Chemistry C3102-2005: Polymers Section

Dr. Edie Sevick, Research School of Chemistry, ANU

2.0 Thermodynamics of an Ideal Chain: Stretching and Squashing

In this section, we begin to investigate the thermodynamics of polymers, starting ﬁrst, with an isolated ideal chain. We begin with the general relation

∆F = ∆U T ∆S (1) − where the first term on the RHS is the change in the free energy, ∆U is the change in the internal energy of the chain, and ∆S is the change in entropy. Examples of interactions which contribute to ∆U range from Lennard-Jones interactions to complex interactions which scientists might arbitrarily propose, all of which are of the form U = f(rij) where rij is a vector of distances between monomers, solvent, or other particle or surface which might be included in the system. By definition, an ideal chain suffers NO interactions and contributes nothing to the internal energy. Consequently, if we take an ideal chain and end-tether it to a phantom wall, or change the solvent, then the change in the internal energy of the system is 0. On the other hand, the chain has considerable disorder, measured by the number of configurations that it can adopt and quantified by the thermodynamic quantity entropy. If we were to hold the two ends of the chain a distance Na apart so that the chain was taut, we would considerably decrease the number of configurations that the chain could adopt. If S1 is the entropy of the ideal chain in its relaxed, or natural state, and S2 the entropy of the chain in its taut state, then S1 >> S2 or

∆S = S2 S1 would be negative, and by the above equation, ∆F , would be positive. That is, − in stretching the chain to its taut state, we would be performing work on the system to increase the free energy of the system. If we were to release the ends of the chain, the chain would spontaneously relax to its natural state that maximises the number of available configurations; that is the system maximises its entropy and minimises its free energy. Thus, in order to describe the thermodynamics of a chain, we need only quantify the entropy of the chain. In this section, we will investigate the thermodynamics of an ideal chain when it is stretched and when it is squashed. The force of stretching and squashing is determined solely by entropy- such forces are referred to as entropic or elastic forces. Boltzmann’s Principle (after Ludwig Boltzmann (1844-1906)) states that the entropy of the system is S = kB ln[p(Σ)], where p(Σ) is the probability distribution of configurations. Imagine that we could actually count the number of configurations of a chain in its natural and fully- stretched states, then the change in entropy would be ∆S = kB ln [Γ/Γ0] where Γ is the number of taut configurations and Γ0 is the larger number of configurations in the chain’s natural state.

1 EXAMPLE PROBLEM

What is the change of entropy associated with an ideal chain of N monomers •• forming a ring?

From Boltzmann’s principle, we ﬁnd that the entropy change for ring formation is

Γring ∆S = Sring S0 = kB ln ( ) (2) − Γ0 where the subscript 0 indicates an FJC chain possessing the same number of monomers, of the same size, but whose ends are not constrained. Γring ad Γ0 are the total number of configurations of the ring and the unconstrained chain, respectively. Following section 1.2, we can express Γ in terms of the number of possible random walks that arrive at a destination m. In 1-dimension, the total number of configurations available to the random walker is 2N , as at each step the walker has 2 possible directions. In order for the walker to form a ring, he must return to his starting point, or in the language of section 1.2, m = 0 and n+ = n−. The number of such ring configurations is N! N N (3) ( 2 )!( 2 )! Therefore the probability of observing a random walker returning to his starting point (in 1-dimension) is 1−D 1 N! 2 1/2 pring = N N N =( ) (4) 2 × ( 2 )!( 2 )! πN where the last term on the right hand side resulted from application of Stirling’s approximation, which is valid in the limit of large N. You should see that this formula also results from eq. 29 in the notes, by replacing m with 0. Likewise, you can transform from discrete destinations to a continuous, end-to-edn distance in three dimensions as achieved on the top of page 7 of the class notes. The result is that the probability of observing a ring in three dimensions is

− 3 p3 D =( )3/2. (5) ring 2πNa2 Therefore, the entropy change upon ring formation of a chain of N statistical monomers, each of size a, is 3k 3 ∆S = k ln p3−D = B ln ( ) B ring 2 2πNa2 3k = B [ln (3) ln(2πNa2)]. 2 − As N is large, ln(3) << ln(2πNa2), and ∆S is negative, that is, entropy is lost in the process of ring formation.

