1. Don’t be in a such a hurry to condemn a person because he doesn’t do what you do, or think as you think. There was a time when you didn’t know what you know today. Malcolm X

2. Systems of variable compositions & Partial molar quantities.
Lecture 4
Systems of variable compositions &
Partial molar quantities.

3. In Lecture 1
The thermodynamic properties of fluids with constant composition was treated.
But in chemical engineering we are faced with processes wherein the composition of the fluids are varying.

4. For example,
In chemical, petroleum or pharmaceuticals industries, multicomponent mixtures, gases or liquids undergo composition changes due to the following:
Mixing and separation processes
Transfer of species from one phase to another
Changes due to chemical reactions
Due to the fact that this type of system depend on composition as well as temperature and pressure, there is the need to develop a foundation theory for the application of thermodynamics to gaseous mixtures and liquid solution.
Department of Chemical/Petroleum Engineering, Afe Babalola University © 2014

5. The theory is introduced via the derivation of fundamental property relation for homogenous solution of variable composition.
These derivations are based on a new property called Chemical Potential.
The chemical potential leads to the introduction of a new thermodynamic property called Partial Property . This is important in the treatment of real gas mixtures and form the basis for the introduction of Fugacity.
Department of Chemical/Petroleum Engineering, Afe Babalola University © 2014

6. Fundamental Property Relation
Why is this fundamental property of importance in solution thermodynamics?

7. the maximum useful work that can be obtained from a system
The Gibbs Energy
the maximum useful work that can be obtained from a system

8. Let us define this fundamental property in mathematical terms

9. 𝜕(𝑛𝐺) 𝜕𝑃 𝑇,𝑛 =𝑛𝑉 𝑎𝑛𝑑 𝜕(𝑛𝐺) 𝜕𝑇 𝑃,𝑛 =−𝑛𝑆
In any closed system, the Gibbs energy is related to temperature and pressure by:
𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃− 𝑛𝑆 𝑑𝑇
Eq. 3.1 may be applied to a single-phase fluid in a closed system where there are no chemical reactions.
For such a system, the composition is constant, hence:
𝜕(𝑛𝐺) 𝜕𝑃 𝑇,𝑛 =𝑛𝑉 𝑎𝑛𝑑 𝜕(𝑛𝐺) 𝜕𝑇 𝑃,𝑛 =−𝑛𝑆

10. 𝑛𝐺=𝑔(𝑃, 𝑇, 𝑛 1 , 𝑛 2 , …, 𝑛 𝑖 ,…) 𝑛 𝑖 =number of species 𝑖
For a single-phase open system that can interchange matter with the surroundings.
The total Gibbs energy 𝑛𝐺 remain functions of temperature and pressure, but now a third function which is the number of moles leaving or entering the system.
A function of the numbers of moles of the chemical species present.
Thus,
𝑛𝐺=𝑔(𝑃, 𝑇, 𝑛 1 , 𝑛 2 , …, 𝑛 𝑖 ,…)
𝑛 𝑖 =number of species 𝑖

11. The total differential of 𝑛𝐺 gives:
𝑑 𝑛𝐺 = 𝜕(𝑛𝐺) 𝜕𝑃 𝑇,𝑛 𝑑𝑃+ 𝜕(𝑛𝐺) 𝜕𝑇 𝑃,𝑛 𝑑𝑇+ 𝑖 𝜕(𝑛𝐺) 𝜕 𝑛 𝑖 𝑃,𝑇, 𝑛 𝑗 𝑑 𝑛 𝑗
The summation is over all species present, and the subscript 𝑛 𝑗 indicates that all mole numbers except the 𝑖th are held constant.
Thus,
The chemical potential of species 𝑖 in the mixture is given by:
𝜇 𝑖 ≡ 𝜕(𝑛𝐺) 𝜕 𝑛 𝑖 𝑃,𝑇, 𝑛 𝑗

