Post on 24-Nov-2014
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
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3. PHASE EQUILIBRIUM - FUNDAMENTALS3. PHASE EQUILIBRIUM - FUNDAMENTALS
Learning Objectives
Student should be able to ;
i. Explain the use of phase equilibrium in chemical engineering processesii. Define and explain the fundamental criteria of phase equilibrium particularly using
Gibbs Free Energy.iii. Define the concept of chemical potential and able to explain its role in defining phase
equilibria.iv. Define the concept of fugacity and the reason for its introduction as a means for
performing calculation related to phase equilibria.v. Define and explain partial molar property concept and able to calculate property
changes of a system as a result of mixing.vi. Derive the Maxwell relations for mixtures.vii. Perform calculation related to phase equilibria and establish the vapour liquid
equilibrium plot (P-x,y or T-x,y) for ideal mixtures.
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
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vapour
liquid
phase changes
Phase Equilibrium is commonly encountered in Chemical Processes, eg. distillation
What happen in the system ?
- equilibrium between liquid and vapour- vaporisation/condensation taking place i.e., phase changes
To make matter more complex, system containsmulti-components.
Recall on what we have learnt before ?
Thermodynamic relationships involving Free Energy function i.e. Gibbs Free Energy in particular - homogenous fluid of pure component/const. composition.
Thermodynamic calculations (enthalpy & saturation vapour pressure) involving phase changes.
Network of Thermodynamic Equations - Maxwell Equations
Now, these thermodynamic property relationships will be extended to systems undergoing phase & composition changes
Application of Phase Equilibrium
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Equilibrium Criteria
Let us review the concept of thermal and mechanical equilibrium….
Thermal Equilibrium Mechanical Equilibrium
Under phase equilibria of pure component,
Phase boundary
Phase
Phase
T = T
Driving Force T
P = P
Driving Force P
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Equilibrium Criteria
Now consider a system of multi component…..
Thermal Equilibrium Mechanical Equilibrium
Phase boundary
Phase
Phase
T = T
Driving Force T
P = P
Driving Force P
Another equilibrium criteria involvingcomposition in the two phase
? = ?
Driving Force
The tendency would be to say composition difference….
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Equilibrium Criteria
Now consider a system of multi component…..
Phase boundary
Phase
Phase
If composition, then the criteria at equilibrium in addition to the pressure and temperature should be x = x . But is it true ?
Air-Water system containing small amount of methanol
Suppose the composition of methanol in water is 0.01
Will the composition of methanol in air is 0.01 too at equilibria?
The answer is no. Therefore, composition is not necessary the driving force…..
We will look at this matter now …
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Equilibrium Criteria
Consider a closed system in terms of changes in matter but not isolated from surroundings in terms of heat and work flow. Let us start with a pure system first….
1st lawdU = dQ + dW = dQ - Pb dV
dS = dQ/Tb + Sp 2nd law
SYSTEM
QW
Sp - entropy production due to irreversibilitysubscipt 'b' - boundary of system
dU = Tb dS - TbSp - Pb dV
combining the two eqn. will yield ;
Since, Sp > 0,
dU < T dS - P dV or dU + P dV - T dS < 0
suppose there is an imbalance of T or P,process leading to equilibrium will continue until T and P for the system boundary equal to surrounding by means of rejecting/accepting heator doing or receiving work.
for the above equation to be true.
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dU < T dS - P dV or dU + P dV - T dS < 0
Using the equation as the basis,
At const. T and P, the eqn. becomes;
Taking the total differential from the equation, we can derive
dG = dU + P dV + V dP - T dS - S dT
Introduce the Gibbs Free EnergyG = H - TS = U + PV - TS We know that H = U + PV
dG T,P = dU + P dV - T dS
Equilibrium Criteria
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dU < T dS - P dV or dU + P dV - T dS < 0
Referring back to the equation,
Hence, at equilibrium we could use this basic criteria
Implication : Any real process occuring spontaneously when system is at const. T and P incurs a decrease in Gibbs Free Energy. And the system will attain its thermodynamic equilibrium when the Gibbs Free Energy value has attained the lowest (minimum) value.
dGT,P = 0
We can derive the criteria below,
dG T,P = dU + P dV - T dS
dGT,P < 0
and comparing it to
Equilibrium Criteria
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For mixtures, we could write the equation in the form below,
Gi = ni gi
+ ni gi
+ ni gi
+ ……. at constant T, P
, and are the respective different phases and ni is the no. of moles of component i.
dGi = d(ni gi
) + d(ni gi
) + d(ni gi
) + ……. < 0
dGi = (ni dgi
+ gi dni
) + (ni dgi
+ gi dni
(ni dgi
+ gi dni
) + ..