EXAMPLE PROBLEM

2 2.1 Stretching a chain

Take an ideal chain, grab each of the free ends, and then extend or stretch the chain by pulling the ends to progressively larger values of R, the end-to-end distance. We want to know the force needed to stretch the chain to half of its contour length. First, lets express the above equation in diﬀerential terms: dF = TdS, where dU = 0 by deﬁnition. Then, since work, w, − is equal to the force, f, integrated over the distance, R dRf, we can write R0 dF dS f = = T (6) −dR dR We already have an expression of the entropy in terms of the end-to-end distance from the Gaussian distribution of end-to-end distance, p(R):

3 3/2 3R2 S = k ln (p(R)) = k ln ( ) exp ( ) (7) B B 2πNa2 −2Na2 We can simplify this to 3 3R2 S = k ln( )3/2 k ( ) (8) B 2πNa2 − B 2Na2 2 Taking the derivative of S with respect to R gives 3kBR/(Na ) as the ﬁrst term on the RHS vanishes as it has no R dependence. Thus, the entropic or elastic force needed to stretch a chain to an end-to-end distance R is R f 3k T (9) ∼ − B Na2 Notice that f is a vector quantity, and in the direction opposite to the end-to-end vector. In other words, f is a restoring force.

EXAMPLE PROBLEMS

An ideal polymer chain is end-tethered to a ﬁxed wall and a weight of mass • m is suspended from the free-end of the chain, as pictured below. The weight is suﬃciently small that the elongation of the chain, x, is between the natural size of the chain and its contour length, (N)a

First, we assume that the force of extension or stretching force is entirely entropic. We can derive the force, f, versus extension, x, relation starting from simple thermodynamics:

E = U T S, − where the elastic force, f, is x dE dS f = = T − dx dx

3 The entropy is given by Boltzmanns principle and the Gaussian probability distribution

SkB ln (P (x)) dS d 3 3/2 3x2 = k ln exp . dx B dx h2πNa2 − 2Na2 i Using the identity ln (ab) = ln a + ln b, the last equation becomes dS d 3 3/2 3x2 = k dx B dxh2πNa2 − 2Na2 i , or dS 3x = k dx − B Na2 . Therefore, the equation for the force versus extension is 3x f = k T − B Na2 Now the force on the chain is simply the mass of the weight times gravitational acceleration, or f = mg, so the sought expression is 3x mg = k T − B Na2 .

Continuing from above, with the weight attached to the free-end of the chain, the • ambient temperature is increased. What happens to the extension of the chain, and why?

From the above equation, you see that the LHS is constant the weight on the free-end of the chain does not change. Thus, if the temperature is increased, then the extension must proportionally decrease. Thus this above result indicates that a stretched chain will contract upon heating. Why? It is primarily an effect due to entropy. You know that entropy rather than interactions dominate because (1) when temperature is increased, the importance of interactions decrease as the magnitude of the interaction, ε, becomes smaller compared with thermal energy,i.e., ε/(k T ) < 1; (2) from E = U TS, you see that increasing the temperature increases the B − role of entropy; (3) the f versus x relation is purely entropic. When temperature is increased, the effect on each statistical monomer is to increase its thermal energy. In other words each statistical monomer becomes more randomised with increased temperature or the entropy of the chain is increased. We know that the conformational entropy of an ideal chain is decreased as its ends are separated apart: or that the ideal chains entropy increases as its end-to-end distance decreases. Thus, with an increase in temperature, there is a randomisation of the statistical monomers, an increase in the chains entropy and a decrease in the end-to-end distance (the chain shortens). Having established that this is an entropic effect, one can analyse this in terms of the exchange of heat and work. Imagine that this chain is now a rubber band. In this problem,

