# Interactions between “particles” to prevent diffusion and to cause phase segregation

The free energy of particles reads, if we assume no internal energy : $$\tilde f(c)= \frac{f(c)}{k_bT}=-c \, ln(c)-(1-c)\, ln(1-c)$$, with $$c$$ the concentration.

If we had a term $$F(c)$$ for the interactions between the particles : $$\tilde f(c)= - c \, ln(c)-(1-c)\, ln(1-c) + F(c)$$, such that $$F(0)=0$$, which is reasonable since it's an interaction potential.

The equilibrium between 2 phases $$1$$ and $$2$$ is determined by the relations : $$\mu_1=\mu_2$$ and $$\pi_1=\pi_2$$.

Now assume you just put your particles that interact with each other in a huge box full of water (an ocean for example). Then, a reasonnable thought is that you can have a phase with only water, and a phase with particles ($$c_0$$) and water ($$1-c_0$$). But the relation $$\mu_1=\mu_2$$ prevents that because that $$\mu(0)=-\infty$$. And I'm really not convinced by that ! If your particles attract each other, they should at some point aggregate, don't they ? In other words, entropy will never let aggregation win totally !!

You are right, the entropy of mixing term will never allow $$c\rightarrow0$$ or $$c\rightarrow1$$.
It is possible to approach those limits very very closely. It depends on whether the interaction term is unfavourable and how unfavourable it is. An example would be $$F(c)=K c(1-c)$$, with some constant $$K>0$$, which vanishes at both extremes and has a maximum (for unfavourable interactions) in the middle. Even in the solid region of two-component systems A+B, where you see almost complete demixing into two solid phases, you can expect a tiny fraction of A atoms dissolved in B, and vice versa. The same idea applies to your solid-liquid example.
The driving force behind this is the gradient of the free energy with respect to $$c$$. If you calculate this from your first equation, you'll see that it diverges at both extremes $$c=0$$ and $$c=1$$, so as to make a small amount of mixing favourable, for any physically reasonable form of the interaction term. In other words, even if $$K$$ is very large, the gradient of the entropy of mixing term will overwhelm the $$F(c)$$ term near the edges of the composition range: the combination $$c\ln c + (1-c)\ln (1-c) + F(c)$$ will have two minima, very close to (but not exactly at) the extremes.