Taking its differential,.
At equilibrium (const. T and P), dgi = 0. Therefore, the equation becomes..
dGi = gi dni
+ gi dni
gi dni
+ ……. < 0In a closed system and considering two phases ( and ), any mass loss in one phase will go to the other phase. Therefore,
dni = - dni
Substituting in the above equation
(gi - gi
) dni < 0
Since at equilibrium dGT,P = 0, therefore….
(gi - gi
) = 0 and gi = gi
Equilibrium Criteria
at constant T, P
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dGT,P = 0
dU < T dS - P dV or dU + P dV - T dS < 0
Also from the same equation shown earlier,
other equilibrium criteria subject to different constraint could be derived.
The other equilibrium criteria consist of ;
dSU,V = 0dAT,V = 0dUS,V = 0dSH,P = 0dHS,P = 0
but the earlier relation involving Gibbs Free Energy is the most popularly used !
Equilibrium Criteria
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The equilibrium criteria established so far seems to be very general. Question is how can we apply such criteria in the calculation for phase equilibria ?
Need to introduce further concepts !
For a constant mass system at equilibrium, we can write the following total differentialeqn. for the Gibbs Free Energy (G)
dG = (G/T)P dT + (G/P)T dP = - S dT + V dP
Recall the total differential
dG = dU + P dV + V dP - T dS - S dT
(G/T)P = -S
(G/P)T = V
Now if we consider a system with variable mass, clearly there should be an additional term to account for this.
dG = (G/T)P dT + (G/P)T dP + (G/n)T,P dn
= - S dT + V dP + dnaccount for changes in G with addition/removal of material = (G/n)T,P where
We now introduce Chemical Potential () which is defined as the rate of changes in G with changes in material amount in the system.
Chemical Potential
Consider the derivation below for pure component system!
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Following from the earlier relation for vapour-liquid equilibrium, we can then write ;
dG + dG = + = 0 =
at const T,P
for pure component system
Chemical Potential
Gi = ni gi
+ ni gi
+ ni gi
+ …….
dG = dG + dG + dG + …… = 0
at equilibrium
Thus, for 2 phase system,
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Chemical potential could also be defined using other functions,
Chemical Potential
= (G/n)T,P Shown earlier
Using the internal energy, U function, it can be defined as;
= (U/n)S,V,n
The eqn. is derived from the total differential eqn. for U.
dU = T dS - P dV + dn
Now, from the definition H = U + PV and dH = dU + P dV + V dP
dH = T dS + V dP + dn
recall the eqn. by combining 1st & 2nd law
dU = TdS - PdV
We could write ;
dH = (H/S)P,n dS + (H/P)S,n dP + (H/n)P,S dn
and by comparing with the equation below;
can equate the two variables
Therefore we could developed another definition for Chemical Potential (H/n)P,S =
j
You can also derive similar relation involving A (Helmholtz Free Energy) at constant T,V, n
dU = (U/S)V,n dS + (U/V)S,n dV + (U/n)S,V dn
Let us attempt another one,
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Equilibrium of Mixtures
Recall the earlier example on a system.
Air-Water system containing small amount of methanol.
Suppose the composition of methanol in water is 0.01
Will the composition of methanol in air is 0.01 too at equilibria?
The answer is no. Therefore, composition is not necessary the driving force…..
If we look carefully at the system, it contains more than one component I.e., exist in the form of mixture.
In separation processes which forms part of a whole chemical process, often mixtures of various components have to be separated either physically or chemically. For a physical based separation, often the understanding of phase equilibria for multicomponent system is important for the design and operation of the process. Distillation, absorption, adsorption etc. are some of the examples of the separation processes.
Earlier, the understanding of phase equilibria was developed around pure component. Using the fundamental developed, we will now explore the multi component system (mixtures).
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Extending the same development to mixtures of components ;
1 dn1 + 2 dn2 + 1 dn1 + 2 dn2 = 0
dn1 = - dn1
dn2 = - dn2
1 = 1
2 = 2
In general terms,
Basis for multi- component VLE
calculation
( i - i ) dni = 0
i = i
i
dG + dG = + = 0
For 2For 2
componentcomponent
Equilibrium of Mixtures
From the earlier equation for phase equilibria for pure component system.
(gi - gi
) = 0 and gi = gi
at constant T, P
This is comparable to the equation developed earlier,,
at constant T, P
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Equilibrium of Mixtures
Chemical potential is an abstract concept. However, it is useful since it provides a simple criterion for equilibria measures. Unfortunately, in application, it turns out that chemical potential has some inconvenient mathematical behaviours. This has led to the introduction of fugacity.