4 heat from the surroundings (δqrev = TdS) is transferred to the rubber band and used to do work (lift the weight as the rubber band contracts) and to increase its temperature. In order for the chain to increase its entropy it must contract as a stretched chain has fewer available conformations than a partially-stretched or unstretched one. In fact, you can write down the entropy change as 2 2 Tf ∆S kB(xi xf )+ Cv ln ∼ − Ti where the ﬁrst term corresponds to the conformational entropy change, from

P (xf ) ∆S kB ln ∼ P (xi) and the second is due to the change in the temperature of the system with Cv being a heat capacity. As xi < xf , the conformational entropy change is increased by a contraction of the chain. This is completely analogous to the entropy associated with the expansion of an ideal gas in a perfectly ﬂexible, conductive container

Vf Tf ∆S nR ln + nCv ln ∼ Vi Ti where the ﬁnal container volume, Vf , is greater than the initial container volume, Vi. An alternative demonstration is to adiabatically stretch/compress the chain; you can achieve this by stretching the chain quickly so that there is no time for heat exchange with the surroundings. To stretch a rubber band, you need to do work on it since there is no heat exchange δq = 0 and ∆S = 0, the work goes into heating the rubber band. This is again analogous to adiabatically compressing an ideal gas: the work done on the gas is manifest as a temperature rise. When the rubber band is retracted or the ideal gas expands, each does work on the surroundings and, as no heat is transferred from the outside, the systems internal energy is used, so it cools down. You could ﬁnd the respective temperature changes by setting ∆S = 0 in the above equations..

EXAMPLE PROBLEM

5 2.2 Squashing a chain

Consider an ideal chain sandwiched between two infinite, impenetrable plates which form an slit of thickness H. We want to know the force needed to compress the plates such that the slit distance is reduced from Na to a. This is an entropic force as there are no interactions between the plates and polymer, except excluded-volume interactions that limit the number of configurations that the chain can adopt. We cannot easily use the above method to determine the force as we do not know how p(R) depends upon the slit distance H. However, we can assume a simple model of a chain on a lattice so that we can “count” the number of configurations available to the chain at any slit thickness. First, assume that the ideal chain must traverse the scaffolding of a lattice. We could pick any lattice, but a square (2-D) lattice, or 4-coordinate lattice, is convenient to draw (as you will see, the choice of lattice does not matter in the derivation). Each statistical monomer must adopt one of four vertices on the lattice. Thus, starting from the first monomer, the chain has 4 possible vertices to choose from, the second monomer has another 4 possible vertices to choose from, etc. Thus, the total number of configurations available to a chain of N statistical monomers on a 4-coordinate lattice is 4N . We can now generalise the lattice to any regular lattice with z vertices or a “z-coordinate” lattice, where zN configurations are available. Now imagine that this lattice is of infinite extent in two directions, but has a thickness H that we will vary. How does the thickness of the scaffolding reduce the number of configurations of the chain? As each statistical monomer has z possible vertices to traverse, a statistical monomer which is located at the boundary has only 1 possible vertex - it must “reverse” its direction. Thus, the number of possible configurations is zN−X where X is the number of monomers which contact the boundaries. If ∆S is proportional to the logarithm of the ratio of the number of configurations in an H-thick slit to that in an infinitely-thick slit, then zN−X ∆S = k ln ( )= k X ln (z) (10) B zN − B × We need to develop an expression that relates X, the number of contacts the chain makes with the boundaries, to H, the separation between the boundaries. If we were to snip the chain at all points of contact with the boundaries, the chain snippets would be on average m monomers long with end-to-end distances of exactly H. As the chains are ideal, the general scaling relationship for chain size holds: H2 = ma2. The number of chain snippets is equal to the number of contact with the boundary, or X. The number of chain snippets is N/m, or replacing m with H2/a2, X = Na2/H2. Therefore, Na2 ∆S = k ln (z) (11) − B H2 × Now the factor ln (z) is just a constant, independent of H, so we will drop this factor as we only care about how the force changes with H and not necessarily its magnitude. The change in free energy in going from an infinite slit to one of thickness H is ∆F = T ∆S k T Na2/H2. − ∼ B × Since F (H = ) = 0, we can equate F (H) = k T Na2/H2. Then force of compression at ∞ B × any given H is f = dF/dH, or − Na2 f k T (12) ∼ B H3

