The Electrostatic Component of Disjoining Pressure

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By Bryan Hassell: AP 255 Fall 11

Keywords: disjoining pressure

Introduction

Investigating the mechanisms of spontaneous charging of interfaces between two phases, one being an electrolyte solution, leads to an equilibrium diffuse atmosphere. If two interfaces are brought together then their ionic atmospheres will overlap and this gives rise to the electrostatic component of the disjoining pressure.

Methods of Calculation of the Disjoining Pressure <math> \Pi_e </math>

There are several approaches for calculating <math> \Pi_e </math>. One by Derjaguin [1] directly determined the electrostatic disjoining pressure acting on the surfaces of parallel plates in an electrolyte solution. He assumed that at equal potentials of the external and internal surfaces of the plate the hydrodynamic pressure is the same and so the disjoining pressure is just the difference between the Maxwell stresses at the external and internal surfaces of the plates:


<math> \Pi_e = \epsilon E_{ex}^2/8\pi - \epsilon E_{in}^2/8\pi </math>


where Ein and Eout are the strength of the electric field at the internal and external surfaces found from the solution of the Poisson-Boltzmann equation with appropriate boundary conditions. Langmuir also proposed [2] a formula based on ionic concentrations in the plane where the potential was a maximum and the electric field is zero. But for the formulation without any assumptions, the calculation of <math> \Pi_e </math> begins with general thermodynamic arguments. First, looking at the differential of Gibbs free energy, taking into account work of external forces that maintain the equilibrium thickness <math> h </math> of the electrolyte layer, and of charge sources which maintain the equilibrium charge densities <math> \sigma_1 </math> and <math> \sigma_2 </math> on the two surfaces:
Figure 1. Model for the derivation of the expression of <math>\Pi_e</math>. Potential distribution in the gap between plates


<math> dG_{T,P,\mu_i} = S\Pi_e dh + S(\psi_1 d\sigma_1 + \psi_2 d\sigma_2) </math>


where <math> \mu_i </math> are the chemical potentials, T is the temperature, P pressure, S is the surface area of the plates and <math> \psi_1 </math> and <math> \psi_2 </math> are the surface potentials as seen in Figure 1. If you then subtract from both sides of the equation the differential of the last term from the previous equation you get:


<math> dG_{T,P,\mu_i} - Sd(\psi_1 d\sigma_1 + \psi_2 d\sigma_2) = -S\Pi_e dh - S(\sigma_1 d\psi_1 + \sigma_2 d\psi_2).</math>


The left hand side of this is a total differential, or


<math>\partial\Pi_e/\partial\psi_{h,\mu_I,\psi_2} = \partial\sigma_1/\partial h|_{\mu_i,\psi_1,\psi_2}.</math>


To get <math> \Pi_e </math> from this we need to integrate the left hand side so we need to transform the right hand side. Referring to the Poisson equation,


<math>{d^2\psi \over dx^2} = \left (\frac{4\pi}{\epsilon} \right) \rho </math>


where the potential <math>\psi</math> is a function of x as seen in Figure 1. The volume charge density is found as


<math> \rho = \sum_{i} ez_i \tilde{n}_i</math>


where the local concentrations of ions are determined by the concentrations of ions in the bulk and the work of their transfer from bulk into the interlayer, <math> w_i = z_i e\psi</math>, therefore the local volume density is a function of local potential, or <math>\rho = \rho(\psi)</math>. In the bulk of the solution where <math>\psi = 0, \tilde{n}_i = n_i</math> the solution is locally electroneutral (volume charge density = 0 and <math>\rho(0) = 0</math>). Multiplying both sides of the Poisson equation by <math> 2{d\psi \over dx}dx</math> and integrating gives:


<math> {\epsilon \over 8\pi} {\left (\frac{d\psi}{dx} \right) }^2 = -\int_{0}^{\psi} \rho(\psi)d\psi + C </math>


where the integration constant C is a function of interlayer thickness h; and as h <math>\rightarrow \infty </math> meaning <math>d\psi/dx \rightarrow 0</math> and <math>\psi \rightarrow 0</math>, then C also goes to 0. For the left plate, the previous equation can be replaced by


