Chapter 16: Polymer Gels

Rubber Networks and Elastomers:

We have discussed the concept of a characteristic relaxation time quite a bit in previous classes and have now mechanically observed different regimes of behavior (viscous liquid vs. elastic solid) based on the relationship between the characteristic relaxation time and the experimentally relevant time over which a perturbation is applied. We largely focused on understanding the extremes of where the characteristic relaxation time is either much greater than or much less than the experimental time so that the material responds as either a liquid or a solid. We also described the intermediate regime of viscoelastic behavior where the material behaves as both a viscous liquid and elastic solid over different timescales. In general, polymers tend to have characteristic relaxation times that are within the viscoelastic regime, where we might expect that despite some elastic behavior, at long enough times it is possible for the polymer to flow and completely relax the stress within the material. However, it is also observed that some polymers exhibit elastic behavior even for very long timescales, much longer than would be expected for a typical polymer melt. These polymers are called elastomers, or equivalently, rubbers. We will describe the molecular characteristics that give rise to rubber elasticity, the physical model used to describe the rubbery response, and finally how we can use this model to find the stress-strain behavior of a elastomeric polymer.

Rubber Elasticity:

Typically when we discuss rubbers and the mechanics of rubbers or elastomers we will talk about networks. And when we deform those networks we are we describe this behavior as rubber elasticity. Microscopically, we envision a rubber network as a large, entwined set of polymer chains that are attached together by crosslinks, referring to points that join otherwise distinct chains. Crosslinks can be chemical or physical in nature - chemical crosslinks refer to actual covalent bonds between chains (vulcanization) , while physical crosslinks refer to non-specific interactions between chains. Examples of physical crosslinks include entanglements in polymer melts, glassy phase separated regions in a rubbery matrix (as in high-impact polystyrene or HIPS), block copolymers, or possibly hydrogen bonding or other types of strong non-covalent bonds between groups that are highly separated along a chain.

The presence of crosslinks joins otherwise disparate chains together such that the entire rubber is connected and individual chains are no longer capable of diffusing completely independently in response to a perturbation. As a result, stressing a crosslinked network leads to a mechanical response that is partially characteristic of an elastic material and partially characteristic of a viscous material, with the important caveat that under a mechanical stress the overall network maintains mechanical robustness. We call materials that have both an elastic and viscous response viscoelastic.

We will quantitatively describe how rubbery networks respond to stress using basic principles of mechanics and a thermodynamic description as well. It is critical to glassy-rubbery behavior. The key concept to recognize is that a rubber is still composed of flexible polymer chains that can easily rearrange in the presence of a stress, allowing large amounts of strain when stressed. However, the presence of crosslinks prevents the chains from completely flowing, and as a result some solid-like mechanical behavior is observed. In the context of the discussion of relaxation time scales from the previous lecture, we could say that the rubber exhibits a longest relaxation time characteristic of a solid (the relaxation time associated with removing crosslinks) while having shorter relaxation times characteristic of liquids (the relaxation time associated with diffusing the polymer chains between crosslinks). We will have multiple relaxation times and see behaviors characteristic of both liquid and solid behavior.

Before describing this behavior and creating a theoretical framework we will be making several key assumptions

1. Gaussian subchains between permanent crosslinks - that is, you can envision the part of a chain between two crosslinked points as a chain itself, which can be described by a Gaussian distribution (hence the number of subchains is always greater than the number of ”regular” chains). This assumption breaks down if the distance between crosslinks is very small.
2. Temperature well above T_g, so that the system has sufficient thermal energy for chains to rearrange.
3. Flexible chains with relatively easy backbone bond rotation potentials, again reflecting the ability of chains to rearrange.
4. Chain deformation occurs by conformational changes, an assumption we have made throughout the course.
5. No relaxation by chain slip - the network appears as fixed on the time scale of experiment, i.e. physical crosslinks can be assumed to be permanent. This means that as we pull the rubber, the chains are not able to slide past each other and thus fall apart.
6. Affine deformation - microscopic deformation by same amount as macroscopic deformation; that is, any given chain’s dimensions change by the same ratio as the change in dimensions of the overall network.
7. No crystallization of the chains at large strains
8. No change in volume of the rubber upon deformation, i.e. incompressibility assumption

Now there are two distinct types of behavior that is well described by our rubber elasticity framework

• Mechanical Deformation: we apply some stress and measure the strain response for a fixed volume
• Gel Swelling: A polymer network swells due to mxing a solvent and the volume increases.

