# 331: Soft Materials

## 1 Catalog Description

Different kinds of polymeric materials. Relationships between structure and physical properties; rubber elasticity, the glassy state, crystallinity in polymers. Lectures, laboratory. Prerequisites: 301 or equivalent; 314 or CHEM 342 1.

## 3 331: Soft Materials

At the conclusion of the course students will be able to:
1. Given the chemical structure of a common polymer, draw the chemical structures of the monomer(s) from which it was made.
2. Given the chemical structure of a monomer (or monomers), draw the chemical structure polymers that can be synthesized from it (or them).
3. Understand, describe and calculate the structural parameters of polymeric materials including monomer units, molecular weight, tacticity, coil dimension, crystallinity, and morphology.
4. Describe the relationship between the above structural parameters and the mechanical and thermal properties of polymeric materials.
5. Describe how the structure and mechanical properties of polymeric materials change at the glass transition temperature and at the melting temperature.
6. Describe how the molecular structure of a surfactant determines its micelle structure.
7. Describe how surfactant molecules and aggregates affect the optical properties, viscosity and surface tension of polymer solutions.
8. Calculate the intermolecular and surface forces for molecules and colloids of different geometries.
9. Design strategies to stabilize or destabilize colloidal systems.
10. Identify a soft material application in daily life, such as in arts, music, sports or food, and explain how material advancement has imporved the application.

## 4 Introduction

Since the title of this book is 'Soft Materials', it makes sense to define what we really mean by 'soft'. Here are two ways to think about it:
1. Soft Materials have Low Elastic Moduli.
By 'low' we mean significantly lower than the moduli of crystalline metals and ceramics. The jellyfish shown in Fig. 4.1 is obviously 'soft' in this sense. Metals and ceramics typically have moduli in the range of 100 MPa (see Fig. 4.2). While the strength of metals can be adjusted by a variety of mechanisms that affect the nature of dislocation motion in these systems, the modulus is set by the nature of the interatomic potentials and there nothing that can really be done to significantly affect the modulus of a given material. Polymers are different, however, and have a much broader range of elastic moduli. The stiffest of these (Kevlar™for example) have elastic moduli in at least one direction that are comparable to the modulus of steel.
2. Thermal Fluctuations Matter in Soft Materials.
Figure 4.1: An example of a 'soft' material.
 Exercise: How high above the earths surface must a single oxygen molecule be lifted in order for its gravitational potential energy to be increased by ${k}_{B}T$? Solution: The gravitational potential energy is $mgh$, where $m$ is the mass of the object, $g$ is the gravitational acceleration (9.8 m/s ${}^{2}$) and $h$ is height. The mass of a single ${O}_{2}$ molecule is obtained by dividing the molecular weight in g/mole by Avogadro's number: $m=32\left(\frac{g}{mole}\right)\left(\frac{mole}{6.02x1{0}^{23}}\right)=5.3x1{0}^{-23}\phantom{\rule{6px}{0ex}}g=5.3x1{0}^{-26}\phantom{\rule{6px}{0ex}}kg$ The conversion to kg illustrates the first point that we want to make with this simple calculation:Don't mix units and convert everything to SI units (m-kg-s) to keep yourself sane and avoid unit errors. Boltmann's constant, ${k}_{B}$ needs to be in SI units as well: ${k}_{B}=1.38x1{0}^{-23}\phantom{\rule{6px}{0ex}}J/K$ Now we just need to equate ${k}_{B}T$ to $mgh$, using something reasonable for the absolute temperature, $T$. I'll use $T$=300 K to get the following: $h=\frac{{k}_{B}T}{mg}=\frac{\left(1.38x1{0}^{-23}\phantom{\rule{6px}{0ex}}J/K\right)\left(300\phantom{\rule{6px}{0ex}}K\right)}{\left(5.3x1{0}^{-26}\phantom{\rule{6px}{0ex}}kg\right)\left(9.8\phantom{\rule{6px}{0ex}}m/{s}^{2}\right)}=8000\phantom{\rule{6px}{0ex}}m=8\phantom{\rule{6px}{0ex}}km$ The full oxygen partial pressure is given proportional to $exp\left(-mgh/{k}_{B}T\right)$, so this value of $h$ is the altitude where the oxygen pressure is a factor of $e$ (2.7) less than the value at sea level.

### 4.1 Synthetic Polymers

Most of this book is devoted to synthetic polymers because of their widespread use and importance, and because they illustrate many of the key concepts that are relevant to natural polymers and other soft materials. We begin with a basic introduction to the synthesis of polymeric materials, and the z component is between properties of these materials. We conclude with some examples of soft materials made up of building blocks that are held together by relative weak forces.

#### 4.1.1 What is a Polymer

• Strong covalent bonds are formed within a molecule (between 'mers').
• Weak Van der Waals or hydrogen bonding are formed between molecules, and cause the materials to condense into a solid or liquid phase.
Figure 4.3: Schematic representation of a polymer

#### 4.1.2 Classification Scheme

Crystallization and glass formation are the two most important concepts underlying the physical properties of polymers. Polymers crystallize at temperatures below ${T}_{m}$ (melting temperature) and form glasses at temperatures below ${T}_{g}$ (glass transition temperature). All polymers will form glasses under the appropriate conditions, but not all polymers are able to crystallize. The
classification scheme
shown in Figure 4.4 divides polymeric materials based on the locations of ${T}_{g}$ and ${T}_{m}$ (relative to the use temperature, $T$) and is a good place to start when understanding different types of polymers.
Figure 4.4: Classification scheme for polymeric materials.

#### Elastomers

Polychloroprene (Neoprene)
Polyisobutylene
Silicone
Figure 4.5: Examples of elastomeric materials.

#### Glassy Polymers

Poly(methyl methacrylate) (PlexiGlas)
Polycarbonate
Poly(phenylene oxide)
Figure 4.6: Examples of glassy polymers.

#### Semicrystalline Polymers

Polyethylene terephthalate.
Poly(tetrafluoroethylene) (Teflon)
Spider Web
Figure 4.7: Examples of semicrystalline polymers.

### 4.2 Understanding Polymer Chemistry

Crystallization and glass formation processes are central to our understanding of polymeric materials, we must eventually address the following question:
• How is a polymer's tendency to crystallize or form a glass determined by is molecular structure?
Before we answer this question, however, we must answer the following question:
• What determines this molecular structure, and how are our choices limited?
In order to answer this question properly, we need to study the processes by which polymeric materials are made. Polymer synthesis involves organic chemistry. After familiarizing ourselves with some of the relevant polymer chemistry, we will be in a position to study the physical properties of polymers. For this reason, our discussion of molecular structure in polymers will include some chemistry.

#### 4.2.1 Covalent Bonding

Polymer molecules consist of atoms (primarily carbon, nitrogen, oxygen and hydrogen) which are covalently bound to one another. The fraction of the periodic table that can form strong covalent bonds is relatively small, corresponding to the ten atoms shown in yellow in Figure 4.8 (H, C, N, O, F, Si, P, S, Cl, Br). It is useful at this point to recall some of the basic principles governing the bonding between these atoms:
• Nitrogen, oxygen, carbon and the other covalent bond forming atoms with $M>6$ (F, Si, P, S, Cl, Br) are surrounded by 8 electrons, included shared electrons.
• Hydrogen atoms are surrounded by 2 electrons, included shared electrons.
• A single bond involves two shared electrons, a double bond involves 4 shared electrons, and a triple bond involves 6 shared electrons.
Figure 4.8: Periodic table, illustrating the ten atoms that make up the covalently-bonded portion of polymeric materials.

#### 4.2.2 Lewis Diagrams

Lewis diagrams (Wikipedia link) provide a convenient way of keeping track of the valence electrons in covalently bonded compounds. Several examples are given here. Note how the rules given on the previous page are followed in each case.
Figure 4.9: Examples of Lewis Diagrams

#### 4.2.3 Bonding

The following principles of covalent bonding in organic materials are very helpful:
• Carbon, Silicon: group 14 (4 valence electrons) - 4 more needed to complete shell -
C, Si form 4 bonds with neighboring atoms
.
• Nitrogen, Phosphorous: group 15 (5 valence electrons) - 3 more needed to complete shell -
N, P form 3 bonds with neighboring atoms
.
• Oxygen, Sulfur: group 16 (6 valence electrons) - 2 more needed to complete shell -
O, S form 2 bonds with neighboring atoms
.
• Hydrogen (group 1) or Fluorine, Chlorine, Bromine (Group 17):
Fl, Cl, Br form 1 bond with neighboring atoms
.
(The situation for P and S is actually a bit more complicated when either of these atoms are bonded to oxygen, but these general rules serve our purpose for now.) The chemical structures throughout this book can be seen to obey these rules.

#### 4.2.4 Shorthand Chemical Notation

Most of the chemical structures illustrated in this text are relatively simple, consisting of single and double bonds between atoms. We generally don't bother to write all of the carbons and hydrogens into to the structure. We use the following common conventions.
• If no element is included, element at junctions between different bonds are assumed to be carbon.
• If atoms are missing, so that the rules given above for the number of bonds attached to each atom type are not satisfied, the missing atoms are hydrogens.
An example of this convention is shown below in the structure for
benzene
: :
Figure 4.10: Examples of Shorthand Notation when drawing chemical structures

## 5 Polymerization Reactions

Polymerization is the process by which small molecules react with one another to form large polymer molecules. Polymerization reactions can be broken up into the following two general categories:
##### Step Growth Polymerizations:
Collections of A and B species react with one another. In linear step growth polymerizations, the ends of molecules react with one another to form longer molecules. A variety of reactions are possible, so you need to know at least a little organic chemistry. We'll focus on just a few of the most common cases.
##### Chain Growth Polymerizations:
Each polymer chain has one reactive site to which additional monomers are added.
The Macrogalleria web site has some excellent, simple descriptions of polymerization reactions. Specific examples are referenced at different points throughout this book.

### 5.1 Step-Growth Polymerizations

Figure 5.1: Polymers produced by step growth polymerization. The red and green circles correspond to 'A' and 'B' monomers that react with one another to form the polymer.
The following pages illustrate some of the specific reactions which take place during the polymerization process. To illustrate the concepts involved, we consider the following polymer types, all of which are produced by step-growth polymerization:

#### 5.1.1 Polyamides

Figure 5.2: Formation of an amide from an amine and a carboxylic acid
In this example, a primary amine reacts with a carboxylic acid to form an amide linkage. Water is liberated during the condensation reaction to form the amide. Primary amines and acid chlorides undergo a similar reaction:
Figure 5.3: Formation of an amide from a primary amine and an acid chloride
Acid chlorides react very rapidly with amines at room temperature, which is very useful for demonstration purposes. Acid chlorides can also react with water to form carboxylic acids, however, and commercial polyamides are generally produced by reaction with carboxylic acids.

### 5.2 Interfacial Polymerizations

Figure 5.4: Schematic representation of the interfacial polymerization off nylon.

#### 5.2.1 Polyesters

Polyesters can be formed by condensation reactions of alcohols with carboxylic acids:
Figure 5.5: Formation of an ester from an alcohol and a carboxylic acid.
Figure 5.6: Ester hydrolysis: The reverse of the esterification reaction.

#### 5.2.2 Polyurethanes

Polyurethanes are formed by the reaction of isocyanates with alcohols, as shown below:
Figure 5.7: Urethane Formation
Note that this is NOT a condensation reaction, since no byproducts are formed during the reaction. Also note that the urethane linkage contains an oxygen atom in the backbone, whereas the amide linkage does not. (The ${R}^{1}$ and ${R}^{2}$ substituents can have different structures, but are always attached to the illustrated linkages by carbon atoms.)

#### 5.2.3 Epoxies

All epoxies involve reactions of epoxide groups (3-membered rings containing an oxygen atom) with curing agents. Amine curing agents are very common, as illustrated Figure , which shows a primary amine reacint with an epoxide group).
The secondary amines which remain can react with additional epoxide groups to form a branched structure as shown in Figure 5.9. The reactive functionality of a primary diamine is 4 when the reaction is with an epoxide (as opposed to its functionality of 2 in the case where the diamine reacts with an acid or acid chloride).
Figure 5.9: Reaction of an epoxide with a secondary amine.

## 6 Molecular Weight Distributions

The repeat unit for this polymer consists of a red and green unit together. The degree of polymerization ( $N$) is the number of repeating units for a single molecule. Because of the statistical nature of the reaction process, not all molecules at any given point in the polymerization will have the same size. This point is illustrated in Figure 6.1, which shows schematic representations of polymer with and $N=10.5$. The molecular weight, $M$, of a polymer is obtained from $N$ by multiplying by ${M}_{0}$, the molecular weight of a repeating unit of a polymer chain:
$\begin{array}{cc}M=N{M}_{0}& \left(6.1\right)\end{array}$
Figure 6.1: Linear, step-growth polymers with $N=3$ (left) and $N=10.5$ (right).
 Some key definitions: $N$: Degree of polymerization - the number of repeat units in a single molecule $M$: Molecular weight - molecular weight of a polymer molecule (g/mole) ${M}_{0}:$ Molecular weight per repeat unit (g/mole)
The complete distribution of molecular sizes can be described by a histogram. There are two different quantities that we can keep track of: ${n}_{N}$ and ${w}_{N}$, which are defined as follows:
• ${n}_{N}$: number of molecules with degree of polymerization = $N$
• ${w}_{n}$: weight of all molecules with degree of polymerization = $N$
The total number of molecules ( $n$) or total weight ( $w$) is obtained by summing over all possible values of $N$:
Note: We've assumed here for simplicity that $N$ takes on integer values. This is true for chain growth polymerizations, but for step growth polymers $N$ can formally have non-integer values (like the value of 10.5 in Figure 6.1). In this case you can just imagine that distribution has divided into 'bins' of width '1'.
Figure 6.2: Histogram of molecular sizes. The same distribution is shown, with the numbers of molecules of each size ( ${n}_{N}$1*2/1tted on the left, and the total weight corresponding to each of the different molecular sizes ( ${w}_{N}$53plotted on the right.

### 6.1 Number Average Molecular Weight

$\begin{array}{cc}{M}_{n}=w/n& \left(6.4\right)\end{array}$
$\begin{array}{cc}{M}_{n}=\frac{{M}_{0}{\sum }_{N}N{n}_{N}}{{\sum }_{N}{n}_{N}}& \left(6.5\right)\end{array}$
$\begin{array}{cc}{N}_{n}=\frac{{\sum }_{N}N{n}_{N}}{{\sum }_{N}{n}_{N}}& \left(6.6\right)\end{array}$

### 6.2 Weight Average Molecular Weight

The
weight average degree of polymerization
, ${N}_{w}$, is obtained simply by dividing ${M}_{w}$ by ${M}_{0}$, i.e. ${N}_{w}={M}_{w}/{M}_{0}.$ The
polydispersity index
, given by ${M}_{w}$/${M}_{N}$ must be greater than or equal to one. Polymers with a polydispersity index close to 1 are said to be monodisperse, with a narrow distribution of molecular weights. While Eq. 6.7 is useful in a formal sense, it's a bit cumbersome to use. Suppose we have a series of molecular weight fractions, with ${w}_{i}$ between the weight of each fraction and ${M}_{i}$ being the weight average molecular weight of each fraction. The weight average molecular weight for this collection of molecules is given by the following summation:
$\begin{array}{cc}{M}_{w}={\sum }_{i}{w}_{i}{M}_{i}& \left(6.8\right)\end{array}$
##### Exercise:
Suppose 3g of a monodisperse polymer with M=50,000 g/mol is blended with 5g of a monodisperse polymer with M=85,000 g/mol. What are the values of ${M}_{n}$, and the polydispersity index for the resultant mixture?
##### Solution:
To get ${M}_{n}$, we need to divide the total weight of polymer (3g + 5g = 8g) by n, the total number of moles of polymer molecules:
$n=\frac{3g}{50,000g/mol}+\frac{5g}{85,000g/mol}=1.9x1{0}^{-4}mol$
${M}_{n}=\frac{8g}{1.19x1{0}^{-4}mol}=67,200g/mol$
The average molecular weight is obtained from the definition of this quantity:
$\begin{array}{cc}{M}_{w}=\frac{{M}_{0}\stackrel{\infty }{\sum _{N=1}}{w}_{N}N}{w}=\frac{1}{w}\stackrel{\infty }{\sum _{N=1}}{w}_{N}{M}_{N}& \left(6.9\right)\end{array}$
In our example, we obtain the following:
${M}_{w}=\frac{1}{8g}\left(3g*50,000g/mol+5g*85,000g/mol\right)=71,875g/mol$
Polydispersity Index = $\frac{{M}_{w}}{{M}_{n}}=\frac{71,875g/mol}{67,200g/mol}=1.07$
##### Exercise:
What if the polymers in the previous example are not monodisperse? Suppose the molecular weights quoted are weight averages, the low molecular weight polymer has a polydispersity of 1.6, and the high molecular weight polymer has a polydispersity of 2.5.
##### Solution:
We divide by the polydispersities to get number average molecular weights, and use these values to get the total number of molecules:
${M}_{n}\left(poly.a\right)=\frac{50kg/mol}{1.6}=31kg/mol$
${M}_{n}\left(poly.b\right)=\frac{85kg/mol}{2.5}=34kg/mol$
$n=\frac{3g}{31,000g/mol}+\frac{5g}{34,000g/mol}=2.44x1{0}^{-4}mol$
${M}_{n}\left(blend\right)=\frac{8g}{2.44x1{0}^{-4}mol}=32,800g/mol$
Due to the form of the expression for the average molecular weight, we can modify the standard expression for the Mw by summing over all different components of the blend, and using the weight average molecular weight for each component:
$\begin{array}{cc}{M}_{w}\left(blend\right)=\frac{1}{w}\sum _{i}{w}_{i}{M}_{wi}& \left(6.10\right)\end{array}$
In our case ${M}_{w}$ is the same as it was in the previous example:
${M}_{w}=\frac{1}{8g}\left(3g*50,000g/mol+5g*85,000g/mol\right)=71,875g/mol$
Polydispersity index = $\frac{{M}_{w}}{{M}_{n}}=\frac{71,875g/mol}{32,800g/mol}=2.19$

### 6.3 Molecular Weight Evolution during Step Growth Polymerization

Suppose ${n}_{a}$ moles of difunctional A molecules with molecular weight ${M}_{a}$ react with ${n}_{b}$ difunctional B molecules with molecular weight ${M}_{b}$ in a step growth polymerization. If ${n}_{a}$ > ${n}_{b}$, then all of the B molecules can react, and ${n}_{a}-{n}_{b}$ molecules will remain at the end of the polymerization:
${n}^{min}={n}_{a}-{n}_{b}$
For condensation reactions, a mole of condensation product with molecular weight ${M}_{c}$ material as a wholeecule of B that reacts. The total weight, $w$, of polymer produced is:
$\begin{array}{cc}w={n}_{a}{M}_{a}+{n}_{b}{M}_{b}-2{n}_{b}{M}_{c}& \left(6.11\right)\end{array}$
The maximum number average molecular weight, ${M}_{n}^{max}$, is obtained by using this value for $w$, an setting $n$ equal to ${n}_{min}$:
$\begin{array}{cc}{M}_{n}^{max}=\frac{w}{{n}^{min}}=\frac{{n}_{a}{M}_{a}+{n}_{b}{M}_{b}-2{n}_{b}{M}_{c}}{{n}_{a}-{n}_{b}}& \left(6.12\right)\end{array}$
For high molecular weight polymers, we can make some additional approximations, since know that ${n}_{a}-{n}_{b}$ << ${n}_{b}$ in this case. We can therefore simplify ${M}_{n}$ to the following:
$\begin{array}{cc}{M}_{n}^{max}\approx \frac{{n}_{b}\left({M}_{a}+{M}_{b}-2{M}_{c}\right)}{{n}_{a}-{n}_{b}}& \left(6.13\right)\end{array}$
The repeat unit molecular weight, ${M}_{0}$ is obtained by adding the molecular weights for the A and B monomers, and subtracting out the molecular weight of the 2 condensation products that are formed (since a repeat unit contains 2 A-B linkages, and therefore involves 2 condensation reactions):
$\begin{array}{cc}{M}_{0}={M}_{a}+{M}_{b}-2{M}_{c}& \left(6.14\right)\end{array}$
From this we obtain the following expression for ${M}_{n}^{max}$ :
The next thing we would like to sort out is how $n$, the total number of molecules, varies with the extent of reaction, $p$. The trick is to realize that every time a reaction takes place, the number of molecules in the system decreases by 1. Each molecule has two groups (one at each end) that can react, and two groups take part in each reaction. For $n$ we obtain the following:
In this equation ${n}_{a}+{n}_{b}$ is the initial number of molecules that we start off with (when $p$=0), and $p\left({n}_{a}+{n}_{b}\right)$ gives the decrease in the number of molecules resulting from all of the reactions that have taken place for some finite value of $p$. The total mass of the polymer that has been produced is $\approx \left({n}_{a}+{n}_{b}\right){M}_{0}/2$ (ignoring the fact that the ends of all the molecules some extra mass that may potentially leave as part of a condensation product). The weight average molecular weight is simply $w/n$, from which we obtain the following:
As mentioned above, the degree of polymerization that can be obtained from a homogeneous mixture of A and B monomers in a linear stop growth polymerization is limited by stoichiometry. Reaction of all A monomers requires the existence of unreacted B monomers, andvice versa. If ${n}_{a}\ne {n}_{b}$, then some of the functional groups on either the A or B monomers must remain unreacted, so that $p$ has a maximum value somewhat less than 1. Let's consider the case where ${n}_{a}-{n}_{b}$, in which case we know there are ${n}_{a}-{n}_{b}$molecules left when all the B groups have reacted. Each of these molecules has an unreacted A group at each end, so there are a total of $2\left({n}_{a}-{n}_{b}\right)$ unreacted groups at the end of the reaction. At the beginning of the reaction (when $p$=0) we had ${n}_{a}+{n}_{b}$ reactive groups, so the total number that have reacted is $2\left({n}_{a}+{n}_{b}\right)-2\left({n}_{a}-{n}_{b}\right)=4{n}_{b}$. This corresponds to a fraction of reacted groups which we refer to as ${p}^{max}$:
Another way to obtain ${M}_{n}^{max}$ is to use Eq. 6.17 for ${M}_{n}$, with $p={p}^{max}$:

$\begin{array}{cc}{M}_{n}^{max}=\frac{{M}_{0}}{2\left(1-{p}_{max}\right)}=\frac{{M}_{0}\left({n}_{a}+{n}_{b}}{2\left(\left({n}_{a}-{n}_{b}\right)}& \left(6.19\right)\end{array}$
which with ${n}_{a}\approx {n}_{b}$ is equivalent to Eq. 6.15.
From the information given so far we know how the number average degree of polymerization evolves with $p$, but in many cases we would like to know what the full molecular weight distribution might look like. To do this we need to account for the statistics, and use the more detailed definitions for ${M}_{n}$ (Eq. ) and ${M}_{w}$ (Eq. ) A molecule formed from a linear step growth polymerization that has a degree of polymerization of $N$ was made from $2N$ molecules ( $N$ of type A and $N$ of type B). Because these $2N$ original molecules are all bifunctional, there are a total of $4N$ reactive groups in the monomers that reacted together to form a polymer of length $N$. Only two of these groups remain unreacted (the end groups). Because each reaction involves two groups, this means that $2N-1$ reactions have taken place ( $4N-2$ groups have reacted).
The key assumption is that all groups react with equal probability. This probability must therefore be equal to the fraction of functional groups which has reacted, which has already been defined as the extent of reaction,p. The probability that any single one of these $2N-1$ reactions has taken place is therefore equal to $p$. We have to multiply all of these probabilities together to get the probability of having a polymer with a degree of polymerization of $N$, so we end up with the following:
$\begin{array}{cc}{n}_{N}\propto {p}^{2N-1}& \left(6.20\right)\end{array}$
The weight, ${w}_{N}$, corresponding to each fraction is proportional to ${n}_{N}N$:
$\begin{array}{cc}{w}_{N}\alpha N{p}^{2N-1}& \left(6.21\right)\end{array}$
The shapes of these distributions for ${n}_{N}$ and ${w}_{N}$ are shown in the plots below, and can be replotted with different values of $p$ in the MATLAB modulue referred to below.
Figure 6.3: Distribution of molecular sizes during an ideal step growth polymerization.
 What happens to the polydispersity index as the degree of polymerization increases? Where are the number and weight averaged degrees of polymerization in comparison to the peak in ${w}_{N}$? How does the shape of the molecular weight distribution change as $N$ gets very large? What sizes of molecules are present in the highest numbers? For what values of p is Nn equal to 0.5, 1, and infinity? Can you develop a simple explanation for these results? What are the values of the polydispersity for these values of p?
Analytic expressions can be derived for number-average and weight average degrees of polymerization for ideal step growth polymerizations that follow the distribution described by Eq. 6.20. Substitution of this expression into Eq. 6.5 gives the following for ${M}_{n}$:
$\begin{array}{cc}{M}_{n}=\frac{{M}_{0}\sum _{N}N{p}^{\left(2N-1\right)}}{\sum _{N}{p}^{\left(2N-1\right)}}& \left(6.22\right)\end{array}$
We assume that our material is isotropic $N$,i.e., $N$=1/2, 1, 3/2, 2, etc. Half integer values of $N$ are allowed because $N$ is the number of repeat units in the molecule. (A value of 3/2 for $N$ would correspond to a molecule with one A monomer between two monomers, for example). Alternatively, we can replace $N$ by $L/2$, where $L$ takes on all positive integral values:
We can follow a similar procedure to get an expression for ${M}_{w}$. Combination of the ideal, linear step growth size distribution (Eq. 6.20) with the definition of ${M}_{w}$ (Eq. 6.7), gives
$\begin{array}{cc}{M}_{w}=\frac{{M}_{0}\sum _{N}{N}^{2}{p}^{\left(2N-1\right)}}{\sum _{N}N{p}^{\left(2N-1\right)}}& \left(6.24\right)\end{array}$
we can again let $L=2N,$ so that $L$ takes on all integer values, so that the expression for ${M}_{w}$ becomes:
$\begin{array}{cc}{M}_{w}=\frac{1}{2}\frac{{M}_{0}\sum _{N}{L}^{2}{p}^{\left(L-1\right)}}{\sum _{N}L{p}^{\left(L-1\right)}}& \left(6.25\right)\end{array}$
These expressions for ${M}_{n}$ and ${M}_{w}$ can be simplified by using the following mathematical formulas:
Substituting the results for these sums into Eqs.6.23 and6.25 results in the following:
$\begin{array}{cc}{M}_{n}=\frac{{M}_{0}}{2\left(1-p\right)};\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}{N}_{n}=\frac{1}{2\left(1-p\right)};& \left(6.27\right)\end{array}$
$\begin{array}{cc}{M}_{w}=\frac{{M}_{0}}{2}\left\{\frac{1+p}{1-p}\right\}& \left(6.28\right)\end{array}$
The expression for ${M}_{n}$ is the same as the result that we got from the previous section, just by keeping track of the total number of molecules in the system. Calculation of ${M}_{w}$ is a bit more complicated though, and requires the statistical approach used in this section. For the polydispersity index we get:
$\begin{array}{cc}\frac{{M}_{w}}{{M}_{n}}=1+p& \left(6.29\right)\end{array}$
Useful, high molecular weight polymers produced by step-growth polymerization are therefore have a polydispersity index that is close to 2.

## 7 Non-Linear Step Growth Polymerizations

Figure 7.1: Nonlinear Step Growth Polymerization

### 7.1 Gelation

Figure 7.2: Molecular schematic of gel network.

