Simple Lattice Model for Fluids. Introduction In this chapter we borrow the idea of lattice...

Simple Lattice Model for Fluids

Introduction

In this chapter we borrow the idea of lattice structures, characteristic of crystals, and apply it to develop lattice-based theories for fluids.

It is not just of academic interest…

Some models widely used in chemical process design can be derived in this way.

2

Introduction

In the lattice, each site will have z nearest neighbors; z is referred to as coordination number.

The value of z depends on the lattice structure and on whether it is a 1-D, 2-D, or 3-D lattice.

In the developments that will follow, we will not need to examine the lattice structure in detail.

3

IntroductionThe volume assigned to each lattice site is denoted as b. The number of lattice sites is M. Therefore, the total volume of the lattice is:

V Mb

Suppose there are N of the M sites are occupied by molecules of a certain substance. The remaining lattice sites may be:

• empty: fluids with many empty sites will have low density. Associating density with the number of empty sites is useful to develop equations of state;

•all occupied by molecules of a second species: useful to obtain models for excess Gibbs energy (activity coefficients).

M N

4

Introduction

Let us recall the connection between entropy and the microcanonical partition function:

, , lnS N V E k

where is the system degeneracy. We will take it to be the number of ways of distributing N molecules of species 1 and (M-N) molecules of species 2 in the lattice. When doing this, we will assume each molecule of species 1 or 2 occupies only one lattice site. Species 2 can be another molecule or a vacancy (i.e., pure fluid 1 with empty spaces between its molecules).

5

Introduction

, , lnS N V E k

The lattice sites are distinguishable from one another. The molecules of each species are undistinguishable. The vacancies are also undistinguishable.

Initially, we will assume that are no intermolecular interactions, or interactions between molecules and vacancies. All configurations will be assumed to have the same energy.

6

Introduction

!

! !

M

N M N

The number of ways of arranging the molecules (and vacancies) in the lattice is the degeneracy:

!ln ln

! !

ln ! ln ! ln !

ln ln ln

ln ln ln

MS k k

N M N

k M N M N

k M M M N N N M N M N M N

k M M N N M N M N

7

Introduction

!

! !

M

N M N

The number of ways of arranging the molecules (and vacancies) in the lattice is the degeneracy:

ln ln lnS k M M N N M N M N

8

Equations of state from lattice theories

, , ,

eng ent

T N T N T N

A U SP T P P

V V V

Let us start with the relationship between the pressure and the Helmholtz energy in classical thermodynamics:

engP

entP

Energetic contribution to the pressure

Entropic contribution to the pressure

9


, , ,

, ,

ln ln ln

ln ln ln

ln ln ln 1

ent

T N T N T N

T N T N

S S MP T T

V M V

M M N N M N M NT S kT

b M b M

M NkT M kT MM M N

b M M N b M N

kT Mb kT V kT Nb

b Mb Nb b V Nb b V

We will consider a lattice with a single species and vacancies:

10

M= total # sitesV = M b ln ln lnS k M M N N M N M N


ln 1ent kT NbP

b V

An interpretation of this result is based on the series expansion:

2 31 1ln 1 ...

2 3x x x x

2 31 1

ln 1 ...2 3

ent kT Nb kT Nb Nb NbP

b V b V V V

2, 1 1

1 ...2 3

entcomp entP V Nb Nb

ZNkT V V

11


Examined in this way, it allows a connection with the virial equation of state. When such connection is made, we observe that the virial coefficients for the entropic contribution to the compressibility factor are temperature independent.

2, 1 1

1 ...2 3

entcomp entP V Nb Nb

ZNkT V V

12


ln 1ent kT NbP

b V

Another interpretation of this result connects it to the van der Waals equation of state. First, retain the first two terms of the entropic contribution to the compressibility factor from the series expansion, i.e.:

, 112

entcomp ent P V Nb

ZNkT V

112

ent NkT NbP

V V

13


If Nb << V, then a reasonable approximation to the entropic contribution to pressure is:

112

2

ent NkT Nb NkTP

bV V V N

The last term in the equation above is similar to the first term in the van der Waals equation of state:

2

RT aP

V b V

14


For the energetic contribution to the properties, we will assume the molecules are randomly distributed in the fluid.