6 EXAMPLE PROBLEMS

The free ends of an ideal chain are end-tethered to two parallel, volume- •• excluding walls. Construct (graph) the force proﬁle associated with changing the separation of the walls from H = a to H = Na.

There are two regimes to consider

Chain squashing over a

at a plate loses a bit of entropy, of order kB, as demonstrated in the lattice model of section 2.2 of the notes. So we simply need to ﬁnd the number times a Gaussian chain

of N monomers contacts the of the chain and multiply this by kB to find the decrease in entropy. Any contiguous segment of a Gaussian chain also obeys Gaussian statistics (as long as the segment is sufficiently long). Thus consider that the squashed chain is severed at each of the contact points and let m be the average number of monomers in the segments. Becuase each segment obeys Gaussian statistics, and the end-to-end distance of each segment is determined by the slit separation, H, we can say that H2 ma2 or ∼ that m scales as (H/a)2. The number of times the chain is severed N/m, or Na2/H2. ∼ Thus, the entropy reduction is k T Na2/H2. Then, B × Na2 ∆F = F (H) F ( ) k T 0 (13) − ∞ ∼ B H2 − where we set F ( ) = 0 since there is no configurational constraints when H . The ∞ →∞ entropic compressing force is then dF (H) Na2 f = k T (14) − dH ∼ B H3 where we have dropped all numerical prefactors of order 1. Note that the free energy of the chain increases as the slit distance decreases; i.e., dF/dH < 0 so that the force is | | positive, pushing the plates apart. This squashing force holds for H such that O(100) < m < N. The lower bound, order of 100, is required so that the segments still obey Gaussian statistics and the upper bound ensures, on average, one contact. This range O(100)

O(100)

7 Scaling predictions of entropic forces for Scaling predictions of entropic forces for a chain of N=400 statistical monomers a chain of N=400 statistical monomers

500 0 0.4 0.4

0.2 0.2 400 -0.2 0 0

300 -0.4 -0.2 -0.2

200 -0.6 -0.4 -0.4

-0.6 -0.6 100 -0.8 -0.8 -0.8

0 -1 -1 -1

-100 -1.2 -1.2 -1.2 1 10 100 1000 10 100

H/a H/a

(a) (b)

Scaling predictions of entropic forces for Scaling predictions of entropic forces for a chain of N=400 statistical monomers a chain of N=400 statistical monomers

0.4 0.4 0.4 0.4

0.2 0.2 0.2 0.2

0 0 0 0

-0.2 -0.2 -0.2 -0.2

-0.4 -0.4 -0.4 -0.4

-0.6 -0.6 -0.6 -0.6

-0.8 -0.8 -0.8 -0.8

-1 -1 -1 -1 10 100 10 100

H/a H/a

Figure 2: Versions of the force profile, fa/kBT versus H/a for squashing (red) and stretching (blue) from (a), (b) draft to final shown in Figure 2(b). In all plots, I adopted N = 400 and I assumed that the order of 1 constant in each expression was unity. In (a) I did not worry about the regimes and I plotted both forces over all H/a but noted on the horizontal coordinate that 10 < H/a < √N = 20 is the squashing regime while √N = 20 < H/a < N = 400 is the stretching regime. In subsequent versions (b)-(d), I simply replot versions to get a proper force profile. (b) Here I’ve changed the horizontal coordinate bounds to 10 < H/a < 400 so that only the Gaussian squashing and stretching regimes are shown. (c) I added constants to each force profile to make the stretching and squashing forces equivalent (and arbitrarily zero) at H = √Na. (d) I eiliminated the stretching points in the squashing regime, and the squashing points in the stretching regime. All done in Kaleidagraph, changing the Kaleidagraph output to .eps format using Illustrator, and inserting them in my Latex document.