12. 𝑆=− 𝜕𝐺 𝜕𝑇 𝑃,𝑥 V= 𝜕𝐺 𝜕𝑃 𝑇,𝑥 𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃− 𝑛𝑆 𝑑𝑇+ 𝑖 𝜇 𝑖 𝑑 𝑛 𝑖 3.3
With this definition and with the first two partial derivatives replaced 𝑛𝑉 and − 𝑛𝑆 , the general equation for the total differential of Gibbs free energy becomes:
𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃− 𝑛𝑆 𝑑𝑇+ 𝑖 𝜇 𝑖 𝑑 𝑛 𝑖
Eq. 3.3 is the fundamental property relation for single phase systems of constant or variable mass and constant or variable composition.
For a special case of one mole of solution, 𝑛=1 and 𝑛 𝑖 = 𝑥 𝑖
𝑑𝐺=𝑉𝑑𝑃−𝑆𝑑𝑇+ 𝑖 𝜇 𝑖 𝑑 𝑥 𝑖
Implicit in Eq. 3.4 is the functional relationship of the molar Gibbs energy to its canonical variables, 𝑇, 𝑃, and 𝑥 𝑖
𝐺=𝐺(𝑃, 𝑇, 𝑥 1 , 𝑥 2 , …, 𝑥 𝑖 ,…)
Eq. 3.4 imply:
𝑆=− 𝜕𝐺 𝜕𝑇 𝑃,𝑥 V= 𝜕𝐺 𝜕𝑃 𝑇,𝑥

13. Chemical Potential and Phase Equilibria
When we have a closed system consisting of two phases in equilibrium, each individual phase is assumed to be an open system free to transfer mass to the other.
The relationship is expressed as:
𝑑 𝑛𝐺 𝛼 = 𝑛𝑉 𝛼 𝑑𝑃− 𝑛𝑆 𝛼 𝑑𝑇+ 𝑖 𝜇 𝑖 𝛼 𝑑𝑛 𝑖 𝛼 a
𝑑 𝑛𝐺 𝛽 = 𝑛𝑉 𝛽 𝑑𝑃− 𝑛𝑆 𝛽 𝑑𝑇+ 𝑖 𝜇 𝑖 𝛽 𝑑𝑛 𝑖 𝛽 b
Where 𝛼 and 𝛽 represent the phases.
At equilibrium both 𝑇 and 𝑃 are uniform throughout the system.
The change in total Gibbs energy of the two phase system is the sum of those equation and each total system property can be expressed by the single equation given as:
𝑛𝑀= 𝑛𝑀 𝛼 + 𝑛𝑀 𝛽

14. 𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃− 𝑛𝑆 𝑑𝑇+ 𝑖 𝜇 𝑖 𝛼 𝑑𝑛 𝑖 𝛼 + 𝑖 𝜇 𝑖 𝛽 𝑑𝑛 𝑖 𝛽 3.7
Hence we have:
𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃− 𝑛𝑆 𝑑𝑇+ 𝑖 𝜇 𝑖 𝛼 𝑑𝑛 𝑖 𝛼 + 𝑖 𝜇 𝑖 𝛽 𝑑𝑛 𝑖 𝛽 3.7
Since the two-phase system is closed, Eq. 3.7 becomes:
𝑑 𝑛𝐺 = 𝑛𝑉 𝑑𝑃− 𝑛𝑆 𝑑𝑇
Comparing Eq. 3.7 and Eq. 3.8 shows that art equilibrium,
𝑖 𝜇 𝑖 𝛼 𝑑𝑛 𝑖 𝛼 + 𝑖 𝜇 𝑖 𝛽 𝑑𝑛 𝑖 𝛽 =
It can therefore be concluded that for multiple phases at the same 𝑇 and 𝑃 in equilibrium the chemical potential of each specie is the same in all phases.

15. 𝑖 𝜇 𝑖 𝛼 − 𝜇 𝑖 𝛽 𝑑 𝑛 𝑖 𝛼 =0 3.10 By law of conservation of mass:
𝛼+𝛽= 𝛼=−𝛽
𝑑 𝑛 𝑖 𝛼 =− 𝑑 𝑛 𝑖 𝛽
𝑖 𝜇 𝑖 𝛼 − 𝜇 𝑖 𝛽 𝑑 𝑛 𝑖 𝛼 =
𝜇 𝑖 𝛼 = 𝜇 𝑖 𝛽 𝑖=1,2,…,𝑁,…
𝜇 𝑖 𝛼 = 𝜇 𝑖 𝛽 = …𝜇 𝑖 𝜋 𝑖=1,2,…,𝑁,…