Assuming again a pure fluid with vacancies, the average number of nearest neighbor sites that are occupied and vacant are respectively equal to:

zN

M

z M N

M

15


We will assume that each molecule interacts with its nearest neighbors. Let u stand for the energy of each interaction. To avoid double counting, a division by 2 will be introduced. Then:

2 2

2 2 2

N zN N z N zbU u u u

M M V

The energetic contribution to the pressure is:

2

2 2

2 2,

,

22

eng

T N

T N

N zbuU N zbu NVP a

V V V V

2a zbu16


Collecting the results of the approximations used for the entropic and energetic contributions, we find:

2

22

2

ent eng NkT N RT aP P P a

b V V B VV N

V Molar volume

This is van der Waals equation of state, a model of great importance in the history of chemical thermodynamics. Van der Waals received the Nobel Prize in 1910 for his work. 17


2

RT aP

V B V

It is cubic in volume. Therefore, given T and P, the model may have either one real root and two complex roots or three real roots. This gives it the ability to represent vapor and liquid phases.

There is an inconsistency: we assumed that the molecules were randomly placed in the lattice; however, their interaction should create non-randomness effects which are not accounted for.

18


Non-randomness affects the entropic and energetic contributions. For the latter, let us assume the probability that, close to a filled site, a nearest neighbor site is occupied:

11

1

u

kTu

kT

c ce c

e

Therefore, the probability that it is not occupied is given by:

u

kTce

c

These two probabilities add to one:

19


We can now return to the expression for the energy:

The energy contribution to the pressure becomes:

2 2

2 2 21 1 1

u u u

kT kT kT

u u u

kT kT kT

N zN e N z e N zb eU u u u

M M Ve e e

2

,

,

2 2

2 2

21

21

u

kTeng

uT N kT

T N

u

kT

u

kT

U N zb eP u V

VVe

N zbu e Na T

V Ve

20


energetic contribution to the pressure:

This changed the energy contribution to the pressure, but the entropy contribution was left unchanged – the algebra becomes complicated.

In practice, the van der Waals equation is seldom used, but its empirical modifications are widely used for chemical process design.

2

2eng N

P a TV

21

u

kT

u

kT

zbu ea T

e

21


Examples:

Redlich-Kwong EOS

Soave-Redlich-Kwong EOS

Peng-Robinson EOS

RT a

PV B TV V B

a TRTP

V B V V B B V B

a TRTP

V B TV V B

22

Activity coefficients for molecules of similar size

A reminder of classical thermodynamics: mixing properties.

, , , , ,imix ii

U T P x U T P x x U T P

, , , , ,imix ii

V T P x V T P x x V T P

, , , , ,imix ii

S T P x S T P x x S T P

All pure substances and mixtures at the same temperature and pressure, and aggregation state (e.g., liquid phase).

23


A reminder of classical thermodynamics: ideal solution properties.

, , ,IMmix ii

i

U T P x x U T P

, , ,IMmix ii

i

V T P x x V T P

, , , lnIMmix ii i i

i i

S T P x x S T P R x x

All pure substances and mixtures at the same temperature and pressure, and aggregation state (e.g., liquid phase). 24


A reminder of classical thermodynamics: excess properties.

, , , , , ,ex IMmixV T P x V T P x V T P x

Real and ideal mixtures at the same temperature and pressure, and aggregation state (e.g., liquid phase).

, , , , , ,ex IMmixU T P x U T P x U T P x

, , , , , ,ex IMmixS T P x S T P x S T P x

25


A reminder of classical thermodynamics: activity coefficients.

Activity coefficients have direct relationships with the excess Gibbs energy and the excess Helmholtz energy:

, , , ,

ln

j i j i

ex ex

ii i

T P n T V n

G RT A RT

n n

26


In previous slides, we analyzed systems with a pure fluid and vacant sites in the lattice. We will now consider fully occupied lattices, i.e., with no vacant sites. For simplicity, we will analyze binary mixtures.

Aspects to consider:

•We will no longer be able to predict the effect of pressure on properties – the number of vacant sites is always zero;

•The total number of sites will be number of sites of component 1 plus the number of sites of component 2 – we will be unable to predict volume changes on mixing. 27


Consider three lattices, all of them with the same spacing of lattice points, same coordination number and volume in each lattice point.

