#### Weak solutions

We now want to derive the dependence of the chemical potential of the solute on its concentration in the dilute limit , i.e. for $$N \ll M$$, $$M$$ being the number of molecules of the solvent (this is the definition of a weak solution). Let $$X$$ and $$Y$$ be the (phase-space) variables of the solute and solvent, respectively. The Hamiltonian of the system reads:

$$H(X,Y) = K_X+K_Y+V(X)+U(Y)+W(X,Y),$$

where the $$K$$'s and $$V(X)$$ and $$U(Y)$$ the kinetic and potential energies of the two components alone, and $$W(X,Y)$$ their interaction. In the dilute limit the interaction between solute molecules can be neglected and we set $$V(X)=0$$. The partition function of the solution then reads:

\begin{align} Z&=\frac{1}{N!}\frac{1}{M!}\int \mathrm{e}^{-\frac{1}{k_BT}\bigl ( K_X+K_Y +U(Y)+W(X,Y)\bigr )} dXdY, \\\\ &= \frac{1}{N!}\int \mathrm{e}^{-\frac{1}{k_BT}\bigl ( K_X + F_Y(X) \bigr )} dX, \end{align}

where $$F_Y(X) = -k_B T\log \left ( \frac{1}{M!} \int \mathrm{e}^{-\frac{1}{k_BT}\bigl ( K_Y + U(Y) + W(X,Y) \bigr )} dY \right )$$ is the free energy of the solvent in the field of the solute, whose molecules are supposed to be kept fixed at positions $$X$$. In the dilute limit $$F_Y(X)-F_Y^\circ=Nf'$$, where $$F^\circ$$ is the free energy of the pure solvent. One has therefore:

$$Z = \mathrm{e}^{-\frac{1}{k_BT}(F^\circ_Y+Nf'+F^\circ_X)},$$

where

\begin{align} F^\circ_X &= -k_BT \log \left ( \frac{1}{N!}\left ( \frac{V}{\lambda^3}\right )^N \right ) \ \\\ &= Nk_BT \log \left ( \frac{n\lambda^3}{e} \right ) \label{eq:Fideal} \end{align}

and $$\lambda = \sqrt{\frac{2\pi\hbar^2}{M_Xk_BT}}$$ are the free energy and the thermal wavelength of an ideal gas at the $$(NVT)$$ conditions of the solute. In the present classical framework, $$\hbar$$ is just an arbitrary constant that fixes the value of the entropy at zero temperature and the units in Eq. \eqref{eq:Fideal}. The free energy of the solution is therefore:

$$F=F^\circ_Y+N\left (f'+ k_BT \log \left ( \frac{n\lambda^3}{e} \right ) \right ).$$

The chemical potential of the solute and its gradiente are therefore:

\begin{align} \mu_X &= {\bar{\mu}}_X + k_B T\log(n), \\ \nabla \mu &= k_B T \frac{\nabla n}{n}. \end{align}