Change in Entropy — Question 6

Question 6

Mixing distinguishable ideal gases adds component expansion entropies

Two distinct ideal gases initially occupy separate compartments at the same temperature and pressure. They contain .250 mol of A and .750 mol of B. A partition is removed in a rigid insulated enclosure. Use $\Delta S_i=n_iR\ln(V_{\mathrm{total}}/V_i)$, $R=8.314$ J/(mol K), and additive component entropies.

1. Find each initial volume fraction and the entropy of mixing.

2. If both compartments instead contain the same gas at the same temperature and pressure, what is the macroscopic entropy change on removing the partition? Explain why treating origin labels as distinct species would be incorrect.

Question 6 - Solution

Mixing distinguishable ideal gases adds component expansion entropies

(a) Method and calculation

$$V_A/V=.250,\quad V_B/V=.750\quad\text{because }V_i\propto n_i.$$$$\Delta S_{\mathrm{mix}}=-R[.250\ln(.250)+.750\ln(.750)]\approx4.67525\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\Delta S_{\mathrm{identical}}=0.$$$$T_f=T_i,\quad p_f=p_i,\quad\text{density and composition are unchanged}.$$For identical gas, there is no new composition variable or distinguishable-species expansion. Merely labeling molecules by their original side does not create a thermodynamic mixing entropy.

Physical check

The distinct-species formula assumes ideal mixing without reaction or a heat of mixing.