8 For a

Chain stretching over √Na

How does the above force profile change if the chain is replaced by a chain of the •• same exact contour length, but is less flexible or more stiff (has a larger persistence length)? Provide a qualitative description

The force and slit distance were appropriately scaled to dimensionless quantities so that the the plots are unchanged by changing the statistical monomer size, which is twice the persistence length. The slope of the force proﬁle, Fig. 1(d), is proportional to 1/N and N, in the stretching and squashing regimes, respectively and independent of the persistence length.

Answer the above question quantitatively (provide a comparative graph) for a •• chain whose persistence length diﬀers by a factor of 5

For a more stiff chain, where the persistence length is 5 times longer, the monomer size would also increase by a factor of 5. Therefore at any H/a, the value of fa/kBT is unchanged, but the force f is 1/5 as large. This is because a stiff chain has fewer conformations, and consequently less entropy to loose. The range of H over which entropic forces are estimated is increased as the chain is larger in size. For a more flexible chain, a would decrease by a factor of 5. Again, at any H/a, the value of fa/kBT is unchanged, but the force f is 5 times as large. This is due to a larger configurational entropy. The range of H over which these entropic forces are estimated is decreased as the chain is small in size. EXAMPLE PROBLEM

9 Summary

In this section, we’ve extended our description of the ideal chain to include thermodynamic predictions of stretching and squashing.

From the Gaussian distribution (or any distribution) of chain sizes, or any way to count the • conﬁgurations of a chain, we can evaluate the thermodynamic entropy using Boltzmann’s principle.

For an ideal chain, ∆U = 0 and the energy of the chain is entirely entropic. • The force of stretching an ideal chain is Hookean; i.e., it is linear with extension. • The force of compression or squashing on an ideal chain scales as the inverse compression • distance, cubed.

10 3.0 The size of chains in good and poor solvent conditions

Obviously, the ideal chain is a simple, first approximate model of an isolated polymer chain. It assumes that the chain is intersectable with itself and inert; i.e., it suffers no energetic interactions with itself or with the surrounding solvent. In this section, we are going to lift these assumptions. But we will do so separately for poor solvent and for good solvent conditions. In general, a poor solvent is one in which the solute (in this case the isolated polymer chain) precipitates. A solute precipitates out of solution so as to avoid energetically unfavourable solute-solvent interactions. In such a case the solute-solute interactions are more favourable than the solute-solvent interactions. Thus precipitation in a poor solvent will lower the internal energy (∆U = U − U − < 0). In contrast, good solvents are those where the solute solute − solute solvent solute-solvent interaction is more favorable than the solute-solute interaction. In this case the solute (or polymer chain) will lower its energy by solubilising in the solvent medium. In this section we ask what effect poor solvent and good solvent have upon the size of a polymer chain.

