The mixing of gases is always accompanied by an increase in entropy. Show that in the formation of a binary mixture of two ideal gases the maximum entropy increase results when X_1 = X_2 = 0.5
For a binary mixture the entropy per mole of the mixture formed is given by
\begin{aligned}\Delta S_{\text {mixing }} & =-R\left[X_{1} \ln X_{1}+X_{2} \ln X_{2}\right] \\& =-R\left[X_{1} \ln X_{1}+\left(1-X_{1}\right) \ln \left(1-X_{1}\right)\right]\end{aligned}For entropy of mixing to be maximum, the first derivative \frac{\delta\left(\Delta \mathrm{S}_{\text {mixing }}\right)}{\delta \mathrm{X}_{1}} should be zero and the second derivative should be negative. Differentiating \Delta S_{\text {mixing }} with respect to X_1 we get
\begin{array}{l}\frac{\delta\left(\Delta \mathrm{S}_{\text {mixing }}\right)}{\delta \mathrm{X}_{1}}=-\mathrm{R}\left[\ln \mathrm{X}_{1}+\frac{\mathrm{X}_{1}}{\mathrm{X}_{1}}+\frac{1-\mathrm{X}_{1}}{1-\mathrm{X}_{1}}(-1)+(-1) \ln \left(1-\mathrm{X}_{1}\right)\right]\\ \\-\mathrm{R}\left(\ln \mathrm{X}_{1}+1-1\right)-\ln \left(1-\mathrm{X}_{1}\right)=0\\ \\\text { or } \ln \mathrm{X}_{1}+1-1-\ln \left(1-\mathrm{X}_{1}\right)=0\end{array}or \ln \frac{\mathrm{X}_{1}}{1-\mathrm{X}_{1}}=0 \quad \text{or} \quad \mathrm{X}_{1}=1-\mathrm{X}_{1}\\
2 \mathrm{X}_{1}=1 \quad \mathrm{X}_{1}=1 / 2=0.5And hence \mathrm{X}_{2}=1-0.5=0.5.