫

p pref p

pref

where: µ(T, pref ) = µ(T ref , pref ) − ∫

D.F. Mendoza (Universidad de Antioquia)

vdp

vdp T

T ref

Phase Equilibrium: Pure Substances

sdT

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Introduction

A way to calculate µ(T, pref ) (or g(T, pref )), using a reference state is: µ(T, pref ) = g(T, pref ) = h(T, pref ) − T s(T, pref ) where: h(T, pref ) = h(T ref , pref ) + [h(T, pref ) − h(T ref , pref )] and s(T, pref ) = s(T ref , pref ) + [s(T, pref ) − s(T ref , pref )]

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Chemical potential of an ideal gas The chemical potential of an ideal gas at T and p, using T ref , pref as reference state is given by: p µig (T, p) = µig (T, pref ) + RT ln ( ref ) p The chemical potential at T and pref is calculated from the values of enthalpy and entropy relative to a reference state µ(T, pref ) = h(T, pref ) − T s(T, pref ) where: h(T, pref ) = ∫

T

cig p dT

and s(T, pref ) = ∫

T

cig p

dT T The chemical potential of an ideal gas with respect to a reference state is: µig (T, p) = ∫

D.F. Mendoza (Universidad de Antioquia)

T ref

T T ref

cig p dT − T ∫

T T ref

cig p T

T ref

dT + RT ln (

Phase Equilibrium: Pure Substances

p ) pref

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Introduction

µig (T, p) for N2(g)

T ref = 298.15 K, pref = 1.01Phase bar.Equilibrium: Pure Substances

D.F. Mendoza (Universidad de Antioquia)

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Introduction

Fugacity The change in µ of a pure substance with respect to the pressure through an isothermal path is p2

∂µ ) dp ∂p T

µ(T, p2 ) − µ(T, p1 ) = ∫

(

µ(T, p2 ) − µ(T, p1 ) = ∫

vdp

p1 p2 p1

Special case: Ideal gas: The variation of µ with pressure keeping T constant is µig (T, p2 ) − µig (T, p1 ) = ∫

D.F. Mendoza (Universidad de Antioquia)

p2 p1

v ig dp = ∫

p2 p1

Phase Equilibrium: Pure Substances

RT p2 dp = RT ln ( ) p p1

26 / 57

Introduction

for any real substance we arbitrarily define a thermodynamic function (fugacity, f ), which preserves the same form of isothermal variation of the chemical potential with pressure for an ideal gas µ(T, p2 ) − µ(T, p1 ) = RT ln (

f2 ) f1

from this definition we note that: • The fugacity of an ideal gas is p, f ig = p • The fugacity of any real substance must be equal to f real = f ig = p as

p approaches to zero. thus, f lim ( ) = 1 p

p→0

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Summarizing • Ideal gas:

µig (T, p2 ) − µig (T, p1 ) = ∫

p2 p1

p2 RT p2 dp = ∫ RT d ln p = RT ln ( ) p p1 p1

• Real substance: (solid, liquid, gas) f2

µ(T, p2 ) − µ(T, p1 ) = ∫

f1

f2 RT f2 df ∫ RT d ln f = RT ln ( ) f f1 f1

It is mandatory that: f lim ( ) = 1 p→0 p for any real substance.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

It will be called the fugacity, represented by the symbol f and defined by the following conditions: 1. The fugacity of a molecular species is the same in two phases when these phases are in equilibrium as regards the distribution of that species. 2. The fugacity of a gas approaches the gas pressure as a limiting value if the gas is indefinitely rarefied. In other words, the escaping tendency of a perfect gas is equal to its gas pressure. That these two conditions are sufficient to define a property of every substance which is not a mathematical, fictitious quantity, but a real physical quantity, capable of experimental determination in every case, must now be shown. 4

4 Lewis G. N. The law of physico-chemical change. Proceedings of the American Academy, 1901. D.F. Mendoza (Universidad de Antioquia)

