#### 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:

\begin{equation} H(X,Y) = K_X+K_Y+V(X)+U(Y)+W(X,Y), \end{equation}

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:

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

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:

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

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

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