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Chemical Engineering II

Lec 2: Phase Equilibria: Thermodynamics of

Dr.-Eng. Zayed Al-Hamamre

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Content

 Learning Objectives  Introduction  Partial Molar Properties  Determination of Partial Molar Properties  Relations Among Partial Molar Quantities

2 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Learning Objectives

 Apply thermodynamics to mixtures. Write the differential for any extensive property, dK, in terms of m + 2 independent variables, where m is the number of species in the .  Define and find values for pure species properties, total properties, partial molar properties, and property changes of mixing.  Define a partial molar property and describe its role in determining the properties of mixtures. Calculate the value of a partial molar property for a species in a mixture from analytical and graphical methods.  Apply the Gibbs–Duhem equation to relate the partial molar properties of different species.

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Introduction

 For a pure species i, all the intermolecular interactions are identical.  The resulting thermodynamic properties—such as —are a manifestation of those interactions.  Mixture contains more than one species and more complex than pure species  Mixture properties are determined only in part by an average of each of the pure species (i-i) interactions.  We must now also take into account how each of the species interacts with the other species in the mixture, that is, the unlike (i-j) interactions  Hence, the properties of a mixture depend on the nature and amount of each of the species in the mixture. The values of the mixture’s properties will be affected not only by how those species behave by themselves but also by how they interact with each other.

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 Upon mixing, o The mass remains constant since it is a conserved quantity o However, the mixture volume is different from the sum of the pure species volume. o The ethanol and water can pack together more tightly than can each species by itself. o This is due to the nature of the hydrogen bonding involved in the structure of the

liquid. 5 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888

Introduction

 When a species becomes part of a mixture, it loses its identity;  Yet it still contributes to the properties of the mixture, since the total solution properties of the mixture depend on the amount present of each species and its resultant interactions

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 Any intensive thermodynamic property can be mathematically described in terms of partial derivatives of two independent, intensive properties.  Since we are now concerned with thermal and mechanical equilibrium, it makes sense to choose T and P as the independent, intensive properties.  For extensive properties, the total number of moles much be also specified.  For mixture, in addition to specifying two independent properties, the number of moles of each species in the mixture should be considered.  Mathematically, we can write the extensive total solution property K in terms of T, P, and the number of moles of m different species:

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Partial Molar Properties

 The differential of K can then be written as the sum of partial derivatives of each of these independent variables, as follows:

8 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Partial Molar Properties

 Define a new thermodynamic function, the partial molar property, as partial derivatives with respect to moles,

K denotes for any extensive properties partial molar property

 The partial molar properties are governed by how a species behaves in the mixture.

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Partial Molar Properties  A partial molar property is always defined at constant and , two of the criteria for phase equilibrium.  Partial molar properties are also defined with respect to number of moles  The number of moles of all other j species in the mixture are held constant; it is only the number of moles of species i that is changed.

 It is a response function, i.e., a measure of the response of total property K to the addition at constant T and P of a differential amount of species I to a

finite amount of solution. 10 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Partial Molar Properties

 Note that

 Because in changing the number of moles of species i, we change the mole fractions of all the other species in the mixture as well, since the sum of the mole fractions must equal 1.

 Using the definition of partial molar properties, the total differential of the variable K becomes:

differentiation at constant composition

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Partial Molar Properties

(k) (k)  Also dK  d(nk)  n  dP  n  dT   Ki dni (11.9)  P T,x  T  P,x

 At constant temperature and pressure, this equation reduces to

Since ni = xin dni xidnndxi

and d(nk)  ndk  kdn k k k k Ki

Rearrange, k k k Ki k Ki

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 k   k  dk-  dP-  dT-Kidxi  0  P T,x  T P,x i  k   k  (I) dk    dP   dT  Kidxi  P T,x  T P,x i

k   xi Ki  0 i

(II) per mole of mixture k   xi Ki i These equations show that the calculation of nk  K  ni Ki (III) i

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Partial Molar Properties  The extensive total solution property K is equal to the sum of the partial molar properties of its constituent species, each adjusted in proportion to the quantity of that species present.  Similarly, the intensive solution property k is simply the weighted average of the partial molar properties of each of the species present.