### 7.2 Prepolymers

Figure 7.3: Example of a Prepolymer
The following factors determine the property of the network:
1. Molecular weight of prepolymer
2. Composition of prepolymer
3. Functionality of curing agent (4 for diamine curing of epoxies)
4. Extent of reaction (number of dangling ends)
5. Network structure (loops, entanglements, etc.)

### 7.3 Corothers Theory of Gelation

We begin with ${n}_{a}$ molecules with a reactive functionality of ${f}_{a}$ that are able to react with ${n}_{b}$ molecules with a reactive functionality of ${f}_{b}$. For illustrative purposes, consider the perfectly stoichiometric case where the total functionality of all the A monomers is equal to the total functionality of all the B monomers:
We define an average reactive functionality, ${f}_{av}$, so that $\left({n}_{a}+{n}_{b}\right){f}_{av}$ is equal to the total number of functional groups:
The number of initial molecules corresponding to a given number of functional groups is inversely proportional to ${f}_{av}$. For large values of ${f}_{av}$, p does not have to reach unity in order for the total number of molecules to be reduced to one. An expression for ${N}_{n}$ can be obtained substituting p ${f}_{av}$/2 for $p$ in Eq. , describing the evolution of ${M}_{n}$ with $p$ for linear step growth polymerization:
$\begin{array}{cc}{N}_{n}=\frac{1}{2-p{f}_{av}}& \left(7.3\right)\end{array}$
The quantity $p{f}_{av}$ describes the average number of times that any given molecule has reacted. The number average molecular weight is predicted to diverge to infinity when this average number of reactions per molecule is 2. This occurs when $p={p}_{gel}$, where ${p}_{gel}$Standardextent of reaction at the endpoint. Setting ${p}_{gel}{f}_{av}=2$ gives:
This simple equation for the gel point is very useful, although it is important to remember that this derivation suffers from the following deficiencies:
1. Gelation actually occurs for finite ${N}_{n}$, since $N=\infty$ only for the largest molecule.
2. The theory neglects loop formation (reactions between two portions of the same molecule). These reactions change the shape of a given molecule, but do not decrease the overall number of molecules.
The first assumption tends to overestimate ${p}_{gel}$, and the second assumption tends to underestimate ${p}_{gel}$. A more accurate description of gelation requires a much more detailed theory than the one presented here.
When one molecule which spans the entire sample, it can no longer flow like a liquid. One the characteristics of non-linear step growth polymerizations is that the viscosity (resistance to flow) of the reaction mixture increases as the extent of reaction increases, and eventually diverges at the gel point. Also, the reaction rate generally decreases as the reaction proceeds and the reactive molecules become larger and larger. These features are illustrated conceptually in Figure 7.4. Here we show schematic representations of the time-dependent extent of reaction, along with the time dependence of the number average degree of polymerization, ${N}_{n}$, and of two physical properties of the material: ${\eta }_{0}$ and ${E}_{\infty }$. Here ${\eta }_{0}$ is the limiting viscosity measured at very low shear rates, and ${E}_{\infty }$ is the elastic modulus measured at very long times. The viscosity is a characteristic of a liquid material, and characterizes the material for values of the extent of reaction, $p$, that are less than the extent of reaction corresponding to the gel point, ${p}_{gel}$. For values of $p$ that exceed ${p}_{gel}$, the material behaves as a solid, and has a finite elastic modulus, ${E}_{\infty }$. At the gel point, the viscosity diverges to infinity, since it is no longer possible for the material to flow. Instead, the material becomes an elastic solid, with a value of ${E}_{\infty }$ that increases from 0 at the gel point.
Note that that materials near the gel point are viscoelastic, and can no longer be described by a single value of the viscosity or modulus. Instead, these properties depend on the timescale of the measurement, as described in more detail on the section on viscoelasticity.
Figure 7.4: Time-dependence of property development in a nonlinear step growth polymerization.
##### Exercise:
A very simple epoxy formulation has bisphenol A diglycidyl ether as the epoxide component and hexamethylene diamine as the crosslinker:
$H2N\left(CH2\right)6NH2$
A) Hexamethylenediamine
B) Bisphenol A diglycidyl ether
1. How many grams of hexamethylene diamine should be added to 1g of Bisphenol A diglycidyl ether in order to optimize the polymerization reaction in a homogeneous solution?
2. Estimate the fraction of amine groups that need to react in order to reach the gel point.
##### Solution:
${n}_{a}=\frac{{n}_{b}{f}_{b}}{{n}_{a}}=\frac{{n}_{b}}{2}$
We have 1g of B, which we divide by it's molecular weight to get the number of moles:
So we know we need about $1.45x1{0}^{-3}$ moles of the hexamethylenediamine. Multiplying by its molecular weight of 116 g/mole gives a total mass of 0.17g. Note that we need a lot less of A than B in this case. If we want our two component epoxy to consist of two parts that we combine in nearly equal volumes, we'll have to add some extra stuff to part A (the diamine part) that doesn't take part in the curing reaction.
To figure out the average functionality, we arrange Eq. 7.2 to give the following for ${f}_{av}$:
${f}_{av}=\frac{{n}_{a}{f}_{a}+{n}_{b}{f}_{b}}{{n}_{a}+{n}_{b}}$
Combining this with Eq. 7.4 for the gel point gives:
${p}_{gel}=\frac{2\left({n}_{a}+{n}_{b}\right)}{{n}_{a}{f}_{a}+{n}_{b}{f}_{b}}$
Substituting our stochiometric condition that ${n}_{a}={n}_{b}/2$ gives ${p}_{gel}=0.75$.

## 8 Chain-Growth Polymerizations

In chain growth polymerization monomers are added one at a time to a reactive site at that typically remains at the end of the monomer that has been most recently added. There are three phases of the reaction that need to be considered in general:
1. Initiation: In this step the a reactive species is formed that is able to form covalent bond with a monomer, reforming the active species at the end of the growing polymer chain.
2. Propagation: Monomers are added one at a time to the growing chain.
3. Termination: The reactive undergoes a reaction of some sort that causes it to lose its reactivity toward other monomers. The polymer chain is 'dead' at this point, and no longer grows in length.
Figure 8.1: Chain growth polymerization.

### 8.1 Chain Growth Polymerization Mechanisms

Chain growth polymerizations generally occur either by addition to a double bond, or by opening a ring. In both cases the overall number of bonds is conserved, and there are no condensation products. If the cases where chain growth occurs by addition to double bond between two carbon atoms, the following takes place:
• A new bond is formed between the active sight and one if the doubly bonded carbons
• The double bond is shifted to a single bond.
Figure 8.2: The two basic types of chain growth polymerizations.
In a ring opening polymerization, the following takes place:
• A new bond is formed between the active species and one of the atoms in a cylic monomer.
• An adjacent bond in the monomer is broken.
• The active site moves to the end of the linear molecule.

### 8.2 Reactive Species for Chain Growth Polymerization

Three of the most common active sites are a free radical, negatively charged anion, and a positively charged cation, illustrated in Figure 8.3 for the addition to the double bond of vinyl polymers. Vinyl polymers have the general structure $C{H}_{2}CHR$, where R is something other than hydrogen. In each of these cases, a single monomer repeat unit is added to the end of the chain, and the reactive site moves to the end of the chain, on the repeat unit that has just been added. Reactions in organic chemistry are all about keeping track of what the bonding electrons are doing, and Lewis diagrams are very helpful in this sense. In the figures designed to illustrate different propagation and termination reactions, we just show the Lewis structures for some of the bonds, to make it simpler to keep track of the situation before and after the reaction has taken place. For our purpose we are not as interested in the detailed reaction mechanism, which would require that we provide a bit more information about the structure of some of the short-lived reaction intermediate.
Anionic Polymerization
Cationic Polymerization
Figure 8.3: Propagation mechanisms for three common chain growth polymerization mechanisms.

### 8.3 Initiation

The previous section outlines the chemistry of some different propagation reactions, where monomers are added to an active site. Initiation is the step by which the active site is produced at the beginning. Here we give two simple examples, one for initiation of a radical polymerization, and the second being the initiation of a an anionic polymerization.

#### 8.3.1 Initiation of a free radical polymerization

In this example, illustrated in Figure 8.4, a peroxide bond between two oxygen atoms splits to form two free radicals. These radicals are able to add to the double bond in a vinyl monomer like styrene to initiate the polymerization reaction.
Figure 8.4: Initiation off a radical polymerization reaction with an epoxide that decomposes into two radical species at elevated temperature.

#### 8.3.2 Initiation of an anionic polymerization

In the example shown in Figure 8.5 is very simple conceptually. The reaction is done in solution, and a small amount of secondary butyl lithium is simply added to a solution of styrene molecule in an appropriate organic solvent. The carbon-lithium bond is very reactive, and has a lot of ionic character. It can really be viewed as an existing anion that is ready to react directly with the styrene monomer.
Figure 8.5: Initiation off a radical polymerization reaction with an epoxide that decomposes into two radical species at elevated temperature.

### 8.4 Termination Mechanisms

Termination is the process by which reactive chain ends become unreactive. Termination reactions can be avoided in 'living polymerizations' like anionic and cationic polymerizations.

Figure 8.6: Termination by radical recombination.
As unpaired electrons, free radicals are very reactive towards one another. As illustrated in this example, two free radicals can readily combine with one another to form a covalent bond. Combination is therefore one type of termination reaction which is very prominent with free radical polymerizations. Note that the number of molecules decreases by one during the combination reaction.

#### 8.4.2 Disproportionation

Figure 8.7: Termination by disproportionation. The circled proton, along with an electron, moves from the molecule on the right, terminating the free radical on the right. The remaining electrons on the molecule on the left rearrange themselves into a double bond.
Plain Laytionation reactions can be viewed as the transfer of a proton andEntanglementsfrom one active molecule to another. The animation here illustrates how the transfer results in the termination of both molecules, with the formation of a double bond. In the final state, the carbon atoms all have eight electrons (including shared electrons) in the valence shell. The molecular weights of the two polymer molecules remain essentially unchanged (with the exception of the transfer of a single proton).

#### 8.4.3 Chain Transfer

Figure 8.8: Chain transfer to toluene.
Chain transfer refers to the migration of the active free radical from one molecule to the other. In this example, the active radical from the growing polymer chain is terminated by the addition of a proton and an electron from the toluene molecule on the right. The net result is that the free radical is transferred from the polymer molecule to the solvent molecule (toluene), which can then initiate the polymerization of additional monomer. Note that the process is very similar to disproportionation, except that the species which donates the proton and electron does not already have a free radical.

### 8.5 Intramolecular Chain Transfer

Figure 8.9: An Animation of Intramolecular Chain Transfer
In this example, the active radical moves from the end of the polymer chain to a different portion on the same polymer chain. Polymerization continues from this radical, resulting in the formation of a short branch, in this case consisting of four carbons. Because the branches are randomly placed along the polymer backbone, they interfere with the polymer's ability to crystallize.
##### Exercise:
1. Chain transfer agents, like toluene in the previous example, can be intentionally added to give some control over the molecular weights of polymers synthesized by free radical polymerizations. In a qualitative sense, how will the addition of chain transfer agents change the number average molecular weight of the polymer?
2. How does your answer to this question change if the chain transfer is always to other sites within the same polymer molecule?
3. Termination reactions involving free radical polymerizations involve the elimination of two radicals to form a new bond. Where does this new bond appear for combination and disproportionation reactions?

### 8.6 Polymerization of Dienes

Dienes are monomers with two double bonds, with a single bond between them. Examples include butadiene, isoprene and chloroprene. The situation for isoprene is illustrated in Figure 8.10. The monomer has 4 carbon atoms in a line, which we number 1 through 4. Double bonds connect carbons 1 to 2 and 3 to 4, and a single bond connects carbon 2 to 3. Carbons 2 and 3 are distinguishable from one another because carbon 2 is bonded to a methyl group that is not present on carbon 2. When a propagating polymer chain interacts with the isoprene monomer, addition can occur in any of 4 different ways, which are illustrated in Figure 8.10:
• 1,2 addition: The active chain end adds across the bond between carbons 1 and 2, just as it would in a normal polymerization.
• 3,4 addition: Like 1,2 addition, but the reaction occurs across the bond between carbons 3 and 4.
• 1,4 addition: Here the reactive site attaches to carbon 1 and the active site moves to carbon 4, with a double bond being formed between carbons 2 and 3. This double bond can exist in a 'cis' conformation or 'trans' configurations. These cis and trans configurations represent chemically different structures, resulting in polymers with different properties.
Natural rubber is a naturally occurring version of polyisoprene that is harvested from certain tropical trees. The excellent elastomeric properties of this material arise from the pure cis 1-4 microstructure that is produced during the natural polymerization of this polymer. Because of the zigzag nature of the polymer backbone, it does not readily crystallize. A segment of the backbone of the polymer is shown in Figure 8.11. This schematic shows a portion of the molecule in a fully extended form. Because of the easy rotation about single bonds, the molecules actually exist in a collection of random configurations.
Figure 8.10: Polymerization of Isoprene

### 8.7 Living Polymerizations

Living polymerizations are chain growth polymerizations that proceed without termination or chain transfer reactions. Relatively monodisperse polymers ( ${M}_{w}$/ ${M}_{N}$ < 1.1) can typically be obtained when the initiation rate is faster than than the propagation rate. Block copolymers are formed by the sequential addition of two (or more) different monomer types in a living polymerization.

### 8.8 Vinyl Polymers and Tacticity

Figure 8.12: Chemical structure of a vinyl monomer.
Figure 8.13: Illustration of different polymer tacticities.
Tacticity is significant because it determines the ability of a polymer to crystallize. The disordered structure of an atactic polymer is inconsistent with the ordered structure of a crystalline polymer. As a result, atactic polymers generally cannot crystallize.Standardons to this rule include polymers where the 'R' group is very small, so that this group can be incorporated into an ordered crystalline array, even if it is randomly placed along the polymer chain. For this reason polyvinyl chloride can be partially crystalline even if the polymer is atactic. Atactic polystyrene and atactic polypropylene, however, are always amorphous.

## 9 Common Polymers

Here we list some common polymer produced by the different synthesis methods introduced in the previous sections.

### 9.1 Chain Growth: Addition to a Double Bond

#### 9.1.1 Polyethylene

The simplest polymer from a structural standpoint is polyethylene, with the structure shown below in Figure 9.1.Dynamicity is not relevant in this case, since there are no substituents other than hydrogen on the carbon backbone.
Figure 9.1: Chemical structure of polyethylene.
##### High Density Polyethylene:
High density polyethylene refers to version with very little chain branching, thus resulting in a high degree of crystallinity. Completely linear polyethylene has a melting point of 138 °C, and a glass transition temperature near -100 °C.
${T}_{g}=\approx -120{\phantom{\rule{6px}{0ex}}}^{○}$C, ${T}_{m}=138{}^{○}$C (perfectly linear)
##### Low Density Polyethylene:
Low density polyethylenes (LDPE's) and high density polyethylenes (HDPE's) are identical in their chemical structure at the atomic level. They are actually structural isomers of one another. Chain branching within low density polyethylene inhibits crystallization, resulting in a material with a melting point lower than 138 degrees C. The decreased crystallinity of LDPE results in a material which is more flexible (lower elastic modulus) than HDPE.
Figure 9.2: Structure of low density polyethylene
The chain branches responsible for inhibiting crystallization in low density polyethylene are typically short. This illustration shows a 3-carbon (propyl) branch, potentially resulting from intramolecular chain transfer during the polymerization reaction.
${T}_{g}=\approx -120{\phantom{\rule{6px}{0ex}}}^{○}$C, ${T}_{m}<138{}^{○}$C (depending on branching)
##### The Importance of Molecular Weight:
Polymeric materials generally have favorable mechanical properties only when the molecular weight is very large - typically hundreds of thousands of g/mol. The point is illustrated with polyethylene:
• M = 16 g/mol: ethylene gas
• M $\cong$200 g/mol: candle wax
• M $\cong 2x1{0}^{5}$- $5x1{0}^{5}$ g/mol: milk jugs, etc.
• M $\cong 3x1{0}^{6}$ - $5x1{0}^{6}$ g/mol: ultrahigh molecular weight polyethylene. This materials as excellent toughness and wear resistance, and is often used as one of the contact surfaces in joint implants.

#### 9.1.2 Polypropylene

Figure 9.3: Structure of polypropylene
The most widely used form of polypropylene is isostatic, with a melting point of 171 °C (for the perfectly isotactic version - a few degrees lower for the actual commercial versions), and a glass transition temperature which is well below room temperature ($\approx 5{\phantom{\rule{6px}{0ex}}}^{○}$C). Single crystals of polypropylene have lower moduli than single crystals of polyethylene along the chain direction, because of the helical structure of propylene. The modulus of isotropic semicrystalline polypropylene is often larger than that of high density polyethylene, however, because of the details of the semicrystalline structure that is formed. The uses of polypropylene and high density polypropylene are similar.

#### 9.1.3 Polybutene-1

This polymer included to illustrate the evolution o the polymer properties when we continue to make the side chain longer.
Figure 9.4: Polybutene-1

#### 9.1.4 Poly(methyl methacrylate):

Poly(methyl methacrylate) (PMMA) is one of the most common materials used to make polymer glass. It is commonly known by the DuPont tradename Plexiglas,™ and has a glass transition temperature between 100 °C and 125 °C, depending on the tacticity. It is also forms the basis for many biomaterials, including dental adhesives.
${T}_{g}=100-125{\phantom{\rule{6px}{0ex}}}^{○}$C, no ${T}_{m}$ (atactic)
Figure 9.5: Poly(methyl methacrylate)

#### 9.1.5 Poly(methyl acrylate)

Poly(methyl acrylate) is not a widely used polymer, primarily because it's glass transition temperature is too low ( $\approx$5 ${}^{○}$C for the atactic polymer) to be useful as a rigid polymer glass, and too high to be useful as an elastomer. It is included here to illustrate the effect that removing the extra methyl group from the polymer backbone has on the glass transition of the polymer.
${T}_{g}=5{\phantom{\rule{6px}{0ex}}}^{○}$C, no ${T}_{m}$ (atactic)
Figure 9.6: Poly(methyl acrylate).

#### 9.1.6 Neoprene

Figure 9.7: Neoprene (polychloroprene).

#### 9.1.7 Polyisobutylene

Polyisobutylene is a common material used to make elastomers, referred to more simply as 'butyl' rubber. It is generally copolymerized by with a small amount of isoprene, so that the resulting double bonds can be used to form a crosslinked material. It is more resistant to solvent penetration than most elastomers, and is often used in applications (like the gloves above) where barrier resistance is needed.
${T}_{g}=-75{\phantom{\rule{6px}{0ex}}}^{○}$C, ${T}_{m}=2{\phantom{\rule{6px}{0ex}}}^{○}$C
Figure 9.8: Polyisobutylene.

#### 9.1.8 Polystyrene

${T}_{g}=100{\phantom{\rule{6px}{0ex}}}^{○}$C, no ${T}_{m}$ (atactic)
Figure 9.9: Structure of polystyrene.

#### 9.1.9 Poly(tetrafluroethylene) (PTFE)Poly(tetrafluoroethylene) (PTFE)

${T}_{g}=130{}^{○}$C (by one report[6]); ${T}_{m}\approx 330{\phantom{\rule{6px}{0ex}}}^{○}$C.
Figure 9.10: Monomer unit of poly(tetrafluroethylene) (Image from[2]).

#### 9.1.10 Poly(vinyl acetate)

Poly(vinyl acetate) is often used as base for chewing gum. It is glassy at room temperature but becomes softer at body temperature, which is just above ${T}_{g}$.
${T}_{g}=30{\phantom{\rule{6px}{0ex}}}^{○}$C, no ${T}_{m}$ (atactic)
Figure 9.11: Monomer unit of poly(vinyl acetate)

#### 9.1.11 Poly(vinyl chloride) PVC

Poly(vinyl chloride) can be partially crystalline even if the material is atactic, because the "R" group in this case is a chlorine atom, which is relatively small. The glass transition temperature of the material is 85 °C, although the addition of small molecules as "plasticizers" can reduce ${T}_{g}$ to below room temperature. When a material is referred to as "vinyl", it is probably PVC. Record albums (before the age of compact disks) and water pipes are commonly made out of poly(vinyl chloride).
${T}_{g}=85{\phantom{\rule{6px}{0ex}}}^{○}$C, no ${T}_{m}$ (atactic)
Figure 9.12: Structure of poly(vinyl chloride)

#### 9.1.12 Poly(vinyl pyridine)

Poly(vinyl pyridine) is very similar to polystyrene, and its physical properties (entanglement molecular weight, etc.) are quite similar to the properties of polystyrene. It exists in one of two forms, poly(2-vinyl pyridine) (P2VP and poly(4-vinyl pyridine) (P4VP), based on the location of the nitrogen in the phenyl ring. Both types interact strongly with metals. The polymers are not used in wide quantities, but have been useful in a range of model studies of polymer behavior, often when incorporated with another material as part of a block copolymer.
${T}_{g}=100{\phantom{\rule{6px}{0ex}}}^{○}$C, no ${T}_{m}$ (atactic)
Figure 9.13: Structure of two types of poly(vinyl pyridine)

### 9.2 Chain Growth: Ring Opening

#### 9.2.1 Poly(ethylene oxide)

Poly(ethylene oxide) (PEO) is generally formed by the ring opening polymerization of ethylene oxide. It is also referred to as polyethylene glycol, although this generally refers to lower molecular weight versions with hydroxyl end groups. PEO is water soluble, and is used in a wide range of biomedical applications, often in a gel form. Lithium salts are also soluble in PEO, and PEO/Li complexes are often used as an electrolyte in battery and fuel cell applications.
${T}_{g}=-65{\phantom{\rule{6px}{0ex}}}^{○}$C; ${T}_{m}\approx 65{\phantom{\rule{6px}{0ex}}}^{○}$C
Figure 9.14: Monomer unit of poly(ethylene oxide)

#### 9.2.2 Polycaprolactam :

${T}_{g}\approx 50{\phantom{\rule{6px}{0ex}}}^{○}$C; ${T}_{m}\approx 220{\phantom{\rule{6px}{0ex}}}^{○}$C
Figure 9.15: Polycaprolactam.

#### 9.2.3 Polycaprolactone :

Polycaprolactone somewhat unique in that is a polyester that is synthesized by ring opening polymerization of a cyclic ester. It can be viewed as a polyester version of the polyamide, polycaprolactam. Contrary to step growth polymerization of polyesters, the ester linkage is not formed during the polymerization reaction, but is already present in the monomer. Polycaprolactone is biodegradable because the polymer slowly degrades by ester hydrolysis over time.
${T}_{g}\approx -50{\phantom{\rule{6px}{0ex}}}^{○}$C; ${T}_{m}=60{\phantom{\rule{6px}{0ex}}}^{○}$C
Figure 9.16: Polycaprolactone.

### 9.3 Step Growth Polymers

A variety of common polymers are discussed briefly in the pages below.

#### 9.3.1 Kevlar ™

${T}_{m}$: above degradation temperature for the polymer.
Figure 9.17: Monomer unit of one version of Kevlar

#### 9.3.2 Polycarbonate

A variety of polycarbonates exist. The most common one, (with the GE trademark of Lexan) has a glass transition temperature of 150 °C. It is used for compact disks, eyeglass lenses, and shatterproof glass. See http://www.pslc.ws/macrog/pcsyn.htm for a good description of the synthesis of polycarbonate via a step growth, condensation reaction involving a phenolic di-alcohol and phosgene $\left(COC{l}_{2}\right).$
${T}_{g}\approx 150{\phantom{\rule{6px}{0ex}}}^{○}$C
Figure 9.18: Polycarbonate.

#### 9.3.3 Polyethylene Terephthalate (PET)

Polyethylene terephthalate (trade names include Mylar and Dacron) is produced in fiber form for textiles, and in film form for recyclable bottles, etc. Its degree of crystallinity is highly dependent on the processing conditions, since it can easily be quenched to a glassy state before crystallization is able to occur.
$=80{\phantom{\rule{6px}{0ex}}}^{○}$C
$=260{\phantom{\rule{6px}{0ex}}}^{○}$C
Figure 9.19: Polyethylene Terephthalate

#### 9.3.4 Poly(phenylene oxide)

Polyphenylene oxide is a high performance polymer has many varied uses, largely because of its excellent performance at high temperatures. $\approx 190{\phantom{\rule{6px}{0ex}}}^{○}$C

#### 9.3.5 Ultem Polyetherimide

Ultem (a trademark of GE) is a form of polyetherimide. It is a high performance polymer that combines high strength and rigidity at elevated temperatures with long term heat resistance ( ${T}_{g}=215{\phantom{\rule{6px}{0ex}}}^{○}$C). The repeat unit is illustrative of the complex chemical structure of many modern, high performance polymers.

#### 9.3.6 Silicones

$=\sim -130ºC$; $=-45ºC$
Figure 9.22: Poly(dimethyl siloxane).
We list PDMS here as a step growth polymer because it can be produced from a self-condensation of silanol (SiOH) groups. The starting point is actually dimethyl-dichlorosilane. In th presence of water the SiCl bonds hydrolyze to SiOH:
Figure 9.23: Hydrolysis of chlorosilane bonds.
The resultant silanol groups than can then condense by the elimination of water:
Figure 9.24: Condensation of silanol groups.
One of the interesting features of silicones is that they can also be synthesized by anionic, ring opening polymerization of cyclic, oligomeric forms of PDMS. Here's one example:
Figure 9.25: Anionic ring opening polymerization of a cyclic silicone oligomer.

Crosslinking (or vulcanization) is the process by which separate molecules in an amorphous polymer are chemically attached to one another, as shown schematically in Figure 10.1. The black lines indicate individual molecules, and the red dots represent crosslink points where these molecules are 'tied' together.
Figure 10.1: Diagram representation of crosslinking.
A representative and commercially important crosslinking system, known by the commercial name of Sylgard 184, consists of the two different parts shown in Figure 10.2. Part 1 consists of poly(dimethyl silioxane) polymer with vinyl groups (double bonds) at each end of the polymer molecules. Part 2 contains a silicone crosslinker with several silane groups (Si-H) that are able to react with the vinyl groups as shown in Figure 10.3. The reaction can be viewed as a step growth reaction involving a component with a reactive functionality of 2 (part 1) and a second component with a reactive functionality that is much larger than 2 (part 2).
Part 1: $n\approx 60$
Part 2: $n\approx 10$
Figure 10.2: A diagram of silicone crosslinking.

## 11 Copolymers

Figure 11.1: Schematic representation of a statistical copolymer.
In
block copolymers
, the different repeating units appear as distinct 'blocks'. A diblock copolymer for example, can be viewed as two different homopolymer molecules that have been covalently joined to one another as shown in Figure 11.2. A wide variety of interesting block copolymer morphologies have been observed. Diblock copolymers are most often synthesized by living, chain growth polymerizations.
Figure 11.2: Schematic representation of a block copolymer.
##### Exercise:
Describe a procedure for making a diblock copolymer of polystyrene and poly(methyl methacrylate).
##### Solution:
The details of how to do this go a bit beyond the scope of this text, but conceptually we know we need to do the following in sequence: Add and appropriate initiator for a living chain growth polymerization, add monomer 'A' until all of it reacts, add monomer B until all of it reacts, terminate the reaction. Here are the details that can work for this system.
1. Add an appropriate initiator. We don't want to use a radical initiator, because we need to avoid the inevitable termination reactions. From the information given in this text, either anionic or cationic polymerization. It turns out that the only anionic polymerization works in this case. We'll choose tetrahydrofuran as a solvent, clean it up very well to get rid of any impurities, and add some secondary butyllithium (see Fig. 8.5) as the initiator. The molar quantity of initiator will be equal to the number of moles of polymer we get at the end of the reaction.
2. Add the first monomer to the initiator solution. The order matters in this case. We need to add the styrene first, and choose reaction conditions so that the initiation is much faster than the subsequent polymerization. This ensures that the polymers all have roughly the same length, since they all started polymerizing a the same time.
3. Once all the styrene reacts, we add the methyl methacyrylate monomer to the solution. Because the polymerization is 'living', the active chain ends from the polystyrene part of the polymerization are still available to react with the methyl methacrylate.
4. Terminate the reaction. This is very easy to do, and can be done by adding a small amount of methanol to the reaction, to produce lithium methoxide and a 'dead' chain after the proton from alcohol trades places with the lithium that is associated with the negatively charged carbon at the end of the polymer chain.

