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 !

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    $\begingroup$ Are you accounting for the entropy change of the atmosphere in your calculation? $\endgroup$ Dec 3 '18 at 12:50
  • $\begingroup$ @probably_someone No , we cant have free expansion when the pressure is less than the atmosperic pressure $\endgroup$ Dec 3 '18 at 12:51
  • $\begingroup$ 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. $\endgroup$ Dec 3 '18 at 12:54
  • $\begingroup$ @Drew ok .. but how this expansion can be possible .. if 10 kpa is less than 100 kpa (1atm)? $\endgroup$ Dec 3 '18 at 13:05
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    $\begingroup$ 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. $\endgroup$
    – GiorgioP
    Dec 3 '18 at 13:21

The question brings up a point that is frequently lost: For a process to be feasible it is not enough for the total entropy generation from initial-to-final state to be zero, it must be zero in every step of the process.

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 $P_A<P_B$ then $\Delta V<0$, i.e., $B$ moves into $A$.

Conclusion The expansion is feasible as long as the high pressure part expands into the low pressure part. This of course we know from simply looking at the forces on the piston. The point is that the second law gives the right answer. So, once we reach the point where pressures are equal, any further expansion would require negative entropy generation and is therefore unfeasible. That's the equilibrium state and once it is reached, the system will remain there.

  • $\begingroup$ Ok.. thanks but I know that it is feasible until equilibrium.. but what if I suppose that the expansion will continue until the pressure in the cylinder falls below the atm pressure .. I know it is impossible process .. but no problem we will neglect that and calulate s gen which should be negative .. but here we will have positive s gen even if we devide the process to 2 processes (one possible and the other is not) because there is no term for the atmospheric pressure in s gen formula ! Does this mean that entropy generation is not ALWAYS a measure for possibility of a process? $\endgroup$ Dec 6 '18 at 14:44
  • $\begingroup$ 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. $\endgroup$
    – Themis
    Dec 6 '18 at 15:30

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