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Introduction

The idea of fugacity is thus evolved from the use of vapor pressure as a measure of escaping tendency. When a substance is in equilibrium with its vapor, the fugacity, in order to fulfil the laws of escaping tendency, must be the same in both. The fugacity of a substance is therefore equal to its vapor pressure if the vapor behaves like a perfect gas. Speaking in terms not very precise, we may say that the fugacity of a substance is equal to the vapor pressure that the substance would have if its vapor were a perfect gas. It has been shown in the preceding paper that for a given substance in a given state the fugacity is a definite property of which the numerical value can in most cases be readily determined, and which is well suited to serve as an exact measure of the escaping tendency.5 5

Lewis G.N. Outlines of a new system of thermodynamic chemistry. Contributions from the research laboratory of physical chemistry of the Massachusetts Institute of Technology, No 17, 1907. D.F. Mendoza (Universidad de Antioquia)

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Introduction

Phase equilibrium and fugacity The phase equilibria criterium between phases α and β is: Tα = Tβ pα = pβ µα = µβ if we compute µα from a reference state ◻ and µβ from a reference state △ (all at T ) fα ) f◻ fβ µβ = µ△ + RT ln ( △ ) f

µα = µ◻ + RT ln (

if we subtract these expressions and arrange terms, we obtain ln (

µα − µβ fα )= =0 β f RT

thus, in the equilibrium f α = f β , no matter what reference state is ◻ or △ D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Phase equilibrium criteria We have two general ways of stating the phase equilibrium criterion: • Using chemical potential µ Tα = Tβ = ⋯ = Tπ pα = pβ = ⋯ = pπ µα = µβ = ⋯ = µπ Characteristics: µ depends on a reference state, numerically not well-behaved. • Using fugacity f Tα = Tβ = ⋯ = Tπ pα = pβ = ⋯ = pπ fα = fβ = ⋯ = fπ Characteristics: f does not depend on a reference state, numerically well-behaved. D.F. Mendoza (Universidad de Antioquia)

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Introduction

The escaping tendency of a given molecular species in a given state is therefore analogous to temperature, and the two laws of escaping tendency are as follows: If the escaping tendency of a given molecular species, X, is the same in two phases, then X will not of itself pass from one phase to the other. If the escaping tendency of X is greater in one phase, it will pass from this phase into the other, when the two are brought together.6

6

Lewis G.N. Outlines of a new system of thermodynamic chemistry. Contributions from the research laboratory of physical chemistry of the Massachusetts Institute of Technology, No 17, 1907. D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Fugacity changes in homogeneous phase: Gas phase To find the change in fugacity due to pressure changes we apply the equivalence between µ and f : RT ∫

ln f2v

ln f1v

p2

d ln f = ∫

v v dp

p1

• v-explicit or z-explicit EoS: (e.g., B truncated virial EoS, compressibility factor EoS) ln (

p2 v v p2 z p r2 z f v (T, p2 ) ) = dp = ∫ dp = ∫ dpr ∫ v f (T, p1 ) p1 RT p1 p p r1 p r

• p-explicit EoS: (e.g., cubic EoS). In this case we change the integration variable: f2

RT ∫

f1

p2 v 2

d ln f = ∫

p1 v 1

v2

v2

dpv − ∫

v1

pdv = p2 v2 − p1 v1 − ∫

pdv

v1

where v1 = v(T1 , p1 ) and v2 = v(T2 , p2 ). ln ( D.F. Mendoza (Universidad de Antioquia)

v2 f v (T, p2 ) 1 ) = pdv + z2 − z1 ∫ f v (T, p1 ) RT v1 Phase Equilibrium: Pure Substances