= + RT ln f0
- chemical potential value at f = unity- function of only T
Fugacity (f) - a measure of volatility of a component in solution
The above relation is derived from ;
d = dg = vdP - sdTd = dg = vdP = RT dln P
T Tin the case of real gas, P is substituted by f
integrate !
f (fugacity) has the unit of pressure (atm, kPa, bar etc)
Extending the relation to mixtures ;
= + RT ln f0
i i i
For ideal gas, the fugacity could be simplified to be equal to the partial pressure exerted by the component in the mixture
f = y P = pi i i
at const. T
The relation between chemical potential and fugacity is described by the equation below;
Fugacity.
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Equilibrium of Mixtures
Writing the eqn. for 2 different phase ( and )
= + RT ln f0
i i i = + RT ln f
0i i i
from equilibrium criteria ;
i = i
it can easily be seen that the following criteria also holds at equilibrium ;
fi = fi
The fugacity of the individual component is equal for the 2 phases.Also basis for
multi- component VLE calculation
Now if we are dealing with vapour liquid equilibria, clearly the two phases will refer to liquid and vapour !
fi = fivl
Fugacity is something easier to handle.Later we will be looking at how to calculate fugacity
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Partial Molar Properties.
When a substance becomes part of a mixture, it losses its identity. However, it still contributes to the mixture properties. For mixtures that do not behave ideally, the mixture properties are not simply the molar-weighted sum of the pure component.
as what you have seen in MEB : Calculation of properties based on Pseudo-Component Approach
Introducing the Partial Molar Properties which has similar appearance in the equation as the chemical potential.
Writing it in a total differential form of M ;
dM = M /P) dP + M/T) dT + M/n ) dnT,n P,n ii
iP,T,n = ij
T,n P,n ii
isum of partial molar properties
dM = M /P) dP + M/T) dT + M dn
The total solution properties depend on the amount present of each species and its resultant interactions. Therefore, need to develop method to enable calculation of the properties in mixture form.
M = ( M/ni ) T,P,nj = 1
Property Changes for Mixing Process
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In respect to the various thermodynamic properties, the partial molar properties could be written as:
dV = (V/T)P,ni + (V/P)T,ni
+ (V/ni)T,P,nj = i dnii = 1
k
dH = (H/T)P,ni + (H/P)T,ni
+ (H/ni)T,P,nj = i dnii = 1
k
For partial volume,
at constant T,P
dV = Vi dnii = 1
k
at constant T,P
dH = Hi dnii = 1
k
Integrate V = Vi ni
k
i = 1
H = Hi ni
k
i = 1Integrate
For partial enthalpy,
or v = vi xi
k
i = 1
(molar volume)
or h = hi xi
k
i = 1
(molar enthalpy)
Property Changes for Mixing Process
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The significance of Partial Molar Properties
UU
consider adding a small portionsof identical mixture to the larger one. What happen is ;
Utot = U + U
By knowing the partial molar properties, we can determine the actual changes in the properties as a result of the addition of material.
Utot = U + U
supposedly from pure molar property
but the actual one
The phenomena could be seen when mixing two species of different molecular sizeeg. benzene with water.
Property Changes for Mixing Process
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
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The Gibbs Duhem EquationThe Gibbs-Duhem equation provides a very useful relationship between the partial molar properties of different species in a mixture. It provides constraints between the partial molar properties of the different species i.e. if the partial molar property of one component is known in a 2-component mixture, the partial molar property of the other component could be calculated.
Integration of the above eqn. at const T and P ;
differentiating the above ;
equating withthe above eqn.
For the condition of constant T and P (under which equilibrium normaly established),
n dM = 0
i i
Gibbs Duhem Eqn.
i = 1
i = 1 i = 1
M = Mi nii = 1
k
M = Mi nii = 1
k
k
dM = Mi dni + ni dMi k k
we can show that ;
M/P) dP + M/T) dT = n dMT,n P,n i ii = 1
k
dM = M /P) dP + M/T) dT + M dn
T,n P,n i i
Consider again the general eqn. for partial molar properties ;
i = 1
k
at constant T and P
Property Changes for Mixing Process
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Alternatively, we start from the eqn. .
T,n P,n ii
idM = M /P) dP + M/T) dT + M dn
At constant T and P
dM = Mi dnii = 1
k
Shown earlier !
