# Difference between revisions of "The Electrostatic Component of Disjoining Pressure"

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− | + | By Bryan Hassell: AP 255 Fall 11 | |

Keywords: [[disjoining pressure]] | Keywords: [[disjoining pressure]] | ||

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− | <math> dG_{T,P,\mu_i} = S\Pi_e dh + S(\ | + | <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> \ | + | 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(\ | + | <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> |

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

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− | <math>{d^2\ | + | <math>{d^2\psi \over dx^2} = {({4\pi \over \epsilon})}\rho </math> |

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

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− | 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\ | + | 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}{({d\ | + | <math> {\epsilon \over 8\pi}{({d\psi \over dx})}^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\ | + | 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}^{\ | + | <math> {2\pi \sigma_{1}^{2} \over \epsilon } = C - \int_{0}^{\psi_1} \rho d\psi</math> |

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− | <math> \sigma_1 = - {\epsilon \over 4\pi}{({d\ | + | <math> \sigma_1 = - {\epsilon \over 4\pi}{({d\psi \over dx})}_{x=0}</math> |

− | and <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}{({\partial\sigma_1 \over \partial h})}_{\ | + | <math> {4\pi\sigma_1 \over \epsilon}{({\partial\sigma_1 \over \partial h})}_{\psi_1} = {({\partial C \over \partial h})}_{\psi_1}</math> |

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− | <math> {({\partial C \over \partial h})}_{\ | + | <math> {({\partial C \over \partial h})}_{\psi_1}{({\partial h \over \partial \psi_1})}_{C}{({\partial \psi_1 \over \partial C})}_{h} = -1 </math> |

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− | <math> \sigma_1 = {\epsilon \over 4\pi}{({\partial \ | + | <math> \sigma_1 = {\epsilon \over 4\pi}{({\partial \psi_1 \over \partial h})}_{C,\psi_2} </math> |

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− | <math> {({\partial\sigma_1 \over \partial h})}_{\ | + | <math> {({\partial\sigma_1 \over \partial h})}_{\psi_1,\psi_2} = -{({\partial C \over \partial\psi_1})}_{h,\psi_2} </math> |

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

## Revision as of 01:10, 7 December 2011

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>

<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} = {({4\pi \over \epsilon})}\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}{({d\psi \over dx})}^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}{({d\psi \over dx})}_{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}{({\partial\sigma_1 \over \partial h})}_{\psi_1} = {({\partial C \over \partial h})}_{\psi_1}</math>

and using the identity

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

and the relation

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

we get finally something we can use

<math> {({\partial\sigma_1 \over \partial h})}_{\psi_1,\psi_2} = -{({\partial C \over \partial\psi_1})}_{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>