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Introduction

Fugacity changes in homogeneous phase: liquid and solid If we assume that liquid and solid phases are incompressible (constant v along the isothermal integral): Liquid phase: ln (

p2 f l (T, p2 ) 1 v l (T )(p2 − p1 ) l ) = v dp = ∫ f l (T, p1 ) RT p1 RT

thus, f l (T, p2 ) = f l (T, p1 ) exp (

v l (T )(p2 − p1 ) ) RT

solid phase: ln (

p2 f c (T, p2 ) v c (T )(p2 − p1 ) 1 v c dp = )= ∫ c f (T, p1 ) RT p1 RT

thus, f c (T, p2 ) = f c (T, p1 ) exp ( D.F. Mendoza (Universidad de Antioquia)

v c (T )(p2 − p1 ) ) RT

Phase Equilibrium: Pure Substances

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Introduction

Fugacity as function of T and p Chemical potential and fugacity are related by the following equation: µ(T, p) − µig (T, p) = RT ln (

f (T, p) ) p

from this equation we obtain ln f (T, p) =

µ(T, p) − µig (T, p) + ln p RT

• Fugacity as function of p: The derivative of fugacity with respect to pressure at constan tempertare is: (

∂ ln f (T, p) 1 ∂µ(T, p) 1 ∂µig (T, p) 1 ) = ( ) − ( ) + ∂p RT ∂p RT ∂p p T T T

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

the partial derivatives of the chemical with respect to the pressure at constant temperature are: (

∂µ(T, p) ) = v(T, p); ∂p T

(

∂µig (T, p) RT ) = v ig (T, p) = ∂p p T

thus, (

∂ ln f (T, p) v(T, p) ) = ∂p RT T

• Fugacity as function of T : The derivative of fugacity with respect to temperature at constant pressure is: (

∂ ln f (T, p) ∂ µ(T, p) ∂ µig (T, p) ) = ( ) − ( ) ∂T ∂T RT ∂T RT p p p

The derivatives at the right-hand side are obtained using the Gibbs-Helmholtz equation.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Gibbs-Helmholtz equation The chemical potential can be expressed as: µ(T, p) = h(T, p) − T s(T, p) dividing by T , µ(T, p) h(T, p) = − s(T, p) T T The derivative with respect to the temperature ∂ µ(T, p) h(T, p) ( ) =− ∂T T T2 p

(

∂ ln f (T, p) h(T, p) − hig (T, p) hR (T, p) ) =− =− 2 ∂T RT RT 2 p

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

• Fugacity as function of T and p: The total derivative of ln f using

T and p as independent variables is: d ln f = (

∂ ln f (T, p) ∂ ln f (T, p) ) dT + ( ) dp ∂T ∂p p T

d ln f = −

hR v dp dT + 2 RT RT

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Fugacity coefficient The fugacity coefficient, φ, is defined as the ratio between the fugacity of a real substance and the fugacity of an ideal gas at the same temperature and pressure: f (T, p) f (T, p) = φ(T, p) = ig f (T, p) p We can obtain this expression from calculating the chemical potentials of the a real and an ideal gas from a very low pressure (p∗ → 0) to p along the isotherm T : µ(T, p) − µ(T, p∗ ) = RT ln (

f (T, p) ) f (T, p∗ ) p µig (T, p) − µig (T, p∗ ) = RT ln ( ∗ ) p

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

under this circumstances µ(T, p∗ ) = µig (T, p∗ ) f (T, p∗ ) = p∗

subtracting µ(T, p) − µig (T, p) = RT ln (

f (T, p) ) = RT ln φ(T, p) p

we can notice that µ(T, p) − µig (T, p) is the residual Gibbs energy g R (T, p) g R (T, p) = RT ln φ(T, p) = ∫

0

D.F. Mendoza (Universidad de Antioquia)

p

(v − v ig )dp = ∫

Phase Equilibrium: Pure Substances

0

p

(v −

RT ) dp p

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Introduction

Fugacity coefficient using virial EoS (gas phase only) z =1+

Bp RT

where: RTc 0 (B + ωB 1 ) pc 0.422 B 0 = 0.083 − T r1.6 0.172 B 1 = 0.139 − T r4.2 B=

integrating

p

RT ln φ(T, p) = RT ∫

D.F. Mendoza (Universidad de Antioquia)

