Diffusiophoresis of a charged particle in a charged cavity with arbitrary electric double layer thickness

Abstract

An analysis is presented for the diffusiophoretic motion of a charged colloidal sphere located at the center of a charged spherical cavity filled with an electrolyte solution at the quasisteady state for the case of arbitrary electric double layers. The electrokinetic equations governing the ionic concentration, electric potential, and velocity distributions in the fluid are linearized with assumption that the system is slightly distorted from equilibrium. These linearized differential equations are solved using a perturbation method with the zeta potentials of the particle and cavity as the small perturbation parameters. An explicit formula for the diffusiophoretic velocity of the particle as a combination of the electrophoretic and chemiphoretic contributions valid for arbitrary values of $\kappa a$ and $a/b$ is obtained by balancing the electrostatic and hydrodynamic forces exerted on it, where $\kappa$ is the Debye screening parameter, $a$ is the radius of the particle, and $b$ is the radius of the cavity. The effect of the charged cavity wall on the diffusiophoresis of the particle is interesting and can be significant. The contributions from the diffusioosmotic (electroosmotic and chemiosmotic) flow taking place along the cavity wall and from the wall-corrected diffusiophoretic force to the particle velocity are comparably important, and this diffusioosmotic flow can reverse the direction of diffusiophoresis. The particle velocity in general increases with an increase in $\kappa a$ and decreases with an increase in $a/b$, but exceptions exist.

Appendix: functions in Eqs. (18)–(20c)

In Eqs. (18) and (19),

$${F_{\mu 00}}(r)={F_{\psi 00}}(r)=\frac{1}{{{\chi _{}}}}\left( {\frac{r}{a}+\frac{{{a^2}}}{{2{r^2}}}} \right),$$

(35)

$$\begin{aligned} F_{{\mu ij}} (r) & = \frac{{Zer}}{{3kTa\chi _{{}} }}\Bigg\{ {\frac{{a^{3} A_{{\mu ij}} (a,b) + 2(1 - \chi )B_{{\mu ij}} (a,b)}}{{2\chi r^{3} }} + \frac{{1 - \chi }}{\chi }\left[ {A_{{\mu ij}} (a,b) + \frac{2}{{a^{3} }}B_{{\mu ij}} (a,b)} \right]} \\ & {\quad + \frac{1}{{r^{3} }}B_{{\mu ij}} (a,r) + A_{{\mu ij}} (r,b)} \Bigg\}, \\ \end{aligned}$$

(36)

$$\begin{aligned} {F_{\psi ij}}(r) & =\frac{1}{{{\kappa ^3}{r^2}}}\Bigg\{ {\frac{{[{g_{1+}}(\kappa a) - {{\text{e}}^{2\kappa a}}{g_{1 - }}(\kappa a)]{A_{\psi ij}}(a,b)+2{g_{1+}}(\kappa a){B_{\psi ij}}(a,b)}}{{{{\text{e}}^{2\kappa a}}{g_{2+}}(\kappa b){g_{1 - }}(\kappa a) - {{\text{e}}^{2\kappa b}}{g_{2 - }}(\kappa b){g_{1+}}(\kappa a)}}} \\ & \quad \times \left[ {{g_{2+}}(\kappa b)\gamma (\kappa r)+\frac{1}{2}\{ {g_{2+}}(\kappa b) - {{\text{e}}^{2\kappa b}}{g_{2 - }}(\kappa b)\} (\kappa r+1){{\text{e}}^{ - \kappa r}}} \right] \\ & \quad +\gamma (\kappa r){A_{\psi ij}}(r,b) - (\kappa r+1){{\text{e}}^{ - \kappa r}}{B_{\psi ij}}(r,b) \Bigg\} , \\ \end{aligned}$$

(37)

where ($i$,$j$) equal to (0,1) or (1,0), $\chi =1+{a^3}/2{b^3}$,

$${A_{\mu ij}}(x,y)=\int_{x}^{y} {\left( {1 - \frac{{{a^3}}}{{{r^3}}}} \right)\frac{{{\text{d}}{\psi _{{\text{eq}}ij}}}}{{{\text{d}}r}}} {\text{d}}r,$$