We have hinted at energetic competition for mechanically deforming a rubber network previously, specifically that stretching polymer chains decreases the conformational entropy of the chain and we had an expression for the energy stretched chain based on the change in the number of microstates

$$G(r)_{stretch} = \frac{3kT}{2} \frac{r^2}{r_0^2} \nonumber$$

Here energy increases if the chain extends and thus we have an entropic spring restoring force that opposes pulling.

Let’s start with the mechanical deformation first.

Rubber Elasticity System:

As usual we want to think about the energetics and thermodynamics associated with changing the state of our system when we apply the mechanical stress to the crosslinked polymer network. The initial state is the unstressed, relaxed polymer network. We assume that we have a large number of subchains which are the chain segments between crosslinked points. We assume that these subchains can be defined as Gaussian with unperturbed end-to-end distance . We also assume that the molecular weight of each subchain, M_x, is the same. Given an initial set of chains, the molecular weight of each subchain will thus decrease the degree of crosslinking is increased.

Upon applying a force (stress) to the network, we assume that the dimensions extend to a new end-to-end distance hr²i. Here, we can invoke the assumption of an affine deformation - the deformation experienced by each individual subchain is equivalent to the deformation experienced by the whole network. Using this assumption, we can relate the microscopic perturbed dimensions to unperturbed dimensions via an extension ratio in each direction x, y, and z which is defined as the ratio between the macroscopic deformed length l_i in that direction and the undeformed length l₀ (assuming a sample with the same l₀ in all 3 directions. These extension ratios are

$$\alpha_x = \frac{l_x}{l_0} \nonumber$$

$$\alpha_y = \frac{l_y}{l_0} \nonumber$$

$$\alpha_z = \frac{l_z}{l_0} \nonumber$$

Because of the affine deformation assumption, we can also write coordinates of an individual subchain in terms of the extension ratios, if we imagine deforming a subchain between one crosslink fixed at (0, 0, 0) and one crosslink fixed at (x, y, z), with deformed coordinates (x’, y’, z’):

$$x' = \alpha _x x; \text{ } y' = \alpha _y y;\text{ } z' = \alpha _z z \nonumber$$

We also then see that the end-to-end distances are defined as

$$r^2 = x'^2 + y'^2 + z'^2 \nonumber$$

$$r_0^2 = x^2 + y^2 + z^2 \nonumber$$

Finally, since we assume that the network is incompressible, the total volume change has to be equal to 0, which means that the product of the deformed lengths has to be equal to the product of the undeformed lengths

$$\alpha_x \alpha_y \alpha_z = 1 \nonumber$$

Therodynamic Perspective of Rubber Elasticity:

Now we need a thermodynamic relationship to derive an expression between force and extension for a general material, which we can then apply to our polymer network. As usual let’s start with Gibbs

$$G = H - TS \nonumber$$

Finding the differential form of G, and including a work term for extension in U gives

$$dG = -SdT + VdP + Fdl \nonumber$$

where G is the Gibbs free energy, l is the extension of the chain, and F is force. By rearranging this equation we can see that F is

$$F = \left ( \frac{\partial G}{\partial l} \right ) _{T,P} \nonumber$$

We can thus take the derivative of our first expression for G with respect to l at constant T and P to get:

$$F = \left ( \frac{\partial H}{\partial l} \right ) _{T,P} - T \left ( \frac{\partial S}{\partial l} \right ) _{T,P} \nonumber$$