## 12 The Glass Transition

### 12.1 Free volume

A very useful way to think about the glass transition involves the concept of unoccupied space, or 'free volume' in a polymer. This concept is illustrated conceptually in Figure 12.1. The red dots connected together by lines represent polymer molecules. In this lattice model of polymers, the red dots are constrained to fit on a lattice. (The lattice is completely artificial, and does not exist in reality). These dots represent the '
occupied volume
' . Empty lattice sites represent
free volume
.
The lattice on the left has no free volume, so there is no way for the molecules to move by the 'hopping' of segments into a small region of unoccupied space. The lattice on the right does have free volume, however, so that molecular motion is relatively easy.
$\begin{array}{cc}\alpha \phantom{\rule{10px}{0ex}}=\frac{1}{V}\frac{dV}{dT}& \left(12.1\right)\end{array}$
Note that the linear thermal expansion coefficient, defined in terms of the linear dimensions of the sample rather than the volume, is equal to α/3. If the fractional change in free volume is small ( $\Delta V/V\ll 1$) then $\alpha$ describes the linear relationship between the change in volume in the change in temperature:
$\begin{array}{cc}\frac{\Delta V}{V}=\alpha \Delta T& \left(12.2\right)\end{array}$
In the glassy phase ( $T<{T}_{g}$), the free volume is assumed to remain constant, so that increases in the volume are governed entirely by increases in the occupied volume:
In the liquid phase ( $T>{T}_{g}$), the free volume and occupied volumes both increase with temperature, and we have:
Combining12.3 and gives the following expression for the free volume in the liquid state:
$\begin{array}{cc}\frac{\Delta {V}_{f}}{V}=\left({\alpha }_{\ell }-{\alpha }_{g}\right)\Delta T& \left(12.5\right)\end{array}$
Now we define a new temperature, ${T}_{\infty }$ which is the theoretical temperature at which the free volume would go to zero, if the material were to continue to follow the liquid behavior below ${T}_{g}$. It is illustrated schematically in Figure 12.2. To determine the fractional free volume at the glass transition, we use Eq. 12.5 with $\Delta T={T}_{g}-{T}_{\infty }$, recognizing that ${V}_{f}/V=0$ at $T={T}_{\infty }$:
Typically, ${T}_{\infty }$ is about 50K below the measured glass transition temperature, and the fractional free volume at ${T}_{g}$ is in the range of a couple percent.
Figure 12.2: The glass transition and the free volume concept.
The concept of free volume can be helpful in sorting out how the glass transition depends on the structure of the polymer, based on the following two guidelines:
1. Changes to the polymer structure that increase the free volume needed in and r for the backbone of the polymer to move will increase ${T}_{g}$. Making the backbone very still (by the incorporation of phenyl groups, as in increase ${T}_{g}$. Also, bulky, 'fat' substituents (like methyl or phenyl groups) tend to increase ${T}_{g}$ when added to the backbone.
2. Changes to the polymer structure that introduce extra free volume will decrease ${T}_{g}$. This is the case for long, thin additions to the backbone, such as alkyl chains $\left(-{\left(C{H}_{2}\right)}_{n}-\right)$.
Figure 12.3: Structure of the repeat units for the materials listed in Table1.
##### Exercise:
Polystyrene has a volume thermal expansion coefficient of $6×1{0}^{-4}\phantom{\rule{6px}{0ex}}{K}^{-1}$ for temperatures above ${T}_{g}$, and a thermal expansion coefficient of $2×1{0}^{-4}\phantom{\rule{6px}{0ex}}{K}^{-1}$ for temperatures below ${T}_{g}$. Estimate the ratio of the free volume to the total volume of sample in the glassy state.Standardthat ${T}_{\infty }$ is 50K below ${T}_{g}$.
##### Solution:
Since the fractional free volume is assumed to be fixed below ${T}_{g}$ equilibr are looking for is simply the value at the glass transition, as given by Eq. 12.5. With ${\alpha }_{\ell }=6×1{0}^{-4}\phantom{\rule{6px}{0ex}}{K}^{-1},$ ${a}_{g}=2×1{0}^{-4}\phantom{\rule{6px}{0ex}}{K}^{-1}$ and ${T}_{g}-{T}_{\infty }=50$ K, we obtain:
$\frac{{V}_{f}}{V}=\left(4×1{0}^{-4}\phantom{\rule{6px}{0ex}}{K}^{-1}\right)\left(50\phantom{\rule{6px}{0ex}}K\right)=0.02$
This value of 0.02 is a typical value for the fractional free volume in a glassy polymer.

### 12.2 Enthalpy and Heat Capacity

Figure 12.4: A graph of enthalpy vs temperature
The heat capacity ( ${c}_{p}$) of the liquid is larger than the heat capacity of the glass. Note that the enthalpy itself is continuous at $T={T}_{g}$. The glass transition is therefore a second order transition - thermodynamic quantities like volume, enthalpy and entropy are continuous at the transition but the derivatives of these quantities with respect to temperature are discontinuous.

### 12.3 Differential Scanning Calorimetry (DSC)

Figure 12.5: Simple schematic representation of a calorimeter.
Figure 12.6: Graph of heat flow vs temperature.

## 13 Structure of Amorphous Polymers

Materials science is the study and application of the relationships between the properties of a material, the structure required to obtain these properties, and the processing methods which can be used to obtain these properties. Our discussion of polymer synthesis was motivated primarily by a need to understand processing methods for polymer materials. We also introduced the structure of different polymers at the atomic level, corresponding to the arrangements of individual atoms in the polymeric repeat units. Our discussion now moves to the structure of polymers on a molecular scale. We begin with amorphous polymers, and follow with a discussion of semicrystalline polymers. Ultimately, we will find that our understanding of many of the important properties of polymers can be related to these structural features by simple, yet remarkably accurate theories. In particular, the theory of rubber elasticity relates the elastic properties of an elastomer to the molecular structure described here.
The squiggly lines in Figure 13.1 represent the backbones of individual molecules in an amorphous polymer. (One molecule is highlighted in red) Our goal is to relate the properties of an amorphous polymer to the distribution of shapes of these molecules. Ultimately, we will show that many of the mechanical properties of amorphous polymers at temperatures above the glass transition are related to this distribution of polymer shapes.
Figure 13.1: A diagram of an amorphous polymer.

### 13.1 End-to-end Vector for Polymer Molecules

Figure 13.2: The end-to-end vector, $\stackrel{\to }{R}$, for a polymer molecule.
The shape of this particular molecule can only be completely specified by describing the path taken by the molecule from one end to the other. Fortunately, the most important quantity is much simpler. This quantity
end-to-end vector
, $\stackrel{\to }{R}$ which is simply the vector spanning the two ends of a given polymer molecule. This vector will in general be different for different polymer molecules in a sample, but the distribution of vectors can be accurately predicted. The shapes of amorphous polymer molecules are random, and random walk statistics can be used to describe the distributions of end-to-end vectors that are obtained.

### 13.2 Random Walk in One Dimension

Figure 13.3: Random walk animation
$\begin{array}{cc}n\left(i\right)=2M{\left(\frac{1}{2\pi {N}_{x}}\right)}^{1/2}exp\left(\frac{-{i}^{2}}{2{N}_{x}}\right)& \left(13.1\right)\end{array}$
We can check that the summation is appropriately normalized. In other words, is the following expression valid?
The summation here is over all possible values of $i$ . Note that for a random walk with an even number of steps, only even values of $i$ are possible. Similarly, for a walk with an odd number of steps, only odd values of $i$ are possible. We can get the summation we want by summing over all integer values of $i$ and then dividing by two:
Functions of the form $f\left(x\right)\propto exp-{\left(x/A\right)}^{2}$ (where $A$ is a constant) are called Gaussian functions, and generally describe random processes. The following integrals of a Gaussian function will be very useful for us:
In our case, $A=\sqrt{2{N}_{x}}$, so ${\sum }_{i}n\left(i\right)=M$ and our expression for the Gaussian distribution of random walks is indeed normalized.

### 13.3 Average of a Function

The entire distribution represented by the Gaussian function will be useful to us, but it is still useful to have some averages. We encountered averages already in our discussion of the number and weight average distribution. In general, if $Pr\left(x\right)$ is the probability that a function has a value of $x$, than the average value of $x$ (referred to as $⟨x⟩$) is obtained from the following expression:
$\begin{array}{cc}⟨x⟩=\sum _{x}xPr\left(x\right)& \left(13.5\right)\end{array}$
where the sum is over all possible values of $x$. Similarly, the average of ${x}^{2}$ is given by: $\begin{array}{cc}⟨{x}^{2}⟩=\sum _{x}{x}^{2}Pr\left(x\right)dx& \left(13.6\right)\end{array}$
The procedure can be generalized to calculate the average of any function of $x$:
$\begin{array}{cc}⟨f\left(x\right)⟩=\sum _{x}f\left(x\right)Pr\left(x\right)& \left(13.7\right)\end{array}$
 Suppose I put the following 10 numbers into a drawer: 1, 4, 8, 12, 19, 25, 28, 33, 37, 45 I randomly pick a number from the drawer and then return it. What is the average value of all the numbers that I pick if I continue with this exercise?
For a function of i, where i is only able to take on discrete values, we have:
$=\sum _{i}f\left(i\right)Pr\left(i\right)$
What if we have a function of a continuous variable, $x$? In this case, we must replacePr(i), with $Pr\left(x\right)dx$, where $Pr\left(x\right)dx$ is the probability that the continuous variable has a value between $x$ and $x+dx$. We must also replace the summation with the appropriate integral to obtain the following result:
 Calculate the average value of ${x}^{2}$, assuming that x is equally likely to take on all values between 1 and 10, and that no values outside this range are possible.

### 13.4 Averages for Random Walks

If the length of each step taken by the random walker is $a$, then the distance between the beginning and the end of the walk is simply $i$ times this step length:
${R}_{x}=ia$
We can substitute ${R}_{x}/a$ for $i$ and $d{R}_{x}/a$ for $di$ in Eq. 13.3 to get an expression for the total number of walks $M.$ We'll generalize a bit further and replace the limits of $-\infty$ and $\infty$ for $i$ to limits of ${R}_{min}$ and ${R}_{max}$ for ${R}_{x}$. We obtain the following for $M\left({R}_{min},\phantom{\rule{6px}{0ex}}{R}_{max}\right)$, the total number of walks with values of ${R}_{x}$ between ${R}_{min}$ and ${R}_{max}$:
$\begin{array}{cc}M\left({R}_{min},\phantom{\rule{6px}{0ex}}{R}_{max}\right)=M\stackrel{{R}_{max}}{\underset{{R}_{min}}{\int }}Pr\left({R}_{x}\right)d{R}_{x}& \left(13.9\right)\end{array}$
with the probability density function, $Pr\left({R}_{x}\right)$ given as follows:
For simplicity, it is helpful to define $A=\sqrt{2{N}_{x}{a}^{2}}$ so that we can write $P\left({R}_{x}\right)$ in the following way:
$\begin{array}{cc}⟨{R}_{x}⟩=\stackrel{\infty }{\underset{-\infty }{\int }}{R}_{x}Pr\left({R}_{x}\right)d{R}_{x}=\frac{1}{\sqrt{\pi }A}\stackrel{\infty }{\underset{-\infty }{\int }}{R}_{x}exp-{\left({R}_{x}/A\right)}^{2}& \left(13.12\right)\end{array}$
$\begin{array}{cc}⟨{R}_{x}^{2}⟩=\stackrel{\infty }{\underset{-\infty }{\int }}{R}_{x}^{2}Pr\left({R}_{x}\right)d{R}_{x}==\frac{1}{\sqrt{\pi }A}\stackrel{\infty }{\underset{-\infty }{\int }}{R}_{x}^{2}exp-{\left({R}_{x}/A\right)}^{2}& \left(13.13\right)\end{array}$
Finally, we use the standard integrals to
$\begin{array}{cc}⟨{R}_{x}⟩=0& \left(13.14\right)\end{array}$
$\begin{array}{cc}⟨{R}_{x}^{2}⟩=\frac{{A}^{2}}{2}={N}_{x}{a}^{2}& \left(13.15\right)\end{array}$
We know that the first result has to be true because the Gaussian distribution function is symmetric: positive and negative values of ${R}_{x}$ with equal magnitudes are equally likely. This means that any contribution to $⟨{R}_{x}⟩$ will be exactly offset by a contribution with the opposite sign. For this reason, the mean-square end-to-end distance, $⟨{R}_{x}^{2}⟩$ , or the root mean square end-to-end distance, $⟨{R}_{x}^{2}⟩$ , is typically used to describe the spatial extent of a random walk.

### 13.5 Random Walks in 3 Dimensions

Now we can extend some of these results to describe the shapes of real polymer molecules. We begin with a discussion of two dimensional random walks. The simplest extension is to give our random walker 4 choices for his movements. In addition to moving to the left and right, he can now move up and down, as illustrated in Figure 13.4a. If we move the constraint that there are only 2 directions in which the walker can move, we get the example shown in Figure 13.4b. These trajectories are now beginning to look like the shapes of actual polymer molecules, which is why this whole exercise is relevant.
Figure 13.4: 2-D Random Walk of 1000 steps on a square lattice (a) and with random bond angles (b). The end-to-end vector, $\stackrel{\to }{R}$, is shown in each case.
The value of ${R}_{x}$ that we studied in the one dimensional random walk case is just the x component of the end-to-end vector, which also has y and z components:
$\begin{array}{cc}\stackrel{\to }{R}={R}_{x}\stackrel{ˆ}{x}+{R}_{y}\stackrel{ˆ}{y}+{R}_{z}\stackrel{ˆ}{z}& \left(13.16\right)\end{array}$
where $\stackrel{ˆ}{x}$, $\stackrel{ˆ}{y}$ and $\stackrel{ˆ}{z}$ are unit vectors in the x, y and z directions, respectively. The probability that a given random walk of N steps in three dimensions has an end-to end vector of with x, y and z compenents of ${R}_{x}$, ${R}_{y}$ and ${R}_{z}$ is given by multiplying the probabilities for the individual components:
$\begin{array}{cc}Pr\left({R}_{x},\phantom{\rule{6px}{0ex}}{R}_{y},\phantom{\rule{6px}{0ex}}{R}_{z}\right)=Pr\left({R}_{x}\right)Pr\left({R}_{y}\right)Pr\left({R}_{z}\right)=\left(2\pi N{a}^{2}/3{\right)}^{-3/2}exp\left(\frac{-3{R}^{2}}{2N{a}^{2}}\right)& \left(13.17\right)\end{array}$
where we have used the following relationships:
$\begin{array}{cc}{R}^{2}={R}_{x}^{2}+{R}_{y}^{2}+{R}_{z}^{2}& \left(13.18\right)\end{array}$
and
$\begin{array}{cc}{N}_{x}={N}_{y}={N}_{z}=N/3& \left(13.19\right)\end{array}$
This relationship can be further simplified by recalling that
$Pr\left({R}_{x},{R}_{y},{R}_{z}\right)d{R}_{x}d{R}_{y}d{R}_{z}$ is the probability that x component of the end-to-end vector is between ${R}_{x}$ and ${R}_{x}+dx$, the y component is between ${R}_{y}$ and ${R}_{y}+dy$, and the z component is between ${R}_{z}$ and ${R}_{z}+dz$. Because the probability only depends on $R$ for an isotropic system (as must be the case for a truly random walk), we can substitute $4\pi {R}^{2}dR$ for $d{R}_{x}d{R}_{y}d{R}_{z}$ to obtain:
$\begin{array}{cc}Pr\left(R\right)=\frac{3\left(6{\right)}^{1/2}{R}^{2}}{{\pi }^{1/2}\left(N{a}^{2}{\right)}^{3/2}}exp\left(\frac{-3{R}^{2}}{2N{a}^{2}}\right)=\frac{4.15{R}^{2}}{{R}_{0}^{3}}exp\left(\frac{-3{R}^{2}}{2{R}_{0}^{2}}\right)& \left(13.20\right)\end{array}$
where $Pr\left(R\right)dR$ is the probability that the magnitude of the end-to-end vector lies between $R$ and $R+dR$, and we have used the following definition of ${R}_{0}$:
$\begin{array}{cc}⟨{R}^{2}⟩=\stackrel{\infty }{\underset{0}{\int }}{R}^{2}Pr\left(R\right)dR=\frac{4.15}{{R}_{0}^{3}}\stackrel{\infty }{\underset{0}{\int }}{R}^{4}exp\left(\frac{-3{R}^{2}}{2{R}_{0}^{2}}\right)={R}_{0}^{2}& \left(13.22\right)\end{array}$
We can define a dimensionless probability density $\stackrel{‾}{P}$, which gives the probability that the normalized quantity $R/{R}_{0}$ lies in some range:
$\begin{array}{cc}\stackrel{‾}{Pr}\left(\stackrel{‾}{R}\right)=4.15\left({\stackrel{‾}{R}}^{2}\right)exp\left(-1.5{\stackrel{‾}{R}}^{2}\right)& \left(13.23\right)\end{array}$
Because $\stackrel{‾}{R}$ is a dimensionless ratio, the probability density, $\stackrel{‾}{Pr}\left(\stackrel{‾}{R}\right)$, is also dimensionless. The shape of this function is shown in Figure 13.5.
Figure 13.5: Probability density.
Note that the distribution is normalized: $\stackrel{\infty }{\underset{0}{\int }}\stackrel{‾}{Pr}d\stackrel{‾}{R}=1$.
##### Exercise:
What fraction of molecules in an amorphous polymer have end to end vectors that are within 10% of ${R}_{0}$?
##### Solution:
We need to integrate the probability density from $R/{R}_{0}$=0.9 to $R/{R}_{0}$ = 1.1. With $x$= $R/{R}_{0}$ we have:
$P{r}_{tot}=4.15\stackrel{1.1}{\underset{0.9}{\int }}{x}^{2}exp\left(-1.5{x}^{2}\right)dx$
This corresponds to the shaded area in the following figure:
Numerical integration of the equation above gives $P{r}_{tot}$ = 0.18, which corresponds to the red shaded area. This is the total fraction of molecules with R between 0.9 ${R}_{0}$ and 1.1 ${R}_{0}$.

## 14 Models of Chain Dimensions

$\begin{array}{cc}\stackrel{\to }{R}=\stackrel{{N}_{b}}{\sum _{i=1}}{\stackrel{\to }{r}}_{i}={\stackrel{\to }{r}}_{1}+{\stackrel{\to }{r}}_{2}+{\stackrel{\to }{r}}_{3}+...+{\stackrel{\to }{r}}_{{N}_{b}}& \left(14.1\right)\end{array}$
Figure 14.1: Vector diagram of a polymer.

### 14.1 General Considerations

We are interested in the average magnitude of the end-to-end vector. As discussed in the previous section, the average of $\stackrel{\to }{R}$ itself is not useful, since it just averages to zero. We are interested in the average value of the square of the magnitude of the end-to end vector. This quantity is equal to the dot product of the end-to-end vector with itself:
$\begin{array}{cc}{R}^{2}=\stackrel{\to }{R}\cdot \stackrel{\to }{R}=\left({\stackrel{\to }{r}}_{1}+{\stackrel{\to }{r}}_{2}+{\stackrel{\to }{r}}_{3}+...+{\stackrel{\to }{r}}_{{N}_{b}}\right)\cdot \left({\stackrel{\to }{r}}_{1}+{\stackrel{\to }{r}}_{2}+{\stackrel{\to }{r}}_{3}+...+{\stackrel{\to }{r}}_{{N}_{b}}\right)& \left(14.2\right)\end{array}$
The dot product has ${N}_{b}^{2}$ terms, which can be represented in the following matrix form:
$\begin{array}{cc}{R}^{2}=\left\{\begin{array}{ccccc}{\stackrel{\to }{r}}_{1}\cdot {\stackrel{\to }{r}}_{1}& +{\stackrel{\to }{r}}_{1}\cdot {\stackrel{\to }{r}}_{2}& +{\stackrel{\to }{r}}_{1}\cdot {\stackrel{\to }{r}}_{3}& +...& +{\stackrel{\to }{r}}_{1}\cdot {\stackrel{\to }{r}}_{{N}_{b}}\\ +{\stackrel{\to }{r}}_{2}\cdot {\stackrel{\to }{r}}_{1}& +{\stackrel{\to }{r}}_{2}\cdot {\stackrel{\to }{r}}_{2}& +{\stackrel{\to }{r}}_{2}\cdot {\stackrel{\to }{r}}_{3}& +...& +{\stackrel{\to }{r}}_{2}\cdot {\stackrel{\to }{r}}_{{N}_{b}}\\ +{\stackrel{\to }{r}}_{3}\cdot {\stackrel{\to }{r}}_{1}& +{\stackrel{\to }{r}}_{3}\cdot {\stackrel{\to }{r}}_{2}& +{\stackrel{\to }{r}}_{3}\cdot {\stackrel{\to }{r}}_{3}& +...& +{\stackrel{\to }{r}}_{3}\cdot {\stackrel{\to }{r}}_{{N}_{b}}\\ +...& +...& +...& +...& +...\\ +{\stackrel{\to }{r}}_{{N}_{b}}\cdot {\stackrel{\to }{r}}_{1}& +{\stackrel{\to }{r}}_{{N}_{b}}\cdot {\stackrel{\to }{r}}_{2}& +{\stackrel{\to }{r}}_{{N}_{b}}\cdot {\stackrel{\to }{r}}_{3}& +...& +{\stackrel{\to }{r}}_{{N}_{b}}\cdot {\stackrel{\to }{r}}_{{N}_{b}}\end{array}\right\}& \left(14.3\right)\end{array}$
The dot product between two vectors, $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ is given in general by $|\stackrel{\to }{A}||\stackrel{\to }{B}|cos\theta$ where $\theta$ is the angle between the two vectors, illustrated below in Figure 14.2. At this point we will make our first simplifying assumption, which is that the lengths of all bonds along the backbone are identical. This assumption is certainly valid for vinyl polymers, and other polymers which have only C-C single bonds along the backbone.
Figure 14.2: Definition of the angle between two vectors.
If this bond length is $\ell$, then ${R}^{2}$ can be represented as follows, where ${\theta }_{ij}$ is the angle between bondi and bondj:
$\begin{array}{cc}{R}^{2}={\ell }^{2}\left\{\begin{array}{ccccc}cos{\theta }_{11}& +cos{\theta }_{12}& +cos{\theta }_{13}& +...& +cos{\theta }_{1{N}_{b}}\\ +cos{\theta }_{21}& +cos{\theta }_{22}& +cos{\theta }_{23}& +...& +cos{\theta }_{2{N}_{b}}\\ +cos{\theta }_{31}& +cos{\theta }_{32}& +cos{\theta }_{33}& +...& +cos{\theta }_{3{N}_{b}}\\ +...& +...& +...& +...& +...\\ +cos{\theta }_{{N}_{b}1}& +cos{\theta }_{{N}_{b}2}& +cos{\theta }_{{N}_{b}3}& +...& +cos{\theta }_{{N}_{b}{N}_{b}}\end{array}\right\}& \left(14.4\right)\end{array}$
The average value of R, will be determined by the average values of θ ${}_{ij}$ In mathematical terms, we have the following expression:
$\begin{array}{cc}⟨{R}^{2}⟩={\ell }^{2}\left\{\begin{array}{ccccc}⟨cos{\theta }_{11}⟩& +⟨cos{\theta }_{12}⟩& +⟨cos{\theta }_{13}⟩& +...& +⟨cos{\theta }_{1{N}_{b}}⟩\\ +⟨cos{\theta }_{21}⟩& +⟨cos{\theta }_{22}⟩& +⟨cos{\theta }_{23}⟩& +...& +⟨cos{\theta }_{2{N}_{b}}⟩\\ +⟨cos{\theta }_{31}⟩& +⟨cos{\theta }_{32}⟩& +⟨cos{\theta }_{33}⟩& +...& +⟨cos{\theta }_{3{N}_{b}}⟩\\ +...& +...& +...& +...& +...\\ +⟨cos{\theta }_{{N}_{b}1}⟩& +⟨cos{\theta }_{{N}_{b}2}⟩& +⟨cos{\theta }_{{N}_{b}3}⟩& +...& +⟨cos{\theta }_{{N}_{b}{N}_{b}}⟩\end{array}\right\}& \left(14.5\right)\end{array}$
In more compact notation we can write:
$\begin{array}{cc}⟨{R}^{2}⟩={\ell }^{2}\stackrel{{N}_{b}}{\sum _{i,j=1}}⟨cos{\theta }_{ij}⟩& \left(14.6\right)\end{array}$
So we have reduced the problem to figuring out what the values of $⟨cos{\theta }_{ij}⟩$ are in the system. This requires a specific model of how the bonds are joined to one another. In the following sections we consider two specific models (the freely jointed and freely rotating models), and a general model where the bond rotation angles are constrained.