For example,

 Where k is the corresponding intensive property to K and xi is the of species i.

14 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Partial Molar Properties

Differentiating Eq (II) dk   xi d K i   K i dxi i i

 Comparison of this equation with Eq. (I) (Subtraction gives) yields

 k   k    dP  dT  xi dKi  0  P T ,x  T P,x i

 This equation must be satisfied for all changes in P, T, and the ki caused by changes of state in a homogeneous phase Gibbs-Duhem equation  As a special case at constant T and P: x d K i  0  i at constant T and P i  Provides a very useful relationship between the partial molar properties of different

species in a mixture 15 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888

Partial Molar Properties

Gibbs-Duhem equation at constant T and P for other thermodynamics properties

 As a solution becomes pure in species i, both properties approach pure species property

lim k  lim K i  Ki xi 1 xi 1

 In the limit of infinite  lim K i  K i  Ki xi 0

16 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Analytical Determination of Partial Molar Properties  Often an analytical expression for the total solution property, k, is known as a function of composition.  In that case, the partial molar property, , can be found by differentiation of the extensive K i expression for K with respect to ni, holding T, P, and the number of moles of the other j species constant, Example

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Example Cont.

But and

Since

and

18 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Example Cont.

xi dMi  0 i dM1 dM2 For binary mixture x1 dM1  x2 dM2  0 x1  x2  0 dx1 dx1

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Example Cont.

20 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Example Cont.

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Graphical Determination of Partial Molar Properties

 Used to calculate the partial (or any other partial molar property) for a binary mixture when we have a graph of the molar volume (or whatever molar property) vs. mole fraction of one component

22 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Graphical Determination of Partial Molar Properties

 For a binary solution A k  x1 K1  x2 K 2

B dk  x1 dK1  K1dx1  x2 dK 2  K 2dx2

 Gibbs-Duhem equation is

x1 dK1  x2 dK 2  0

 Dividing by dx1, we have the Gibbs-Duhem equation in derivative forms

dK1 dK 2 C x1  x2  0 dx1 dx1

Because x1  x2 1 dx1  dx2 23 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888

Graphical Determination of Partial Molar Properties

Eq. B becomes dk  K1  K 2 dx1 dk From Eq A and D dk K1  k  x2 K 2  k  x1 dx1 dx1  These equations can be used to obtain partial molar properties from solution property. dk dk  Or k  K1  x2 k  K 2  x1 dx1 dx1

intercept slope  Similarly,

24 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Graphical Determination of Partial Molar Properties

 K 1

k1 k k  K 2

k2

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Graphical Determination of Partial Molar Properties

 In the figure, the slope of the line at point k is

dk k  I2 dk   I1  I2 dx1 x1 dx1

dk dk I2  k  x1 I1  k  (1 x1) dx1 dx1

 Comparing these equation with dk dk K1  k  x2 K 2  k  x1 dx1 dx1

K 2  I K1  I1 2

26 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Graphical Determination of Partial Molar Properties

 The limiting cases of the partial molar volume of species 1 in terms of composition of 1.

 In the limit as x1 goes to 1, we have all 1 and no 2 in the mixture.  In this case, the partial molar volume just equals the pure species molar volume:

lim k  lim K i  Ki xi 1 xi 1

 In the limit of infinite dilution

 lim K i  K i  Ki xi 0

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Example The need arise in a laboratory for 2000 cm3 of an antifreeze solution consisting of 30 mol % methanol in water. What volumes of pure methanol and of pure water at 25 C must be mixed to form the of antifreeze, also at 25 C? Partial molar volumes for methanol and water in a 30 mol % methanol solution and their pure-species molar volume, both at 25 C , are: Methanol (1) and water (2):

3 1 3 1 V1  38.632 cm mol V1  40.727 cm mol 3 1 3 1 V2  17.765 cm mol V2  18.068 cm mol

28 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Example Cont.