Based on definition of partial molar property,
M = Mi ni
We differentiate the equation,
i = 1
k
dM = Mi dni + ni dMi k
i = 1
k
i = 1
For both equation to be true, ni dMi must be equal to 0i = 1
k
n dM = 0
i iGibbs Duhem Eqn.
i = 1
k
at constant T and P
Property Changes for Mixing Process
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Now let us apply the derivation using Gibbs Free Energy as the basis.
G = i ni which on a molar basis is written as (divide by n),
G = Gi ni
Using Gibbs Free Energy
by differentiation,
ni di = 0This is the Gibbs-Duhem Eqn. using Chemical Potential as the Extensive Property.
G/n = i yi = Gi yi = g
we can prove that at const T and P,
i = 1
k
i = 1 i = 1
k k
By definition of partial molar property
i = 1
k
dG = Gi dni + ni dGi = i dni + ni dii = 1 i = 1 i = 1 i = 1
k k k k
through comparison to eqn :
- S dT + V dP - i dni = 0 at equilibrium dG = i = 1
k
Property Changes for Mixing Process
i = 1
k
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Experimental Determination of Partial Molar PropertiesExperimental Determination of Partial Molar Properties
We know now that all thermodynamic properties must be either measured directlyor calculated from experimentally determined data. We will now focussed on determining property changes on mixing and partial molar properties.
For the property changes, we can write a general eqn ;
M = M - nimi
property of the pure component
mixture property
property changes
M = Mi ni
Using the eqn.,
Lead to M = niMi - nimi = niMi
partial molar property change
partial molar property
G = H - TS = i ni
For Gibbs Free Energy changes on forming a liquid solution, we can write,
G = niGi - nigi = niGi
alternatively from eqn. we can write,
G = nii - nii = nii
Chem & Process Therm. by B.G. Kyle (pg 378-379)
i ii
i i i
Additional Note !
i = 1
k
i = 1
k
i = 1
k
i = 1
k
i = 1
k
i
To simplify writing the eqn., let us assume,
= i = 1
k
i
Property Changes for Mixing Process
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Consider the equation again ;
M = niMi - nimi = niMi
After conducting few experiments, we can have a relation of V against some combinations of compositions.
If a plot is to be generated for a binary mixture ;
V
x1 1
i i i
let say we do it for volume (V)
V = niVi - nivi = niVi i i i
0
Positive deviation !
Property Changes for Mixing Process
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From the eqn.,
we can write ;
divide by n1 + n2
V / (n1 + n2) = v = x1V1 + x2V2
differentiate v wrt x1 ;
In accordance with eqn. above,the last 2 terms can be written as
x1dV1/dx1 - x1dv1/dx1 + x2 dV2/dx1 - x2dv2/dx1
as v1 and v2 are property of pure components, their derivatives are 0. And from Gibbs Duhem eqn.,the terms x1dV1/dx1 + x2dV2/dx2 is also equal to 0.
Thus, the remaining of the eqn. becomes ;
dv/dx1 = V1 - V2
eliminating V2 from the eqn.,
V1 = v + (1-x1)dv/dx1 (continue next page)
i i i
ni dM i = 0
Gibbs Duhem eqn.
i
V = n1V1 + n2V2
V = niVi - niVi = niVi
dv/dx1 = V1 - V2 + x1 dV1/dx1 + x2 dV2/dx1
(terms appear in mol fraction)
Property Changes for Mixing Process
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V
x1
A
G
E
F D
C
B
K
V1 = v + (1-x1)dv/dx1The eqn.
can be viewed graphically using the figure below. Using point A as the reference ;
v = AG or BD, (1-x1) = GD or AB, and the slope dm/dx1 = CB/AB or KC
V1 (at point A) = BD + AB(CB/AB) = BD + CB = CD
and also we can write the eqn.;
V2 = V1 - dv/dx1 = CD - KC/1 = KD = EF
slope
Thus from experimental data, we could find the respective value for V1 and V2 which shows the differences between partial molar volume to pure species volume with the respective compositions.
Property Changes for Mixing Process
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To demonstrate the applicability of the equations, lets suppose we have experimental data representing volume changes with respect to composition for a binary system.
Experimental data of the molar volume for a binary system of components 1 and 2.