### 14.2 Freely Jointed Chain Model

The simplest (and most unrealistic) model of chain dimensions is the freely jointed chain model, where all bond orientations are equally likely. In the freely jointed chain model $⟨cos{\theta }_{ij}⟩=0$ for $i\ne j$, because orientations giving positive and negative values of $cos{\theta }_{ij}$ are equally likely. The only exception is for the ${N}_{b}$ diagonal terms for which $i=j$. These represent the projection of a vector onto itself, and all have $\theta =0$ and $cos\theta =1$.
$\begin{array}{cc}⟨{R}^{2}⟩={N}_{b}{\ell }^{2}& \left(14.7\right)\end{array}$

### 14.3 Freely Rotating Chain Model

In this model, bond angles are fixed, but rotation about bonds is possible. Rotation about a bond sweeps out a cone as shown below. The bond rotation angle is defined as $\phi$, and in the freely rotating model all values of $\phi$ are assumed to be equally likely.
Figure 14.3: Freely Rotating Chain Models with views perpendicular to the axis of the rotating bond and along the axis of the rotating bond
In the freely rotating chain model, the angle between adjacent bonds is fixed at certain angle, referred to here simply as $\theta$ (without the subscripts). Adjacent bonds,i.e. those with $i-j=1$ or $i-j=-1$, have $⟨cos{\theta }_{ij}⟩=cos\theta$. It can be shown that terms with $i-j$=2 or -2 have $⟨cos{\theta }_{ij}⟩={cos}^{2}\theta$, terms with $i-j$=3 or -3 have $⟨cos{\theta }_{ij}⟩=cos\theta$, etc. The matrix of terms making up the summation to give $⟨{R}^{2}⟩$ therefore has the following form:
$\begin{array}{cc}⟨{R}^{2}⟩={\ell }^{2}\left\{\begin{array}{ccccc}1& +cos\theta & +{cos}^{2}\theta & +...& +{cos}^{\left({N}_{b}-1\right)}\theta \\ +cos\theta & +1& +cos\theta & +...& +{cos}^{\left({N}_{b}-2\right)}\theta \\ +{cos}^{2}\theta & +cos\theta & +1& +...& +{cos}^{\left({N}_{b}-3\right)}\theta \\ +...& +...& +...& +...& +...\\ +{cos}^{\left({N}_{b}-1\right)}\theta & +{cos}^{\left({N}_{b}-2\right)}\theta & +{cos}^{\left({N}_{b}-3\right)}\theta & +...& +1\end{array}\right\}& \left(14.8\right)\end{array}$
Of the ${N}_{b}^{2}$ terms in the summation, ${N}_{b}$ terms (the diagonal terms with $i=j$) are equal to 1. There are ${N}_{b}-1$ terms with $i-j$=1, and an additional ${N}_{b}$-1 terms with $i-j$=-1. There are therefore 2( ${N}_{b}$-1) terms with a magnitude equal to $cos\theta$. Similarly, there are 2( ${N}_{b}$ -2) terms equal to $cos{}^{2}\theta$, 2( ${N}_{b}$ -3) terms equal to $cos{}^{3}\theta$, etc. The expression for $⟨{R}^{2}⟩$ can therefore be written as follows:
$⟨{R}^{2}⟩={\ell }^{2}\left\{{N}_{b}+2\left({N}_{b}-1\right)cos\theta +2\left({N}_{b}-2\right)\left(cos\theta {\right)}^{2}+...+2{\left(cos\theta \right)}^{{N}_{b}-1}\right\}$
In the more compact notation, we have:
Equation 14.9 is an exact expression that is valid for any value of ${N}_{b}$. Unfortunately, it's not a very useful equation. Fortunately, we can get a greatly simplified expression that is very nearly exact for any reasonably large value of ${N}_{b}$. We are able to do this because $cos\theta$ is less than one, so that $\left(cos\theta {\right)}^{i}$ rapidly decreases asi increases. For example, for $\theta =7{1}^{○}$ (the value corresponding to C-C single bonds), $\left(cos\theta {\right)}^{i}$ = $1.3x1{0}^{-5}$ for $i=10$. For large values of ${N}_{b}$, we only need to consider contributions from values of i which are much smaller than ${N}_{b}$. A very good approximation is obtained by substituting ${N}_{b}$ for ${N}_{b}-i$ and extending the sum to $i=\infty$, which leads to the following:
$\begin{array}{cc}⟨{R}^{2}⟩={N}_{b}{\ell }^{2}\left\{1+2\underset{i=1}{\sum ^{\infty }}{\left(cos\theta \right)}^{i}\right\}& \left(14.10\right)\end{array}$
At this point we are in a position to use the following standard sum, valid for any value of $x$ with an absolute magnitude less than 1:
$\begin{array}{cc}\stackrel{\infty }{\sum _{i=1}}{x}^{i}=\frac{x}{1-x}& \left(14.11\right)\end{array}$
Use of this expression with $x=cos\theta$ gives (after a little algebraic rearrangement):
$\begin{array}{cc}⟨{R}^{2}⟩={N}_{b}{\ell }^{2}\left\{\frac{1+cos\theta }{1-cos\theta }\right\}& \left(14.12\right)\end{array}$
For singly bonded carbon along the backbone, $\theta =7{1}^{○}$ and $⟨{R}^{2}⟩=2{N}_{b}{\ell }^{2}$.

### 14.4 Bond Angle Restrictions

The freely rotating model is overly simplistic, because all bond rotation angles cannot be equally likely. At certain angles, functional groups attached to adjacent carbon atoms are too close to one another, as illustrated here.
We know that the freely rotating chain model must also be overly simplistic, because some bond rotation angles (those corresponding to the gauche and trans configurations, for example) are more likely than others. In this case $⟨{R}^{2}⟩$ is given by the following expression, which we present here without proof.

### 14.5 Characteristic Ratio

##### Exercise:
What would the characteristic ratio be for polystyrene if 70% of the bonds were in trans configurations and the remaining 30% were in gauche configurations?
##### Solution:
We use Eq. 14.15 for ${C}_{\infty },$with $\theta =7{1}^{○}$, the correct bond angle for singly bonded carbon, which we have along the polystyrene backbone. Gauche bonds have $\phi =±12{0}^{○}$ ( $cos\phi =-0.5\right)$ trans bonds have $\phi =0$ ( $cos\phi =1\right)$. With our assumed distribution of gauche and trans bonds we obtain the following for $⟨cos\phi ⟩$:
$⟨cos\phi ⟩=0.7\left(1\right)+0.3\left(-0.5\right)=0.55$
This gives ${C}_{\infty }=6.9$, which is actually a pretty reasonable value. Most polymers with singly bonded carbons along the backbone have a value for ${C}_{\infty }$ that is between 6 and 10.
##### Exercise:
How is the statistical segment length of a vinyl polymer related to ${C}_{\infty }$?
##### Solution:
The characteristic ratio and statistical segment length are both defined in terms of an expression for R2. All we need to do is equate these two expressions (Eqs.13.21 and14.14):
$\begin{array}{cc}{C}_{\infty }{N}_{b}{\ell }^{2}=N{a}^{2}& \left(14.16\right)\end{array}$
The repeat unit for a vinyl polymer contains two C-C single bonds along the backbone, so ${N}_{b}=2N$. Also, single bonds between carbon atoms have a bond length of 1.54 Å. The statistical segment length is therefore related to ${C}_{\infty }$ as follows:
$\begin{array}{cc}a=2.18Å\sqrt{{C}_{\infty }}& \left(14.17\right)\end{array}$

### 14.6 Self-Avoiding Random Walks

Figure 14.6: Random walk diagram
An isolated polymer molecule in a solvent no longer obeys random walk statistics. If the polymer is surrounded by solvent molecules, these surroundings no longer look the same as other parts of the same molecule. The chain will adopt shapes that avoid direct polymer/polymer contacts, and the end-to-end distance will be larger than the value predicted by random walk models. The rms end-to-end distance increases with the 0.6 power of the chain length, instead of the 0.5 power that was obtained for random walk statistics:
Note: The value of ${a}_{s}$ is not exactly the same as the statistical segment length describing chain dimensions in pure polymer melts, but it will be close to it.

### 14.7 Deformation of a Single Polymer Molecule

Figure 14.7: Single strand deformation
$\begin{array}{cc}\Omega \left(\stackrel{\to }{R}\right)=Cexp\left(-1.5{\left(R/{R}_{0}\right)}^{2}\right)& \left(14.19\right)\end{array}$
The entropy is obtained from the number of possible molecular shapes:
$\begin{array}{cc}{S}_{d}\left(\stackrel{\to }{R}\right)={k}_{B}ln\Omega \left(\stackrel{\to }{R}\right)={k}_{B}lnC-\frac{3{k}_{B}{R}^{2}}{2N{a}^{2}}& \left(14.20\right)\end{array}$
The free energy is assumed to be entirely dominated by this entropic contribution:
Note: In reality, there is a small enthalpic contribution to the stretching free energy as well, because stretching the chain increases the relative proportion of higher energy gauche bonds, but we're not going to worry about that correction here. Because we are only every interested in changes in the free energy, constant terms that don't depend on $R$ are going to drop out when we compare free energies, so the detailed value of this constant is not going to matter.
Figure 14.8: Free energy of deformation
Figure 14.9: Force of deformation
Figure 14.10: Generalized deformation
If the end-to-end vector is not directed purely along the x-axis, we need to consider the effects of the extensions in other directions as well. Because the free energy scales with the square of the separation, only the magnitude of the end-to-end vector matters. This expression for the elastic free energy of an individual molecule in terms of its end-to-end distance is remarkably simple. It is one of the most important and widely used results in polymer science. In the following section, we will see how it is used to obtain estimates for the elastic modulus of a crosslinked elastomer.

## 15 Elasticity

In this section, we discuss the mechanical properties of polymers, defined as the relationships which are observed between an applied stress state and the resulting strain. Are focus in this course is on elasticity and viscoelasticity. We begin with a refresher on the fundamental definitions of stress and strain, and then discuss the three most important experimental geometries for mechanical measurements of soft materials.

### 15.1 Fundamental Definitions

#### 15.1.1 Specification of the State of Stress

The two dimensional stress state is illustrated below in Figure 15.1. In two dimensions there are 3 independent components of the stress tensor. Two of these are normal forces where the stress is perpendicular to the relevant surface. These are ${\sigma }_{xx}$ and ${\sigma }_{yy}$, with forces acting in the x and y directions, respectively. There is also a shear stress, ${\sigma }_{xy}$, which act parallel to each of the 4 surfaces of a square chunk of material. Note that the magnitudes of the shear forces acting on the 4 different surfaces are all equal to one another, a fact imposed by the requirement that there must be no net force or torques acting on this overall square of material (otherwise the material would be accelerating in response to this net force). The values of these three stress components depend on the way we define the orientation of the coordinate system. It is always possible to define a coordinate system so that the shear components disappear, and we are only left with normal stresses. These are the
principal stresses
, which in two dimensions are ${\sigma }_{1}^{p}$ and ${\sigma }_{2}^{p}$. The same holds for a 3-dimensional stress state: we can always find a coordinate system for any given stress state so that only normal stresses. In three dimensions we have three principal stresses, ${\sigma }_{1}^{p}$, ${\sigma }_{2}^{p}$ and ${\sigma }_{3}^{p}$. These act in the three, mutually perpendicular principal stress directions as illustrated in Figure 15.2. In summary, any state of stress is completely specified by the three principal stresses ( ${\sigma }_{1}^{p}$, ${\sigma }_{2}^{p}$, ${\sigma }_{3}^{p}$), and the orientation of the coordinate system used to define these principal stresses.
Figure 15.1: Representation of a two-dimensional stress state.

#### 15.1.2 Extension Ratios and the Strain Ellipsoid

A similar concept of principal axes applies to the strain state as well. For polymeric systems (and elastomers in particular), we need a description that can be used to describe both small and large strains. We the concept of the
strain ellipsoid
, illustrated in Figure 15.3. An undeformed sphere of radius ${r}_{0}$ is deformed into an ellipsoid with principal radii of ${r}_{0}{\lambda }_{1}$, ${r}_{0}{\lambda }_{2}$ and ${r}_{0}{\lambda }_{3}$. Here ${\lambda }_{1}$, ${\lambda }_{2}$ and ${\lambda }_{3}$ are the
principal extension ratios
. The values of these principal extension ratios, along with the orientation of the principal axes of the strain ellipsoid, completely specify the state of strain.

#### 15.1.3 Uniaxial Extension or Compression

A common mechanical testing geometry corresponds to the application of a single normal force to the sample, as illustrated in Figure 15.4. For our purposes we define the coordinate system so that the normal force is applied in the z direction, and we refer to this force as ${P}_{z}$. We also assume that our sample has isotropic mechanical properties and is in the shape of a cylinder. The result of the applied force is to increase the length of of the sample from ${\ell }_{0}$ to $\ell$, and to decrease it's diameter from ${d}_{0}$ to $d$.
$\begin{array}{cc}{P}_{z}d\ell =d\left(\Delta {F}_{d}\right)& \left(15.1\right)\end{array}$
We can write this force in terms of ${\lambda }_{z}$, the extension ratio in this direction:
$\begin{array}{cc}{P}_{z}=\frac{d\left(\Delta {F}_{d}\right)}{d\ell }=\frac{d\left(\Delta {F}_{d}\right)}{d{\lambda }_{z}}\frac{d{\lambda }_{z}}{d\ell }& \left(15.2\right)\end{array}$
With ${\lambda }_{z}=\ell /{\ell }_{0}$, we have $\frac{d{\lambda }_{z}}{d\ell }=1/{\ell }_{0}$. We also have ${V}_{0}={A}_{0}{\ell }_{0}$, so and we obtain the following for the engineering stress, ${\sigma }_{eng}$:
$\begin{array}{cc}{\sigma }_{eng}=\frac{{P}_{z}}{{A}_{0}}=\frac{d}{d{\lambda }_{z}}\left(\frac{\Delta {F}_{d}}{{V}_{0}}\right)& \left(15.3\right)\end{array}$
Here ${A}_{0}$, ${\ell }_{0}$ and ${V}_{0}$ are the cross sectional area, length and volume of the undeformed material.
The deformation free energy normalized by the undeformed volume is the deformation free energy density, $\Delta {f}_{d}$:
$\begin{array}{cc}\Delta {f}_{d}=\frac{\Delta {F}_{d}}{{V}_{0}}& \left(15.4\right)\end{array}$
With these definitions the following simple expression gives the engineering stress in terms of the deformation free energy density, for uniaxial deformation along the z axis:

#### 15.1.4 Simple Shear

In a simple shear deformation, two parallel surfaces within a material remain at a fixed separation as they slide past one another, as illustrated in Figure 15.5. In this example an applied shear force, ${P}_{xy}$ is applied to the two surfaces, which each have an area, $A.$ The separation between the two surfaces is $h$, and the relative displacement of the two surfaces is $u$. For these quantities we obtain the shear sress, ${\sigma }_{xy}$, the engineering shear strain, ${e}_{xy}$ or $\gamma$, and the shear modulus, $G$:
Note that ${e}_{xy}$ and $\gamma$ are both commonly used symbols to represent the shear strain.
The force is again given by the derivative of the free energy with respect to the displacement in the direction of the applied force:
$\begin{array}{cc}{P}_{xy}=\frac{d\left(\Delta {F}_{d}\right)}{du}=\frac{d\left(\Delta {F}_{d}\right)}{d\gamma }\frac{d\gamma }{du}=\frac{1}{h}\frac{d}{d\gamma }\left(\Delta {F}_{d}\right)& \left(15.6\right)\end{array}$
The shear stress is then obtained by dividing by the cross-sectional area (note: $hA=V$, ${\nu }_{el}$ = ${n}_{el}$/V, ${\sigma }_{xy}$ = ${P}_{xy}/A$:
One complicating factor of simple shear in comparison to uniaxial extension or compression is that the strain ellipsoid is rotated continuously during the shear experiment. As a result we cannot connect the principal axes of the coordinate system in a straightforward way. The situation for shear in the x-y plane is illustrated in Figure 15.6. Shear is a two-dimensional stress state, so the material is undeformed in the third direction, in this case the z direction, so ${\lambda }_{z}=1.$ The principal strains in the x-y plane are ${\lambda }_{1}$ and ${\lambda }_{2}$. While the relationship between the orientation of the strain ellipsoid is a complicated one, the shear strain itself is given by the difference between ${\lambda }_{1}$ and ${\lambda }_{2}$:
$\begin{array}{cc}\gamma ={e}_{xy}={\lambda }_{1}-{\lambda }_{2}& \left(15.8\right)\end{array}$
Eq. 15.8 will enable us to determine the deformation free energy of a material undergoing a shear deformation, and is used below in Section15.2.5.
Figure 15.6: Extension ratios for simple shear.

#### 15.1.5 Torsion

An important geometry for characterizing soft materials is the torsional geometry shown in Figure 15.7. In this material a cylindrical or disk-shaped material is twisted about an axis of symmetry. The material could be a long, thin fiber (Figure 15.7a) or a flat disk sandwiched between two plates (Figure 15.7b). We obtain the shear modulus by looking at the torsional stiffness of mateiral, $i.e.$, the Torque, $T$ required to rotate the top and bottom of the fiber by an angle ${\theta }_{0}$.
Figure 15.7: Fiber torsion.
We define a cylindrical system with a z axis along the fiber axis. The other axes in this coordinate system are the distance $r$ from this axis of symmetry, and the angle $\theta$ around the z axis. The shear strain in the $\theta -z$ plane depends only on $r$, and is given by:
$\begin{array}{cc}{e}_{\theta z}=r\frac{d\theta }{dz}=r\frac{{\theta }_{0}}{\ell }& \left(15.9\right)\end{array}$
The corresponding shear stress is obtained by multiplying by the shear modulus, $G$:
$\begin{array}{cc}{\sigma }_{\theta z}=Gr{\theta }_{0}/\ell & \left(15.10\right)\end{array}$
We integrate the shear stress to give the torque, $T$:

### 15.2 Rubber Elasticity

Crosslinked rubbers are unique in that thermodynamic arguments can be used to predict their elastic moduli with remarkable accuracy. Our starting point will be a description of the free energy of an elastomer as a function of its deformation. This free energy is dominated by entropic contributions arising from restrictions on the number of different conformations (or shapes) that polymer strands are able to adopt. The detailed descriptions of polymer chain statistics given earlier were developed so that we would be in a position to describe the mechanical properties of rubbery materials. The pages that follow are proof that statistics and thermodynamics are actually useful!!

#### 15.2.1 Molecular Deformation

Different types of molecular deformations, and the characteristic lengths and forces, are illustrated here.
Figure 15.8: Various types of molecular deformation

#### 15.2.2 Free energy of a stretched rubber

When a rubber is stretched, the free energy increases. In general the free energy increase has enthalpic and entropic contributions ( ${F}_{d}={H}_{d}-T{S}_{d}$). The basic assumption of rubber elasticity theory is that the free energy increase due to deformation is dominated by the decrease in the entropy. In other words $|T\Delta {S}_{d}|>>|\Delta {H}_{d}|$. The second assumption we will make is that the deformation at a microscopic level mimics the deformation at a macroscopic level. In other words, relative changes in the spacings between crosslink points are identical to relative changes in the overall sample demonstrations. This assumption is referred to as the affine deformation assumption, and is illustrated on the following page.

#### 15.2.3 Affine Deformation Assumption

Figure 15.9: A chunk of rubber
The block to the left represents a chunk of rubber. The black lines are individual network strands , and the red circles are crosslink points that tie these strands together. If the molecules are deforming "affinely", the distances between crosslink points change by ratios that are identical to the extension ratios of the material as a whole.

#### 15.2.4 Free energy change due to deformation

Figure 15.10: Contrasting free energies before and after deformation
$\begin{array}{cc}{F}_{0}=-{k}_{B}TlnC+\frac{3{k}_{B}T}{2N{a}^{2}}\left\{{R}_{x}^{2}+{R}_{y}^{2}+{R}_{z}^{2}\right\}& \left(15.12\right)\end{array}$
After deformation, the values of ${R}_{x}$, ${R}_{y}$ and ${R}_{z}$ are each multiplied by the respective extension ratio ( $\stackrel{\to }{R}={\lambda }_{x}{R}_{x}\stackrel{ˆ}{x}+{\lambda }_{y}{R}_{y}\stackrel{ˆ}{y}+{\lambda }_{z}{R}_{z}\stackrel{ˆ}{z}$), so that ${R}^{2}={\lambda }_{x}^{2}{R}_{x}^{2}+{\lambda }_{y}^{2}{R}_{y}^{2}+{\lambda }_{z}^{2}{R}_{z}^{2}$. The deformed free energy, ${F}_{def}$ is:
$\begin{array}{cc}{F}_{def}=-{k}_{B}TlnC+\frac{3{k}_{B}T}{2N{a}^{2}}\left\{{\lambda }_{x}^{2}{R}_{x}^{2}+{\lambda }_{y}^{2}{R}_{y}^{2}+{\lambda }_{z}^{2}{R}_{z}^{2}\right\}& \left(15.13\right)\end{array}$
The free energy change due to deformation of the molecule is given as follows:
$\begin{array}{cc}\Delta {F}_{d}={F}_{def}-{F}_{0}=\frac{3{k}_{B}T\left\{\left({\lambda }_{x}^{2}-1\right){R}_{x}^{2}+\left({\lambda }_{y}^{2}-1\right){R}_{y}^{2}+\left({\lambda }_{z}^{2}-1\right){R}_{z}^{2}\right\}}{2N{a}^{2}}& \left(15.14\right)\end{array}$
This result is for the deformation of a single polymer molecule, which for a crosslinked elastomer corresponds to a segment that connects crosslink points. There are a huge number of these segments in a macroscopic chunk of rubber. To get the free energy change for the material as a whole, we need to replace ${R}_{x}^{2}$ by ${n}_{el}⟨{R}_{x}^{2}⟩$, where ${n}_{el}$ is the total number of nework strands and $⟨{R}_{x}^{2}⟩$ is the average value of ${R}_{x}^{2}$ for these segments. We need to make similar substitutions for ${R}_{y}^{2}$ and ${R}_{z}^{2}$ to obtain the following:
$\begin{array}{cc}\Delta {F}_{d}=\frac{3{k}_{B}T{n}_{el}\left\{\left({\lambda }_{x}^{2}-1\right)⟨{R}_{x}^{2}⟩+\left({\lambda }_{y}^{2}-1\right)⟨{R}_{y}^{2}⟩+\left({\lambda }_{z}^{2}-1\right)⟨{R}_{z}^{2}⟩\right\}}{2N{a}^{2}}& \left(15.15\right)\end{array}$
Now we assume that the material was isotropic when it was crosslinked, so that $⟨{R}_{x}^{2}⟩=⟨{R}_{y}^{2}⟩=⟨{R}_{z}^{2}⟩=⟨{R}^{2}⟩/3$. With this assumption, and with ${R}_{0}^{2}-N{a}^{2}$, we obtain the following result:
$\begin{array}{cc}\Delta {F}_{d}=\frac{{k}_{B}T{n}_{el}⟨{R}^{2}⟩\left\{{\lambda }_{x}^{2}+{\lambda }_{y}^{2}+{\lambda }_{z}^{2}-3\right\}}{2{R}_{0}^{2}}& \left(15.16\right)\end{array}$
Note that $⟨{R}^{2}⟩$ is the mean square end-to-end distance of the polymer strands that span the crosslinks, and that ${R}_{0}^{2}$ is the value of $⟨{R}^{2}⟩$ when the polymer strands obey random walk statistics. The relationship between these two quantities depends on the conditions of the crosslinking reaction.
It is often useful to work in terms of intensive free energy changes (free energy per unit volume). The free energy of deformation per unit volume $\Delta {f}_{d}$ is obtained very simply from $\Delta {F}_{d}$ by dividing by $V$, the volume of a sample. We retain the same expression as shown above, but with the strand concentration, ${\nu }_{el}$ substituted for the number of strands, ${n}_{el}$:
$\begin{array}{cc}\Delta {f}_{d}=\frac{\Delta {F}_{d}}{V}=\frac{{k}_{B}T{v}_{s}\beta \left\{{\lambda }_{x}^{2}+{\lambda }_{y}^{2}+{\lambda }_{z}^{2}-3\right\}}{2}& \left(15.17\right)\end{array}$
where ${\nu }_{el}$ and $\beta$ are defined as follows:
$\begin{array}{cc}{\nu }_{el}=\frac{{n}_{el}}{V};\phantom{\rule{6px}{0ex}}\phantom{\rule{6px}{0ex}}\beta =\frac{⟨{R}^{2}⟩}{{R}_{0}^{2}}& \left(15.18\right)\end{array}$

#### 15.2.5 Shear deformation of an elastomer

$\begin{array}{cc}\Delta {f}_{d}=\frac{{k}_{B}T{v}_{s}\beta \left\{{\lambda }_{1}^{2}+{\lambda }_{2}^{2}-2\right\}}{2}=\frac{{k}_{B}T{v}_{s}\beta \left({\lambda }_{1}-{\lambda }_{2}{\right)}^{2}}{2}=\frac{{k}_{B}T{v}_{s}\beta {\gamma }^{2}}{2}& \left(15.20\right)\end{array}$
The shear stress is obtained by differentiation of the strain energy function with respect to $\gamma$ (Eq. 15.7):
$\begin{array}{cc}{\sigma }_{xy}=\frac{d}{d\gamma }\left(\Delta {f}_{d}\right)=G\gamma & \left(15.21\right)\end{array}$
where $G$ is the shear modulus, given as follows:
$\begin{array}{cc}G={k}_{B}T{v}_{s}\beta & \left(15.22\right)\end{array}$
Note that the shear stress is proportional to the shear strain, even for large values of the strain.

#### 15.2.6 Uniaxial deformation of an elastomer

For uniaxial extension or compression the deformation is applied along one axis, which we define as the z axis. We assume that our material is isotropic, so that the extensions in the x and z directions are identical to one another, i.e. ${\lambda }_{x}={\lambda }_{y}$. In addition, the material is assumed to be incompressible, so ${\lambda }_{x}{\lambda }_{y}{\lambda }_{z}$ = 1. We therefore have ${\lambda }_{x}={\lambda }_{y}={\lambda }_{z}^{-1/2}$ . We can therefore write the free energy of a deformed elastomer as a function of the single extension ratio, ${\lambda }_{z}$:
$\begin{array}{cc}\Delta {f}_{d}=\frac{G}{2}\left\{{\lambda }_{z}^{2}+\frac{2}{{\lambda }_{z}}-3\right\}& \left(15.23\right)\end{array}$
$\begin{array}{cc}{\sigma }_{eng}=\frac{d}{\partial d\lambda }\left(\Delta {f}_{d}\right)=G\left\{{\lambda }_{z}-\frac{1}{{\lambda }_{z}^{2}}\right\}& \left(15.24\right)\end{array}$
Young's modulus (E) is defined as the derivative of the stress with respect to strain, evaluated at low strain ( ${\lambda }_{z}=1$):
$\begin{array}{cc}E\equiv \frac{\partial {\sigma }_{eng}}{\partial {\lambda }_{z}}{|}_{{\lambda }_{z}=1}=G{.\left\{1+\frac{2}{{\lambda }_{z}^{3}}\right\}|}_{{\lambda }_{z}=1}=3G=3{v}_{s}\beta {k}_{B}T,& \left(15.25\right)\end{array}$
The concentration of network strands ( ${v}_{s}$) is the inverse of the volume per strand. This can be calculated from the molecular weight and density. Defining ${M}_{s}$ as the number average molecular weight of a network strand (molecular weight between crosslinks) gives:
$\begin{array}{cc}{v}_{s}\left(\frac{strands}{volume}\right)=\frac{\rho \left(mass/volume\right)}{{M}_{s}\left(mass/mole\right)}\cdot {N}_{av}\left(strands/mole\right)& \left(15.26\right)\end{array}$
where ${N}_{av}$ is Avogadro's number ( $6.02x1{0}^{23}$). With $R={k}_{B}{N}_{av}=8.314\phantom{\rule{6px}{0ex}}J/mole\cdot K$ the expression
$E=\frac{3\rho \beta RT}{{M}_{s}}$
Finally, note that 1 J/m ${}^{3}$ = 1 Pa. Stresses and elastic moduli have units of force/area or energy/volume.
##### Exercise:
Calculate the expected value of Young's modulus for a crosslinked polyisoprene that has a number average molecular between crosslinks of 4000 g/mole. The density of polyisoprene is 0.9 g/cm ${}^{3}$. Assume that the polymer was crosslinked under equilibrium conditions in the melt state.
##### Solution:
Because the polymer was crosslinked under equilibrium conditions, the network strands obey random walk statistics, with  = 1. To keep everything in SI units, we need the density in units of kg/m ${}^{3}$ (1 g/ $c{m}^{3}$ = 1000 kg/ ${m}^{3}$), and ${M}_{s}$ in kg/mole. In our example, $\rho$ = 900 kg/ ${m}^{3}$ and ${M}_{s}$ = 4 kg/mole. We'll also assume that we are interested in the elastic modulus near room temperature (T $\approx$ 300 K).
$E=\frac{3\left(900\phantom{\rule{6px}{0ex}}kg/{m}^{3}\right)\left(8.314J/mole-K\right)\left(300K\right)}{4kg/mole}=1.6x1{0}^{6}\phantom{\rule{6px}{0ex}}J/{m}^{3}.$

## 16 Viscoelasticity

Many of the concepts of rubber elasticity still apply under certain situations to materials that are not actually crosslinked. Consider, for example, the behavior of Silly-Putty. Silly-putty is based on silicones that are not actually crosslinked. Nevertheless, this material still bounces like an ordinary crosslinked rubber ball. This is because over very short times it behaves elastically. The deformation energy stored in the material as it comes into contact with the surface and deforms is available to propel the material back into the air as the deformation relaxes and this strain energy is converted to kinetic energy. If we let the sample sit on a surface for a long period of time, however, it eventually flows and behaves as a liquid.
##### Solids:
In this case the stress is proportional to the strain, and the rate at which the strain is applied does not matter. As a result the time dependence of the stress looks just like the time dependence of the strain. The slopes of the time-dependent stress and strain curves (the top two curves in Figure 16.2) are related to one another by the shear modulus of the material.
##### Liquids:
In liquids the stress is proportional to the rate at which the strain is applied, and is independent of the current strain. Liquids do not store any strain energy, and as soon as the strain stops changing, the stress drops back to zero. The stress for the time dependent strain shown in Figure 16.2 is constant while the stress is increasing and is zero otherwise. The stress is given by the shear viscosity , $\eta$, as follows:
$\begin{array}{cc}{\sigma }_{xy}=\eta \frac{d\gamma }{dt}=\eta \stackrel{˙}{\gamma }& \left(16.1\right)\end{array}$
##### Viscoelastic Materials:
Viscoelastic materials have characteristics of both solids and liquids. The shear stress depends on both the shear stress and the strain rate, and is not a simple function of either one. The shear stress actually depends on the details of the previous strain history, as described in more detail below.