V  x1V1  x2V2  (0.3)(38.632)  (0.7)(17.765) V  24.025 cm3mol1 V t 2000 cm3 n    83.246 cm3 V 24.025 cm3mol1

n1  x1n  (0.3)(83.246 )  24.974 mol

n2  x2n  (0.7)(83.246 )  58.272 mol

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Example Cont. Solution The line drawn tangent to the V-x1 curve at x1=0.30, illustrates the values of

V1=40.272 cm3 mol-1 and V2=18.068 cm3 mol-1.

t 3 1 V1  (24.497 mol)(40.727 cm mol ) t 3 V1  1017 cm t 3 1 V2  (58.272 mol)(18.068 cm mol ) t 3 V2  1053 cm

30 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Example

Find the partial molar volume of species b in a binary solution when we know the partial molar volume of species a, , as a function of composition

Applying Gibbs-Duhem equation at constant T and P

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Summary of the Different Types of Thermodynamic Properties

32 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Summary of the Different Types of Thermodynamic Properties

Vi ,Ui , Hi ,Si , Ai ,Gi these properties must be intensive

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Relations Among Partial Molar Quantities

 Partial molar properties can be related to each other, considering the equation

 Since the pressure is constant,

 Applying the definition of a partial molar quantity

34 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Relations Among Partial Molar Quantities

 Similarly,

 What is the most important property ? ……G……….  For pure component; G = G (T, P)

 Also, for closed system: no mass transfer across boundary or in a single-phase fluid in a closed system wherein no chemical reactions occur

 When there is no change in composition, dG  VdP SdT 35 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888

Relations Among Partial Molar Quantities

 For a homogeneous mixture e.g. containing i components mixture;

G = G (T, P, n1, n2, …, ni)

ni is the number of moles of species i

(G) (G) (G) d(G)    dP    dT     dni  P T ,n  T  P,n i ni  T ,P,n j

all mole numbers held constant all mole numbers except ni held constant

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 Also,

 G  G  G    V    S     Gi     P T ,n  T  P,n ni  P,T ,n j

 Application of the criterion of exactness

 V   S   Gi  (S)           T ni  T P ,n  P T ,n  P ,n   P,T ,n j

 G   i     Si  T P ,x

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Relations Among Partial Molar Quantities

 Gi  (V )  G      i         Vi  P  ni T ,n   P,T ,n j  P T ,x

38 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Relations Among Partial Molar Quantities  Thus, every equation that provides a linear relation among thermodynamic properties of a constant-composition solution has as its counterpart an equation connecting the corresponding partial properties of each species in the solution. For example,

dG  VdP  SdT dGi  VidP  SidT

H U  PV Hi Ui  PVi

G  H TS Gi  Hi TSi

 G   Gi    V    Vi  P T ,x  P T ,x 39 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888

Relations Among Partial Molar Quantities

 Also, for closed system: no mass transfer across boundary or in a single-phase fluid in a closed system wherein no chemical reactions occur

 For a mixture,

40 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Relations Among Partial Molar Quantities

dG  VdP SdT  Gi dNi i

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Relations Among Partial Molar Quantities

42 Chemical Engineering Department | University of Jordan | Amman 11942, Jordan Tel. +962 6 535 5000 | 22888 Partial Properties in Binary

 For a binary (For two components) solution

dH  TdS VdP(μ1 )S,Pdn1  (μ2 )S,Pdn2

dU  TdS  PdV (μ1 )V,Sdn1  (μ2 )V,Sdn2

dGVdP -SdT(μ1 )T,Pdn1  (μ2 )T,Pdn2

dA PdV -SdT(μ1 )T,V dn1  (μ2 )T,V dn2  H   U   G   A          G1  1             n1 n1 n1 n1  S,P,n2  V ,S,n2  T ,P,n2  T ,V ,n2

 Keep in mind that,

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Excursion

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