2000
1750
1500
1250
1000
0 10.2 0.4 0.6 0.8
v = x1 V1 + x2 V2 Using the equation :
v = (1-x2) V1 + x2 V2
Property Changes for Mixing Process
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Differentiating the eqn wrt x2 and applying the Gibbs Duhem Eqn.,
v = (1-x2) V1 + x2 V2
x2 (dv/dx2)= -x2 V1 + x2 V2
x2 (dv/dx2)= - V1 + ( x1V1 + x2 V2 )
= vv = V1 + x2 (dv/dx2)
2000
1750
1500
1250
1000
0 10.2 0.4 0.6 0.8
V2
V1
v1
v2
X2 mix = 0.7
vmix
At x2 = 0.7
Property Changes for Mixing Process
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Let’s try with another example involving molar enthalpy
Develop an expression for the partial molar enthalpy of sulphuric acid in water at 21 oC using equation;
HH2SO4 = h - xH20 (dh/dxh20)
Given the enthalpy expression
h = 1.596 xH2SO4 + 1.591xH20 - 74.40 xH2SO4
xH20 ( 1 - 0.561 xH2SO4 ) kJ/kmol
Differentiate the eqn.,
dh/dxH20 = 1.591 – 74.4 [ (xH2SO4 - xH20) (1 - 0.561 xH2SO4
) + 0.561 xH2SO4 xH20 ] kJ/kmol
Substitute the two eqn. above into the top equation,
HH2SO4 = 1.596 – 74.40 xH20
2 + 83.48 xH2SO4 xH20
2 kJ/kmol
Applying Gibbs Duhem Equation
Property Changes for Mixing Process
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Analytical Determination of Partial Molar PropertiesAnalytical Determination of Partial Molar Properties
Often an analytical expression for total solution property is known as a function of composition. An example would be the expression developed using virial equation for mixtures (will look at the expression in detail later!).
Additional Note !
v = (RT/P) [ 1 + (Bmix) / RT)] = (RT/P) + y12 B11 + 2y1y2 B12 + y2
2 B22
for 2 component
V = ( n1 + n2) v = ( n1 + n2) (RT/P) + [ ( n12 B11 + 2n1n2 B12 + n2
2 B22 ) / (n1 + n2) ]
Molar volume
System volume
Taking the differentiation,
V1 = (V/n1)T,P,n2 = RT + 2n1
B11 + 2n2 B22 - n1
2 B11 + 2n1n2 B12 + n22 B22
P (n1 + n2) (n1 + n2) 2
Simplifying the eqn,
Partial molar volume of species 1
V1 = RT + ( y12 + 2y1y2
) B11 + 2 y22 B12 - y2
2 B22
P
Property Changes for Mixing Process
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Additional Note !
Similarly, the expression for species 2 could be derived.
The molar volume for pure species 1 is obtained by setting y2 = 0
V2 = RT - y12 B11 + 2y1
2 B12 + ( y22 + 2 y1y2 ) B22
P
v1 = RT/P + B11
The volume change for mixing is given by,
vmix = v - ( y1v1 + y2v2 )
Substituting for v, v1 and v2 and simplifying the eqn. will yield the expression ;
vmix = 2 y1y2 [ B12 - ( B11 + B22 ) / 2 ]
Property Changes for Mixing Process
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A property changes of mixing describes the extent of property changes as a result of mixing. It is defined as the difference between the total solution property in the mixture and the sum of the pure species properties of its constituents, each in proportion to how much is present in the mixture.
Property Changes for Mixing Process
Mmix = M - ni Mii = 1
k
Hmix = H - ni Hii = 1
k
Writing it for the respective properties,
Vmix = V - ni Vii = 1
k
Smix = S - ni Sii = 1
k
The properties of the pure component are determined at the set T and P.
For cases where partial molar properties could be used to determine the total mixture property eg. volume,
Vmix = niVi - ni Vii = 1
k
OVERALL VIEW
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Property Changes for Mixing Process
When involving partial molar properties, the equations can also be written in the form,
Vmix = niVi - ni Vii = 1
k
i = 1
k
Vmix = ni ( Vi - Vi )i = 1
k
Hmix = niHi - ni Hii = 1
k
i = 1
k
Hmix = ni ( Hi - Hi )i = 1
k
and
For volume
For enthalpy
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At a molecular level, the property change of mixing reflects how the interactions between unlike species in the mixture compare to the like interactions of the pure species they replaced.
Property Changes for Mixing Process
For the case of volume changes as a result of mixing between different species,
Vmix = niVi - ni Vii = 1
k
The volume change as a result of mixing reflects the differences in how closely species in a mixture can pack together in comparison to how they pack as pure species. A negative Vmix when the unlike interaction “pull” the species in the mixture closer together while a positive Vmix when the unlike interaction “push” the species apart when they are mixed. However, if the Vmix is zero, then the different species behave identically in the mixture as the pure species.
For the case of enthalpy changes as a result of mixing between different species,
Hmix = niHi - ni Hii = 1
k
The enthalpy change as a result of mixing reflects the energetic interactions between the different species. A negative Vmix reflects more stable presence of the species in the mixture as compared to at their pure state while a positive Vmix reflects less stable presence of the species in the mixture as compared to at their pure state . However, if the Vmix is zero, there is no changes in the stability of their presence whether in mixture or pure state.