28



The first lattice contains M molecules of species 1

29



The first lattice contains M molecules of species 1.

The second lattice contains (N-M) molecules of species 2

30




The second lattice contains (N-M) molecules of species 2.

The third lattice is formed by joining the first two lattices and mixing the molecules of the two species.

It contains a total of N molecules distributed in its N sites. In this mixing process: 0mixV

31


Let us compute the entropy in each of these cases.

First lattice with M molecules of species 1 in M sites.

There is only one way of distributing M identical, undistinguishable molecules in the M sites.

1 1ln ln1 0S k k

Second lattice with (N-M) molecules of species 2 in (N-M) sites.

2 2ln ln1 0S k k

32


Let us compute the entropy in each of these cases.

Third lattice with M molecules of species 1 and (N-M) molecules of species 2 in a total of N sites.

12

!ln ln

! !

NS k k

M N M

!

, , ln! !mix

NS T P x k

M N M

33


1 2 1 1 2 2

!, , ln

! !

ln ln ln

ln ln ln

ln ln ln ln

ln ln

ln ln ln ln

mix

NS T P x k

M N M

k N N N M M M N M N M N M

k N N M M N M N M

k M N N M N M M N M N M

N MMk M N M

N N

k M x N M x kN x x x x

34


On a molar basis:

1 1 2 2, , ln lnmixS T P x kN x x x x

1 1 2 2, , ln lnmix S T P x R x x x x

0exS

35


Considering the same three lattices, let us now evaluate the energy of mixing.


1 11

1

2U Mzu

Second lattice with (N-M) molecules of species 2 in (N-M) sites.

2 22

1

2U N M zu

36


12 1 11 2 22

2 12 1 12

1 11 2 12 1 12 2 22

1 1 11 2 12 2 1 12 2 22

2 21 11 1 2 12 2 22

1 1

2 21 1

2 2

2

2

22

U Mzx u N M zx u

Mzx u N M zx u

zM x u x u N M x u x u

zNx x u x u x x u x u

zNx u x x u x u

Third lattice with M molecules of species 1 and (N-M) molecules of species 2 in a total of N sites.

37


12 1 2mixU U U U

2 21 11 1 2 12 2 22 11 22

2 21 11 1 11 1 2 12 2 22 2 22

1 1 11 1 2 12 2 2 22

1 2 11 1 2 12 2 1 22

1 2 12 11 22 1 2 1 2

1 12

2 2 2

22

1 2 12

22

22 2

mix

zNU x u x x u x u Mzu N M zu

zNx u x u x x u x u x u

zNx x u x x u x x u

zNx x u x x u x x u

z zNx x u u u Nx x Nx x

38


12 1 2mixU U U U

1 2 1 22

exmixU z U

x x x xN N

12 11 222u u u Exchange energy

Summary:

1 2

exUx x

N 0exS 0exV

39


ex ex ex exG U TS PV

1 2 1 20 0exG Nx x Nx x

1 2 1 2ex

Av mG x x N x x

1 21 2

1 2

1 1ex

m m

n nGnx x

RT RT RT n n

2

21 2

1, ,

lnex

m

T P n

G RTx

n RT

1

22 1

2, ,

lnex

m

T P n

G RTx

n RT

40


The model developed here is called Regular Solution Model.

Consistent with the assumptions made for its development, this model works best for:

•Mixtures with molecules of similar size;

•Intermolecular interactions in the pure components and in the mixture are similar, so that deviations from random distribution of molecules are small.

41

Flory-Huggins model

There are many cases of practical interest in which the molecules have very different sizes.

Polymer-solvent solutions are an extreme example.

The final expression of the Flory-Huggins model is amazingly simple, yet it captures the essential effect of size differences on thermodynamic properties.

Paul Flory received the Nobel Prize in Chemistry in 1974.

42

Flory-Huggins model

We will assume that:

•The mixture is binary: polymer + solvent;•All sites of the lattice are occupied;•There are p equal polymer molecules;•Each polymer molecule occupies c lattice sites;•There are s equal solvent molecules;•Each solvent molecule occupies 1 lattice site;•The lattice coordination number is z.