### 16.1 Relaxation Modulus

$\begin{array}{cc}G\left(t\right)\equiv \frac{{\sigma }_{xy}\left(t\right)}{{\gamma }_{0}}& \left(16.2\right)\end{array}$The shear modulus is independent of ${\gamma }_{0}$, provided that ${\gamma }_{0}$ is sufficiently small. This low strain regime, referred to as the linear viscoelastic regime is the focus of our remaining discussion of viscoelasticity.
Figure 16.3: Application of an instantaneous shear strain.

### 16.2 Boltzmann Superposition Principle

In the linear viscoelastic regime the effects of strains applied at different times are additive, a concept known as the
Boltzmann superposition principle
. A simple example is illustrated in Figure 16.4, where the following step strains are applied to the sample:
• A strain of $\Delta {\gamma }_{1}$ applied at time ${t}_{1}$
• A strain of $\Delta {\gamma }_{2}$ applied at time ${t}_{2}$
• A strain of $\Delta {\gamma }_{3}$ applied at time ${t}_{3}$
Figure 16.4: Material response to a series of step strains.
The stress contribution from these different strains is obtained by multiplying by the shear relaxation modulus, evaluated at the time that has passed since the strain was applied. If strain was applied at ${t}_{1}$, for example, the stress at some later time, $t$, is obtained by multiplying by $G\left(t-{t}_{1}\right)$. Adding up the contributions from the three different step strains in our example leads to the following::
$\begin{array}{cc}{\sigma }_{xy}=\Delta {\gamma }_{1}G\left(t-{t}_{1}\right)+\Delta {\gamma }_{2}G\left(t-{t}_{2}\right)+\Delta {\gamma }_{3}G\left(t-{t}_{3}\right)& \left(16.3\right)\end{array}$
This expression can easily be extended to include an arbitrary number of step strains:
$\begin{array}{cc}{\sigma }_{xy}=\sum _{i}\Delta {\gamma }_{i}G\left(t-{t}_{i}\right)& \left(16.4\right)\end{array}$
By taking a sum of very small strain increments, we can generalize this expression to account for a continuously changing strain. We begin by writing $\Delta {\gamma }_{i}$ in the following way:
$\begin{array}{cc}\Delta {\gamma }_{i}=\frac{d\gamma }{dt}\left({t}_{i}\right)\Delta {t}_{i}=\stackrel{˙}{\gamma }\left({t}_{i}\right)\Delta {t}_{i}& \left(16.5\right)\end{array}$
This substitution leads to the following expression for the stress:
$\begin{array}{cc}{\sigma }_{xy}=\sum _{i}\stackrel{˙}{\gamma }\left({t}_{i}\right)G\left(t-{t}_{i}\right)\Delta {t}_{i}& \left(16.6\right)\end{array}$
We can write this in integral form by taking $\Delta {t}_{i}\to 0$ and replacing the summation by an integral over all times less than the current time:
$\begin{array}{cc}{\sigma }_{xy}\left(t\right)=\stackrel{t}{\underset{-\infty }{\int }}\stackrel{˙}{\gamma }\left({t}_{i}\right)G\left(t-{t}_{i}\right)d{t}_{i}& \left(16.7\right)\end{array}$
This expression gives the current stress that remains as a result of all of the strains introduced at different times in the past. It is often convenient to change variables so that $s=t-{t}_{i}$, $d{t}_{i}=-ds$. Note that s = $\infty$ when ${t}_{i}=-\infty$, and $s=0$ when ${t}_{i}=t$. The integral can therefore be rewritten as follows:
Note that because $s$ represents time, $G\left(s\right)$ and $G\left(t\right)$ represent the same time-dependent relaxation modulus. We are just using different variables ( $s$ and $t$) to represent time.
The simplest application of Eq. is 'steady shear', where a constant strain rate is applied. Because $\stackrel{˙}{\gamma }$ is independent of time in this case, it can be brought outside the integral, giving the following:
$\begin{array}{cc}{\sigma }_{xy}=\stackrel{˙}{\gamma }\stackrel{\infty }{\underset{0}{\int }}G\left(t\right)dt& \left(16.9\right)\end{array}$
The following expression is obtained for the viscosity:

### 16.3 Idealized Relaxation Curves

One thing to keep in mind when looking at the viscoelastic properties of materials is that processes occur over a very large range of time scales. To capture all of these time scales we typically plot the time on a logarithmic scale. The same is true for the frequency-domain experiments discussed below. The relaxation modulus can often vary of several orders of magnitude, so that we also plot the relaxation modulus itself on a logarithmic scale.
$\begin{array}{cc}G\left(t\right)={G}_{g}exp\left(-t/{\tau }_{g}\right)& \left(16.11\right)\end{array}$This relaxation time could, for example, describe the decay in the modulus from a glassy value of $\approx 1{0}^{9}$ Pa. This relaxation time can be viewed as a 'glass transition time', and its existence is a clue that some equivalence between time and temperature must exist. The relaxation time characterizing the glass transition depends very strongly on the temperature, but is independent of the polymer molecular weight. If the polymer molecular weight is very high, however, the polymers become entangled with one another, and behave elastically for times that are too short for these entanglements (shown schematically in Figure 16.6) to relax by molecular diffusion. In addition to ${\tau }_{g}$, there is a second transition time, ${\tau }_{e}$, determined by the lifetime of these molecular entanglements. Addition of this second relaxation results in the following expression for the relaxation modulus:
$\begin{array}{cc}G\left(t\right)={G}_{g}exp\left(-t/{\tau }_{g}\right)+{G}_{e}exp\left(-t/{\tau }_{e}\right)& \left(16.12\right)\end{array}$
Here ${G}_{e}$ is the plateau modulus (often referred to as ${G}_{N}^{0}$ in the literature). Entanglements behave like crosslinks, but they have finite lifetimes as the molecules diffuse and the entanglements are eliminated and reformed elsewhere. The plateau modulus is given by the concentration of entanglements, ${v}_{e}$, or equivalently, by the number average molecular weight between entanglements, ${M}_{e}$ (referred to as the entanglement molecular weight):
$\begin{array}{cc}{G}_{e}={v}_{e}{k}_{B}T=\frac{\rho RT}{{M}_{e}}& \left(16.13\right)\end{array}$
$\begin{array}{cc}G\left(t\right)=\stackrel{{N}_{r}}{\sum _{i=1}}{G}_{i}exp\left(-t/{\tau }_{i}\right)& \left(16.14\right)\end{array}$
Inclusion of a sufficiently large number of relaxation processes (large ${N}_{r}$) enables very complicated relaxation behavior to be modeled accurately.

### 16.4 Temperature Dependence

A characteristic of single-component polymeric materials is that the temperature dependence of all of the different relaxation times is the same. The concept is illustrated in Figure 16.7, which shows the relaxation modulus at two different temperatures for an idealized material with two different relaxation times. The relaxation times scale with temperature according to the same temperature dependent shift factor, ${a}_{T}$. The shift factor is equal to one at a reference temperature, ${T}_{ref}$, and increases as the temperature is decreased. Because both relaxation times are multiplied by the same factor, the curve is shifted linearly along the logarithmic x axis.
Figure 16.7: Temperature dependence of a material with two relaxation times.
Because all of the relaxation times are multiplied by ${a}_{T}$ when the temperature is changed, data obtained at different temperatures superpose when plotted as a function of $t/{a}_{T}$. As a result the relaxation behavior over a large range of times can be obtained by measuring the relaxation spectrum at different times, and shifting the data to the reference temperature.
$\begin{array}{cc}log\left({a}_{T}\right)=A+\frac{B}{T-{T}_{\infty }}& \left(16.15\right)\end{array}$
The factor A is determined from the requirement that ${a}_{T}=1$ at $T={T}_{ref}$:
$\begin{array}{cc}A=-\frac{B}{{T}_{ref}-{T}_{\infty }}& \left(16.16\right)\end{array}$
The relaxation times diverge to $\infty$ at $T={T}_{\infty }$, which in free volume theory is the temperature at which the free volume of the equilibrium liquid goes to zero.

### 16.5 Dynamic Moduli

Figure 16.8: Dynamic mechanical spectroscopy.
Experimentally, a wide variation in time scales is accessed by oscillating or vibrating the sample and measuring the frequency response of the material. Consider, for example, an oscillatory shear strain:
$\begin{array}{cc}\gamma ={\gamma }_{0}sin\left(\omega t\right)& \left(16.17\right)\end{array}$
The strain rate is also sinusoidal:
$\begin{array}{cc}\stackrel{˙}{\gamma }\left(t\right)=\frac{d\gamma \left(t\right)}{dt}=\omega {\gamma }_{0}cos\left(\omega t\right)=\omega {\gamma }_{0}sin\left(\omega t+\pi /2\right)& \left(16.18\right)\end{array}$
Note that the strain and the strain rate are out of phase by $\pi$/2 (90 ${}^{○}$). This concept of a phase difference is very important in understanding the frequency dependent dynamic moduli. In this case we use Boltzmann superposition to obtain an expression for the stress:
$\begin{array}{cc}{\sigma }_{xy}\left(t\right)=\stackrel{\infty }{\underset{0}{\int }}\stackrel{˙}{\gamma }\left(t-s\right)G\left(s\right)ds=-\omega {\gamma }_{0}\stackrel{\infty }{\underset{0}{\int }}cos\left\{\omega \left(t-s\right)\right\}G\left(s\right)ds& \left(16.19\right)\end{array}$
Now we make use of the following trigonometric identity:
$\begin{array}{cc}cos\left(a-b\right)=sin\left(a\right)sin\left(b\right)+cos\left(a\right)cos\left(b\right)& \left(16.20\right)\end{array}$
We can therefore write the time dependent stress in the following way.
We define ${G}^{\prime }$ (storage modulus) and ${G}^{\prime \prime }$ (loss modulus) such that ${G}^{\prime }$ describes the component of the stress that is in phase with the strain (the elastic component) and ${G}^{\prime \prime }$ describes the component of the stress that is in phase with the strain rate (the viscous component).
By comparing Equations 16.21 and 16.22, we obtain the following for ${G}^{\prime }$ and ${G}^{\prime \prime }$:
$\begin{array}{cc}\begin{array}{c}{G}^{\prime }\left(\omega \right)=\omega \stackrel{\infty }{\underset{0}{\int }}G\left(t\right)sin\left(\omega t\right)dt\\ {G}^{\prime \prime }\left(\omega \right)=\omega \stackrel{\infty }{\underset{0}{\int }}G\left(t\right)cos\left(\omega t\right)dt\end{array}& \left(16.23\right)\end{array}$
In is useful to consider the behavior of ${G}^{\prime }$ and ${G}^{\prime \prime }$ in limiting cases where the system is a perfectly elastic solid with no viscous character, and where the material is a Newtonian liquid with no elastic character:
##### Perfectly Elastic System:
In this case the shear modulus is independent of both time and frequency. For $\gamma \left(t\right)={\gamma }_{0}sin\left(\omega t\right)$ we have:
$\begin{array}{cc}{\sigma }_{xy}\left(t\right)=G\gamma \left(t\right)=G{\gamma }_{0}sin\left(\omega t\right)& \left(16.24\right)\end{array}$
Comparing to Eq. 16.22 gives:
$\begin{array}{cc}\begin{array}{c}{G}^{\prime }=G\\ {G}^{\prime \prime }=0\end{array}& \left(16.25\right)\end{array}$
##### Perfectly Viscous System:
We again have $\gamma \left(t\right)={\gamma }_{0}sin\left(\omega t\right)$, but this time the shear stress depends only on the strain rate:
$\begin{array}{cc}{\sigma }_{xy}\left(t\right)=\eta \frac{d\gamma \left(t\right)}{dt}=\eta \omega {\gamma }_{0}cos\left(\omega t\right)& \left(16.26\right)\end{array}$
Comparing to Eq. 16.22 gives:
$\begin{array}{cc}\begin{array}{c}{G}^{\prime }=0\\ {G}^{\prime \prime }\left(\omega \right)=\omega \eta \end{array}& \left(16.27\right)\end{array}$

#### 16.5.1 Phase Angle and Loss Tangent

${G}^{\prime }$ and ${G}^{\prime \prime }$ can be viewed as the real and imaginary components of a a complex modulus, ${G}^{*}$:
$\begin{array}{cc}{G}^{*}\left(\omega \right)={G}^{\prime }\left(\omega \right)+i{G}^{\prime \prime }\left(\omega \right)& \left(16.28\right)\end{array}$
As with any complex number, we can represent ${G}^{*}$ by its magnitude, which we refer to as $|{G}^{*}|$ and its phase angle, which we refer to as $\delta$:
$\begin{array}{cc}\begin{array}{c}|{G}^{*}\left(\omega \right)|={\left\{{\left({G}^{\prime }\left(\omega \right)\right)}^{2}+{\left({G}^{\prime \prime }\left(\omega \right)\right)}^{2}\right\}}^{1/2}\\ \delta ={tan}^{-1}\left(\frac{{G}^{\prime \prime }\left(\omega \right)}{{G}^{\prime }\left(\omega \right)}\right)\end{array}& \left(16.29\right)\end{array}$
The significance of $|{G}^{*}|$ and $\delta$ are illustrated by writing the sinusoidal displacement and the resulting stress in the following way:
$\begin{array}{cc}\frac{\gamma \left(t\right)}{{\gamma }_{0}}=sin\left(\omega t\right)& \left(16.30\right)\end{array}$
$\begin{array}{cc}\frac{{\sigma }_{xy}\left(t\right)}{{\gamma }_{0}}=|{G}^{*}|sin\left(\omega t+\delta \right)& \left(16.31\right)\end{array}$
$|{G}^{*}|$gives the ratio of the stress amplitude to the strain amplitude, and $\delta$ is the phase lag between the stress and strain. The following expressions relate $|{G}^{*}|$, ${G}^{\prime }$, ${G}^{\prime \prime }$ and $\delta$:
$\begin{array}{cc}\begin{array}{c}{G}^{\prime }=|{G}^{*}|cos\left(\delta \right)\\ {G}^{\prime \prime }=|{G}^{*}|sin\left(\delta \right)\\ tan\left(\delta \right)=\frac{{G}^{\prime \prime }}{{G}^{\prime }}\end{array}& \left(16.32\right)\end{array}$

#### 16.5.2 Spring and Dashpot Models

As mentioned above, a generalized Maxwell model, where the time-dependent relaxation is described by a sum of exponential relaxations, is often used to describe the viscoelastic behavior of soft materials. A way of visualizing the effect of different relaxation times is to imagine a series of springs (elastic elements) and dashpots (viscous elements) that are linked together in some way. Because it is easy to visualize the extension of springs in a tensile experiment than it is to visualized them in a shear experiment, we consider the time dependent behavior of the tensile modulus, $E$ (Young's modulus in a traditional, static experiment). A single maxwell element (Figure 16.9) has a spring of modulus $E$ in series with a dashpot of viscosity $\eta$, and gives and gives an exponential relaxation with a relaxation time, $\tau$, given by $\eta /E$. In a generalized Maxwell the stress is given by the sum of the contributions from all the individual Maxwell elements. In Figure 16.10 we include a spring with modulus ${E}_{r}$ that is not in series with a dashpot. This represents the 'relaxed' modulus, and describes the stress remaining in the system for $t\to \infty$. The overall stress after the instantaneous application of a tensile strain, $e$, is:
Figure 16.9: Maxwell model.
$\begin{array}{cc}\frac{\sigma }{e}={E}_{r}+{\sum }_{j}{E}_{j}exp\left(-t/{\tau }_{j}\right),& \left(16.33\right)\end{array}$with ${\tau }_{j}={\eta }_{j}/{E}_{j}$. In order to obtain ${G}^{\prime }$ and ${G}^{\prime \prime }$ we use Eq. 16.23. The sine and cosine transforms of an exponential function can be determined analytically, and we end up with the following for ${G}^{\prime }\left(\omega \right)$ and ${G}^{\prime \prime }\left(\omega \right)$ if $G\left(t\right)={G}_{0}exp\left(-t/\tau \right)$:
(Note: A MATLAB script that demonstrates the use of the use of MATLAB's symbolic algebra capability to determine these integrals is given in Appendix 1). Plots of these values for ${G}^{\prime }$ and ${G}^{\prime \prime }$ are shown in Figure 16.11. Note that for $\omega =1/\tau$, ${G}^{\prime }={G}^{\prime \prime }$. For $\omega >1/\tau$ the material behaves as a liquid, with ${G}^{\prime \prime }>{G}^{\prime }$. For angular frequencies above $1/\tau$, the opposite is true, and the material has solid-like behavior with ${G}^{\prime }>{G}^{\prime \prime }$.
Figure 16.10: Spring-dashpot representation of a generalized Maxwell model.
Because the generalized Maxwell model is just the sum of individual Maxwell elements, we can apply Eq. 16.34 to each of these individual elements, and sum the result to get the storage and loss moduli for the general case where many different relaxation times are needed to fully characterize the linear viscoelastic response of the material:
$\begin{array}{cc}\begin{array}{c}{G}^{\prime }\left(\omega \right)={E}_{r}+{\sum }_{j}\frac{{G}_{j}\omega {\tau }_{j}}{1+{\left(\omega {\tau }_{j}\right)}^{2}}\\ {G}^{\prime \prime }\left(\omega \right)={E}_{r}+{\sum }_{j}\frac{{G}_{j}{\left(\omega {\tau }_{j}\right)}^{2}}{1+{\left(\omega {\tau }_{j}\right)}^{2}}\end{array}& \left(16.35\right)\end{array}$
##### Zero Shear Viscosity
If ${E}_{r}=0$ and $\omega <<1/{\tau }_{0}$, where ${\tau }_{0}$ is the longest relaxation time of the material, then the response will be liquid like, with ${G}^{\prime \prime }=\eta \omega$. The zero shear viscosity is obtained from the response in this low frequency regime:
$\begin{array}{cc}{\eta }_{0}\equiv \underset{\omega \to 0}{lim}\frac{{G}^{\prime \prime }\left(\omega \right)}{\omega }& \left(16.36\right)\end{array}$

#### 16.5.3 Time-Temperature Superposition of Dynamic Mechanical Data

Figure 16.12: An interactive showing dynamic mechanical data
##### Measurement of ${G}_{N}^{0}$ and ${M}_{e}$
Figure 16.13: Plot showing temperature shifted curve
In this plot we show the temperature shifted master curve forG', from the PTBA data. The specific value of ${G}_{N}^{0}$ is generally taken as the inflection point in G', or as the value of G' at the frequency at which $\delta$ is minimized. Once G' is known, the entanglement molecular weight, Me can be determined from therelevant expression.
##### Viscocity from Rheological Data
The zero shear viscosity, ${\eta }_{0}$, at the temperature where ${a}_{T}$ = 1 is determined from the low frequency data, in the regime where G' is proportional to $\omega$. The full temperature dependence of n is then given by the measured temperature dependence a dependence of ${a}_{T}$.
Figure 16.14: Rheological data

## 17 Semicrystalline Polymers

Many polymers of commercial importance are able to crystallize to a certain extent. The most widespread crystallizable polymers are polyethylene and isotactic polypropylene. As we will see, crystallization is never complete, and there is always a certain amorphous fraction in any polymer. For this reason, many of the concepts which apply to amorphous polymers are still valid for semicrystalline polymers. For example, semicrystalline polymers are often quite brittle at temperatures below the glass transition temperature of the amorphous fraction.
In the following pages, we introduce some common semicrystalline polymers, and discuss their basic structural feature polymers, in addition to kinetic issues that are important in their processing.

### 17.1 Structural Hierarchy in Semicrystalline Polymers

The structure of semicrystalline polymers is much more complex than the structure of amorphous polymers, and our discussion in the following pages is not as quantitative as our discussion of the random walk configurations of polymer chains in amorphous polymers. Instead, we illustrate the features of semicrystalline polymers by considering the following basic structural features:
1. Helices formed by individual polymer molecules
2. Perfect crystals formed by the lateral packing of these helices
3. Lamellar crystallites formed by folded helices
4. Semicrystalline regions formed by stacking of lamellar crystallites
5. Spherulitic morphologies formed by radial growth of lamellar crystallites

### 17.2 The Structural Repeat Unit

Exercise:What are the structural repeats for each of the following polymers:
1. Poly(tetrafluoroethylene)
2. Syndiotactic Polypropylene
3. Isotactic Polypropylene
4. Atactic Polystyrene
Solution:
1. The repeating unit here is simply $C{F}_{2}$ since I can reconstruct the polymer simply by connecting these units to one another.
2. The repeating unit here contains 4 backbone carbons and two methyl groups, since there is a methyl group on every other carbon, and these methyl groups alternate from one side of the backbone to the other.
3. In this case there are two backbone carbons and a single methyl group (a single proplyne unit).
4. Since the polymer is atactic, it has an inherently disordered strucutre, with no structural repeat. This is why it is always amorphous.

### 17.3 Helix Formation

Figure 17.1: Diagram of a planar zigzag (left), with the specific geometry for tetrahedrally bonded carbon (right). Note that the bond angles appear distorted in the leftmost portion of the figure because the horizontal and vertical scales are not necessarily equivalent.)
The planar zigzag configuration is one special case of the shapes of polymers along the chain axis. In the more general case, the backbone carbon atoms all lie along the a helix that is formed about some central axis. Helices of vinyl polymers form by rotation about the C-C single bonds which make up the backbones of the molecules. The curved line in the side view of the planar zigzag in Figure 17.1 represents the helix that passes through each of the carbon atoms in a single molecule in the polyethylene crystal structure.
The general notation describing the helical structure is $x*y/z$ where $x$, $y$ and $z$ are defined as follows:
• $x$: the number of backbone carbons in the structural repeat. This is the only number that we can get by looking at the molecular structure of the polymer itself. We need to know the crystal structure to know what $y$ and $z$ are.
• $y$: the number of structural repeats per crystallographic repeat. This means that the product $xy$ gives the number of carbon atoms that repeat along the chain direction within one unit cell.
• $z$: the number of turns of the helix per crystallographic repeat.
The planar zigzag can be formally viewed as a 1*2/1 helix, with 3 turns of the helix shown in Figure 17.1. From the top view, we see that all of the backbone atoms of this helix are in the same plane. More complicated helices form because of steric interactions between the other substituents that are placed on the carbon backbones. It's easy for polyethylene molecules to adopt an all-trans configuration because the hydrogen atoms are small and don't interfere with each other when the polymer chain adopts this particular shape. Fluorine atoms are larger than hydrogens, however, so poly(tetrafluorethylene) (Teflon ${}^{TM}$) adopts a more complicated, 13/7 helix, which is is shown below in Figure 17.2.
Exercise: What is length of a polyethylene molecule with M = 10,000 g/mole, if the molecule is in all-trans (planar zigzag) configuration.
Solution: For bond angle, $\theta$, for singly bonded carbon atoms are 71 ${}^{○}$, and the C-C bond length is 1.54 Å. From the drawing below, we see that the projected length of this bond along the direction of the zigzag is equal to 1.25 Å. We just need to multiply this length by the number of carbons in the polyethylene chain, which we get by dividing the molecular weight of 10,000 g/mole by the molecular weight of a single $C{H}_{2}$ unit (14 g/mole). This gives the following for the total length, $L$:
$L=1.25\phantom{\rule{6px}{0ex}}Å\left(\frac{10,000\phantom{\rule{6px}{0ex}}g/mole}{14\phantom{\rule{6px}{0ex}}g/mole}\right)=960\phantom{\rule{6px}{0ex}}Å$
Figure 17.2: Representation of a 13/7 helix.

### 17.4 Crystalline unit cells

Figure 17.3: Diagram of a crystalline unit cell.
The perfect crystal structure of a polymer is defined by the helix type, unit cell dimensions and by the packing of the helices in the unit cell. Polyethylene can exist in multiple crystal structures. One of these, referred to as polyethylene I, is illustrated in Figure 17.4. This crystal structure has two 1*2/1 helices packed into an othorhombic unit cell ( $\alpha$ = $\beta$ = $\gamma$ = 90 ${}^{○}$). As is often the case, the helix axis corresponds to the $c$ direction of the unit cell.
$\alpha =\beta =\gamma =9{0}^{○}$
$a$= 4.946 A
$b$= 7.418 A
$c$= 2.546 A
2 helices per unit cell
Figure 17.4: Structure of polyethylene I.
 Polymer Helix Type a, b, c (Å) $\alpha$, $\beta$, $\gamma$ structural units/cell polyethylene I 1*2/1 7.41, 4.95, 2.55 90, 90, 90 4 polyethylene II 1*2/1 8.09, 4.79, 2.53 90, 107.9, 90 4 poly(tetrafluoroethylene ) I 1*13/6 5.59, 5.59, 16.88 90, 90, 113.3 13 polypropylene (syndiotactic) 4*2/1 14.50, 5.60, 7.40 90, 90, 90 8 polypropylene (isotactic) 2*3/1 6.66, 20.78, 6.50 90, 99.6, 90 12 PET 12*1/1 4.56, 5.96, 10.75 98.5, 118, 112 Polyisoprene (cis 1,4) 8*1/1 12.46, 8.86, 8.1 90, 90, 90 8
Exercise:Calculate the density of perfectly crystalline polyethylene, given assuming the crystal structure from the previous page.
Solution: Density is mass/volume, so we just need to calculate the volume of a unit cell, and the total mass of material included in the unit cell:
1. Unit cell volume: $a\cdot b\cdot c$= $9.341×1{0}^{-23}\phantom{\rule{6px}{0ex}}c{m}^{3}$
2. Mass: 14 g/mole = $2.326×1{0}^{-23}$ g per $C{H}_{2}$ unit. Since there are 4 $C{H}_{2}$ units in a unit cell, mass per unit cell = $4\left(2.326×1{0}^{-23}\phantom{\rule{6px}{0ex}}g\right)=9.302×1{0}^{-23}\phantom{\rule{6px}{0ex}}g$.
3. Density = mass/volume = 0.996 g/cm ${}^{3}$.
Note that the density is close to 1 g/cm ${}^{3}$, which is generally true for most polymers.