Remember the physical interpretation, the equation above tells us the the force associated with a change in chain extension is related to the derivative of the enthalpy, which measures bond interactions, and entropy. For ideal rubbers, we assume that the change in enthalpy is 0 as the rubber network is stretched. So then

$$F = - T \left ( \frac{\partial S}{\partial l} \right ) _{T,P} \nonumber$$

Here we see the restoring force of rubber elasticity is driven by entropy. We assume there is NO change in enthalpy when a rubber is stretched in part because of the fixed volume assumption - since the volume is not changing, the net number of interactions is assumed to be the same on average, so there is no net enthalpy change. This assumption of no net enthalpy change is one the unique characteristics of polymers - for example, we could not apply the same reasoning to metals because of strong enthalpic changes associated with stretching bonds, etc. Even if we relax the assumption of a zero enthalpy change, we would still find that for polymers the change in entropy is the dominant term, allowing us refer to the corresponding force as an entropic spring force.

Since the elastic force that arises from stressing a rubber network is related to the derivative of the entropy, we need to calculate the entropy based on the extension ratios defined previously. Entropy is related to the number of microstates available to the polymer, which is related to the probability distribution of the Gaussian subchains. So we can define the entropy change for a single chain as

$$S_2 - S_1 = \Delta S = k \ln \frac{\Omega_2}{\Omega_1} \nonumber$$

$$\Omega_1 = \text{const} \cdot \exp \left ( \frac{-3r_0^2}{2nl^2} \right ) \nonumber$$

$$\Omega_2 = \text{const} \cdot \exp \left ( \frac{-3(\alpha_x^2x^2+\alpha_y^2y^2+\alpha_z^2z^2)}{2nl^2} \right ) \nonumber$$

Here we assume that Ω₁ ∝ P(r₀) and Ω₂ ∝ P(r) where P(r) is the Gaussian distribution, and r₀ and r are the relaxed and stressed dimensions defined above. This then gives

$$\Delta S = k\ln \frac{\Omega_2}{\Omega_1} = \frac{-3k}{2nl^2} \left [ (\alpha_x^2-1)x^2+(\alpha_y^2-1)y^2+(\alpha_z^2-1)z^2 \right ] \nonumber$$

Again, we are able to state that the entropy of a single chain is related to the extension ratios associated with the macroscopic sample because of our assumption of affine/homogeneous deformation. Right now we have the entropy in terms of the relaxed coordinates x,y,z. In this state, where the overall chain dimensions can be described by the Gaussian distribution, then we can expect that there is no distinction between a random walk in either of the 3 directions, so the average contribution to from x², y², and z² is the same so

$$\langle x^2 \rangle = \langle y^2 \rangle = \langle z^2 \rangle = \frac{\langle r_0^2 \rangle}{3} \nonumber$$

Substituting in these values and knowing that nl² = r₀ gives

$$\Delta S = \frac{-k}{2} \left ( \alpha_x^2+\alpha_y^2+\alpha_z^2 - 3 \right ) \nonumber$$

Finally F is

$$\Delta F_i=-T\left ( \frac{\partial \Delta S(\alpha_i)}{\partial l_i} \right)_{T,P} \nonumber$$

where l_i denotes the force in a given direction i = x,y,z.

Uniaxial Deformation of a Rubber Network:

Now that we have an expression for the change in entropy of a chain given some deformation and have a thermodynamic relation that yields the force associated with that deformation, we can start analyzing different scenarios. Let’s start simple with uniaxial deformation and find the resulting elastic force. Let’s take deformation along the x axis so that

$$\alpha_x = \frac{l_x}{l_0} \nonumber$$

$$\alpha_y = \frac{1}{\sqrt{\alpha_x}} \nonumber$$

$$\alpha_z = \frac{1}{\sqrt{\alpha_x}} \nonumber$$

Here, we calculate α_y and α_z from the incompressibility condition α_xα_yα_z = 1. We can then write ∆S in terms of just α_x

$$\Delta S (\alpha _x)= -\frac{k}{2}(\alpha _x^2 + \frac{2}{\alpha_x} -3) \nonumber$$

