Entropy generation

Does entropy generation fully define if the process is possible or not? For example if I have a piston cylinder device expands freely and isothermally from 200kpa to 10kpa - which is not possible because of the atmosheric pressure - but here we find entropy generation is positive !

• Are you accounting for the entropy change of the atmosphere in your calculation? Dec 3 '18 at 12:50
• @probably_someone No , we cant have free expansion when the pressure is less than the atmosperic pressure Dec 3 '18 at 12:51
• Entropy generation fully defines if a process is reversible or not. You can have a reversible isothermal expansion, but you're right that it won't be reversible if happens freely. Dec 3 '18 at 12:54
• @Drew ok .. but how this expansion can be possible .. if 10 kpa is less than 100 kpa (1atm)? Dec 3 '18 at 13:05
• The question does not specify whether the piston is in contact with the atmosphere (at the sea level) or it is in a vacuum or in contact with a gas at a pressure of 10 kPa. In the last two cases, I do not see any problem. Dec 3 '18 at 13:21

Let's see why the experiment you propose is not feasible. We start with a rigid insulated box divided into parts: $$A$$ is the system, $$B$$ is the surroundings. If $$B$$ is very large it becomes the atmosphere, but this is not important. Both systems are at the same temperature but at different pressures. Now we move the piston isothermally by a small amount so that the volume of part $$A$$ changes by $$\Delta V$$ and the volume of $$B$$ by $$-\Delta V$$. During this step the energy of part $$A$$ changes by $$\Delta U$$ and part $$B$$ by $$-\Delta U$$ (the box is an isolated system).
We will calculate entropy changes using the entropy equation $$dS = \frac{dU}{T} + \frac{P dV}{T}$$ For small $$\Delta V$$: $$\Delta S_A = \frac{\Delta U}{T} + \frac{P_A \Delta V}{T}$$ $$\Delta S_B = -\frac{\Delta U}{T} - \frac{P_B \Delta V}{T}$$ The entropy generation is $$S_\text{gen} = \Delta S_A + \Delta S_B =(P_A-P_B)\frac{\Delta V}{T}\geq 0$$ This must be positive, or zero at most. This means that if $$P_A>P_B$$ then $$\Delta V>0$$, i.e., $$A$$ moves into $$B$$; if $$P_A then $$\Delta V<0$$, i.e., $$B$$ moves into $$A$$.
• You are asking a good question: What if the overall entropy generation along a path that connects the initial and final states is positive but the entropy generation for some portion along this path is negative? Then it is possible to construct some other path to reach the final state, such that $dS_\text{gen}$ is positive everywhere along that path. Thermodynamics cannot tell you which path that is, only that it exists. For example, you might use weights, springs etc. to store some of the energy and use it later to reach a pressure below atmospheric. Dec 6 '18 at 15:30