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

Expert verified

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

Step by step solution

01

Given information

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

02

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:

ΔSmix=-nR(xln(x)+(1-x)ln(1-x))

03

Explanation

Take the derivative of ΔSmixto show that the slope of the entropy with respect to x is zero at x = 0 and x = 1.

ddxΔSmix=-nRx×1x+ln(x)-(1-x)×11-x-ln(1-x)ddxΔSmix=nR(ln(1-x)-ln(x))ddxΔSmix=nRln1-xx

At x=0, we have

ddxΔSmixx=0=nRln10=nRln()=ddxΔSmixx=0=

At x=1, we have

ddxΔSmixx=1=nRln01=nRln(0)=ddxΔSmixx=1=

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