Question 15.7.3: The mixing of gases is always accompanied by an increase in ......

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

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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=0.5

And hence \mathrm{X}_{2}=1-0.5=0.5.

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