Therefore, the number of lattice sites that the mixture occupies is:

N p c s 43

Flory-Huggins model

As done when we developed a model for activity coefficients in mixtures of molecules of similar size, we will adopt the microcanonical ensemble:

The notation here is:

ln , ; ln ; ln ;mix S k p s p c s k p p c k s s

, ;p s p c s # of ways of placing p polymer molecules and s solvent molecules in pxc+s lattice sites.

;p p c # of ways of placing p polymer molecules in pxc lattice sites (pure polymer).

;s s # of ways of placing s solvent molecules in s lattice sites (pure solvent). 44

Flory-Huggins model

Pure solvent:

;s s # of ways of placing s solvent molecules in s lattice sites (pure solvent).

; 1s s

Remember: the molecules are indistinguishable.

45

Flory-Huggins model

Pure polymer:


This result will be a particular situation of the polymer + solvent case. So, we will come back to the pure polymer later.

46

Flory-Huggins model

Polymer + solvent mixture:

We will proceed as follows:

•Assume the lattice is initially empty;•First place the polymer molecules in the lattice, one at a time;•After that, place the solvent molecules (this will be very simple).

, ;p s p c s # of ways of placing p polymer molecules and s solvent molecules in pxc+s lattice sites.

47

One arrangement of an 8-mer polymer on a lattice

48

Flory-Huggins model


The lattice has N sites. Since it is initially empty, there are N ways to place the first segment of the first polymer molecule: N

All other lattice positions are empty, but the second segment of the chain has to be in one of the z (coordination number) nearest neighbor sites. Therefore, the number of ways to place the first two segments is: Nz

To place the third segment, only (z-1) positions are available. The number of ways to place the first three segments is: 1Nz z

49

Flory-Huggins model


At this point, it is tempting to try to extrapolate this result for the remaining segments of the polymer chain and propose that the number of ways to place the first chain is:

21

cNz z

50

Flory-Huggins model



It is tempting, but it is wrong!

21

cNz z

51

Flory-Huggins model



It is tempting, but it is wrong!

For example, if the chain folds, not all (z-1) nearest neighbor sites may be available to put the next polymer chain segment. Also, when placing the second chain, some sites may be occupied by segments of the first polymer molecule and so on.

21

cNz z

52

Flory-Huggins model


To improve our counting strategy, let us assume we will place 4-segment polymer chains starting from an empty lattice. The number of ways to place the first segment is:

NWe know z nearest neighbors are available for the next segment, since the lattice is empty, except for the first segment already placed. However, as part of a general strategy, let us assign a probability that it is empty, based on the number of segments already placed. Then, the number of ways of placing the first two segments becomes: 1N

NzN

53

Flory-Huggins model


Following the same strategy for the third and fourth segments, the number of ways to place the four segments of the first chain is:

2

2

1 2 31 1

11 2 3

1 !

4 !

N N NNz z z

N N N

z zN N N N

N N

z z N

N N N

54

Flory-Huggins model


Having placed the first chain, it is time to place the second one in the lattice, taking into account the fact the first is already there. The number of ways of doing that is:

2

2

5 6 74 1 1

14 5 6 7

4 !1

8 !

N N NN z z z

N N N

z zN N N N

N N

Nz z

N N N

55

Flory-Huggins model


For the third chain:

2

2

9 10 118 1 1

18 9 10 11

8 !1

12 !

N N NN z z z

N N N

z zN N N N

N N

Nz z

N N N

56

Flory-Huggins model


The number of ways of placing the first three chains is the product of the results of the previous slides:

2 2 2

3 2 3

4 ! 8 !1 ! 1 1

4 ! 8 ! 12 !

1 !

3 4 !

N Nz z N z z z z

N N N N N N N N N

z z N

N N N

Note above that 3 is the number of polymer molecules and 4 is the number of segments per molecule.

57

Flory-Huggins model


The number of ways of placing all p polymer molecules, each with c segments is:

2 2 2

3 4 2 3

4 ! 8 !1 ! 1 1

4 ! 8 ! 12 !

1 !

3 4 !

N Nz z N z z z z

N N N N N N N N N

z z N

N N N

21 !