3.1 A chain in poor solvent

We start here with a FJC chain in a poor solvent. In this case, monomer-monomer interactions are favoured over monomer-solvent interactions. Thus, the chain will adopt a configuration that minimises contact with the poor solvent and maximise self-contact. Very simply, a long, flexible chain will “collapse” upon itself, forming a dense monomer mass that excludes solvent molecules from its interior. In its collapsed state, the chain cannot completely avoid contact with the solvent, but if it adopts a spherical collapsed state, then it will minimise its solvent-monomer interactions. In terms of the free energy of the chain, ∆F = ∆U T ∆S, you can see that − the collapse of the chain will lower ∆U simply because Umonomer−solvent > Umonomer−monomer. However, what will happen to the chain’s entropy? In going from a solvated, fully flexible chain to a collapsed chain with a dense core, the chain will lose entropy, i.e., ∆S < 0. Thus there are two competing effects: the interactions with the solvent will promote collapse (i.e., the interactions contribute to a lowering of the free energy), but the loss of entropy will actually suppress the chain collapse (i.e., entropy contributes to an increase in the free energy). Indeed, the balance of these two effects is determined by the temperature. Given that the temperature is low, such that the chain adopts a collapsed configuration, how does the size of the chain vary with the number of monomers in the chain? The volume of a chain is simply the contour length, Na, times the cross-sectional area of the chain, which is πr2 where r is the cross sectional radius of the chain. The value of r will be on the order of a, so we can say that the volume of the chain is V Na3, where the prefactor depends upon chain ∼ the cross-sectional area of the chain contour. In its collapsed state, a single chain will form a 3 spherical globule of volume Vsphere = (4/3)πR where R is the radius of the sphere. If the chain collapses so as to expel all monomer, then this spherical volume is comprised of polymer only, or V = V . That is, Na3 R3, or the size of the collapsed chain scales with the cubed chain sphere ∼ root of the number of monomers: R aN 1/3 (18) ∼ This geometrical argument works fairly well for flexible chains: implicitly, it lifts the simplifying

11 approximation of a self-intersectable chain. If a chain were intersectable, it would intersect itself over and over again to form the smallest volume (smallest surface area) which is a point. However this analysis does not work for a rigid rod chain, as it has NO flexibility and will change its shape minimally when in the presence of a poor solvent. What about a semi-flexible chain like DNA? Recall that we said such chains were like a garden hose - imagine collapsing such a hose into a dense ball - it would be very difficult. Indeed, experiments have shown that when subjected to poor solvent conditions, DNA coils up, very much like a neatly wound garden hose, forming a doughnut (or donut), or more mathematically, a torus. This torus is the shape that minimises the solvent contact with the semi-flexible DNA chain.

3.2 A chain in a good solvent

A chain in a good solvent will adopt a configuration that maximises contact with the solvent. That is, the chain will adopt a random walk configuration, with the exception that it will not self-intersect itself. This is called a self-avoiding random walk or SAW. There are a handful of ways of deriving the scaling relationship for the size of a chain in good solvent. Here we will follow the so-called “Flory” derivation (after Paul Flory, American Chemist and Nobel Laureate)¿ Flory used a simple thermodynamic argument: the chain will adopt an average size that will minimise its free energy: F = U TS (19) − We need to cast each term on the RHS in terms of the chain size. The entropy term is rather straightforward. We said that the solvent-swollen chain is very similar to a random walk chain with the exception that it does not have self-intersections. As you might imagine, the entropy of a SAW of N steps is very similar to a self-intersecting random walk - and consequently, we will simply apply the ideal chain description of entropy, derived in the previous section. Thus, the only difference in a Flory good-solvent chain and an ideal chain is ∆U. We can write down a virial expansion of U as

U = Vk T (βη2 + γη3 + δη4 + ) (20) B ··· where β, γ, δ are the second, third, fourth virial coeﬃcients which are fully deter- ··· ··· mined by the form of the energetic interactions (e.g., LJ interactions). We will ignore all but the leading term in this expansion, which represents the energetics of all two-body interactions. Higher order terms represent three-body interactions, four-body interactions, etc. If the monomer-monomer interactions are unfavourable, or repulsive, then the second virial coeﬃ- cient is positive, β > 0. If the monomer-monomer interactions are favourable, or attractive, then β < 0. For the good solvent case, β > 0. η is the concentration of statistical monomers, η N/V where V is the volume. Thus, we can recast the free energy in terms as ≡ 3R2 F = k T (V βη2 + ). (21) B 2Na2 This expressions states that the energy of the chain is comprised of all binary interactions amongst the monomers and the ideal elastic contribution of a random walker. We need to get

