Regular Solution

Let \(w_{ij} = w_{ji}\) and \(w_{ii} =0\). \begin{equation} G = RT \sum_{i=1}^{n} x_i \ln x_i +\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} w_{ij}x_i x_j \end{equation} \begin{align} \frac{\partial G}{\partial x_k} &= RT(\ln x_k+1) + \frac{1}{2} \sum_{j=1}^{n} w_{kj}x_j +\frac{1}{2} \sum_{i=1}^{n} w_{ik}x_i \nonumber \\ &=RT(\ln x_k+1) +\sum_{i=1}^{n} w_{ik}x_i \quad \quad \quad \quad (k=1,\cdots,n) \end{align} Chemical potential of the component \(k\) is \begin{align} \mu_k =& G-\sum_{j=1}^{n} x_j \frac{\partial G}{\partial x_j} +\frac{\partial G}{\partial x_k} \nonumber \\ =&RT \sum_{i=1}^{n} x_i \ln x_i +\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} w_{ij}x_i x_j-\sum_{j=1}^{n} x_j[RT (\ln x_j+1)+\sum_{i=1}^{n} w_{ij}x_i] \nonumber \\ &+RT (\ln x_j+1)+\sum_{i=1}^{n} w_{ik}x_i \nonumber \\ =&RT \ln x_k -\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} w_{ij}x_i x_j+\sum_{i=1}^{n} w_{ik}x_i \quad \quad \quad \quad (k=1,\cdots,n) \label {A4} \end{align} Let \(w_{ij}=m_{ij}RT_{C}\). Then, Eq.\eqref{A4} could be written as \begin{align} \mu_k &=RT \ln x_k -\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} m_{ij}RT_{C}x_i x_j+\sum_{i=1}^{n} m_{ik}RT_{C}x_i \nonumber\\ &=RT_{C}\left[\tau \ln x_k -\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} m_{ij}x_i x_j+\sum_{i=1}^{n} m_{ik}x_i \right] \quad \quad \quad \quad (k=1,\cdots,n) \end{align} where \(\tau = \frac{T}{T_C}\) is the reduced temperature. As long as every \(m_{ij}\) is not a function of \(T\), all the phase diagrams should be identical for the same given value of the reduced temperature, \(\tau\). This statement is also true even if the interaction parameter \(w_{ij}\) is a function of composition, e.g. a subregular solution model.