The 3-dimensional problem has now broken down into only knowing the extension in a single dimension. We can take the derivative with respect to l_x to find F, using the chain rule

$$F_x = -T\frac{\partial}{\partial l_x}\Delta S (\alpha_x) \nonumber$$

$$= -T\frac{\partial}{\partial \alpha_x}\left ( \frac{\partial \alpha_x}{\partial l_x} \right )\Delta S (\alpha_x) \nonumber$$

$$= \frac{kT}{2l_0} \frac{\partial}{\partial \alpha_x}\left ( \alpha_x^2 + \frac{2}{\alpha_x} - 3 \right ) \nonumber$$

$$F_x = \frac{kT}{l_0}\left ( \alpha_x - \frac{1}{\alpha_x^2} \right ) \nonumber$$

To put this in more familiar terms, we can find the stress by dividing by the cross sectional area (in the y − z plane), which we’ll call A₀. So far we’ve derived F_x for a single chain only, and would ideally like to know the stress required to deform a network of z total subchains per volume V gives

$$\sigma_x = \frac{zF_x}{A_0} = \frac{zkT}{A_0l_0}(\alpha_x - \frac{1}{\alpha_x^2}) = N_{xlinks}kT(\alpha_x - \frac{1}{\alpha_x^2}) \nonumber$$

Here, we simplify by letting A₀l₀ = V and $\frac{z}{V} = N_{xlinks}$, where N_xlinks is the number of crosslinks per unit volume. Typically the restoring force is written in terms of the molecular mass of crosslinks instead of their number, so we can rearrange to find the crosslink molecular mass in terms of the density as

$$M_x = \frac{\text{mass of polymer/volume}}{\text{number of subchains per volume}} = \frac{\rho}{N_{xlinks}/N_A} \nonumber$$

$$\sigma_x = \frac{\rho N_A kT}{M_x}\left ( \alpha_x - \frac{1}{\alpha_x^2} \right ) \nonumber$$

where N_A is Avogadro’s number. We can write the stress as a function of strain (the typical measurement of material deformation) using the relation:

$$\alpha_x = \frac{l_x}{l_0} = \frac{l_0 + \Delta l}{l_0} = 1 + \frac{\Delta l}{l_0} = 1 + \epsilon_x \nonumber$$

$$\sigma_x = \frac{\rho N_A kT}{M_x}\left (1 + \epsilon_x - \frac{1}{(1 + \epsilon_x)^2} \right ) \nonumber$$

In the limit of small strains, we see that the stress scales linearly with the strain, and hence we can define the Young’s modulus

$$E = \frac{\rho N_A kT}{M_x} \nonumber$$

Let’s stop here for a moment and appreciate this result. We see here that for a rubber network the Young’s modulus increases with temperature and decreases with the molecular weight of crosslinks. This implies that if we increase the density of crosslinks, then the molecular weight of the crosslinks decreases and the Young’s modulus goes up, leading to a greater elastic restoring force. However, do recall that if the density gets too high, the assumption of Gaussian subchains will eventually break down, and the derivation will no longer be valid.

Since we have shown that the Young’s modulus depends on crosslink molecular mass (and hence crosslink density), and we have stated that entanglements can act as physical crosslinks at short times (since at longer times the entanglements will eventually diffuse away), then we can measure the elastic response of an entangled melt at some known temperature to approximate the number of statistical segments between entanglements N_E. In principal, this type of experiment would also allow us to identify the critical molecular weight for crossing over between the Rouse and reptation regimes since the onset of the reptation regime would be associated with the onset of elastic behavior at small timescales.