(38a)

$${B_{\mu ij}}(x,y)=\int_{x}^{y} {({r^3} - {a^3})\frac{{{\text{d}}{\psi _{{\text{eq}}ij}}}}{{{\text{d}}r}}} {\text{d}}r,$$

(38b)

$${A_{\psi ij}}(x,y)={\kappa ^2}\int_{x}^{y} {{{\text{e}}^{ - \kappa r}}(\kappa r+1){W_{ij}}(r){\text{d}}r} ,$$

(39a)

$${B_{\psi ij}}(x,y)={\kappa ^2}\int_{x}^{y} {\gamma (\kappa r){W_{ij}}(r){\text{d}}r} ,$$

(39b)

$${g_{1 \pm }}(x)=2 \pm 2x+{x^2},$$

(40a)

$${g_{2 \pm }}(x)=1 \pm x,$$

(40b)

$$\gamma (x)=\sinh x - x\cosh x,$$

(41)

$${W_{ij}}(r)=\left[ {{F_{\mu ij}}(r)+\frac{{Ze}}{{kT}}{\psi _{{\text{eq}}ij}}(r){F_{\mu 00}}(r)} \right].$$

(42)

In Eqs. (20a), (20b) and (20c),

$${F_{00r}}(r)={C_{001}}+{C_{002}}\frac{a}{r}+{C_{003}}{\left( {\frac{a}{r}} \right)^3}+{C_{004}}{\left( {\frac{r}{a}} \right)^2},$$

(43a)

$${F_{00\theta }}(r)= - {C_{001}} - \frac{{{C_{002}}}}{2}\frac{a}{r}+\frac{{{C_{003}}}}{2}{\left( {\frac{a}{r}} \right)^3} - 2{C_{004}}{\left( {\frac{r}{a}} \right)^2},$$

(43b)

$${F_{p00}}(r)={C_{002}}{\left( {\frac{a}{r}} \right)^2}+10{C_{004}}\frac{r}{a},$$

(43c)

$$\begin{aligned} {F_{ijr}}(r) & ={C_{ij1}}+{C_{ij2}}\frac{a}{r}+{C_{ij3}}{\left( {\frac{a}{r}} \right)^3}+{C_{ij4}}{\left( {\frac{r}{a}} \right)^2}+\frac{1}{5}{\left( {\frac{a}{r}} \right)^3}J_{{ij}}^{{(5)}}(r) \\ & \quad - \frac{a}{r}J_{{ij}}^{{(3)}}(r)+J_{{ij}}^{{(2)}}(r) - \frac{1}{5}{\left( {\frac{r}{a}} \right)^2}J_{{ij}}^{{(0)}}(r), \\ \end{aligned}$$

(44a)

$$\begin{aligned} {F_{ij\theta }}(r) & = - {C_{ij1}} - \frac{{{C_{ij2}}}}{2}\frac{a}{r}+\frac{{{C_{ij3}}}}{2}{\left( {\frac{a}{r}} \right)^3} - 2{C_{ij4}}{\left( {\frac{r}{a}} \right)^2}+\frac{1}{{10}}{\left( {\frac{a}{r}} \right)^3}J_{{ij}}^{{(5)}}(r) \\ & \quad +\frac{a}{{2r}}J_{{ij}}^{{(3)}}(r) - J_{{ij}}^{{(2)}}(r)+\frac{2}{5}{\left( {\frac{r}{a}} \right)^2}J_{{ij}}^{{(0)}}(r), \\ \end{aligned}$$

(44b)

$${F_{pij}}(r)={C_{ij2}}{\left( {\frac{a}{r}} \right)^2}+10{C_{ij4}}\frac{r}{a} - 2\frac{r}{a}J_{{ij}}^{{(0)}}(r) - {\left( {\frac{a}{r}} \right)^2}J_{{ij}}^{{(3)}}(r),$$