0

z−1 dp = Bp p

Phase Equilibrium: Pure Substances

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Introduction

Fugacity coefficient using Cubic EoS (gas and liquid phases) The general form of a CEoS is: p=

a(T ) RT − v − b (v + δ1 b)(v + δ2 b)

where: a(T ) = a(Tc )f (T );

a(Tc ) = Ωa (

R2 Tc2 ); pc √

f (T ) = [1 + m(1 −

EoS SRK PR

δ1 1√ 1+ 2

δ2 0√ 1- 2

D.F. Mendoza (Universidad de Antioquia)

Ωa 0.42748 0.45724

Ωb 0.08664 0.0778

b = Ωb (

RTc ) pc

2

Tr )]

m 0.48 + 1.574 ω - 0.176 ω 2 0.37464 + 1.54226 ω - 0.26992 ω 2

Phase Equilibrium: Pure Substances

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Introduction

using change of variable from dp to dv, we get: p

RT ln φ = ∫

0

(v −

RT ) dp p v

RT ln φ = RT (z − 1) − ∫

∞

(p −

RT ) dv v

This expression is integrated using p given by the general CEoS to obtain: RT ln φ(T, p) = RT ln (

v a(T ) δ1 b + v )+ ln ( ) − RT ln(z) + RT (z − 1) v−b (δ2 − δ1 )b δ2 b + v

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Fugacity coefficient using corresponding states principle7 pr

RT ln φ = RT ∫

(z − 1) 0

dpr pr

7 Fugacity coefficient (zc = 0.27). Hougen, Watson, Ragatz. Chemical process principles, part. II, Wiley, 1959. D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Lee-Kesler corresponding states8 ln φ = ln φ(0) + ω ln φ(1)

8

Koretsky M. Engineering and chemical thermodynamics, 2 ed., US: Wiley, 2013.

D.F. Mendoza (Universidad de Antioquia)

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Introduction

Lee-Kesler corresponding states9

9

Koretsky M. Engineering and chemical thermodynamics, 2 ed., US: Wiley, 2013.

D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

Calculation of fugacities • Gas phase:

f v (T, p) = φv (T, p)p

where φ is calculated using any suitable EoS (e.g. virial, cubic, corresponding states) • Liquid phase: 1 Using any EoS able to model gas and liquid phases (e.g. cubic, corresponding states): f l (T, p) = φl (T, p)p 2

Incompressible fluid model: f l (T, p) = φv,sat (T, psat )psat (T ) exp (

v l (T )(p − psat ) ) RT

• Solid phase: In this case we assume the incompressible solid model f c (T, p) = φsub,sat (T, psub )psub (T ) exp ( D.F. Mendoza (Universidad de Antioquia)

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Introduction

Note that. . . When we use the incompressible (fluid or solid) model: • φv,sat is calculated from the EoS that describes the vapor phase and psat (T ) is calculated from a vapor pressure correlation (e.g. Antoine, Wagner, etc.) • φsub,sat is calculated from the EoS that describes the gas phase and psub (T ) is calculated from a sublimation pressure correlation. When we use a cubic EoS to calculate liquid and vapor phases: • φv (T, p) is calculated using the gas (vapor) volume given by the CEoS at T and p. • φl (T, p) is calculated using the liquid volume given by the CEoS at T and p.

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Introduction

Example Calculate the gas-phase fugacity of propane at 85 bar and 444 K using: • Virial EoS • SRK EoS • Corresponding states

Data: Tc = 369.8 K, pc = 42.5 bar, ω = 0.153, zc = 0.27.