Limits of Rubber Elasticity Model:

This model works very well at applied strains that are relatively low however in regimes where the chain extension is high the model begins to break down

• Strain-induced Crystallization: at large extensions the polymer chains can pack tightly and form crystalline regions (shish-kebob spherulites) and thus the Young’s modulus will be larger than predicted by theory
• Subchain Length Reaches Contour Length: at this point we are pulling on covalent bonds and the Young’s modulus will approach that of metals
• Bond Rotation Away from Trans State at High Extension: (see discussion below)
• Dangling Ends: experimental Young’s modulus is lower than theory because there are regions between the end of the chain and cross link that can form entanglements. This increases the crosslink density of the network somewhat artificially as the the physical cross links disappear when deformation occurs.
• System Temperature Drops Below T_g: Recall that we are implicitly assuming that the rubbery network is above its glass transition temperature, and thus chains are freely able to deform and hence explore a full conformational space. If the network is cooled below the glass transition temperature, the entropic spring arguments will fail because the network no longer acts like a Gaussian chain, but instead will give an elastic response consistent with the kinetic barriers to chain rearrangement
• Too Many Crosslinks: Major assumption of this model is that every subchain in the rubbery network can be treated as Gaussian chain, and thus loses a significant amount of conformational entropy upon stretching. If the crosslink density is high, however, the number of segments between crosslinks will correspondingly become very low and the subchains may no longer be considered Gaussian.
• Too Few Crosslinks: If the chains are too long entanglements can occur between subchains and the entanglements can act like crosslinks on short timescales, so we could imagine the effective crosslink density being much higher than anticipated due to the presence of crosslinks. Note that even in the case of moderate crosslinking density, there will typically be chain ends which are capable of forming crosslinks and may result in spurious results. Write problem on this

Gels:

Our framework for rubber elasticity can be utilized to describe gels which are a highly crosslinked network of polymer chains and then swollen in solution with a solvent. You can also start with a polymer solution and then crosslink the polymer in the presence of solvent via UV radiation, chemical reagents, or enzymes.

In this system there are again two major competing thermodynamic forces this time it is the mixing of solvent with the polymer chains via enthalpic/entropic driving forces (Flory-Huggins theory) and the networked chains to resist mechanical deformation due to the loss of conformational entropy. By balancing these two contributions to the system free energy we can calculate the optimal swelling ratio for a given gel.

Flory-Rehner Theory of Gel Swelling:

For the swelling of a crosslinked, elastic gel we can imagine solvent mixing in an unperturbed rubber, leading to a reduction of free energy due to mixing. As the solvent mixes with the rubbery gel, however, it will tend to expand the volume of the rubber, causing the chains to stretch and exert a restoring force. The basic idea here is to combine our understanding of Flory-Huggins theory and rubber elasticity to derive a total free energy change. We first assume that the free energy of mixing and the free energy of elastic stretching are additive

$$\Delta G = \Delta G_{mix} + \Delta G_{elastic} \nonumber$$

Again, the competition here is between the favorable free energy of mixing, driven by favorable translational mixing entropy and either a favorable or unfavorable χ interaction, and the unfavorable free energy of elastic stretching, which emerges from the stretching of entropic springs in the network. Let’s define some variables inour system

• V₀ = initial volume of gel

• V = final volume of gel

• hr₀i = unperturbed end to end distance of subchains

• hri = stretched end to end distance of subchains

• x = unperturbed dimension of overall gel (not subchain; entire network). Assume cube-like network geometry (see slides for picture)

• x⁰ = perturbed dimension of gel.