12 this expression in terms of one variable, say α which we define as the swelling or expansion coefficient, R α = (22) R0 where R0 is the natural size of the chain, R0 = √Na. This swelling coefficient becomes larger than unity when the chain exceeds the natural size of an ideal chain, and is smaller than one when the chain collapses to sizes less than that of an ideal chain. We want to know what value α takes on when the solvent is good; clearly α will be greater than 1. The concentration, η = N/V 3 3 is expressed in terms of R and R0 using V α R . In terms of the swelling coefficient, the free ∼ 0 energy is then F β√N 3 = + α2 (23) kBT a3α3 2 As you can see from the first term on the RHS, the interactions with the good solvent promote swelling the of the chain; i.e., as α becomes larger than 1, the first term becomes smaller and the free energy is thus smaller. On the other hand, a swollen chain has less entropy and therefore the second term on the RHS suppresses swelling; i.e., as α increases beyond 1, the second term becomes larger and the free energy increases. Thus, the energetic interactions with the solvent and the entropic elasticity are in opposition. At low swelling, the free energy will be high due to the energetic interactions, while at high swelling, the free energy will also be high due to the entropic elasticity of the chain. There will be an intermediate or optimal value of swelling at which the two opposing driving energies will be equal and the free energy will then be minimal. You can determine an equation for the optimal swelling coefficient α by solving dF/dα = 0, and checking to ensure that d2F/dα2 > 0. The equation for α at which the free energy is minimal is β α ( )1/5N 1/10 (24) ∼ a3 which, when inserted into the definition of α, R = αR0, yields β R ( )1/5N 3/5a, (25) ∼ a3 or, R N 3/5a. ∼ This derivation is very simplistic, but the end result is effectively correct. A chain in a good solvent is often referred to as a “Flory chain” after Paul Flory who first wrote down this analysis. Of course, there are several simplifications and Pierre DeGennes (French Physicist and Nobel Laureate) carried out a more precise and very complicated calculation in 1972 and effectively got the same result. The Flory treatment effectively assumes that the only difference between a good solvent chain and an ideal chain is in the incorporation of (repulsive) pairwise interactions. Let’s explore this further. First let us summarise the ideal chain, investigating the occurence of self-intersections with itself. The ideal chain has natural size R √Na, the volume circumscribing the chain is ∼ V R3 = N 3/2a3, and the volume fraction of segments in that circumscribed volume is ∼ −1 φ = Nν/V (ν/a3)√N . Now the number of segments on the ideal chain that intersect ∼ another segment of the chain is N φ as φ is, to first order, the probability of finding a × −1 segment at any locations within V . Because φ √N , the number of 2-body interactions or ∼ 13 single-intersections of the chain scales with √N. Thus, the number of intersections increases with the square root of the N. N is usually large, so the number of binary self-intersections is appreciable. How many intersections are there that are 3-body ior 3-way (i.e. the chain self-intersects three times at one point)? That number is N φ2, and does not scale with N, × that is, we can find a three-way intersection on the order of once each chain, irrespective of the number of statistical monomers. The number of a four-way intersection is N φ3 and scales −1 × as √N - and again, as N is usually large, the number of four-body intersection vanishes. So in a random walk, two-way intersections are frequent, we might find a three-way intersection, but we will not find higher-order intersections. Now contrast that with the Flory chain. In the Flory chain, an ”intersection” corresponds to a close enough approach of the monomers that they feel the (repulsive) monomer-monomer interaction strongly. Whether the monomers intersect or not is determined by the form of the interaction potential. For example, in a Lennard Jones-type interaction, there is a strong repulsion at close distances that prohibits overlap. The size of the good-solvent chain is R ∼ N 3/5a, the volume circumscribing the chain is V R3 = N 9/5a3, and the volume fraction of ∼ segments in that volume is φ = N/ν/V (ν/a3)N −4/5. The number of 2-way self-intersections ∼ is Nφ N 1/5, and the number of three-way intersections is Nφ2 N −13/5. For large N, ∼ ∼ the number of a three-way intersection is vanishingly small, and for higher-order intersection probabilities, it is even smaller. Consequently, in both the ideal chain and the Flory chain, the occurence of a three-way intersection or interaction is not a dominant contribution to either chain’s configuration. How- ever, the probability that a chain intersects itself in a two-way intersection is quite large for an ideal chain, but binary interactions in a Flory chain are smaller in number. Indeed, this we already know: the solvent acts to expand the chain, thereby decreasing the probability that two monomers, on arbitrary locations on the contour, approach closely. We can additionally state that it is these two-body interactions which effectively increase the size of the chain under good solvent conditions.

