# Regular Solution

Let $$w_{ij} = w_{ji}$$ and $$w_{ii} =0$$. $$G = RT \sum_{i=1}^{n} x_i \ln x_i +\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} w_{ij}x_i x_j$$ \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.