16. Partial Properties 𝑀 𝑖 = 𝜕 𝑛𝑀 𝜕 𝑛 𝑖 𝑃,𝑇, 𝑛 𝑗 3.12 𝜇 𝑖 ≡ 𝐺 𝑖
The molar partial properties of species 𝑖 is defined as:
𝑀 𝑖 = 𝜕 𝑛𝑀 𝜕 𝑛 𝑖 𝑃,𝑇, 𝑛 𝑗
𝑈 𝑖 , stand for the partial molar internal energy
𝐻 𝑖 , the partial molar enthalpy
𝑆 𝑖 , the partial molar entropy
𝐺 𝑖 , the partial molar Gibbs energy
It is a response function which represent the change of total property 𝑛𝑀 due to addition at constant 𝑇 and 𝑃 of a differential amount of species 𝑖 to a finite amount of solution.
Comparing the equation with the definition of chemical potential With Eq. 3.12:
𝜇 𝑖 ≡ 𝐺 𝑖

17. Equation Relating Molar & Partial Molar Properties
The definition of partial molar partial property in Eq provides the means for calculation of partial properties from solution-property data.
The derivation of the equation begins with the observation that the thermodynamics properties of a homogeneous phase are functions of temperature, pressure and the number of moles of the individual; species which comprises the phase.
Hence for the thermodynamic property 𝑀:
𝑛𝑀=Μ(𝑃, 𝑇, 𝑛 1 , 𝑛 2 , …, 𝑛 𝑖 ,…)
The total differential of 𝑛𝑀 is:
𝑑 𝑛𝑀 = 𝜕 𝑛𝑀 𝜕𝑃 𝑇,𝑛 𝑑𝑃+ 𝜕 𝑛𝑀 𝜕𝑇 𝑃,𝑛 𝑑𝑇+ 𝑖 𝜕 𝑛𝑀 𝜕 𝑛 𝑖 𝑃,𝑇, 𝑛 𝑗 𝑑 𝑛 𝑖
Department of Chemical/Petroleum Engineering, Afe Babalola University © 2014

18. 𝑑 𝑛𝑀 =𝑛 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+𝑛 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇+ 𝑖 𝑀 𝑖 𝑑 𝑛 𝑖 3.14
Where the subscript 𝑛 indicates that all mole numbers are held constant and 𝑛 𝑗 indicated that all mole numbers except 𝑛 𝑖 are held constant.
Because the first two partial derivatives on the right are evaluated at constant 𝑛 and
Because the partial derivative of the last term on the right is given by Eq. 3.12,
Eq. 3.13, the total differential of 𝑛𝑀 becomes:
𝑑 𝑛𝑀 =𝑛 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+𝑛 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇+ 𝑖 𝑀 𝑖 𝑑 𝑛 𝑖
Where
Subscript 𝑥 denotes differentiation at constant composition.

19. 𝑑𝑀= 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+ 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇+ 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖
In application,
The choice of a system of any size as represented by 𝑛 can be made, and
The choice of any variation in the size of a system as represented by 𝑑𝑛 can also be made
Thus 𝑛 and 𝑑𝑛 are independent and arbitrary.
Also, for Eq to be equal to zero, each term in the expression must be zero, hence;
𝑑𝑀= 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+ 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇+ 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖
3.16
And,
𝑀= 𝑖 𝑀 𝑖 𝑥 𝑖

20. 𝑖 𝑥 𝑖 𝑑 𝑀 𝑖 + 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖 = 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+ 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇+ 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖
Multiplying Eq by 𝑛 gives:
𝑛𝑀= 𝑖 𝑛 𝑖 𝑀 𝑖
Differentiating Eq. 3.17, we have:
𝑑𝑀= 𝑖 𝑥 𝑖 𝑑 𝑀 𝑖 + 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖
Comparing Eq with Eq. 3.19, we have:
𝑖 𝑥 𝑖 𝑑 𝑀 𝑖 + 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖 = 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+ 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇+ 𝑖 𝑀 𝑖 𝑑 𝑥 𝑖