• α_x = swelling ratio = $\frac{x'}{x}$- this is akin to extension ratios defined for rubber elasticity

• φ₂ = volume fraction of polymer =$\frac{V_0}{V}$ - note that this is because the increase in volume is due to solvent, while V₀ is always the volume of polymer, so as V increases the volume fraction goes down

• N = number of subchains in network = $\frac{\rho V_0 N_A}{M_x}$ as defined in rubber elasticity

To simplify the system, we can assume isotropic swelling, meaning that we assume all 3 dimensions swell by α_x = α_s. The final volume, initial volume, swelling ratio, and polymer volume fraction are then

$$V = V_0 \alpha_s^3 \nonumber$$

$$\phi_2 = \frac{V_0}{V} \nonumber$$

$$\alpha_s^3 = \frac{1}{\phi_2} \nonumber$$

Having defined some variables, let’s define the chemical potential difference for solvent in gel versus pure solvent, since the condition of equilibrium will be that the chemical potential difference is 0. The chemical potential difference is

$$\mu_1 - \mu_1^0 = \left ( \frac{\partial \Delta G_{mix}}{\partial n_1} \right )_{T,P,n_2} + \left ( \frac{\partial \Delta G_{elastic}}{\partial \alpha_s} \right ) \left ( \frac{\partial \alpha_s}{\partial n_1} \right )_{T,P,n_2} \nonumber$$

This expression defines the chemical potential difference as the change in the free energy (change) as a function of adding molecules of solvent. We have expression for the first term from Flory-Huggins which gave us

$$\left ( \frac{\partial \Delta G_{mix}}{\partial n_1} \right )_{T,P,n_2} = RT \left [ \ln \Phi_1 +(1-\frac{1}{x_2})\phi_2 +\chi_{12} \Phi_2^2 \right ] \nonumber$$

In Flory-Huggins theory we derived this expression for the chemical potential for a set of chains with degree of polymerization given by x₂ - however, now we are talking about a network of covalently linked polymers, which have an effective molecular weight much higher than that of independent chains. We can thus reasonably approximate the degree of polymerization as infinite since the network is effectively one huge polymer chain, so we can simplify this as

$$\left ( \frac{\partial \Delta G_{mix}}{\partial n_1} \right )_{T,P,n_2} = RT \left [ \ln \Phi_1 + \phi_2 +\chi_{12} \Phi_2^2 \right ] \nonumber$$

We can now turn our attention to the elastic term. Recall that one of our assumptions of rubber elasticity was that there was no enthalpy change on stretching, so we can write the free energy change strictly in terms of the conformational entropy

$$\Delta G_{elastic} = -T\Delta S_{elastic} \nonumber$$

We derived a form for ∆S in terms of extension ratios, assuming that the rubber was incompressible. However, we have now explicitly stated that the volume of the network DOES change, so we have to include a new term related to this increase in volume to our change in entropy, since increasing the volume will increase the number of accessible microstates available to the chain and hence increase its final entropy. Our new entropy change for N subchains is

$$\Delta S_{elastic} = -\frac{k}{2} N \left [ (3\alpha_s^2 - 3) - \ln \frac{V}{V_0} \right ] \nonumber$$

Note the form we chose for the change in volume - we have taken the ratio of the volume of the final gel to the volume of the initial, then taken the logarithm; this form is identical to the change in conformational entropy of N ideal gas particles moving from a volume V₀ to V , and reflects the assumption of ideal polymer chains.

We now have a form for the change in entropy, and given that rubbery stretching does not involve an enthalpic term we also then have an expression for the free energy. To find the chemical potential, we take the first derivative of the free energy change with respect to the number of solvent molecules

$$\Delta G_{elastic} = -T\Delta S_{elastic} \nonumber$$

$$= \frac{NkT}{2} \left [ (3\alpha_s^2 - 3) - \ln \frac{V}{V_0} \right ] \nonumber$$

$$\left ( \frac{\partial \Delta G_{elastic}}{\partial n_1} \right )_{T,P,n_2} = \left ( \frac{\partial \Delta G_{elastic}}{\partial \alpha_s} \right ) \left ( \frac{\partial \alpha_s}{\partial n_1} \right )_{T,P,n_2} \nonumber$$

$$= \frac{NkT}{2} \frac{\partial}{\partial \alpha_s} \left ( 3\alpha_s^2 - 3 - \ln \frac{V}{V_0} \right ) \left ( \frac{\partial \alpha_s}{\partial n_1} \right )_{T,P,n_2} \nonumber$$

$$= \frac{NkT}{2} \left ( 6\alpha_s - \frac{3}{\alpha_s} \right ) \left ( \frac{\partial \alpha_s}{\partial n_1} \right )_{T,P,n_2} \nonumber$$

