Rearranging these formulas

Question

The entropy of mixing formula is $\Delta S_{\mathrm{mix}} = R(x_1\ln(x_1) + x_2\ln(x_2))$ where $x$ is the mole fraction $x_1 = \frac{n_1}{n_1+n_2}$ and $x_2 = \frac{n_2}{n_2+n_1}$ where $n$ is the number of moles and the subscript denotes which compound it is. I am asked to express this entropy of mixing formula in terms of the mass fraction which is $y_1 = \frac{m_1}{m_1+m_2}$ where $m$ is the mass of the substances. now $m_1 = n_1M_1$ and $m_2 = n_2M_2$ where $M$ is the molar mass, I have tried to rearrange them a lot and could not manage to express $x$ in terms of $y$. Could anyone help me please?

Answer 1

Given

\begin{align} \Delta S_{\mathrm{mix}} &= R(x_1\ln(x_1) + x_2\ln(x_2)) \tag{1}\label{1} ,\\ x_1 &= \frac{n_1}{n_1+n_2} \tag{2}\label{2} ,\\ x_2 &= \frac{n_2}{n_1+n_2} \tag{3}\label{3} ,\\ y_1 &= \frac{m_1}{m_1+m_2} \tag{4}\label{4} ,\\ m_1 &= n_1M_1 \tag{5}\label{5} ,\\ m_2 &= n_2M_2 \tag{6}\label{6} . \end{align}

From \eqref{4}-\eqref{6} we have \begin{align} y_1 &= \frac1{1+\displaystyle\frac{n_2}{n_1}\cdot \frac{M_2}{M_1}} \tag{7}\label{7} , \end{align}

from \eqref{2}-\eqref{3} follows

\begin{align} x_1+x_2&=1 \tag{8}\label{8} ,\\ \frac{n_2}{n_1} &= \frac{x_2}{x_1} =\frac{1-x_1}{x_1} \tag{9}\label{9} , \end{align}

so \eqref{9} combined with \eqref{7}, \eqref{8} gives the desired:

\begin{align} y_1 &= \frac1{1+\frac{x_2}{x_1}\cdot \tfrac{M_2}{M_1}} \tag{10}\label{10} ,\\ y_1 &= \frac1{1+\frac{1-x_1}{x_1}\cdot \tfrac{M_2}{M_1}} \tag{11}\label{11} ,\\ y_1 &= \frac1{1+\left(\frac1{x_1}-1\right)\cdot \tfrac{M_2}{M_1}} \tag{12}\label{12} ,\\ \frac1{y_1} &= {1+\left(\frac1{x_1}-1\right)\cdot \tfrac{M_2}{M_1}} \tag{13}\label{13} ,\\ \frac1{y_1}-1 &= {\left(\frac1{x_1}-1\right)\cdot \tfrac{M_2}{M_1}} \tag{14}\label{14} ,\\ \left(\frac1{y_1}-1\right)\cdot \tfrac{M_1}{M_2} &= \frac1{x_1}-1 \tag{15}\label{15} ,\\ 1+\left(\frac1{y_1}-1\right)\cdot \tfrac{M_1}{M_2} &= \frac1{x_1} \tag{16}\label{16} , \end{align}

\begin{align} x_1 &= \frac 1{1+\displaystyle\frac{1-y_1}{y_1}\cdot \tfrac{M_1}{M_2}} \tag{17}\label{17} ,\\ x_2 &= \frac 1{1+\displaystyle\frac{y_1}{1-y_1}\cdot \tfrac{M_2}{M_1}} \tag{18}\label{18} . \end{align}