(44c)

where ($i$,$j$) equals (0,1), (1,0), (0,2), (1,1), and (2,0),

$${C_{001}}=\frac{1}{\omega }a(4{a^4}+4{a^3}b+4{a^2}{b^2}+9a{b^3}+9{b^4}),$$

(45a)

$${C_{002}}= - \frac{6}{\omega }b({a^4}+{a^3}b+{a^2}{b^2}+a{b^3}+{b^4}),$$

(45b)

$${C_{003}}=\frac{2}{\omega }{b^3}({a^2}+ab+{b^2}),$$

(45c)

$${C_{004}}= - \frac{3}{\omega }{a^3}b(a+b),$$

(45d)

$$\begin{aligned} {C_{ij1}} & ={\left( {\frac{b}{a}} \right)^2}{C_{004}}J_{{ij}}^{{(0)}}(b)+\frac{1}{\omega }b{\text{(}}9{a^4}+9{a^3}b+4{a^2}{b^2}+4a{b^3}+4{b^4}{\text{)}}J_{{ij}}^{{(2)}}(b) \\ & \quad +\frac{a}{b}{C_{002}}J_{{ij}}^{{(3)}}(b)+{\left( {\frac{a}{b}} \right)^3}{C_{003}}J_{{ij}}^{{(5)}}(b), \\ \end{aligned}$$

(46a)

$${C_{ij2}}={C_{003}}J_{{ij}}^{{(0)}}(b)+{C_{002}}J_{{ij}}^{{(2)}}(b)+{C_{001}}J_{{ij}}^{{(3)}}(b)+{C_{004}}J_{{ij}}^{{(5)}}(b),$$

(46b)

$$\begin{aligned} {C_{ij3}} & = - \frac{6}{{5\omega }}{b^5}J_{{ij}}^{{(0)}}(b)+{C_{003}}J_{{ij}}^{{(2)}}(b)+{\left( {\frac{b}{a}} \right)^2}{C_{004}}J_{{ij}}^{{(3)}}(b) \\ & \quad - \frac{1}{{5\omega }}{a^3}(4{a^2} - 5ab - 5{b^2})J_{{ij}}^{{(5)}}(b), \\ \end{aligned}$$

(46c)

$$\begin{aligned} {C_{ij4}} & =\frac{1}{{5\omega }}{b^3}(5{a^2}+5ab - 4{b^2})J_{{ij}}^{{(0)}}(b)+{C_{004}}J_{{ij}}^{{(2)}}(b) \\ & \quad +{\left( {\frac{a}{b}} \right)^3}{C_{003}}J_{{ij}}^{{(3)}}(b) - \frac{6}{{5\omega }}{a^5}J_{{ij}}^{{(5)}}(b) , \\ \end{aligned}$$

(46d)

$$\omega ={{\text{(}}a - b{\text{)}}^3}{\text{(}}4{a^2}+7ab+4{b^2}{\text{),}}$$

(47)

$$J_{{ij}}^{{(n)}}(r)=\int_{a}^{r} {{{\left( {\frac{r}{a}} \right)}^n}{G_{ij}}(r){\text{d}}r} ,$$

(48)

$${G_{01}}(r)=\frac{{\varepsilon {\kappa ^2}{a^4}}}{{3Zer}}{F_{\mu 00}}(r)\frac{{{\text{d}}{\psi _{{\text{eq}}01}}}}{{{\text{d}}r}},$$

(49a)

$${G_{10}}(r)=\frac{{\varepsilon {\kappa ^2}{a^4}}}{{3Zer}}{F_{\mu 00}}(r)\frac{{{\text{d}}{\psi _{{\text{eq}}10}}}}{{{\text{d}}r}},$$

(49b)

$${G_{02}}(r)= - \frac{{\varepsilon {\kappa ^2}{a^4}}}{{3Zer}}{W_{01}}(r)\frac{{{\text{d}}{\psi _{{\text{eq}}01}}}}{{{\text{d}}r}},$$

(50a)

$${G_{11}}(r)= - \frac{{\varepsilon {\kappa ^2}{a^4}}}{{3Zer}}\left[ {{W_{01}}(r)\frac{{{\text{d}}{\psi _{{\text{eq}}10}}}}{{{\text{d}}r}}+{W_{10}}(r)\frac{{{\text{d}}{\psi _{{\text{eq}}01}}}}{{{\text{d}}r}}} \right],$$

(50b)

$${G_{20}}(r)= - \frac{{\varepsilon {\kappa ^2}{a^4}}}{{3Zer}}{W_{10}}(r)\frac{{{\text{d}}{\psi _{{\text{eq}}10}}}}{{{\text{d}}r}}.$$

(50c)