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Introduction

Example Calculate the liquid fugacity of propane at 253.15 K and 10 bar and using: • Virial EoS and incompressible fluid model. • SRK EoS • Corresponding states

Data: Tc = 369.8 K, pc = 42.5 bar, ω = 0.153, zc = 0.27, pvap (253.15 K) = 2.444 bar, v l,sat. (253.15 K) = 1.802 × 10−3 m3 /kg.

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Phase Equilibrium: Pure Substances

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Introduction

Example A compound has a solid molar volume of 85 cm3 /mol and at a pressure of 0.25 bar, it sublimates at T = 275 K. Estimate the fugacity of this compound in the solid phase at T = 275 K and each of the following pressures, 1

1 bar

2

10 bar

3

100 bar

4

1000 bar

Assume: φsub (275 K, 0.25 bar) ≈ 1 and incompressible solid model. R = 83.1452 bar cm3 /(mol K)

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Introduction

Phase equilibrium calculation According to the Gibbs’ phase rule the solution10 of a two-phase problem has one degree of freedom. This degree of freedom can be either T sat or psat . Let’s suppose we are interested in calculating the equilibrium properties of phases α and β. 1. Case 1: Given T α = T sat , calculate: T β , pα , pβ using the following equations: Tα = Tβ pα = pβ f α (T α , pα ) = f β (T β , pβ ) if we replace the first two equations into the third one, we only need to solve the third equation f α (T sat , psat ) = f β (T sat , psat ) where psat is the only unknown, since pα = pβ = psat , and T α = T β = T sat 10 by solution we mean finding the intensive properties in the equilibrium, i.e. T and p of each phase. D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

2. Case 2: Given pα = psat , calculate: T α , T β , pβ using the following equations: Tα = Tβ pα = pβ f α (T α , pα ) = f β (T β , pβ ) if we replace the first two equations into the third one, we only need to solve the third equation f α (T sat , psat ) = f β (T sat , psat ) where T sat is the only unknown, since pα = pβ = psat , and T α = T β = T sat

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Introduction

In summary 1

Case 1: Given T sat find psat , solving: f α (T sat , psat ) = f β (T sat , psat ) then use the equality in temperature and pressure to assign values to temperature and pressure in each phase: T α = T β = T sat pα = pβ = psat

2

Case 2: Given psat find T sat , solving: f α (T sat , psat ) = f β (T sat , psat ) then use the equality in temperature and pressure to assign values to temperature and pressure in each phase: T α = T β = T sat pα = pβ = psat

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Introduction

Application: Vapor-liquid equilibrium Example Find T vap of benzene at 20 bar using: 1

Virial equation (vapor phase) and incompressible fluid model (liquid phase).

2

Generalized fugacity coefficent charts (Corresponding states principle) for both phases.

3

PR EoS to describe liquid and vapor phases.

Data: Tc = 562.2 K, pc = 48.9 bar, zc = 0.27. ln pvap (mmHg) = 15.9008 −

2788.51 ; T (K) − 52.36

(280K ≤ T ≤ 377K)11

11 Note that this expression is not suitable to solve our problem since pvap (20 bar) = 494.5 bar. D.F. Mendoza (Universidad de Antioquia)

Phase Equilibrium: Pure Substances

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Introduction

To find T

vap

we need to solve the following phase equilibrium equation f v (T vap , 20 bar) = f l (T vap , 20 bar)

The way we calculate fugacity depends on how we model each phase: 1

Virial equation (vapor phase) and incompressible fluid model (liquid phase). φv (T vap , 20 bar) ⋅ 20 bar = φv (T vap , 20 bar) ⋅ pvap (T vap ) thus, we nee to find T vap , such that: pvap (T vap ) − 20 bar = 0

2

Corresponding states principle and PR EoS follow the same approach to describe liquid and vapor phases. φv (T vap , 20 bar) ⋅ 20 bar = φl (T vap , 20 bar) ⋅ 20 bar thus, we nee to find T vap , such that: φv (T vap , 20 bar) − φl (T vap , 20 bar) = 0

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