### 17.5 Stiffness of Polymer Single Crystals

Figure 17.5: Macromol. 23, 2365-2370 (1990).
Single crystal polyethylene has the following calculated values for the modulus along the 3 crystal axes:
• E=315 GPa along c axis
• E=8.0 GPa along a axis
• E=9.9 GPa along b axis
Amorphous fraction: E=0.001 GPa
The actual modulus is almost always much lower than this because of imperfect crsytal alignment, chain folding and the presence of an amorphous fraction. See the following pages for details.

### 17.6 Chain Folding

Figure 17.6: Diagram of polymer chain folding

#### 17.6.1 Models of chain folding

Figure 17.7: Extreme views of chain folded crystals
These drawings illustrate two extreme views of the structure of chain folded crystals. In the adjacent reentry model, the molecules are assumed to fold back on themselves in a very well-organized way. In the random switchboard model, chains which leave the crystal at the tops and bottom surfaces reenter the crystal at random positons. Reality lies somewhere between these two extremes.

#### 17.6.2 Amorphous Fraction and Tie Molecules

Figure 17.8: Lamellar crystallites

#### 17.6.3 Percent Crystallinity for Lamellar Crystallites

Figure 17.9: Diagram of percent crystallinity of lamellar crystallites

### 17.7 Spherulitic morphologies

Figure 17.10: A representation of a spherulite
Figure 17.11: Formation of a spherulite
During crystallization, the spherulite grows radially outward, with each branch extending in the growth direction while maintinaing a constant crystal thickness, ${\lambda }_{c}$.
Figure 17.12: Geometric representation of a spherulite
This is the geometric requirement for addition of new lamellae; the additional space taken up by the spherulite as a whole is filled by new lamellae.
Figure 17.13: Video of the formation of a spherulite

### 17.8 Birefringence

Figure 17.14: Interactive model of birefringence
 Light emerging from the sample for $\psi =45º$ and $\Delta \phi =90º$ is said to be circularly polarized. Why is this a sensible description of the polarization state you observe under these circumstances? For what values of $\psi$ is no light at all transmitted through the analyzer? The maximum light transmission through the analyzer is obtained for $\varphi =45º$ and $\Delta \phi =180º$ . What fraction of the light emerging from the sample is transmitted under these circumstances?

#### 17.8.1 Birefringence and Radial Symmetry

Figure 17.15: Diagram showing birefringence
If the object between the polarizer and analyzer has spherical symmetry, as is the case with these liquid crystalline droplets, the'Maltesecross ' pattern shown here will be obtained. Dark patches correspond to $\Psi$= 0 and 90°, and the bright patches are obtained at $\Psi$= ± 45°.

### 17.9 Growth of a Lamellar Crystallite

The overall free energy change ( $\Delta F$) associated with the formation of a small crystallite has two components:
1. A surface energy component, which is always positive.
2. A bulk free energy component, which is negative for temperatures below the equilibrium melting temperature, ${T}_{m}^{0}$.
The bulk component can be written in terms of the undercooling ( ${T}_{m}^{0}-T\right)$ by writing the bulk free energy change per unit volume ( $\Delta f$ ${}_{c}$) in terms of its entropic and enthalpic components:
$\begin{array}{cc}\Delta {f}_{c}=\Delta {h}_{c}-T\Delta {s}_{c}& \left(17.1\right)\end{array}$
• $\Delta {h}_{c}$ is the enthalpy required to melt the crystal, and is always a positive number.
• From its definition, $\Delta T$ is positive whenever the temperature is below the equilibrium melting temperature.
• $\Delta {f}_{c}$ is the free energy difference between the crystal and the amorphous material, and is negative for temperatures below the equilibrium melting temperature.
Figure 17.16: Schematic representation of a growing lamellar crystallite.
If we account for the surface energies associated with the new crystal/amorphous interface that is created, the increase in free energy for adding $n$ "strands" to the face of a lamellar crystallite is given by the following expression:
$\begin{array}{cc}\Delta F=2a\text{'}{\lambda }_{c}{\gamma }_{s}+2na\text{'}b\text{'}{\gamma }_{e}+na\text{'}b\text{'}{\lambda }_{c}\Delta {f}_{c}& \left(17.3\right)\end{array}$
If we assume that $n$ is large we can neglect the term involving ${\gamma }_{s}$, so that $\Delta F<0$ when $\Delta {f}_{c}<\frac{2{\gamma }_{e}}{{\lambda }_{c}}$. With $\Delta {f}_{c}$ given by Eq. 17.2, we can rewrite the condition that $\Delta F<0$ in the following form:
$\begin{array}{cc}{\lambda }_{c}>\frac{2{\gamma }_{e}{T}_{m}^{0}}{\Delta {h}_{c}\Delta T}& \left(17.4\right)\end{array}$
A crystal will grow by the addition of additional strands when this criterion is met. Thick crystals (high ${\lambda }_{c}$) are thermodynamically favorable, but thin crystals (low ${\lambda }_{c}$) are kinetically favorable. In this model, the crystal thickness which is actually obtained is the lowest, thermodynamically possible value. The inequality from the previous page therefore becomes an approximate equality:
$\begin{array}{cc}{\lambda }_{c}\approx \frac{2{\gamma }_{e}{T}_{m}^{0}}{\Delta h\Delta T}& \left(17.5\right)\end{array}$
Because thick crystals are thermodynamically more stable than thin crystals, the thick crystals will melt at a slightly higher temperature than thin crystals. One therefore expects a relationship between the temperature at which the crystals will form, and the temperature at which the crystals will melt. Crystals crystallized at higher temperatures will also melt at higher temperatures. At an undercooling of zero the actual melting temperature, ${T}_{m}$, will be equal to the equilibrium melting temperature. For very low undercoolings, however, the crystallization rate becomes very slow, so crystallization will never take place at the exact equilibrium melting temperature. Values of the equilibrium melting temperature are obtained from measurements of ${T}_{m}$ at higher undercoolings, and extrapoloating to $\Delta T=0$ as illustrated in the following graph:
Figure 17.18: Melting temp ( ${T}_{m}$) as a function of the crystallization temperature, ${T}_{c}$.

### 17.10 Density of Semicrystalline Polymers

The percent crystallinity in a polymer sample is generally determined by measuring the density of the material. A linear relationship between density and percent crystallinity is assumed, so that the following relationship holds:
$\begin{array}{cc}%Crystallinity=\frac{\rho -{\rho }_{amorphous}}{{\rho }_{crystalline}-{\rho }_{amorphous}}x100& \left(17.6\right)\end{array}$
Figure 17.19: Density vs percent crystallinity

## 18 Liquid Crystals

Liquid crystals are materials that that are intermediate between true traditional crystalline materials, and liquids. The molecules in a liquid crystalline phase are not organized into a full, 3- dimensionally periodic structure, but they do have some degree of orientational or positional order. Examples of molecular organization in some liquid crystalline phases are shown below.
Figure 18.1: Structure of liquid crystals

### 18.1 Nematic Liquid Crystals

Figure 18.2: Structure of nematic liquid crystals

### 18.2 Nematic Liquid Crystals - Texture

Figure 18.3: Image of a nematic liquid crystal
This photograph was taken between crossed polarizers, uing a microscope geometry like that described in the description of Section 17.8.

### 18.3 Liquid Crystal Displays

Figure 18.4: Structure of an LCD

## 19 Solutions and Blends

We now move from our discussion of the shapes of individual polymer molecules in a homogeneous, one component system, to the behavior of solutions and blends consisting of different types of molecules that have been mixed together. We are interested in studying these systems for many reasons, including the following:
1. Characterization: many of the properties of polymer molecules are determined by the properties of polymer solutions.
2. Processing: many applications of polymers involve require that the polymer molecules be dissolved in an appropriate solvent.
3. Improved materials: new materials with improved properties can often be formed by blending or "alloying" different polymers.
4. Recycling: polymers need to be separated before they can be recycled because the different polymers do not mix favorably.
The key quantity when discussing thermodynamics of polymer solutions (polymer + small-molecule solvent) and polymer blends (polymer A + polymer B) is the free energy of mixing. We therefore begin with a discussion of polymer solution thermodynamics. In general, we consider the case where we mix ${n}_{a}$ A molecules with a degree of polymerization ${N}_{a}$ with ${n}_{b}$ B molecules with a degree of polymerization ${N}_{b}$. Polymer solutions will correspond to the case where one of the degrees of polymerization is very small (typically 1). We will make the assumption that the sizes of the repeat units for the components of the mixture are identical. This assumption is not as restrictive as it would initially appear, since one can always define "effective" repeat units with the desired volume. In this sense, ${N}_{a}$ and ${N}_{b}$ are determined by the relative molecular volumes of the components in the mixture, and are defined in terms of some reference volume.
Exercise: The room temperature densities of polystyrene and toluene are 1.05 $g/c{m}^{3}$and 0.87 $g/c{m}^{3}$, respectively. What reference volume, ${V}_{0}$, should we use to characterize the thermodynamics of polystyrene solutions in toluene? What degrees of polymerization would characterize a solution where the polystyrene molecular weight is 100,000 g/mole?
Solution: The obvious reference volume in this case is the volume of a toluene molecule, which we take as having N=1. This volume is obtained from the density and molecular weight of toluene. Toluene ( ${C}_{6}{H}_{8}$) has a molecular weight of 92 g/mole.
${V}_{0}=\left(\frac{92\phantom{\rule{6px}{0ex}}g}{mole}\right)\left(\frac{c{m}^{3}}{0.87\phantom{\rule{6px}{0ex}}g}\right)=106\phantom{\rule{6px}{0ex}}c{m}^{3}/mole$
Note that we have used the molar volume for ${V}_{0}$. This convention will be followed throughout our discussion.
The value of $N$ for the polymer molecules will be given by the molecular volume, divided by the reference volume:
${N}_{polystyrene}=\left(\frac{100,000\phantom{\rule{6px}{0ex}}g}{mole}\right)\left(\frac{c{m}^{3}}{1.05\phantom{\rule{6px}{0ex}}g}\right)\left(\frac{mole}{106\phantom{\rule{6px}{0ex}}c{m}^{3}}\right)=898$
In our discussion of the thermodynamics of mixtures, we will refer to N as the number of "segments", defined in terms of a reference volume as described here. In some cases this reference volume will correspond to a repeat unit of the polymer, but this does not need to be the case in general.

### 19.1 Chemical Potentials and Free Energy of Mixing

$\begin{array}{cc}F={n}_{a}{\mu }_{a}+{n}_{b}{\mu }_{b}& \left(19.1\right)\end{array}$
The derivative with respect to ${n}_{a}$ is taken with ${n}_{b}$ held constant, and vice versa. Note that $F$ is the total extensive free energy of the system, since it increases as the numbers of A and B molecules ( ${n}_{a}$ and ${n}_{b}$) increases.
Figure 19.1: Animation of the chemical potential of a molecule
The chemical potentials of A and B molecules of the pure components are defined as ${u}_{a}^{0}$ and ${u}_{b}^{0}$. The free energy of the two components before mixing is therefore given by:
$\begin{array}{cc}{F}_{unmixed}={n}_{a}{\mu }_{a}^{0}+{n}_{b}{\mu }_{b}^{0}& \left(19.3\right)\end{array}$
The free energy of mixing is equal to the difference in the free energies of the mixed and unmixed systems:
$\begin{array}{cc}\Delta {F}_{mix}=F-{F}_{unmixed}={n}_{a}{\mu }_{a}+{n}_{b}{\mu }_{b}-\left({n}_{a}{\mu }_{a}^{0}+{n}_{b}{\mu }_{b}^{0}\right)={n}_{a}\left({\mu }_{a}-{\mu }_{a}^{0}\right)+{n}_{b}\left({\mu }_{b}-{\mu }_{b}^{0}\right)& \left(19.4\right)\end{array}$
Because we are only interested in changes in chemical potentials due to mixing, we can set the chemical potentials of the pure, unmixed components to zero ( ${\mu }_{a}^{0}={\mu }_{b}^{0}=0$).
By defining the chemical potentials of the pure components as zero, we obtain:
Equation 19.5 is completely general, and valid for mixtures of molecules of any size. It's not that useful, however, because we generally don't write down values of ${n}_{a}$ and ${n}_{b}$ when we want to specify the composition of a multicomponent mixture. For mixtures of small molecules (metal alloys, for example), we commonly use mole fractions, ${X}_{a}$ and ${X}_{b}$, to specify the composition:
$\begin{array}{cc}\begin{array}{c}{X}_{a}=\frac{{n}_{a}}{{n}_{a}+{n}_{b}}\\ {X}_{b}=\frac{{n}_{b}}{{n}_{a}+{n}_{b}}\end{array}& \left(19.6\right)\end{array}$
This makes sense for metals because the molar volumes of the different components are not that different from one another. In polymer systems, however, we are often dealing with systems where the components have molar volumes that can differ by orders of magnitude. This is the situation if we are interested in the solubility of a high molecular weight polymer (which can have a molecular weight of several hundred thousand g/mole) in a solvent (which will typically have a molecular weight of $\approx 100$ g/mole). Organic solutions are typically specified by the weight fractions of the different components. In discussions of polymer solution thermodynamics it is more common to work in terms of volume fractions of the different components in the systems.
The concepts in this section can be easily extended to systems with more than two components, but we confine ourselves here to binary solutions of just two components: A molecules of length ${N}_{a}$ and B molecules of length ${N}_{b}$. In our discussion of polymer thermodynamics it is important to keep in mind ${N}_{a}$ ${N}_{b}$ are not necessarily true degrees of polymerization as defined earlier in this text. Instead they are obtained by dividing the molar volumes of the different components by some reference volume, ${V}_{0}.$ For polymer/solvent mixtures this reference volume is typically taken as the molar volume of the solvent, and for polymer/polymer mixtures it is typically taken as the volume per repeat unit for one of the polymers. The total volume, $V$, and volume fractions, ${\phi }_{a}$ and ${\phi }_{b}$ are given by the following expressions:
The free energy of mixing per unit volume, $\Delta {f}_{mix}$, is obtained from $\Delta {F}_{mix}$ by dividing by the volume of the system:
$\begin{array}{cc}\Delta {f}_{mix}=\frac{{n}_{a}{\mu }_{a}+{n}_{b}{\mu }_{b}}{{n}_{a}{N}_{a}{V}_{0}+{n}_{b}{N}_{b}{V}_{0}}=\frac{{\mu }_{a}{\phi }_{a}}{{N}_{a}{V}_{0}}+\frac{{\mu }_{b}{\phi }_{b}}{{N}_{b}{V}_{0}}& \left(19.9\right)\end{array}$
By rearranging the equation at the bottom of the previous page we obtain:
The quantities $\Delta {f}_{mix}$, ${\mu }_{a}/{N}_{a}$ and ${\mu }_{b}/{N}_{b}$ all depend on the composition of the material. The relationship between these quantities is illustrated by the tangent construction shown in Figure 19.2, where the composition dependence of $\Delta {f}_{mix}$ is plotted as a function of ${\phi }_{b}$ for a hypothetical material. For a given composition (denoted ${\phi }_{b}^{\prime }$ in Figure 19.2), the chemcial potentials are obtained by first drawing a tangent to the free energy curve at ${\phi }_{b}={\phi }_{b}^{\prime }$. The value of ${\mu }_{a}/{N}_{a}$ is determined by extrapolating this tangent line to ${\phi }_{b}=0$ and ${\mu }_{a}/{N}_{a}$ is obtained by extrapolating the tangent line to ${\phi }_{b}=1.$ This result is consistent with Eq. 19.10, which shows that normalized free energy of mixing at a certain composition is determined by the weighted average of the normalized chemical potentials. As a result a straight line drawn between the normalized chemical potentials on the left and right axes must hit the free energy curve at ${\phi }_{b}={\phi }_{b}^{\prime }$.
To fully validate the tangent construction illustrated in Figure 19.2 we still need to show that the straight line between the chemical potentials must be equal to the slope of the free energy curve. This can done in formal terms by combining Eqs. 19.2 and 19.8, which after simplification leads to the following relationship:
$\begin{array}{cc}\frac{\partial \left(\Delta {f}_{mix}{V}_{0}\right)}{\partial {\phi }_{b}}=\frac{{\mu }_{b}}{{N}_{b}}-\frac{{\mu }_{a}}{{N}_{a}}=-\frac{\partial \left(\Delta {f}_{mix}{V}_{0}\right)}{\partial {\phi }_{a}}& \left(19.11\right)\end{array}$
The fact that the tangent must give the difference between ${\mu }_{b}/{N}_{b}$ and ${\mu }_{a}/{N}_{b}$ makes sense conceptually, because the only way changing the composition by some amount requires that an A segment be replaced by a B segment.
Figure 19.2: Relationship of the free energy of mixing to the chemical potentials.

### 19.2 Ideal Entropy of Mixing for Polymers

For small molecules the ideal free energy of mixing is given by the following expression:
$\begin{array}{cc}\frac{\Delta {s}_{ideal}{V}_{0}}{R}=-{X}_{a}ln{X}_{b}-{X}_{b}ln{X}_{b}& \left(19.12\right)\end{array}$
The quantity ${s}_{ideal}{V}_{0}$ is the entropy of mixing per mole of molecules. If the molar volumes of the A and B molecules are the same, than we can simply replace the mole fractions, ${X}_{a}$ and ${X}_{b}$ with the corresponding volume fractions, ${\phi }_{a}$ and ${\phi }_{b}$. The entropy of mixing per molecule is independent of the size of the molecule. Therefore, the entropy of mixing per repeat unit is smaller than the entropy of mixing per molecule by a factor of $N$, where $N$ is the degree of polymerization. In general, we have the following expression for the ideal free energy of mixing per volume for polymers with molar volumes of ${N}_{a}{V}_{0}$ and ${N}_{b}{V}_{0}$:
$\begin{array}{cc}\frac{{s}_{ideal}{V}_{0}}{R}=-\frac{{\phi }_{a}ln{\phi }_{a}}{{N}_{a}}-\frac{{\phi }_{b}ln{\phi }_{b}}{{N}_{b}}& \left(19.13\right)\end{array}$

### 19.3 Idealized Enthalpy of Mixing

The key result is that the entropy of mixing for a given volume is much smaller for large molecules than it is for small molecules. This result explains why different types of polymers almost never mix with one another in the liquid state. The driving force for two liquids to mix is normally the entropy of mixing, but this driving force is very low for mixtures of very large polymer molecules. In fact, the entropy of mixing for polymer mixtures is sometimes negative. While the ideal entropy of mixing must be positive, it is very small, and can be overwhelmeed by non-ideal contributions to the entropy of mixing which can be positive or negative. Before discussing these effects, however, we will continue in our derivation of the Flory Huggins equation for the free energy of mixing. The Flory-Huggins theory combines the ideal free energy of mixing with a very simple form for the enthalpy of mixing. Unlike the entropy of mixing, the enthalpy of mixing is assumed to be independent of the size of the molecules. We can therefore use small molecules to illustrate the origins of the equation for the enthalpy of mixing.
Figure 19.3: An animated simulation of mixing
Figure 19.4: Pictoral representations of mixing
In this example, the blue square is placed into an environment where one of its four nearest neighbors is blue, and three are red. Red-blue contacts have an energy (or enthalpy) of ${E}_{ab}$, red-red contacts have an energy of ${E}_{aa}$, and blue-blue contacts have an energy of ${E}_{bb}$. The enthalpy change, $\Delta$H, associated with the removal of the square from a pure blue phase on the left (4 blue neighbors) to the mixture on the right (1 blue neighbor and 3 red neighbors) is therefore given by the following formula:
$\begin{array}{cc}\Delta H=3{E}_{ab}+{E}_{bb}-4{E}_{bb}=3\left({E}_{ab}-{E}_{bb}\right)& \left(19.14\right)\end{array}$
Figure 19.5: Diagram showing contacts
How do we calculate the energy of a randomly mixed collection of A and B molecules? We begin by assuming that the energy of an A-A contact is ${E}_{aa}$, the energy of an B-B contact is ${E}_{bb}$, and the energy of an A-B contact is ${E}_{ab}$. Also assume that each molecule $z$ nearest neighbors ( $z=4$ for the 2-dimensional array of squares shown above.) The number of A-B contacts in a mixture of ${n}_{a}$ A molecules and ${n}_{b}$ B molecules is given by:
$\begin{array}{cc}\begin{array}{c}{n}_{aa}={n}_{a}z{p}_{a}/2\\ {n}_{bb}={n}_{b}z{p}_{b}/2\\ {n}_{ab}={n}_{a}z{p}_{b}\end{array}& \left(19.15\right)\end{array}$
where ${p}_{a}$ is the probability that a nearest neighbor is an A molecule. The factor of 2 in the expressions for ${n}_{aa}$ and ${n}_{bb}$ is needed to avoid double counting. (For example, if ${n}_{a}$ =2, there is only one interaction between the two molecules, not 2.) Now we invoke the mean field (or random mixing) approximation, which is that ${p}_{a}={\phi }_{a}$, and ${p}_{b}={\phi }_{b}$. We obtain the follwing.
$\begin{array}{cc}\begin{array}{c}{n}_{aa}={n}_{a}z{\phi }_{a}/2\\ {n}_{bb}={n}_{b}z{\phi }_{b}/2\\ {n}_{ab}={n}_{a}z{\phi }_{b}\end{array}& \left(19.16\right)\end{array}$
The energy of the mixed state is obtained by multiplying the numbers of different contacts by their energy:
$\begin{array}{cc}{E}_{mixed}={E}_{aa}{n}_{aa}+{E}_{bb}{n}_{bb}+{E}_{ab}{n}_{ab}& \left(19.17\right)\end{array}$
Substitution for ${n}_{aa}$, ${n}_{bb}$ and ${n}_{ab}$ gives the following:
$\begin{array}{cc}{E}_{mixed}={E}_{aa}{n}_{a}z{\phi }_{a}/2+{E}_{bb}{n}_{b}z{\phi }_{b}/2+{E}_{ab}{n}_{a}z{\phi }_{b}& \left(19.18\right)\end{array}$
The energy of the unmixed state is obtained in a similar fashion, remembering that in this state all A's are surrounded by other A's, and all B's are surrounded by other B's:
$\begin{array}{cc}{E}_{unmixed}={E}_{aa}{n}_{a}z/2+{E}_{bb}{n}_{b}z/2& \left(19.19\right)\end{array}$
The energy change due to mixing is obtained by subtracting the energy of the mixed state from the energy of the mixed state:
$\begin{array}{cc}\Delta {E}_{mix}={E}_{mixed}-{E}_{unmixed}=\frac{{E}_{aa}{n}_{a}z{\phi }_{a}}{2}+\frac{{E}_{bb}{n}_{b}z{\phi }_{b}}{2}+& \\ {E}_{ab}{n}_{a}z{\phi }_{b}-\frac{{E}_{aa}{n}_{a}z}{2}-\frac{{E}_{bb}{n}_{b}z}{2}& \left(19.20\right)\end{array}$
This expression can be further simplified by recalling that ${n}_{a}$ = ${\phi }_{a}n$ and ${n}_{b}$ = ${\phi }_{b}n$, where $n$ is the total number of molecules ( $n={n}_{a}$ + ${n}_{b}$):
$\begin{array}{cc}\Delta {E}_{mix}=nz\left[\frac{{E}_{aa}{\phi }_{a}^{2}}{2}+\frac{{E}_{bb}{\phi }_{b}^{2}}{2}+{E}_{ab}{\phi }_{a}{\phi }_{b}-\frac{{E}_{aa}{\phi }_{a}}{2}-\frac{{E}_{bb}{\phi }_{b}}{2}\right]& \left(19.21\right)\end{array}$
After some rearrangement we obtain the following expression:
$\begin{array}{cc}\Delta {E}_{mix}=nz{\phi }_{a}{\phi }_{b}\left[{E}_{ab}-\frac{{E}_{aa}+{E}_{bb}}{2}\right]& \left(19.22\right)\end{array}$
$\begin{array}{cc}\chi \equiv \frac{z}{{k}_{B}T}\left[{E}_{ab}-\frac{{E}_{aa}+{E}_{bb}}{2}\right]& \left(19.23\right)\end{array}$
With this definition of $\chi$, the energy of mixing for A and B molecules (with the same size) is:
$\begin{array}{cc}\Delta {H}_{mix}=\Delta {E}_{mix}=RT\chi n{\phi }_{a}{\phi }_{b}& \left(19.24\right)\end{array}$
Strictly speaking, the 'enthalpy' and the 'energy' of a system are not the same thing. The Gibbs free energy involves the enthalpy, and is the appropriate thermodynamic quantity at a fixed temperature and pressure. The Helmholtz free energy involves the energy, and is the appropriate thermodynamic quantity at fixed temperature and volume. For compressible systems the two quantities differ slightly. We use the expression from the previous page for the extensive enthalpy of mixing, and divide by the volume of the system $\left(n{V}_{0}\right)$ to obtain the intensive enthalpy of mixing:
$\begin{array}{cc}\Delta {h}_{mix}=\frac{\Delta {H}_{mix}}{n{V}_{0}}=\frac{RT\chi {\phi }_{a}{\phi }_{b}}{{v}_{0}}& \left(19.25\right)\end{array}$
Normalizing as we did for the ideal entropy of mixing gives:
$\begin{array}{cc}\frac{\Delta {h}_{mix}{V}_{0}}{RT}=\chi {\phi }_{a}{\phi }_{b}& \left(19.26\right)\end{array}$

### 19.4 Flory-Huggins Free Energy of Mixing

The most commonly used expression used to describe the free energy of mixing for polymer/polymer and polymer/solvent mixtures combines mean-field enthalpy of mixing with the ideal combinatorial entropy of mixing:
$\begin{array}{cc}\Delta {f}_{mix}=\Delta {h}_{mix}-T\Delta {s}_{ideal}& \left(19.27\right)\end{array}$
Combining these forms for the enthalpy and entropy gives the following expression, referred to as the Flory-Huggins free energy of mixing:
It is important to recognize that many assumptions went into the derivation of this equation, and that they can't possibly all be true. Nevertheless, this equation is very commonly used in polymer science, and needs to be understood. In fact, this equation is often used as a definition for $\chi$. Based on our earlier definition of $\chi$ in terms of energetic parameters, one would expect that it's magnitude would decrease as the temperature increases. In fact, the actual free energy of mixing can have very different behavior, and the free energy of mixing is generally accurately described by the Flory- Huggins expression only when $\chi$ is allowed to vary with temperature and composition. We begin on the next page by constructing phase diagrams, with the assumption that the free energy of mixing is adequately described by a value of $\chi$ that is independent of composition. Free energy curves for ${N}_{a}={N}_{b}=1$ are shown in Figure 19.6 for $\chi$ ranging from 0 to 4. For ${N}_{a}={N}_{b}=1,$ the critical value of $\chi$ is equal to 2. When $\chi$ is larger than this critical value there will be a region of compositions where the free energy curve has a negative curvature in the free energy curve.
Figure 19.6: Regular solution form or for the free energy of mixing of small molecules.