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The earlier shown example demonstrate the applicability of the equation below for volume changes as a result of mixing;
Property Changes for Mixing Process
Vmix = niVi - ni Vi
k
2000
1750
1500
1250
1000
0 10.2 0.4 0.6 0.8V1
V2
v1
v2
X2 mix = 0.7
vmix
At x2 = 0.7
Experimental data of the molar volume for a binary system of components 1 and 2.
x2
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Recall on the development of Maxwell eqn. for single component (pure) system .
dG = V dP - S dT
(G/T)P = -S(G/P)T = V
(G/PT)P = (V/T)P2 (G/TP)T = -(S/P)T
2
Equal
eg.
The same procedure could be applied to equations representing a mixture.
dG = - S dT + V dP + dn
From the earlier derivation involving pure system, we could show that the following applies to mixture system.
= (G/n)T,P where
(G/T)P,n = -S (G/P)T,n = V
( G/TP)n = - (S/P)T,n
2 ( G/PT)n = (V/T)P,n
2
Equal
Maxwell Equations for Mixtures.
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Following the same procedure but using other properties (H, U and A) will lead to the following Maxwell Equations.
- (S/P)V,n = (V/T)S,n
(S/V)P,n = (P/T)S,n
(S/V)T,n = (P/T)V,n
Now if n is not being held constant, we could derive
dG = - S dT + V dP + dn = (G/n)T,P where
(G/T)P,n = -S (G/n)T,P =
( G/Tn)P = - (S/n)T,P
( G/nT)P = (/T)P,n
22
Equal
More Maxwell equations could be derived using other properties (H,U and A)
Maxwell Equations for Mixtures.
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
39
Phase Rule – A Review
Consider a system at phase equilibrium
Vapour & Liquid in equilibrium
Conditions at equilibrium ;
Tl = Tv
l = v
Pl = Pv
l = f(Tl, Pl, xi)
v = f(Tv, Pv, yi)2 phases
From phase rule ;
F = C - P + 2F - degree of freedomC - No. of componentP - No. of phases
For the above case, clearly F =1 for a single component
As we add-in additional component, we will have 3 additional eqns and 4 variables !
l = v
l = f(Tl, Pl, xi)
v = f(Tv, Pv, yi)
l , v
xi , yi
1 extra degrees of freedomi.e., 1 extra spec. on system variable
No of spec. on system variables to be made in orderto fully specified the system.
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Ideal Mixtures
There are 2 classes of extensive properties
1. They are functions of entropy (S,A,G). Their partial molar properties is not equal to the pure molar properties even for ideal mixtures.
2 They are not function of entropy (V,U,H). Their partial molar properties is equal to the pure molar properties for ideal mixtures.
The free energy function i.e. Gibbs (or Helmholtz) free energies, belong to the first group.
G = ini = Gi ni ( = Gi ni )
partial molarGibbs Free Energy
pure molarGibbs Free Energy
Total Gibbs Free Energyof a mixture system
iii
M = xi mii = 1
k
mi – pure molar properties
To simplify writing the eqn., let us assume,
= i = 1
k
i
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G = i ni + RT ni ln fi
recall the formula ;
The Gibbs free energy equation can be written in terms of fugacities using the eqn. below ;
For an ideal mixtures, we can write the equation (using partial pressures) ;
G = i ni + RT ni ln pi
For a number of ideal gas mixed at temperature T and initial partial pressure pio we can write ;
G = i ni + RT ni ln pio sum of the ideal gas components G prior to mixing
0
0
0
mixed all the ideal gas components together at const. T and P. The changes in the free energy resulted from mixing is ;
Gmix = RT ni ln ( pi / pio )
i.e. the difference between the two eqn. above.
i
i
i
i
i
i
i = + RT ln f0
i i i
Ideal Mixtures
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Consider the mixing of pure gases at same initial pressure (p io) together to form mixture which will also be at the same pressure.
pio = P pi = yi P
Therefore
initial pressure = final pressure partial pressure (final)
Using the 2 above relation and yi = ni / n into the equation,
Gmix = RT ni ln ( pi / pio )
Gmix = RT ni ln yiwhich can be derived into the form
Note that the answer will always be negative. Why ?
Remember when a process took place spontaneously, the Gibbs Free Energy will always seek for a minimum value shown earlier.
