Irreversible Processes and Entropy — Question 8

Question 8

Reversing a gas state does not erase the entropy made in a free expansion

One mole of ideal gas at 300 K first expands freely and adiabatically from $V$ to $2V$. It is then compressed reversibly and isothermally back to $V$ by a 300 K reservoir and a work source. Use $R=8.314$ J/(mol K) and $\Delta S=nR\ln(V_f/V_i)$ for an isothermal ideal-gas change.

1. Find gas entropy changes for both legs and the net gas change.

2. Find net reservoir entropy change and work input over the complete protocol. Has the whole environment returned to its starting state?

Question 8 - Solution

Reversing a gas state does not erase the entropy made in a free expansion

(a) Method and calculation

$$\Delta S_{g,1}=R\ln2\approx5.76283\,\mathrm{J/K},\quad\Delta S_{g,2}=-R\ln2.$$$$\Delta S_{g,\mathrm{cycle}}=0.$$

(b) Interpretation and further calculation

$$W_{\mathrm{on,2}}=RT\ln2\approx1728.85\,\mathrm J,\quad Q_{r,2}=RT\ln2.$$$$\Delta S_r=Q_{r,2}/300=R\ln2>0,\quad S_{\mathrm{gen,total}}=R\ln2.$$The work source loses energy and the reservoir gains heat. Restoring the gas alone does not reverse every change in the composite.

Physical check

A cyclic subsystem has zero net entropy change; its surroundings need not.