### 19.5 The Coexistence (binodal) and Spinodal Curves

The critical value of $\chi$ decreases as ${N}_{a}$ and ${N}_{b}$ increase. In addition, the free energy curves are no longer symmetric about ${\phi }_{b}=0.5$ when ${N}_{a}\ne {N}_{b}$. Both of these points are illustrated in Figure 19.7, where we plot the Flory-Huggins free energy of mixing (Eq. 19.28) for ${N}_{a}=200$, ${N}_{b}=500$ and $\chi =$0.012. A two-phase region will exist for average compositions of the alloy that are between the compositions between ${\phi }_{1}$ and ${\phi }_{2}$, the two compositions defined by the common tangent construction in Figure 19.7. Because the tangent to the curve at these two compositions are identical, they have the same values of ${\mu }_{a}$ and ${\mu }_{b,}$ which is a requirement for equilibrium. In addition, the average free energy for a mixture of these two phases (dashed line in Figure 19.7) is less than the free energy for a homogenous phase of this average composition (solid line in Figure 19.7). By plotting these two composition as a function of $\chi$, we obtain the binodal curve shown as the solid line in Figure 19.8. The binodal curve is also referred to as the coexistence curve because it shows the compositions of the phases that are in equilibrium with one another in the two-phase region of the phase diagram.
Figure 19.7: Free energy of mixing for a system with asymmetric molecular sizes in the two phase regime.
The spinodal curve (or stability limit ) represents the boundary in the phase diagram where the free energy of mixing changes from a positive curvature to a negative curvature. It occurs when $\chi ={\chi }_{s}$, where ${\chi }_{s}$ defines the spinodal curve. For ${N}_{a}=200$ and ${N}_{b}=300$ the spinodal curve is shown as the dashed line in Figure 19.8. The change in curvature from positive to negative values occurs when the second derivative of the free energy with respect to composition is equal to zero. We obtain the following expression for ${\chi }_{s}$ by setting ${\phi }_{b}=1-{\phi }_{a}$ in the free energy expression:
$\begin{array}{cc}\frac{{\partial }^{2}}{\partial {\phi }_{a}^{2}}\left(\frac{{\phi }_{a}ln{\phi }_{a}}{{N}_{a}}+\frac{\left(1-{\phi }_{a}\right)\left(ln1-{\phi }_{a}\right)}{{N}_{b}}+{\chi }_{s}{\phi }_{a}\left(1-{\phi }_{a}\right)\right)=0& \left(19.29\right)\end{array}$
Evaluating the second derivative leads to the following:
$\begin{array}{cc}\frac{1}{{\phi }_{a}{N}_{a}}+\frac{1}{\left(1-{\phi }_{a}\right){N}_{b}}-2{\chi }_{s}=0& \left(19.30\right)\end{array}$
which can be rearranged to give the following for ${\chi }_{s}$:
$\begin{array}{cc}{\chi }_{s}=\frac{1}{2{\phi }_{a}{N}_{a}}+\frac{1}{2{\phi }_{b}{N}_{b}}& \left(19.31\right)\end{array}$

### 19.6 Critical Point

The phase diagrams obtained from the free energy of mixing expression show under what conditions polymeric liquids are completely miscible, and under what conditions phase separation will occur. The critical point is the point on the phase diagram where a liquid is just beginning to undergo phase separation. Mathematically, the critical point corresponds to the point on the spinodal curve where ${\chi }_{s}$ has the lowest possible value. We obtain this point by setting the derivative of ${\chi }_{s}$ to zero:
$\begin{array}{cc}\frac{\partial }{\partial {\phi }_{a}}\left(\frac{1}{2{\phi }_{a}{N}_{a}}+\frac{1}{2\left(1-{\phi }_{a}\right){N}_{b}}\right)=0& \left(19.32\right)\end{array}$
The solution to this equation is given by ${\phi }_{a}={\phi }_{a,crit}$, at which point ${\chi }_{s}$ is equal to ${\chi }_{crit}$:
Exercise:The value of for blends of polystyrene (PS) and poly(methyl methacrylate) (PMMA) is 0.037 at 170 °C, based on a reference volume equal to the reference volume of a polystyrene repeat unit. Suppose monodisperse samples of PS and PMMA of equal molecular weights are mixed with one another. What range of molecular weights will be completely miscible with one another at 170 °C? (Assume the densities of the two polymers are identical).
Solution: First we calculate ${\chi }_{crit}$:
$\begin{array}{cc}{\chi }_{crit}=\frac{{N}_{a}+{N}_{b}+2\sqrt{{N}_{a}{N}_{b}}}{2{N}_{a}{N}_{b}}=\frac{2}{N}& \left(19.35\right)\end{array}$ for ${N}_{a}={N}_{b}=N$
The polymers are completely miscible for $\chi$< ${\chi }_{crit}$, which occurs when N < 2/ $\chi$. With $\chi$= 0.037, we have complete miscibility for N < 54. The repeat unit molecular weight for polystyrene is 104 g/mol, so the critical molecular weight below which the polymers are completely miscible is 54*104 g/mol = 5616 g/mol

### 19.7 Spinodal Decomposition

"Spinodal Decomposition" is the mechanism of phase separation where small perturbations in the composition of the single phase grow because this decreases the overall free energy. This happens inside the spinodal curve defined previously. The microstructure is generally characterized by a well defined wavelenth, as illustrated below.
Figure 19.9: Characteristic microstructures for phase separation by spinodal decompositon (left) and nucleation and growth (right).

### 19.8 Phase Diagrams in Temperature Space

Actual values of $\chi$ for real systems are generally obtained by fitting the Flory Huggins expression to the free energy of mixing to actual experimental data. This procedure can give actual values of $\chi$ which either increase or decreaes with increasing temperature. Smaller values of $\chi$ (where mixing is favored) can therefore be obtained at either high or low temperatures, giving the following two types of phase diagrams:
Figure 19.10: Phase diagrams in temperature space.

### 19.9 Chemical Potentials

Recall that the Section 19.1 of species A is determined by differentiating the extensive free energy with respect to the number of A molecules in the system. The extensive free energy is obtained by multiplying the free energy density, f, by the volume of the system, V. For a binary mixture of A and B polymers we have:
$\begin{array}{cc}\Delta {f}_{mix}=\frac{RT}{{v}_{0}}\left\{\frac{{\phi }_{a}ln{\phi }_{a}}{{N}_{a}}+\frac{{\phi }_{b}ln{\phi }_{b}}{{N}_{b}}+\chi {\phi }_{a}{\phi }_{b}\right\}V& \left(19.38\right)\end{array}$
Using the expressions for ${\phi }_{a}$ and ${\phi }_{b}$ (Eq. 19.8) and $V$ (Eq. 19.7) we have:
$\begin{array}{cc}\frac{\Delta {F}_{mix}}{RT}=\left\{{n}_{a}ln{\phi }_{a}+{n}_{b}ln{\phi }_{b}+\chi {n}_{a}{N}_{a}{\phi }_{b}\right\}& \left(19.39\right)\end{array}$
We are now in a position to obtain expressions for the chemical potentials by differentiating the extensive free energy of mixing. We make the usual assumption that the chemical potentials of the pure components are equal to zero (activities of the pure components equal to one):
$\begin{array}{cc}{\mu }_{a}=\frac{\partial \Delta {F}_{mix}}{\partial {n}_{a}}=RT\frac{\partial }{\partial {n}_{a}}\left({n}_{a}ln{\phi }_{a}+{n}_{b}ln{\phi }_{b}+\chi {n}_{a}{N}_{a}{\phi }_{b}\right)& \left(19.40\right)\end{array}$
In this way we obtain the following result:
$\begin{array}{cc}\frac{{\mu }_{a}}{RT}=ln{\phi }_{a}+{\phi }_{b}\left(1-\frac{{N}_{a}}{{N}_{b}}\right)+\chi {N}_{a}{\phi }_{b}^{2}& \left(19.41\right)\end{array}$
When discussing polymer/solvent mixtures we replace the 'a' and 'b' subscripts with 'p' (for polymer) and 's' (for solvent). In addition, we take ${N}_{s}$ = 1. Formally, ${N}_{p}$ is the ratio of the polymer volume to the solvent volume. With these substitutions, we obtain the following expression for ${\mu }_{s}$, the solvent chemical potential, which is valid for all concentrations:
$\begin{array}{cc}\frac{{\mu }_{s}}{RT}=ln\left(1-{\phi }_{p}\right)+{\phi }_{p}\left(1-\frac{1}{{N}_{p}}\right)+\chi {\phi }_{p}^{2}& \left(19.42\right)\end{array}$

#### 19.9.1 Limiting case for high molecular weight polymer

For ${N}_{p}\to \infty$ the solvent chemical potential reduces to ${\mu }_{s}^{\infty }$:
In addition the critical point from Eqs. 19.33 and 19.34 occurs at ${\phi }_{p}=0$ and $\chi =0.5.$ This means that for $\chi >0.5$ pure solvent will be in equilibrium with a phase that contains both polymer and solvent. The polymer volume fraction in this phase is determined from the requirement that the solvent chemical potential must be zero. In other words, if a chunk of high molecular weight polymer is immersed in a solvent, solvent diffuses into the polymer until ${\phi }_{p}$ decreases to the point where ${\mu }_{s}^{\infty }=0.$ In general, this solution must be solved numerically. In Figure 19.11 we plot the concentration dependence of ${\mu }_{s}^{\infty }$ for $\chi =0$, $\chi =0.5$ and $\chi =1$.
Figure 19.11: Concentration dependence of the solvent chemical potential from Eq.19.43.

#### 19.9.2 Limiting case for dilute solutions

For small values of ${\phi }_{p}$ the log term can be expanded as follows:
$\begin{array}{cc}ln\left(1-{\phi }_{p}\right)=-{\phi }_{p}-\frac{{\phi }_{p}^{2}}{2}-...& \left(19.44\right)\end{array}$
Retaining only the first two terms in the expansion gives the following for the solvent chemical potential:
##### Exercise:
What is the polymer volume fraction for a polymer solution with a concentration of 0.05 g/cm ${}^{3}$? Assume the solvent has a density, ${\rho }_{s}$, of 0.85 g/cm ${}^{3}$ and the polymer has a density, ${\rho }_{p}$, of 1.1 g/cm ${}^{3}$ and there is no volume change on mixing.
##### Solution:
We just need to calculate the total weight of solution in a volume of 1 $c{m}^{3}$:
$\begin{array}{cc}c={\phi }_{p}{\rho }_{p}=0.05\left(1.1\phantom{\rule{6px}{0ex}}g/c{m}^{3}\right)=0.055g/c{m}^{3}& \left(19.46\right)\end{array}$
Note: Actual solution concentrations are generally reported as weight of polymer per total volume of solution - the quantity that we refer to as c. From a theoretical standpoint, it is generally more convenient to use the polymer volume fraction, ${f}_{c}$. These quanties are linearly related to one another through the polymer density, as illustrated here. The solvent density is not needed for this calculation, although it is needed to determine the ratio of polymer to solvent

### 19.10 Osmotic Pressure

The concept of osmotic pressure can be illustrated by considering the operation of the membrane osmometer illustrated conceptually in Figure 19.12. The device consists of a polymer/solvent mixture that is separated from a bath of pure solvent by a membrane that is permeable to solvent but impermeable to the polymer. The chemical potential of the pure solvent is zero, by definition. Because the solvent chemical potential in the polymer/solvent mixture is less than zero, there is a driving force for solvent to move into this side of the membrane from the pure solvent side. Solvent will continue to move into this side of the device until the hydrostatic pressure generated by the increased depth of the liquid layer provides counteracts the driving force associated with the chemical potential balance. This pressure is the osmotic pressure, and is obtained by dividing the chemical potential difference by the solvent volume:
$\begin{array}{cc}\Pi =-{\mu }_{s}/{V}_{s}& \left(19.47\right)\end{array}$
One important aspect of an osmotic pressure measurement is that it can be used to provide a direct measure of the number average molecular weight of the polymer. Osmotic pressure effects are important in many other situations as well. For example, the osmotic pressure is the minimum pressure required to purify a solution by forcing it through a membrane. The osmotic pressure of aqueous solutions is also important in a wide variety of situations in biology. The height of many trees is limited by the maximum osmotic pressure that can be sustained in the leaves, since this is the driving force that draws water to the upper reaches of the tree. Here we explore these phenomena in more detail, and develop some quantitative descriptions of the osmotic pressure.
By combining the Flory-Huggins expression for the solvent chemical potential (Eq. 19.42) with the definition of the osmotic pressure (Eq. 19.47), we obtain the following expression for $\Pi$:
$\begin{array}{cc}\Pi =\frac{RT}{{V}_{s}}\left(-ln\left(1-{\phi }_{p}\right)-{\phi }_{p}\left(1-\frac{1}{{N}_{p}}\right)-\chi {\phi }_{p}^{2}\right)& \left(19.48\right)\end{array}$
If ${N}_{p}$ is large the osmotic pressure depends only on the polymer concentration, and not on the degree of polymerization. The following result is obtained for ${N}_{p}$ >>1:

#### 19.10.1 Osmometry for Molecular Weight Determination

The number average molecular weight can be obtained by measuring the osmotic pressure of a dilute polymer solution. Using the dilute solution form of the Flory-Huggins expression for the solvent chemical potential (Eq. 19.45) gives the following for $\Pi$:
$\begin{array}{cc}\frac{\Pi }{RT}=\frac{{\phi }_{p}}{{N}_{p}{V}_{s}}+\left(1/2-\chi \right)\frac{{\phi }_{p}^{2}}{{V}_{s}}& \left(19.50\right)\end{array}$
The polymer volume fraction and overall solution volume (V) are given by the following expressions:
$\begin{array}{cc}{\phi }_{p}=\frac{{n}_{p}{N}_{p}{V}_{s}}{V}& \left(19.51\right)\end{array}$
Use of this expression for ${\phi }_{p}$ gives the following for $\pi$:
$\begin{array}{cc}\frac{\Pi }{RT}=\frac{{n}_{p}}{V}+\left(0.5-\chi \right){N}_{p}^{2}{V}_{s}{\left(\frac{{n}_{p}}{V}\right)}^{2}+...& \left(19.52\right)\end{array}$
Here it is helpful to rewrite ${n}_{p}/V$ in the following way:
$\begin{array}{cc}\frac{{n}_{p}}{V}=\left(\frac{w}{V}\right)\left(\frac{{n}_{p}}{w}\right)=\frac{C}{{M}_{n}}& \left(19.53\right)\end{array}$
where $c$ is simply the polymer concentration (kg/m ${}^{3}$ in SI units, g/cm ${}^{3}$ in the units that are more commonly used) and ${M}_{n}$ is the number average molecular weight (kg/mole in SI units). The osmotic pressure is therefore related to the number average molecular weight and concentration as follows:

#### 19.10.2 Scaling Theory of Osmotic Pressure

The previous derivation of the expressions for the osmotic pressure are useful, but they are not packed with physical insight regarding the actual meaning of osmotic pressure. Some simple physical arguments are more useful in this sense. We can start by making an analogy to the ideal gas law, which must be valid in the dilute regime. We start with the familiar ideal gas law:
$\begin{array}{cc}PV=nRT& \left(19.56\right)\end{array}$
$\begin{array}{cc}\Pi =\frac{CRT}{{M}_{n}}& \left(19.57\right)\end{array}$
In the dilute regime, molecular collisions are rare, and the pressure and free energy are dominated by the entropic penalty associated with confinement of the molecules into a fixed volume. Virial coefficients quantify the deviations from this ideal, dilute limit.
Below the overlap concentration, we can specify the polymer concentration by $\xi$, the average distance between polymer molecules in solution, as illustrated in Figure 19.14. The concentration of polymer molecules (molecules per volume) is $1/{\xi }^{3}$ and the osmotic pressure is given by the following expression:
$\begin{array}{cc}\Pi =\frac{RT}{{\xi }^{3}}& \left(19.59\right)\end{array}$
##### Exercise:
What is the overlap concentration poly(methyl methacrylate) with M = 100,000 g/mole in toluene?
##### Solution:
We can combine Eq. 19.58 for the overlap concentration with Eq. 14.18 for ${R}_{0}$ to obtain the following:
$\begin{array}{cc}{C}^{*}=\frac{M}{{R}_{0}^{3}{N}_{av}}=\frac{M}{{a}_{s}^{3}{N}^{1.8}{N}_{av}}=\frac{{M}_{0}}{{a}_{s}^{3}{N}^{0.8}{N}_{av}}& \left(19.60\right)\end{array}$
Note that we have used the self-avoiding walk expression for ${R}_{0}$, which is the relevant expression for good solvent conditions ( $\chi <0.5\right).$ PMMA has ${M}_{0}$ = 100 g/mole, so $N=1000$ for the polymer in our example. With these values of $N$ and ${M}_{0}$, and an assumed value of 7 Å ( $7×1{0}^{-8}$ cm) for ${a}_{s}$, we obtain ${c}^{*}=1.9×1{0}^{-3}\phantom{\rule{6px}{0ex}}g/c{m}^{-3}$. The density of toluene is slightly less than 1 g/cm ${}^{3}$, so this corresponds to a polymer weight fraction of a couple tenths of a percent.
The basic assumption of scaling theory of osmotic pressure is that $\Pi$ is still given by Eq. 19.59 at concentrations above the overlap concentration, but that $\xi$ has a different meaning in this regime. The approach works in the 'semidilute' solution regime where the overall concentration is still low ( ${\phi }_{p}$ < 0.25), but where there is substantial overlap between polymer molecules. Each overlap point represents a constraint that contributes an entropic contribution of ~ ${k}_{B}T$ to the free energy of the system. The osmotic pressure is therefore ~${k}_{B}T$ times the concentration of overlap points, or ${k}_{B}T/{\xi }^{3}$, where $\xi$ is the correlation length , or average distance between overlap points. As the solution gets more concentrated, the average distance between polymer contacts decreases., as shown schematically in Figure 19.15.
Figure 19.15: Change in correlation length with solution concentration.
To get the concentration dependence of the correlation length, we can start with the postulate that the corrections to the low-conentration form of the osmotic pressure must depend only on $c/{c}^{*}$, so that the osmotic pressure has the following form:
$\begin{array}{cc}\frac{\Pi M}{CRT}=1+f\left(C/{C}^{*}\right)& \left(19.61\right)\end{array}$
$\begin{array}{cc}\xi ={a}_{0}{\phi }_{p}^{-3/4}& \left(19.62\right)\end{array}$
Here ${a}_{0}$ is a typical monomer size, comparable in magnitude to the cube root of the molecular volume, or to the statistical segment length. This value for $\xi$ can then be combined with Eq. 19.59 to give the following expression for the osmotic pressure in the semidilute concentration regime:
$\begin{array}{cc}\Pi =\frac{{k}_{B}T}{{a}_{0}^{3}}{\phi }_{p}^{9/4}& \left(19.63\right)\end{array}$
The full expression for the osmotic pressure for dilute and semidilute concentrations is obtained by using this value of $\xi$ for $C>{C}^{*}$and adding the osmotic pressure that dominates in the dilute limit:
$\begin{array}{cc}\Pi =\frac{CRT}{M}+\frac{{k}_{B}T}{{a}_{0}^{3}}{\phi }_{p}^{9/4}& \left(19.64\right)\end{array}$
While derived in completely different ways, Eqs. 19.49 and 19.63 actually behave similarly in the concentration regime where they both apply. This is illustrated in Figure 19.16., where these two expressions are compared for a specific combination of ${a}_{0}$ and $\chi$.
Figure 19.16: Concentration dependence of the Flory-Huggins form for the osmotic pressure (Eq. 19.49) with the semidilute scaling form (Eq. 19.63), with $\chi =0.4$ and ${a}_{0}^{3}=4{v}_{s}$.

#### 19.10.3 Typical Magnitude of the Osmotic Pressure

The fundamental pressure scale for water is the thermal energy, $RT$ divided by the solvent molar volume, ${V}_{s}$. For water, we have (at 25 ºC):
$\begin{array}{cc}{V}_{s}=\frac{18\phantom{\rule{6px}{0ex}}g}{mole}\frac{1{0}^{-6}{m}^{3}}{1g}=1.8x1{0}^{-5}\phantom{\rule{6px}{0ex}}{m}^{3}=18\phantom{\rule{6px}{0ex}}c{m}^{3}& \left(19.65\right)\end{array}$
$\begin{array}{cc}\frac{RT}{{V}_{s}}=\frac{\left(8.314\phantom{\rule{6px}{0ex}}J/mole-K\right)\left(298\phantom{\rule{6px}{0ex}}K\right)}{3.0×1{0}^{-29}\phantom{\rule{6px}{0ex}}{m}^{3}}=1.4×1{0}^{8}\phantom{\rule{6px}{0ex}}Pa& \left(19.66\right)\end{array}$
This pressure is relatively large. The actual osmotic pressure is obtained by multiplying by a prefactor involving the Flory-Huggins interaction parameter and the polymer volume fraction. Osmotic pressures of many atmospheres (1 atm = $1{0}^{5}$ Pa) can easily be obtained.
Exercise: What is the osmotic pressure of seawater?
Solution: The osmotic pressure arises from all of the dissolved salts in seawater. Typical concentrations are tabulated below.[4] The total concentration of all dissolved salts is 1.12 moles/kg, which after multiplying by the density of seawater (1025 kg/m ${}^{3}$) gives an overall salt concentration of 1.15 moles/m ${}^{3}$. The osmotic pressure is obtained by multiplying this concentration by $RT$:
$\Pi =\left(8.314\phantom{\rule{6px}{0ex}}J/mol\cdot K\right)\left(300\phantom{\rule{6px}{0ex}}K\right)\left(1025\phantom{\rule{6px}{0ex}}kg/{m}^{3}\right)=2.6x1{0}^{6}\phantom{\rule{6px}{0ex}}Pa$
 Species moles/kg ${H}_{2}O$ 53.6 $C{l}^{-}$ 0.546 $N{a}^{+}$ 0.469 $M{g}^{2+}$ 0.0528 $S{O}_{4}^{2-}$ 0.0282 $C{a}^{2+}$ 0.0103 ${K}^{+}$ 0.0102 $B{r}^{-}$ 0.000844 $S{r}^{2+}$ 0.000091 ${F}^{-}$ 0.000068 Total Ions 1.12

### 19.11 Equilibrium Swelling of a Neohookean Material

$\begin{array}{cc}\Delta {f}_{d}=\left(\frac{{V}_{dry}}{{V}_{wet}}\right)\frac{{G}_{dry}\left\{{\lambda }_{x}^{2}+{\lambda }_{y}^{2}+{\lambda }_{z}^{2}-3\right\}}{2}& \left(19.67\right)\end{array}$
The The factor of ${V}_{dry}/{V}_{wet}$ accounts for the fact that as the volume of the gel increases due to solvent swelling, the concentration of network strands decreases accordingly. The extension ratios appearing in Eq. 19.67 are referenced to the undeformed, dry elastomer. To calculate the shear modulus of the solvent-swollen elastomer we need to consider a deformation process that occurs in two steps: istropic expansion of the gel to an equilibrium solvent-swollen state and shear deformation of this isotropically deformed gel.

#### 19.11.1 Equilibrium Swelling of the Gel

The geometry of an experiment used to measure the solvent swelling is shown in Figure 19.17. The polymer sample of interest is placed in a sealed container containing wither pure solvent (solvent activity, ${a}_{s},$equal to 1), or a solvent diluted with soluble polymer to give ${a}_{s}<1.$ When water is the solvent the solvent activity is simply the relative humidity. Elasticity of the gel introduces an elastic pressure, ${p}_{el}$ that limits the amount of solvent that is able to diffuse into the gel. To calculate this pressure we begin with the deformation free energy given in Eq. 19.67, with the following values for the extension ratios and relative volumes of the dry and wet states:
$\begin{array}{cc}{\lambda }_{x}={\lambda }_{y}={\lambda }_{z}\equiv {\lambda }_{s}& \left(19.68\right)\end{array}$
$\begin{array}{cc}{V}_{dry}/{V}_{wet}={\phi }_{p}={\lambda }_{s}^{-3}& \left(19.69\right)\end{array}$
Here ${\phi }_{p}$ is the volume fraction of polymer in the swollen gel.
Figure 19.17: Experimental geometry for obtaining the equilibrium swelling.
A general expression for the equilibrium swelling is obtained by writing the deformation free energy for the swelling (step 1 in the previous subsection) in terms of the volume:
The elastic pressure obtained directly from this expression by differentiating with respect to the sample volume:
$\begin{array}{cc}{\mu }_{s}=RTln{a}_{s}=RT\left\{ln\left(1-{\phi }_{p}\right)+{\phi }_{p}+\chi {\phi }_{p}^{2}\right\}+{p}_{el}{V}_{s}& \left(19.72\right)\end{array}$
The last term accounts for the fact that the pressure is larger inside the gel than outside the gel. This term involves the shear modulus of the dry material, equal to RT multiplied by the concentration of elastic strands in the network (Eq. 15.22, where we take $\beta =1$). This concentration can be expressed in terms of the average molar volume of a network strand, ${V}_{el}$. We further define ${N}_{el}$, an effective degree of polymerization of an elastic strand, as the ratio of ${V}_{el}$ to ${V}_{s}$:
$\begin{array}{cc}{G}_{dry}={\nu }_{el}RT=\frac{RT}{{V}_{el}}=\frac{RT}{{N}_{el}{V}_{s}}& \left(19.73\right)\end{array}$Now we can combine Eqs. 19.71, 19.72 and 19.17 to obtain an expression relating the solvent activity to ${\phi }_{p}$:
$\begin{array}{cc}ln{a}_{s}=ln\left(1-{\phi }_{p}\right)+{\phi }_{p}+\chi {\phi }_{p}^{2}+\frac{{\phi }_{p}^{1/3}}{{N}_{el}}& \left(19.74\right)\end{array}$
Figure 19.18: Example swelling curves for a fixed value of $\chi$ and different values of ${N}_{el}$ (part a) and for a fixed value of ${N}_{el}$ and different values of $\chi$ (part b).

#### 19.11.2 Shear Deformation of the Swollen Elastomer

Suppose that we now apply a shear strain to the swollen gel. The , $\gamma ={\lambda }_{1}-{\lambda }_{2}$, with ${\lambda }_{1}{\lambda }_{2}=1$ and ${\lambda }_{3}={\lambda }_{s}$ (see Section 15.2.1). For the swollen and sheared state we have ${\lambda }_{x}={\lambda }_{s}{\lambda }_{1}$, ${\lambda }_{y}={\lambda }_{s}{\lambda }_{2}$ and ${\lambda }_{3}={\lambda }_{s}.$
With these extension ratios we have the following expressions for volume ratio the free energy change from state 1 (the isotropically swollen state) to state 2 (the swollen and subsequently sheared state):
$\begin{array}{cc}\Delta {f}_{1\to 2}=\frac{{G}_{dry}}{2{\lambda }_{s}}\left\{{\lambda }_{1}^{2}-{\lambda }_{2}^{2}+2\right\}=\frac{{G}_{dry}}{2{\lambda }_{s}}\left\{{\lambda }_{1}^{2}-{\lambda }_{2}^{2}\right\}=\frac{{G}_{dry}}{2{\lambda }_{s}}{\gamma }^{2}& \left(19.75\right)\end{array}$
We differentiate twice with respect to $\gamma$ (see Section 15.2.5) to obtain the shear modulus of the swollen elastomer:
$\begin{array}{cc}{G}_{wet}=\frac{{d}^{2}}{d{\gamma }^{2}}\left(\Delta {f}_{1\to 2}\right)=\frac{{G}_{dry}}{{\lambda }_{s}}={G}_{dry}{\phi }_{p}^{1/3}& \left(19.76\right)\end{array}$
Note that the elastic swelling pressure calculated from the previous section is equal to the shear modulus of the solvent-swollen polymer.