3.3 θ-solvency conditions

It is important to understand that the sign of the second virial coeﬃcient, β, indicates the type of solvent. If β < 0, the solvent is poor, if β > 0, the solvent is good. It is indeed possible for β to take on values that are very close to zero: in that case there is not energetic preference between monomers and solvent molecules and the system has eﬀectively no change in its internal energy, i.e., ∆U = 0. This condition is called “theta” solvency conditions and under these conditions, the chain size does indeed scale as R √Na, as predicted by the ideal ∼ chain.

EXAMPLE PROBLEMS

14 Describe an isolated, single polymer chain in (a) good solvent, (b) theta solvent, • and (c) poor solvent conditions.

good solvent: In a good solvent, an isolated chain is swollen by the solvent. Thermody- namically we understand the size of the chain as a balance between the solvent-polymer interactions that promote swelling and the entropy (or stretching penalty) that suppresses swelling. The chain is ﬂuctuating in size and shape. The average size of the chain scales as R N 3/5a, where N is the number of statistical segments in the chain and a is the ∼ size of a statistical segment.

theta solvent In a theta solvent, the monomer-monomer interactions are identical to the monomer-solvent interactions so that the chain is eﬀectively inert to itself and the solvent. The chain is referred to as an ideal chain, as it is similar to the ideal gas model: it can intersect itself and has no interaction energy. We model such a chain as a random walk that can intersect itself. The probability of ﬁnding a chain of size R obeys a Gaussian distribution for R

/bulllet What happens to the moderately extended chain if the solvent is changed from a theta to a good solvent?

The chain extension will not change appreciably, extending only slightly. As an ideal chain is stretched, it will have fewer self-intersections, and non-contiguous monomers will rarely be close. In a good solvent, a chain swells because non-contiguous monomers that are close will feel a repulsion. However, if that chain is stretched and there are few close monomer pairs, then there will be minimal swelling of the chain. You might detect a small increase in the extension in good solvents. You would however see a bigger eﬀect if the solvent were made poor: in this case the chain will collapse, and depending upon the weight, you might get a complete collapse of the chain, pulling the weight up to the tether point, or you might simply see the ball-chain conﬁguration, where there is a slight decrease in the overall extension of the chain.

Consider that you have a chain that is doubly end-tethered to opposite walls, •• separated at such a distance that a small force is required to keep the walls from sandwiching together. The wall separation is signiﬁcantly smaller than the contour length of the chain. The surrounding solvent is a good one for the chain. Qualita- tively describe what happens to the chain and the force needed to keep the walls ﬁxed when the solvent is changed from good to poor. EXAMPLE PROBLEM

15 Summary

In this ﬁnal section, we used geometrical reasoning and thermodynamics to explore non-ideal chain, namely chains which interact favourably and unfavourably with their solvent.

The size of a chain, collapsed upon itself in poor solvent is R N 1/3a • ∼ The size of a chain, swelled in good solvent, is R N 3/5a • ∼ As the second virial coeﬃcient decreases from positive vlaues, to zero, then to negative • values, the chain is in a good solvent, theta solvent, than then poor solvent.