21. 𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+ 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇− 𝑖 𝑥 𝑖 𝑑 𝑀 𝑖 =0
On simplification we have:
𝜕𝑀 𝜕𝑃 𝑇, 𝑥 𝑑𝑃+ 𝜕𝑀 𝜕𝑇 𝑃, 𝑥 𝑑𝑇− 𝑖 𝑥 𝑖 𝑑 𝑀 𝑖 =0
3.20
Eq is known as the Gibbs-Duhem Equation
(Pierre-Maurice-Marie Duhem, French Physicist ( ))
This equation must be satisfied for all changes in 𝑃, 𝑇 and the 𝑀 𝑖 caused by changes of state in a homogenous phase.
At constant T and P, Eq becomes:
𝑖 𝑥 𝑖 𝑑 𝑀 𝑖 =0
3.21

22. Property relations are the same in the form on either basis:
General Comments:
The constituents of a solution are in fact intimately intermixed and owing to molecular interactions cannot have private properties of their own.
Nonetheless, partial molar properties as defined by Eq have all characteristics of properties of individual species as they exist in solution.
For practical purposes, they may be assigned as property values to the individual species.
The symbol 𝑀 may express solution properties on a unit-mass basis as well as on a mole basis.
Property relations are the same in the form on either basis:
By replacing with 𝑛 representing number of moles (partial molar properties)
By replacing with 𝑚 representing mass (partial specific properties)

23. General Comments:
In solution thermodynamics, three kinds of properties are used and are distinguished by the following symbols:
Solution properties 𝑀, for example: 𝑈, 𝐻, 𝑆, 𝐺
Partial properties 𝑀 𝑖 for example: 𝑈 𝑖 , 𝐻 𝑖 , 𝑆 𝑖 , 𝐺 𝑖
Pure-species properties 𝑀 𝑖 for example: 𝑈 𝑖 , 𝐻 𝑖 , 𝑆 𝑖 , 𝐺 𝑖

24. Partial Properties in Binary Solutions
For binary systems, the summability relation, Eq. 3.17
𝑀= 𝑀 1 𝑥 1 + 𝑀 2 𝑥
Differentiating 𝑀, we have:
𝑑𝑀= 𝑥 1 𝑑 𝑀 𝑀 1 𝑑 𝑥 1 + 𝑥 2 𝑑 𝑀 𝑀 2 𝑑𝑥
When 𝑀 is a known function of 𝑥 1 at constant 𝑇 and 𝑃, the appropriate form of the Gibbs-Duhem equation is:
𝑥 1 𝑑 𝑀 1 + 𝑥 2 𝑑 𝑀 2 =
Since 𝑥 1 + 𝑥 2 =1
Then d 𝑥 1 =−𝑑 𝑥
Hence substituting Eq into Eq & combining with Eq. 3.24
𝑑𝑀= 𝑥 1 𝑑 𝑀 𝑀 1 𝑑 𝑥 1 + 𝑥 2 𝑑 𝑀 2 − 𝑀 2 𝑑𝑥 1

25. Partial Properties in Binary Solutions Cont’d
𝑑𝑀= 𝑥 1 𝑑 𝑀 1 + 𝑥 2 𝑑 𝑀 𝑀 1 𝑑 𝑥 1 − 𝑀 2 𝑑𝑥 1
𝑑𝑀= 𝑀 1 𝑑 𝑥 1 − 𝑀 2 𝑑𝑥 1
𝑑𝑀 𝑑 𝑥 1 = 𝑀 1 − 𝑀 2
𝑀 2 = 𝑀 1 − 𝑑𝑀 𝑑 𝑥 (A)
Substituting (A) into the summability relation of Eq. 3.22
𝑀= 𝑀 1 𝑥 1 + 𝑥 2 𝑀 1 − 𝑑𝑀 𝑑 𝑥 1
𝑀= 𝑀 1 𝑥 1 + 𝑥 2 𝑀 1 − 𝑥 2 𝑑𝑀 𝑑 𝑥 1