Gases will always tend to mix and not unmix unless being forced to be separated. And this is the natural direction of the process which is towards the minimum value for the Gibbs Free Energy
i
Ideal Mixtures
i = 1
k
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The entropy changes for the mixing proces could be derived from the definition of G = H - TS
Gmix = Hmix - T Smix
Smix = - Gmix / T
For ideal mixture, there are no changes in enthalpy due to mixing (Hmix = 0) as H is not a function of entropy.Therefore, their partial molar properties is equal to the pure molar properties for ideal mixture.
Hence,
Smix = - R ni ln yiWill always be
positive
The logical explanation is that for any spontaneous process, the entropy for the system will always increase. Principles of increasing entropy
This also help to explain that for entropy dependent partial molar properties (S, A, G), they have different values from the pure molar properties (S, A, G) even for ideal mixtures
On the contrary to the Gibbs Free Energy, the answer for entropy changes due to mixing will always be positive. Why ?
Ideal Mixtures
i = 1
k
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Using similar method of derivation but applying it to ideal liquid solution mixtures will lead to the following results :-
Smix = - R ni ln xi
Gmix = RT ni ln xi
For Gibbs Free Energy of mixing ideal solutions
For entropy changes from the mixing process of ideal solution
Ideal Mixtures
i = 1
k
i = 1
k
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Ideal Mixtures
Example : Find the ideal change in G and S when 3 mol of Argon (1), initially at 1 atm, 298 K are allowed to mix with 2 mol of Xenon (2) at the same initial condition.
The total moles are 3 + 2 = 5
Calculating for GmixGmix = RT ni ln xi
i = 1
k
= 8.314 . 298 [ 3 ln (3/5) + 2 ln (2/5) ]= - 8339 J
Calculating for Smix Smix = - R ni ln xii = 1
k
= - 8.314 [3 ln (3/5) + 2 ln (2/5) ]= + 28 J/K
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
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Vapour Liquid Equilibria for Ideal Mixtures
Recall the concept of fugacity of each component in a mixture.
= + RT ln f0
i i i
d = dg = v dP – s dT
dT = dgT = v dP = RT dln P
The relation was derived from
= + RT ln f / f
Introduce fugacity and substitute it into the equation,
0 0
Taking the f 0 at atmospheric pressure as the reference f 0 = P = 1and writing the equation for components in mixtures ;
= + RT ln f0
i i i
Reviewing the derivation for fugacity ….
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Vapour Liquid Equilibria for Ideal Mixtures
Now, the fugacity of each component in a simplest liquid mixture could be calculated by ;
fi = fi xi0
fi - fugacity of pure liquid at T & P of mixture0
At low pressure (under ideal gas condition), fi0 = Pi
0
where Pi0 is the partial pressure of the pure liquid at T & P of the mixture
and under the condition where the vapour phase is an ideal gas mixture, we can write
fi = P yi
Following the equilibrium criteria where
fi = fil v
we can derive that ;
yi P = Pi xi0
This is known as Raoult's Law
fi = fi
The fugacity of the individual component is equal for the 2 phases.
Recall the basis for equilibrium
Pi0 = saturation vapour pressure for component I at specified T,P
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The Raoults law (equation) enable a vapour liquid equilibria relation be established for an ideal liquid/gas system mixture.
For ideal mixture, can be calculated from : Raoult's Law
yi P = xi Pi
o
only need information on,- Total Pressure- Vapour Pressure Pi
can determine x-y diagram
x
yWe can also go
directly from left to right ?
P
x,y
liquid
vapour
IsothermalVLE diagram
P1
o
P2
o
o
Vapour Liquid Equilibria for Ideal Mixtures
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T
x,yFor ideal mixture, can be calculated from :
Raoult's Law
x
y
liquid
vapourIsobaricVLE diagram
T1
o
T2
o
yi = xi Pi / Po
- have to calculate Pi
for T value in between boiling point.
o
Use Antoinne
Eqn.
Can we go directly from left to right ?
can determine x-y diagram
Vapour Liquid Equilibria for Ideal Mixtures
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Vapour Liquid Equilibria for Ideal Mixtures
Example : Find the vapour composition above a liquid mixture of 50 mol% benzene(1) and hexane(2) at 25 C if it is assumed that both the liquid and vapour form ideal solutions. The vapour pressure of benzene and hexane at 25 C are 93.9 and 149.2 mmHg respectively.
yi P = Pi0 xi
Using equation,
Calculate for y1 and y2.
y1 = 0.5 . 93.9 / P y2 = 0.5 . 149.2 / P
P = y1 P + y2 P = x1 P10 + x2 P2
0 = 121.5 mmHg
The composition in vapour could then be calculated
y1 = 0.386 y2 = 0.614
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Vapour Liquid Equilibria for Ideal Mixtures
Example : Develop a T-x,y diagram for a system containing n-pentane(1) and n-heptane(2) at pressure 101.3 kPa. Given the Antoinne eqn. for saturation vapour pressure for the 2 components as below;
Component 1 : ln P10 = 13.818 – 2477.07 / (T + 233.21)
Component 2 : ln P20 = 13.8587 – 2911.32 / (T + 216.64)
First, find the two temperature limit I.e., the boiling points at specified pressure.