!

p c pz z N

N N N p c

58

Flory-Huggins model


With all p polymer molecules in the lattice, it is time to place the s solvent molecules there. There are s lattice sites available for that. The number of ways of placing s indistinguishable solvent molecules in s lattice sites is equal to 1. Therefore, the number of ways of placing the p polymer molecules and s solvent molecules is:

2

2

1 !, ;

!

!1

!

p c p

p c p

z z Np s p c s

N N N p c

p c sz z

p c s p c s s

59

Flory-Huggins model

Pure polymer:


Using, the formula for the polymer + solvent case:

2

2

!1,0; 0

0!

1!

p c p

p c p

p cz zp p c

p c p c

z zp c

p c p c

60

Flory-Huggins model

These expressions can now be substituted in:

ln , ; ln ; ln ;mix S k p s p c s k p p c k s s

Please refer to the textbook for the sequence of algebraic steps. The result is:

ln ln ln

ln ln ln ln

ln ln

ln ln

mix

P s

S k p s p c s p p c s s

k p p c s s p c s p p c s s

p c sk p s

p c s p c s

k p s

61

Flory-Huggins model

These are the fractions of the lattice volume occupied by each species. On a molar basis:

P

p c

p c s

s

s

p c s

ln lnmix P P s sS R x x

The entropy of mixing to form an ideal solution is:

ln lnIMmix P P s sS R x x x x

62

Flory-Huggins model

The excess entropy is:

ln lnex IM sPmix mix P s

P s

S S S R x xx x

64

Flory-Huggins model

Energy of mixing

Energy of the pure solvent:

1

2S ssU szu

65

Flory-Huggins model

Energy of mixing

Energy of the pure polymer: for long chains, only a small fraction of the segments will be at the end of a chain. Most will be internal segments, each with 2 other segments attached to it. Here, we will count interactions between non-bonded segments. Therefore, for each segment, the energy is:

2 ppz u

The total energy of the pure polymer is:

12

2p ppU pc z u 66

Flory-Huggins model

Energy of mixing

Energy of the mixture:

In this expression, we have used that the probability of having a neighbor of a certain type depends on the volume fraction.

1 12

2 21 1

22 2

ps P pp s ss

s ps p ps

U pc z u sz u

pc z u sz u

67

Flory-Huggins model

Energy of mixing

1 1 12 2

2 2 21 1 1

22 2 2

2 21

22

mix ps s p

P pp s ss s ps

p ps ss pp

P pp s ss s ps

p ps ss pp

U U U U

pc z u sz u pc z u

sz u szu pc z u

pc s pcz u z u z u

pc s pc s pc spc s

s s pcz u zu z u

pc s pc s pc s

68

Flory-Huggins model

Energy of mixing

2 21

2 2

12 1 1 2 2

21

2 2 221

22

P P pp s s ss P s ps

mix

s p ps s ss P pp

pp P P ss s s P s ps

P s ps pp ss

P s ps pp ps ss

z u z u z uU pc s

z u zu z u

pc s z u zu u z

pc s u z z u zu

pc s z u u z u u

69

Flory-Huggins model

Energy of mixing

12

2mix ps pp ps ss P s

P s

U pc s z u u z u u

pc s kT

Energy per molecule:

mixP s P s P s

U pc skT kT cx x

p s p s

Energy per mol:

ex exmix P s P sU RT cx x U H

70

Flory-Huggins model

Excess Molar Gibbs Energy

ln lnex ex ex

sPP s P s P s

P s

G H TScx x x x

RT RT x x

Activity coefficients:

21ln ln 1s

s P Psx c

2ln ln 1pP s s

p

c cx

71

Flory-Huggins model

Comments:

•Widely used for polymer solutions;•Internally inconsistent – the interaction energies create non-random spatial distributions, neglected during model derivation;•The polymer chain was assumed to be linear, but many polymer molecules are branched.•Staverman-Guggenheim model (used in UNIQUAC and UNIFAC) is a refinement:

1 1

ln ln2

ex C Ci i

i i ii ii i

S zx x q

R x

72

Staverman-Guggenheim model

1 1

ln ln2

ex C Ci i

i i ii ii i

S zx x q

R x

1

i ii C

j jj

x r

x r

1

i ii C

j jj

x q

x q

,i ir q Volume and surface area parameters, respectively

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