26. 𝑀+ 𝑥 2 𝑑𝑀 𝑑 𝑥 1 = (𝑥 1 + 𝑥 2 ) 𝑀 1 𝑀 1 =𝑀+ 𝑥 2 𝑑𝑀 𝑑 𝑥 1 3.26
𝑀+ 𝑥 2 𝑑𝑀 𝑑 𝑥 1 = (𝑥 1 + 𝑥 2 ) 𝑀 1
𝑀 1 =𝑀+ 𝑥 2 𝑑𝑀 𝑑 𝑥 1
3.26
Similarly, from (A)
𝑀 1 = 𝑀 2 + 𝑑𝑀 𝑑 𝑥 (B)
Substituting (B) into the summability relation of Eq. 3.22
𝑀= 𝑀 1 𝑥 1 + 𝑥 2 𝑀 2
𝑀= 𝑥 1 ( 𝑀 2 + 𝑑𝑀 𝑑 𝑥 1 )+ 𝑥 2 𝑀 2
Department of Chemical/Petroleum Engineering, Afe Babalola University © 2014

27. 𝑀= 𝑥 1 𝑀 2 + 𝑥 1 𝑑𝑀 𝑑 𝑥 1 + 𝑥 2 𝑀 2 𝑀− 𝑥 1 𝑑𝑀 𝑑 𝑥 1 =( 𝑥 1 + 𝑥 2 ) 𝑀 2
𝑀= 𝑥 1 𝑀 2 + 𝑥 1 𝑑𝑀 𝑑 𝑥 1 + 𝑥 2 𝑀 2
𝑀− 𝑥 1 𝑑𝑀 𝑑 𝑥 1 =( 𝑥 1 + 𝑥 2 ) 𝑀 2
𝑀 2 =𝑀− 𝑥 1 𝑑𝑀 𝑑 𝑥 1
3.27
For binary systems, the partial properties are readily calculated directly from an expression for the solution property as a function of composition at constant T and P.

28. Working Session 3

29. The need arises in a laboratory for 2,000 cm3 of an antifreeze solution consisting of 30-mol% methanol in water. What volumes of pure methanol and of pure water at 25⁰C must be mixed to form the 2000cm3 of antifreeze also at 25⁰C? Partial molar volumes for methanol and water in a 30-mol% methanol solution and their pure species molar volumes, both at 25⁰C are: methanol (1): 𝑉 1 = cm3mol-1 𝑉 1 = cm3mol-1 Water (2): 𝑉 2 = cm3mol-1 𝑉 2 = cm3mol-1

30. Working Session 3
Solution

31. 𝑥 1 = 𝑥 2 =0.7
𝑉 = 𝑥 1 𝑉 1 + 𝑥 2 𝑉 2
𝑉 = =24.025
But 𝑉 𝑡 =2000
Total moles, 𝑛= 𝑉 𝑡 𝑉 = = 𝑚𝑜𝑙
Of this 30% is methanol and 70% is water
𝑛 1 = 𝑥 1 𝑛= =24.974
𝑛 2 = 𝑥 2 𝑛= =58.272
Volume of pure species, 𝑉 𝑖 𝑡 = 𝑛 𝑖 𝑉 𝑖
𝑉 1 𝑡 = 𝑛 1 𝑉 1 = =1017cm3
𝑉 2 𝑡 = 𝑛 2 𝑉 2 = =1053cm3
Department of Chemical/Petroleum Engineering, Afe Babalola University © 2014

32. The molar volume (cm3mol-1) of a binary liquid mixture at T and P is given by:
𝑉=120 𝑥 𝑥 𝑥 1 +8 𝑥 2 𝑥 1 𝑥 2
Find the expressions for the partial molar volume of species 1 and 2 at T and P
Show that when these expressions are combined in accordance with Eq (V= 𝑖 𝑉 𝑖 𝑥 𝑖 ), the given equation for 𝑉 is recovered.
Show that these expressions satisfy Eq ( 𝑖 𝑥 𝑖 𝑑 𝑉 𝑖 =0) (Gibbs-Duhem equation).
Show that 𝑑 𝑉 1 /𝑑 𝑥 1 𝑥 1 =1 = 𝑑 𝑉 2 /𝑑 𝑥 1 𝑥 1 =0 =0
Plot values of 𝑉, 𝑉 1 , and 𝑉 2 calculated by the given equation for 𝑉 and by the equation developed in (1) vs. 𝑥 1 . Label points 𝑉 1 , 𝑉 2 , 𝑉 1 ∞ and 𝑉 2 ∞ and show their values.