Derive an expression from the Gibbs-Thompson relation

[Jacob Daniel]

Homework Statement

The Gibbs-Thompson relation provides a correlation between solubility and radius of curvature. Consider two spherical particles of radius r and R present in a solvent, with R >> r. The rate of dissolution of the smaller particles (the rate of decrease in its radius with time) is directly proportional to the difference in the solubility of the two particles times its surface area. Derive an expression of how the radius of the smaller particle evolves with time. Both particles are made of the same material which has an atomic volume of Ω, and have the same surface energy of γ.

Homework Equations

ln(S_c/S_infinity) = [(γ*Ω)/(K*T)]*[(1/R₁)+(1/R₂)]

This is the Gibbs-Thompson relation. For a sphere, R₁=R₂

The Attempt at a Solution

So I think what this question is asking is to find an equation which equates the radius of the smaller particle with time. This has to do with an effect called Ostwald ripening where material from smaller particles is transferred to larger particles for some reason.

Let -d_r/d_t be the rate at which the the radius of the smaller particle changes with time. Apparently it is directly proportional to the difference in the solubility of the two particles times its surface area (I assume it means the surface area of the smaller particle).

So uh... I think you need to solve for solubility of each particle and then find the difference. Multiply that difference by the SA of the smaller particle. Set that equal to the rate at which the radius of the smaller particle decreases. And then integrate to find an expression which relates r to t.

 

[Nino MMP]

But I'm really confused about how to solve for the solubility of each particle. Any help would be appreciated!