## 20 Surfaces and Interfaces

### 20.1 Equilibrium Contact Angle

The shape of a liquid droplet on a solid surface is determined by a balance of forces where the solid, liquid and ambient environment (typically air) come into contact. These three phases meet at the contact line. The surface energy of the liquid ( ${\gamma }_{\ell }$), the surface energy of the solid ( ${\gamma }_{s}$) and the interfacial energy of the solid/liquid interface ( ${\gamma }_{sl}$) are related to one anther by the following expression, obtained by balancing horizontal forces at the contact line.
${\gamma }_{s}={\gamma }_{sl}+{\gamma }_{l}cos{\theta }_{e}$
Figure 20.1: Diagram of the equilibrium contact angle

### 20.2 Polymer/polymer Interfacial Free Energy

The interface between two immiscible polymers is characterized by a finite interfacial width, $w$. The actual value of $w$ is determined by a balance of enthalpic considerations, which favor a small value of $w$, and entropic considerations, which favor a large value of $w$. Below we obtain an expression for the interfacial energy as a function of $w$, and minimize this with respect to $w$ to get expressions for both $w$ and the interfacial free energy between immiscible A and B polymers.
The enthalpic contribution to the interfacial free energy is obtained by integrating the enthalpy per unit volume, $\Delta h$, associated with contacts between A and B repeat units in the interfacial zone. Here we use the same regular solution form that forms the basis of the Flory-Huggins expression for the free energy of mixing:
$\begin{array}{cc}\Delta h\left(z\right)=\frac{RT}{{V}_{0}}\chi {\phi }_{a}\left(z\right){\phi }_{b}\left(z\right)& \left(20.1\right)\end{array}$
$\begin{array}{cc}{\phi }_{a}\left(z\right)=1-z/w;& \left(20.2\right)\end{array}$ $\begin{array}{cc}{\phi }_{b}\left(z\right)=z/w& \left(20.3\right)\end{array}$
Figure 20.3: Linear approximation to the concentration profile across an interface between immiscible polymers .
These expressions result in the following for ${\gamma }_{\chi }$, the enthalpic contribution to the interfacial free energy. We refer to this contribution as ${\gamma }_{\chi }$ because it includes all of the thermodynamic interactions that are included in $\chi$. In addition to all the enthalpy of mixing, this parameter also includes the non-ideal entropy of mixing:
$\begin{array}{cc}{\gamma }_{\chi }=\stackrel{w}{\underset{0}{\int }}\Delta h\left(z\right)dz=\frac{RT\chi }{{V}_{0}}\stackrel{w}{\underset{0}{\int }}\left(\frac{z}{w}\right)\left(1-\frac{z}{w}\right)dz=\frac{RTw\chi }{6{V}_{0}}& \left(20.4\right)\end{array}$
A simple argument for the entropic contribution every time a random walk goes from one side of the interface to the other, it must turn around, so that we use one half of the available molecular configurations. If the number of molecular configurations is $\Omega$, the configurational entropy of associated with these configuration is ${k}_{B}ln\Omega$. If the number of configurations decreases by a factor or 2, then the entropy decreases by ${k}_{B}ln\left(2\right)$ and the free energy increases by ${k}_{B}Tln\left(2\right)$. Because our theory is already approximate, we'll assume that this free energy increase per forced turnaround of a polymer molecule is simply ${k}_{B}T$.
Our calculation of the entropic contribution to the free energy proceeds in the four steps outlined below:
1. Determine ${N}_{w}$, number of repeat units in random walk with ${R}_{0}=w$.
$a\sqrt{{N}_{w}}=w;$ ${N}_{w}={\left(w/a\right)}^{2}$
2. Calculate ${v}_{w}$, the volume of each of these random walks:
${v}_{w}={N}_{w}{v}_{0}={v}_{0}{\left(w/a\right)}^{2}$
3. Calculate ${\Gamma }_{w}$, the number of these random walks per unit area in the interfacial region:
${\Gamma }_{w}=\frac{w}{{v}_{w}}=\frac{{a}^{2}}{w{v}_{0}}$
4. multiply by ${\Gamma }_{i}$ by ${k}_{B}T$ to get ${\gamma }_{s}$, the entropic contribution to the interfacial free energy:
${\gamma }_{s}=\frac{{k}_{B}T{a}^{2}}{w}$
The overall interfacial free energy, ${\gamma }_{ab}$, is obtained by adding ${\gamma }_{\chi }$ and ${\gamma }_{s}$:
${\gamma }_{ab}={\gamma }_{\chi }+{\gamma }_{s}=\frac{{k}_{B}T}{{v}_{0}}\left(\frac{\chi w}{6}+\frac{{a}^{2}}{w}\right)$
The actual value of $w$ is obtained by minimizing $\gamma$ with respect to $w$. We obtain the following:
$w=a\sqrt{\frac{6}{\chi }}$
${\gamma }_{ab}=\frac{2a{k}_{B}T}{{v}_{0}}\sqrt{\chi /6}=\frac{2aRT}{{V}_{0}}\sqrt{\chi /6}$
This approximate approach gives the correct scaling ( $w\propto a{\chi }^{-1/2}$; ${\gamma }_{ab}\propto \frac{a{k}_{B}T}{{v}_{0}}{\chi }^{1/2}$), but the exact numerical prefactors are not obtained with quantitative accuracy with this approach. A more complete treatment results in the following for $w$ and $\gamma$:
$\begin{array}{cc}w=\frac{2a}{\sqrt{6\chi }}& \left(20.5\right)\end{array}$

### 20.3 Block Copolymer Morphologies

Block copolymers consisting of two different polymers characterized by a sufficiently large value of $\chi$ form a wide variety of interesting morphologies. In these cases two different polymer types that prefer to separate into different phases are covalently linked to one another along a single molecule. The morphologies observed for simple diblock copolymer molecules (a sequence of A repeat units linked to a sequence of B repeat units) is shown in Figure 20.4. If the length of the A block is much longer than the length of the B block, then the B blocks segregate into isolated spheres in a continuous matrix of A polymer formed from the A blocks. If the relative length of the B block is increased, the spheres transition to a hexagonal array of cylinders, and then to a morphology where the B cylinders form an interconnected, 3-dimensional network within the A matrix. If the A and B block lengths are similar in length, then the morphology is a simple alternating array of A and B lamellae, as shown schematically in Figure 20.5. An electron micrograph of a lamellar diblock copolymer morphology is shown in Figure 20.6.
Figure 20.4: Diblock copolymer morphologies.
The free energy per molecule of a lamellar-forming diblock copolymer is the sum of a chain stretching free energy, ${F}_{s}$ , and a free energy associated with the interface between the A and B domains. We refer to this interfacial free energy per chain as ${F}_{\gamma }$. For our purposes we assume that the A and B blocks have the same statistical segment length, $a$. The overall volume fractions of the two different blocks, ${f}_{a}$ and ${f}_{b}$, are given by the relative degrees of polymerization of the two blocks:
$\begin{array}{cc}\begin{array}{c}{f}_{a}=\frac{{N}_{a}}{{N}_{a}+{N}_{b}}=\frac{{N}_{a}}{N}\\ {f}_{a}=\frac{{N}_{b}}{{N}_{a}+{N}_{b}}=\frac{{N}_{b}}{N}\end{array}& \left(20.7\right)\end{array}$
Here $N$ is the overall degree of polymerization of the molecule,i.e., $N={N}_{a}+{N}_{b}$. The overall width of the A region of the block copolymer is ${f}_{a}\ell$, and the width of the B region is ${f}_{b}\ell$. The stretching free energy is obtained from the requirement that each block of the molecule must have an rms end-to-end distance equal to half the width of the respective region:
$\begin{array}{cc}\begin{array}{c}{R}_{0}^{a}={f}_{a}\ell /2\\ {R}_{0}^{b}={f}_{b}\ell /2\end{array}& \left(20.8\right)\end{array}$
We define ${f}_{a}$ and If the degrees of polymerization of the two blocks are ${N}_{a}$ and ${N}_{b}$, and the statistical segment length for each of the two different repeating units is $a$, we have:
The interfacial free energy is obtained by multiplying the interfacial free energy by the area per molecule along the A/B interface, which we refer to as $\Sigma$:
This area per molecule is given by dividing the molecular volume, $N{V}_{0}$, by the length per molecule, $\ell /2$:
The overall free energy per chain, ${F}_{tot}$, is given by the sum of ${F}_{s}$ and ${F}_{\gamma }$. Combination of Eq. 20.6 for ${\gamma }_{ab}$ with Eqs. 20.9, 20.10 and 20.11 gives:
$\begin{array}{cc}{F}_{tot}=\frac{3}{8}\frac{RT{\ell }^{2}}{N{a}^{2}}+\frac{2aNRT}{\ell }\sqrt{\chi /6}& \left(20.12\right)\end{array}$
The equilibrium value of $\ell$, which we refer to as ${\ell }_{eq}$ is obtained as the value of $\ell$ for which ${F}_{tot}$ is minimized. Setting $d{F}_{tot}/{d}_{\ell }$ to 0 gives:
$\begin{array}{cc}{\ell }_{eq}={\left(\frac{8}{3\cdot \sqrt{6}}\right)}^{1/3}a{N}^{2/3}{\chi }^{1/6}=1.03a{N}^{2/3}{\chi }^{1/6}& \left(20.13\right)\end{array}$
We can use the fact that ${R}_{0}={N}^{1/2}$a to write the expression for ${\ell }_{eq}$ in the following way:
$\begin{array}{cc}\frac{{\ell }_{eq}}{{R}_{0}}=1.03{\left(\chi N\right)}^{1/6}& \left(20.14\right)\end{array}$
We can now substitute ${\ell }_{eq}$ for $\ell$ in Eq. 20.12 to obtain the following expression for the equilibrium free energy per molecule in the lamellar microstructure:
$\begin{array}{cc}{F}_{eq}=1.19RT{\left(\chi N\right)}^{1/3}& \left(20.15\right)\end{array}$
Equation 20.15 gives us the molar free energy for a lamellar phase, but it still does not tell us if this lamellar phase has a lower free energy for the disordered phase, where the A and B blocks remain completely mixed. In order to do this we need to compare to the molar free energy in the disordered phase, where the A and B blocks remain completely mixed with one another. To do this we use the Flory-Huggins free energy of mixing (Eq. 19.28). Because the block copolymer is a one component system, there is no entropy of mixing, so we don't include the ideal entropy of mixing terms. We just have the term involving $\chi$, which accounts for the interactions between the A and B repeat units in the mixed, disordered phase. The free energy per molecule is obtained by multiplying the free energy per unit volume by the molecular volume, $N{V}_{0}$. We get the following simple expression for ${F}_{dis}$, the molar free energy of the disordered phase:
$\begin{array}{cc}{F}_{dis}=RT\chi {\phi }_{a}{\phi }_{b}=RT\chi {f}_{a}{f}_{b}& \left(20.16\right)\end{array}$
For a symmetric diblock copolymer with ${f}_{a}={f}_{b}=0.5$, ${F}_{dis}={F}_{eq}$ when $\chi N$ is equal to 10.4. When $\chi N$ is larger than this value, the ordered phase free energy given by Eq. 20.15 is lower than the disordered phase free energy given by Eq. 20.16. As a result we expect that an ordered lamellar phase will be formed for ${f}_{a}=0.5$ when $\chi N$ is greater than 10.4. A disordered phase will be formed for $\chi N<10.4$. This value of 10.5 is remarkably similar to the value of 10.5 that is obtained by a much more detailed calculation. These more detailed calculations are necessary in order to understand how the preferred geometry of the ordered phase (spheres, cylinders, etc.) depends on the ${f}_{a}$. A full calculation of the phase diagram is shown in Figure 20.7.
Figure 20.7: Block copolymer phase diagram (adapted from ref.[5]).

## 21 Case Study: Thermoreversible Gels from Triblock Copolymer Solutions

### 21.1 Introduction

Many of the principles of this course are illustrated by the behavior solutions of acrylic triblock copolymers in higher alcohols. The specific molecular structures of the polymer and the solvent are shown in Figure 21.1 and consist of a midblock of poly(n-butyl acrylate) (PnBA) between two blocks of poly(methyl methacrylate) (PMMA). The value of $\chi$ for the PnBA/solvent system is less than 0.5 for all temperatures of interest, but the $\chi$ for the PMMA/solvent system has the following temperature dependence:
$\begin{array}{cc}\chi =1.45-0.0115T& \left(21.1\right)\end{array}$
with $T$ in ${}^{○}$C.
The following points are illustrated by this example:
1. Polymers are soluble in solvents when $\chi <0.5.$
2. For $\chi >0.5,$ the solvent content within the polymer can be obtained from the chemical potential expression.
3. To understand the temperature dependence of the solubility you need to know the temperature dependence of $\chi$.
4. Polymer liquids have higher heat capacities than the corresponding polymer glasses.
5. Calorimetry can be used to detect the enthalpy recovery peak from an aged polymer glass.
6. Gelation occurs when the average functionality is $\approx$2.
7. Time-temperature superposition works when the structure does not change appreciably with Temperature.
8. The viscosity is obtained by integrating $G\left(t\right)$ or from the response at very low frequencies.
From Eq. 21.1 we see that the value of $\chi$ characterizing the PMMA/solvent interaction is greater than 0.5 for $T<83{\phantom{\rule{6px}{0ex}}}^{○}$C. At temperatures above 83 ${}^{○}$C the PMMA and PnBA blocks are both in good solvent conditions, and the polymer behaves as a normal polymer solution. At lower temperatures, however, the PnBA midblock remains in good solvent conditions, but the PMMA endblocks are no longer soluble. As a result these endblocks aggregate to form the micellar structures illustrated schematically in Figure 21.2. As a solution is cooled, $\chi$ increases and solvent is expelled from these aggregates. When sufficient solvent has been expelled from the PMMA aggregates, they become glassy, and the system behaves as a solid.
Figure 21.2: Schematic representation of the temperature and concentration dependence of the structure of the triblock copolymer solutions.

### 21.2 Thermoreversible Gelcasting of Ceramics

The transition from 'liquid-like' to 'solid-like' behavior as the temperature is changes enable these materials to be used in a variety of applications, including the ceramics processing application illustrated in Figures21.3 and21.4. In this application ceramic particles are added to the triblock copolymer at a relatively high volume fraction (around 50% by volume). The thermoreversible nature of the triblock copolymer solution allows the suspension to be poured into a silicone mold at high tempertures (typically $\approx 80{\phantom{\rule{6px}{0ex}}}^{○}$C). When the system is cooled to room temperature, the triblock copolymer forms a solid gel and the material can be removed from the mold. A ceramic material with the shape of the mold is obtained by letting the solvent evaporate from the suspension, and firing the object at elevated temperature to burn off the polymer and sinter the ceramic particles together.
Figure 21.3: Schematic representation of the alumina particles dispersed within a thermoreversible triblock copolymer gel.

### 21.3 Quantifying the Solid/Liquid Transition

The temperature dependence of the mechanical properties of the triblock copolymer solutions are quantified by the frequency dependence of the storage and loss moduli, ${G}^{\prime }\left(\omega \right)$ and ${G}^{\prime \prime }\left(\omega \right)$. It is conceptually simpler to think in terms of the time-dependence of the relaxation modulus, $G\left(t\right)$, which can be written as a sum of exponential relaxations:
$\begin{array}{cc}G\left(t\right)={\sum }_{i}{G}_{i}exp\left(-t/{\tau }_{i}\right)& \left(21.2\right)\end{array}$
In the frequency domain, the values storage and loss moduli are given by the following expressions:
The most important thing for us is the viscosity, which we can obtain in terms of the values of ${G}_{i}$ and ${\tau }_{i}$, or in terms of the loss modulus at very low frequencies:
$\begin{array}{cc}\eta ={\sum }_{i}{G}_{i}{\tau }_{i}={lim}_{\omega \to 0}\frac{{G}^{\prime \prime }}{\omega }& \left(21.5\right)\end{array}$
We generally obtain data as a master plot (where the frequency is multiplied by the temperature shift factor, ${a}_{T}$). In this case we obtain the temperature dependence of the viscosity by multiplying by ${a}_{T}$:
$\begin{array}{cc}\eta ={a}_{T}{lim}_{{a}_{T}\omega \to 0}\frac{{G}^{\prime \prime }}{{a}_{T}\omega }& \left(21.6\right)\end{array}$
We can also obtain an expression for the limiting modulus at high frequency (or low temperature):
$\begin{array}{cc}{G}_{0}={\sum }_{i}{G}_{i}& \left(21.7\right)\end{array}$
Finally, we define an average relaxation time, ${\tau }_{av}$ in the following way:
$\begin{array}{cc}{\tau }_{av}=\frac{\eta }{{G}_{0}}=\frac{\sum {G}_{i}{\tau }_{i}}{\sum {G}_{i}}& \left(21.8\right)\end{array}$
Values of the storage and loss moduli for the triblock copolymer solutions are shown in Figure 21.5, along with the values of ${G}_{i}$ and ${\tau }_{i}$ used to fit the data. The temperature dependent viscosity for a solution with ${\phi }_{p}=0.15$ is shown in Figure 21.6, and the temperature dependent relaxation times for a variety of solution concentrations are shown in Figure 21.7.
Figure 21.5: a) Master curves for ${G}^{\prime }$ and ${G}^{\prime \prime }$ (symbols) with fits to Eqs.21.3 and21.4. b) Values of ${G}_{i}$ and ${\tau }_{i}$ used to fit the data from part a (the relaxation times, ${\tau }_{i}$ are referred to by ${\lambda }_{i}$ in the figure). The reference temperature (defined as the temperature where ${a}_{T}=1$) is 65 ${}^{○}$C for these data.

### 21.4 Characterizing the Glass Transition in PMMA domains

The glass transition is typically observed as step change in heat flow as a sample is cooled through the glass transition. In our case the PMMA domains go through a glass transition during cooling, but the PMMA concentration is so low that the glass transition is not clearly observed during cooling or when when the sample is immediately reheated after cooling (see Figure 21.8). However, a large enthalpy peak is observed after the sample has been left to 'age' at room temperature for a long time before reheating the sample. As shown in Figure 21.9, the enthalpy peak gets larger and moves to higher temperature as the aging time at room temperature is increased. This peak has the appearance of a melting peak associated with crystallization, but in this case it arises from the glassy PMMA domains.
The origins of the enthalpy peak observed during the aging experiments can be understood by realizing that the glass transition separates an equilibrium, liquid regime at temperatures above ${T}_{g}$ from a non-equilibrium, glassy regime at temperatures below ${T}_{g}$. At temperatures below the glass transition, the enthalpy content of the sample is higher than the equilibrium enthalpy content defined by the extrapolation of the liquid behavior (the dashed line in Figure 21.10). As a result, the enthalpy slowly decreases toward this equilibrium line as the sample is aged below the glass transition. The decrease in enthalpy during this aging process is labeled as $\Delta {H}_{a}$ in Figure 21.10. When the sample is reheated, the enthalpy increases with temperature according to the glassy heat capacity, ${C}_{p}^{g}$. As a result the enthalpy content eventually crosses the equilibrium line and becomes less than the equilibrium enthalpy content. At a temperature somewhere just above ${T}_{g}$, the sample is able to equilibrate, and the enthalpy increases by an amount $\Delta {H}_{r}$ in order to catch up to the equilibrium value. This is the enthalpy corresponding to the peaks in Figure 21.9. As the aging time decreases, $\Delta {H}_{a}$ and $\Delta {H}_{r}$ both decrease, and the temperature at which the enthalpy is recovered moves closer to ${T}_{g}$. The actual value of the ${T}_{g}$ can be estimated by measuring the area of the enthalpy recovery peak and plotting against the location of this peak. Doing this for the data shown in Figure 21.9 results in an estimate for ${T}_{g}$ of the PMMA domains of 35 ${}^{○}$C.
Figure 21.8: Measured heat flow during cooling, immediate reheating, and reheating after room temperature storage for 130 days.
As a final check on all of this, we can see if it makes sense that the glass transition of the PMMA domains should be around 35 ${}^{○}$C. The glass transition of PMMA is about 125 ${}^{○}$C, which is certainly a lot higher than the value of 35 ${}^{○}$C that we are claiming here. However, we need to remember that the PMMA domains contain a lot of solvent, and this solvent will depress ${T}_{g}$ significantly. How much solvent do we expect there to be in the PMMA domains? From Eq. 21.1 we obtain $\chi$=1.05 at $T=35{\phantom{\rule{6px}{0ex}}}^{○}$C. The relationship between $\chi$ and ${\phi }_{s}$, the solvent volume fraction in the PMMA cores of the micelles is obtained by setting ${N}_{p}$ to $\infty$ and setting the solvent chemical potential (from Eq19.43) to 0:
$\begin{array}{cc}\begin{array}{c}ln\left({\phi }_{s}\right)+1-{\phi }_{s}+\chi {\left(1-{\phi }_{s}\right)}^{2}=0\\ ln\left(1-{\phi }_{p}\right)+{\phi }_{p}+\chi {\phi }_{p}^{2}=0\end{array}& \left(21.9\right)\end{array}$
This equation needs to be solved numerically to obtain ${\phi }_{s}$ (or ${\phi }_{p}$) from a specified value of $\chi$. A simple MATLAB script used to do this is given in Section25.2. A value of 1.05 for $\chi$ gives ${\phi }_{s}$=0.289. Independent measures of the glass transition temperature of bulk PMMA samples with solvent loadings comparable to this do in fact show measured glass transition temperatures close to 35 ${}^{○}$C. The relationship between $\chi$ and ${\phi }_{p}$ given by Eq. 21.9 is plotted in Figure 21.11.
Figure 21.11: Relationship between ${\phi }_{p}$ and $\chi$ obtained from Eq. 21.9.

### 21.5 Concentration Dependence of the Gel Modulus

Elasticity of the triblock copolymer solutions at low temperatures arises from the fact that the middle, PnBA blocks of the copolymer can span different PMMA aggregates, thereby linking the whole structure together. The PMMA aggregates behave as physical crosslinks with a functionality given by the number of midblocks that that bridge different aggregates. This functionality is obtained by multiplying the aggregation number,i.e. the number of PMMA endblocks in a single aggregate, by the probability $f$ that a PnBA midlbock spans two different PMMA aggregates (we also have to divide by two to account for the fact that there are two PMMA blocks on each triblock copolymer molecule). Aggregation numbers depend on the polymer concentration, and are typically very large, as illustrated in Figure 21.12. For high molecular weight polymers the percolation threshold, where the average functionality of a micelle is 2, is quite low, corresponding to ${\phi }_{p}\approx 0.035.$ Above this percolation threshold, the shear modulus is given by the following expression:
$\begin{array}{cc}{G}_{0}=\nu {k}_{B}T\frac{{D}^{2}}{{R}_{0}^{2}}& \left(21.10\right)\end{array}$
Here $D$ is the average distance between micelle cores, which can be measured experimentally by x-ray scattering. We also have ${R}_{0}={N}^{1/2}a$, where $N$ is the midblock degree of polymerization and $a$ is the statistical segment length for the midblock. The quantity $\nu$ is the conentration of 'load bearing strands', which in this case is the concentration of triblock copolymer chains with bridging midblocks:
$\nu =\frac{f{\phi }_{p}\rho {N}_{av}}{M}$
where $M$ is the molecular weight of the triblock copolymer molecule and $\rho$ is the polymer density. The modulus is strongly concentration dependent because of the concentration dependence of the bridging fraction, $f$, which is shown in Figure 21.13. Measured and calculated values of ${G}_{0}$ are shown in Figure 21.14.
Figure 21.12: Concentration dependence of aggregation number for three different PMMA-PnBA-PMMA block copolymer solutions. The three data sets correspond to three differnt triblock copolymers, with different molecular weights for the PMMA and PnBA blocks.

### 21.6 Hydrogels: Water as the Solvent

As a final illustration of what can be done with these sorts of triblock copolymer gels, we consider materials where the midblock is replaced with poly(acrylic acid), a polymer that is water soluble at neutral pH. The structure of these polymer is shown in Figure 21.15. This figure also shows a scheme for forming gels from these materials. Instead of adjusting $\chi$ between the solvent and the PMMA endblocks by changing temperature, we do this by adding a small amount of water to the solvent (which is initially dimethyl sulfoxide). Addition of just a small amount of water increases the effective value of $\chi$ characterizing the solvent/PMMA interaction. The result is that the relaxation times for the triblock copolymer solution increase dramatically, as illustrated by the rheological data in Figure 21.16. The effects of solvent composition can be illustrated by introducing a shift factor, ${a}_{s}$ that depends on the composition of the solvent. It's use is illustrated in Figure 21.17. The viscosity at a given temperature is proportional to ${a}_{s}$, so we see that small increases in the water content of the solvent result in an increase in the solution viscosity by several orders of magnitude. This occurs because water induces the aggregation of PMMA blocks in to discrete domains, just as reducing temperature did for the case where alcohol was used as the solvent.
Figure 21.15: Formation and ionic crosslinking of PMMA-PMAA-PMMA triblock copolymer hydrogels (PMAA = poly(methacrylic acid).
A useful feature of the hydrogels formed from the triblock copolymers with the poly(acrylic acid midblocks) is that the properties can be changed dramatically by immersing the gels in solutions containing divalent cations like Ca ${}^{2+}$ or Zn ${}^{2+}$. These ions add additional crosslinks to the material, increasing the modulus by a factor of 100 or more (Figure 21.18). The materials also have high toughness, and can be extended to several times their original length prior to failure (Figure 21.19).
Figure 21.18: Elastic modulus for triblock copolymer gels that have been ionically crosslinked to varying degrees.

## 22 Summary

### 22.1 Classification Scheme

Figure 22.1: Summary of polymer properties and classes

### 22.2 Molecular Weight Averages

N: Degree of polymerization
M: Molecular weight
${M}_{0}$Molecular weight per repeat unit
${M}_{n}=w/n$

${M}_{w}=\frac{1}{w}\sum _{i}{w}_{i}{M}_{i}=$Weight Average Molecular
w = total weight of polymer
n = total number of polymer molecules
${w}_{i}$ = weight of fraction i (grams)
${M}_{i}$ = molecular weight of fraction i

### 22.3 Polymerizations

#### 22.3.1 Linear step growth polymerization

${M}_{n}=\frac{{M}_{0}}{2\left(1-p\right)}$
$\frac{{M}_{w}}{{M}_{n}}=1+p$
$n={n}_{a}-{n}_{b}$
(p = extent of reaction)

#### 22.3.2 Nonlinear step growth polymerization

${M}_{n}=\frac{{M}_{0}}{2-p{f}_{av}}$
$\left({n}_{a}+{n}_{b}\right){f}_{av}={n}_{a}{f}_{a}+{n}_{b}{f}_{b}$
${f}_{a}$, ${f}_{b}$ = functionality of A and B monomers
${n}_{a}$, ${n}_{b}$ = numbers of A and B monomers
$p{f}_{av}$ = 2 at the gel point

#### 22.3.3 Chain Growth Polymerization

Active site is typically a free radical, anion, or cation.
Propagation corresponds to addition to a double bond, or opening of a ring.
Termination can occur by Disproportionation or combination.
Intermolecular chain transfer reduces molecular weight.
Intramolecular chain transfer results in chain branching.

### 22.4 Chain Dimensions

#### 22.4.1 Structure of Amorphous Polymers

The probability density distribution for the end to end vector, R (with components ${R}_{x}$, ${R}_{y}$ and ${R}_{z}$) is:


Here a is the statistical segment length, defined by the following expression:


#### 22.4.2 Structure of amorphous vinyl polymers

${R}_{0}^{2}={C}_{\infty }{N}_{b}{l}^{2}=N{a}^{2}$