Component 1 : T = 36.04 C
Component 2 : T = 98.42 C
T
x-y
36.04
98.42
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Vapour Liquid Equilibria for Ideal Mixtures
To solve for the composition at the intermittent temperature between the boiling temperatures, use equation;
y1 = x1 P10 / P
x2 = 1 – x1
P = y1 P + y2 P = x1 P10 + x2 P2
0
Solve for x1
Then
T
x-y
36.04
98.42
T x1 y1
40 0.86 0.98370 0.250 0.70080 0.142 0.517
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Solubility in an Ideal Solution
Gas solubility in a liquid solvent is determined by equating vapour and liquid fugacities of the species.
fi = pi = yi P = xi Pi0
this is only for ideal mixture !
For solid solubility, the equation used for determining solid solubility in ideal solution is :
( ln xi / T)P = hm/RT2
enthalpy of melting
Note : The equation is independent of solvent properties (In actual fact, it is not true for actual solution)
If we integrate with assumption that the enthalpy remain constant, the result is
ln xT = (hm/R) (dT/T ) = (hm/R) [ 1/Tm - 1/T ] T
Tm
2
Tm - melting temperature of the soluteT - temperature of solvent
The principle of equilibrium
Now, we got ourself an equation to determine solid solubility in ideal solution.
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
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Freezing point depressionFreezing point depression
The extend of lowering in the melting temperature. Example : Addition of salt to water
The effect is to lower the activity of the solvent in the liquid phase primarily by loweringits mole fraction from unity to some lower value.
The equation used to estimates the lowering in melting temperature
T / ln x2 = T / ln (1-x1) = RT / hm2
solute composition in solvent solvent composition after mixing with solute
Integrating the above equation
ln (1-x1) = hm/T) . [ 1/Tm - 1/T ]
This is almost equal to -x1
as its value is very small
rearrange to give : T = x1 RTm / hm 2
refer to solvent
refer to solvent
Thus, we can predict the loweringof freezing point for a solution dueto addition of solute.
Solubility in an Ideal Solution
CAB 2023 : CHEMICAL THERMODYNAMICS July 2007
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Boiling Point ElevationBoiling Point Elevation
A similar form of analysis can be done to determine the extent of boiling point elevation
eg : salt dissolved in water will result in higher boiling temp.
A similar form of equation could be derived
ln (1-x1) = hvap/T) . [ 1/T - 1/Tvap ]
solute composition in solvent x1 << 1
T = x1 RTvap / hvap
After rearranging,
2
refer to solvent
This is almost equal to -x1
as its value is very small
Thus, we can predict the increasein boiling point for a solution dueto addition of solute.
Solubility in an Ideal Solution
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Conclusions and Review
The importance of phase equilibria in Chemical Eng. Separation Processes
Fundamental of Phase Equilibria
Fundamental Relation using Gibbs Free Energy
dG T,P = dU + P dV - T dS
dGT,P = 0
Deriving fundamental criteria at equilibrium
extended to mixture…
Introduce chemical potential to provide more sense…
ii And relate them through the equation involving fugacity
fRTii ln0 Thus allowing for the criteria to be expressed in fugacity
ii ff v
ili ff
Easier to perform calculation !
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Conclusions and Review
Property variation in mixture as composition changes
Chemical engineers have to deal mainly with mixtures
Introduce partial molar property
1,,
jnPTin
MM To account for property changes as a result of
composition change
Equation used for computing the changes (for volume and enthalpy) ;
i
k
iimix VnVV
1
i
k
iimix HnHH
1
i
k
iimix vxvv
1
i
k
iimix hxhh
1
Molar property (per mol basis)
Could be determine from pure molar property using the eqn. derived from Gibbs Duhem relation.
221dx
dMxMM for binary mixture
Has also established for Gmix and smix.
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Conclusions and Review
Demonstration on the application of the phase equilibria
Consider ideal mixture, where Raoult’s law is used to simplify the phase equilibria complexity
0iii PxPy
Fixing the system pressure and knowing the saturation vapour pressure at respective temperature, would allow for VLE composition to be computed with respect to the temperature range between the two boiling points.
T
x,y
yx