We can find $\left ( \frac{\partial \alpha_s}{\partial n_1} \right )$ by writing the change in volume in terms of the molar volume of solvent added, since the total volume after swelling has to be equal to the initial volume (volume of polymer) plus the volume of solvent. We can now write

$$\alpha_s^3 = \frac{V}{V_0} = \frac{V_0 + n_1 v_1}{V_0} \nonumber$$

$$\alpha_s = \left( 1 + \frac{n_1 v_1}{V_0} \right)^{1/3} \nonumber$$

where n₁ is the number of moles of solvent and v₁ is the volume per mole of solvent, so that n₁v₁ is the total volume of solvent. From this definition we can take the derivative to get

$$\left ( \frac{\partial \alpha_s}{\partial n_1} \right ) = \frac{1}{3\alpha_s^2}\frac{v_1}{V_0} \nonumber$$

Combining these terms gives

$$\left ( \frac{\partial \Delta G_{elastic}}{\partial n_1} \right ) = \frac{NkT}{2} \left ( 6\alpha_s - \frac{3}{\alpha_S} \right ) \frac{1}{3\alpha_s^2}\frac{v_1}{V_0} \nonumber$$

We can rearrange this result in terms of $\phi_2 = \frac{1}{\alpha_s^3}$ and M_x to simplify to

$$\left ( \frac{\partial \Delta G_{elastic}}{\partial n_1} \right ) = NkT\frac{v_1}{V_0} \left ( \phi_2^{1/3} - \frac{\phi_2}{2} \right ) = \frac{RT\rho v_1}{M_x} \left ( \phi_2^{1/3} - \frac{\phi_2}{2} \right ) \nonumber$$

where we substituted in the relation for $N=\frac{\rho V_0 N_A}{M_x}$ and kN_a = R. Finally, we then add in the term from Flory-Huggins theory to give the change in chemical potential as

$$\mu_1-\mu_1^0 = RT[\ln (1-\phi_2) + \phi_2 +\chi_{12}\phi_2^2] + \frac{RT\rho v_1}{M_x} \left ( \phi_2^{1/3} - \frac{\phi_2}{2} \right ) \nonumber$$

We now set the change in chemical potential to 0, as we stated was the condition for equilibrium. Note that this is essentially the same as saying the change in free energy from mixing has to be equal to the change in free energy from elastic energy. We then write as a final equation

$$\boxed{\ln (1-\phi_2) + \phi_2 +\chi_{12}\phi_2^2 = -\frac{\rho v_1}{M_x} \left ( \phi_2^{1/3} - \frac{\phi_2}{2} \right )} \nonumber$$

This equation has no simple solution, but can be used to relate the swelling ratio at equilibrium (here related to φ₂) to the properties of the rubbery network (M_x) and the enthalpic interactions between the solvent and polymer (χ).

Again, the most important thing to remember here is the qualitative balance between free energy terms - there is a driving force from the mixing entropy between solvent and polymer that encourages swelling. As swelling occurs, the volume increase elicits a mechanical response from the rubbery network that tends to oppose swelling. Finally, the value of the χ parameter could either oppose or favor swelling depending on its sign.

On the basis of the equation just derived, we can calculate several important parameters from swelling experiments. For instance, if we know ρ and v₁, which should generally be known for a given solvent and polymer, we can measure φ₂ from the swelling ratio and could then measure M_x from stretching a dry rubber, then measure χ₁₂ by adding solvent to the rubber and measuring the swelling. Similarly, if we know χ₁₂ from membrane osmometry, we can use a single swelling experiment to measure M_x if the same parameters as above are known.