<math> {2\pi \sigma_{1}^{2} \over \epsilon } = C - \int_{0}^{\psi_1} \rho d\psi</math>


and by Gauss's theorem,


<math> \sigma_1 = - {\epsilon \over 4\pi} {\left (\frac{d\psi}{dx} \right)}_{x=0}</math>


and <math>\psi(0) = \psi_1</math>. If the potential of the right hand plate is considered constant or <math>\psi_2 =</math> const. then by differentiating the equation for the left hand plate we get


<math> {4\pi\sigma_1 \over \epsilon} {\left (\frac{\partial\sigma_1}{\partial h} \right)}_{\psi_1} = {\left (\frac{\partial\C}{\partial h} \right)}_{\psi_1}</math>


and using the identity


<math> {\left (\frac{\partial C}{\partial h} \right)}_{\psi_1} {\left (\frac{\partial h}{\partial \psi_1} \right)}_{\C} {\left (\frac{\partial\psi_1}{\partial C} \right)}_{h} = -1 </math>


and the relation


<math> \sigma_1 = {\epsilon \over 4\pi} {\left (\frac{\partial\psi_1}{\partial h} \right)}_{C,\psi_2}</math>


we get finally something we can use


<math> {\left (\frac{\partial\sigma_1}{\partial h} \right)}_{\psi_1,\psi_2} = -{\left (\frac{\partial\C}{\partial \psi_1} \right)}_{h,\psi_2} </math>


Substituting this into the original equation we were trying to solve for <math>\Pi_e</math> and integrating we get


<math> \Pi_e(h) = -C </math>


where we've taken into account that <math>\Pi_e(\infty) = 0</math> and <math> C(\infty) = 0 </math>. Taking the integration constant we found before, we finally get the key equation:


<math> \Pi_e(h) = -\int_{0}^{\psi}\rho(\psi)d\psi - {\epsilon E^2 \over 8\pi} </math>


This is the most general formula for <math>\Pi_e</math> which holds for both symmetrical and asymmetrical double layers in electrolyte solutions of arbitrary composition and doesn't have any assumptions on the relationship between local values of volume charge density to potential. This relationship depends on conditions of electrochemical equilibrium for all the charged dissolved components in the layer and is a complicated problem.

Another derivation of <math>\Pi_e</math> may be found by using hydrostatics and electrostatics. Without going into too much detail we first start with the equality which holds in equilibrium electrolyte layers:


<math> grad p + \rho grad \psi + {1 \over 8\pi} grad \left (E^2 \frac{\partial\epsilon}{\partial\rho_l}\rho_l \right) = 0 </math>


As it is seen, this is a balance of the hydrostatic pressure, electric force acting on the volume charge and the last term which is electrostriction. Writing this in differential form, then noting that in the bulk of solution <math>p\equiv p_o</math>, <math>\psi\equiv 0</math> and <math>E\equiv 0</math> and in the interlayer <math>p\not\equiv p_o</math>, <math>\psi\not\equiv 0</math> and <math>E\not\equiv 0</math> we can get the following equation:


<math> p_o = p(x,h) + \int_{0}^{\psi_{x,h}}\rho d\psi + \frac{1}{8\pi}E^2(x,h)\frac{\partial\epsilon}{\partial\rho_l}\rho_l</math>


Substituting the Poisson equation into the equilibrium equation we see that the pressure normal to the surfaces in any transverse cross section of the film are:


<math>p_1 \equiv p(x,h) - \frac{\epsilon E^2(x,h)}{8\pi}+\frac{E^2(x,h)}{8\pi}\frac{\partial\epsilon}{\partial\rho_l}\rho_l</math>


which is constant, only a function of h independent of x. This is compose of hydrostatic pressure, Maxwell stress, and striction pressure. The difference between <math>p_1(h)</math> and <math>p_o</math> is exactly the disjoining pressure so by subtracting the last two equations from each other, you get the equation derived before with the electrostriction term gone. This result shows that the effect of electrostriction is just that hydrodynamic pressure will be a function of local electrostatic potential and local field. Integration of <math>\Pi_e</math> over a distance h between the interfaces of diffuse layers makes it possible to obtain an important characteristic of overlapping diffuse atmospheres: their Gibbs free energy of interaction.