Question

Prove that the entropy of mixing of an ideal mixture has an infinite slope, when plotted vs. x, at x = 0 and x= 1.

Short answer

Therefore, it is proved that the entropy of mixing of an ideal mixture has an ideal slope.

Step-by-step solution

Given information

The entropy of mixing of an ideal mixture has an infinite slope, when plotted vs. x, at x = 0 and x= 1.

Concept

When two ideal gases are allowed to mix at the same pressure and temperature and the total volume remains unaltered, the change in entropy is:

$\Delta S_{\text{mix}}=-nR\left(x\ln \right(x)+(1-x\left)\ln \right(1-x\left)\right)$

Explanation

Take the derivative of $\Delta S_{mix}$to show that the slope of the entropy with respect to x is zero at x = 0 and x = 1.

$$\begin{gathered} \frac{d}{dx}\Delta S_{mix}=-nR\left(x\times \frac{1}{x}+\ln \left(x\right)-(1-x)\times \frac{1}{1-x}-\ln (1-x)\right) \\ \frac{d}{dx}\Delta S_{mix}=nR\left(\ln \right(1-x)-\ln (x\left)\right) \\ \frac{d}{dx}\Delta S_{mix}=nR\ln \left(\frac{1-x}{x}\right) \end{gathered}$$

At x=0, we have

$$\begin{gathered} {\left.\frac{d}{dx}\Delta S_{mix}\right|}_{x=0}=nR\ln \left(\frac{1}{0}\right)=nR\ln (\infty )=\infty \\ {\left.\frac{d}{dx}\Delta S_{mix}\right|}_{x=0}=\infty \end{gathered}$$

At x=1, we have

$$\begin{gathered} {\left.\frac{d}{dx}\Delta S_{mix}\right|}_{x=1}=nR\ln \left(\frac{0}{1}\right)=nR\ln \left(0\right)=\infty \\ {\left.\frac{d}{dx}\Delta S_{mix}\right|}_{x=1}=\infty \end{gathered}$$