This question is inspired by Reif Problem 5.5. Note that it is not a homework problem and, even if it were, my question only loosely relates to it.
A vertical cylinder contains $N$ molecules of a monatomic ideal gas and is closed off at the top by a piston of mass $M$ and area $A$. The acceleration due to gravity is $g$. The heat capacities of the piston and cylinder are negligibly small, and any frictional forces between the piston and the cylinder walls can be neglected. The whole system is thermally insulated. Initially, the piston is clamped in position so that the gas has a volume $V$ and a temperature $T$. The piston is now released and after some oscillations, comes to rest in a final equilibrium situation corresponding to a larger volume of the gas. Does the entropy increase?
Now the answer, intuitively, is yes, since the system finds a new equilibrium precisely because that equilibrium is entropy-maximizing (has the most microstates).
But consider the following reasoning (and please tell me why it's wrong). While the particular process described is not quasi-static, we can imagine a corresponding quasistatic process which takes us to the prescribed final state. Since entropy is a function of state, the change in entropy will be precisely the same. Now for such a quasistatic process, we are now able to write that $dS=dQ/T=0$ because the system is thermally insulated. Thus we seem to conclude that there is no change in entropy.
I have two questions then. The first is about the aforementioned system, and the second is about "calculating with quasistatic processes" more generally.
(1) Was my reasoning flawed in that $dq\neq0$ in the imagined quasistatic process case? That is to say, is it the case that the imagined quasistatic process which takes us to our final equilibrium cannot have $dQ=0$ (i.e. there is no quasistatic, thermally insulated process with which we can reach our final state)? If so, how can I see this?
(2) In general, when am I allowed to calculate with quasistatic processes? Are there general guidelines for constructing valid quasistatic processes which take us to a desired final state?
Edit: after thinking a little more, I think it is fair to conclude that the gas itself gains no entropy (as evidenced by my argument), but that the universe itself does gain entropy (the entropy of the environment increases). This still begs the question though: if $Q=0$ here, and the environment is a heat reservoir so that $\Delta S = Q/T$ then the environment should also not gain entropy?
Best Answer
Yes. Because the process is not quasi static, entropy increases. But the increase is not due to entropy transfer which can only occur when there is heat transfer. It is instead due to entropy generated by the sudden drop in pressure and the ensuing rapid expansion and oscillations.
The quasi static process you imagine cannot be adiabatic because a process that is both quasi static (and frictionless) and adiabatic is by definition isentropic (constant entropy), whereas the process you are replacing has generated entropy. So in order for your quasi static process to connect the two same states there needs to be entropy transfer to the system in the form of heat from the surroundings equal to the entropy generated in the irreversible adiabatic process it is replacing.
Yes it was flawed. The imagined quasi static process needs to involve heat transfer from the surroundings. Such heat transfer cannot occur for any adiabatic process.
You are allowed to use any reversible (quasi static and frictionless) between two states to calculate the change in entropy between the two states, since entropy is a state function. For helpful guidelines on how to construct a valid process, see the following: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
There can be no increase in entropy of the environment (surroundings) unless there is heat transfer from the system to the surroundings. Only heat can transfer entropy. Work does not transfer entropy. This means there is never a change in entropy of the surroundings as a result of any adiabatic system process (quasi static or otherwise). Likewise an increase in entropy of an adiabatic system can never be the result of entropy transfer from the surroundings. The increase in entropy of an adiabatic system has to be all generated entropy.
Bottom Line: Entropy is generated between the initial and final equilibrium state in the non quasi static process you presented. In order to return to the initial state this entropy needs to be transferred to the environment so that the total change in system entropy is zero. Entropy can only be transferred to the surroundings in the form of heat. Therefore no adiabatic path exists that can return the system to its initial state.
As a final comment it can be shown, without an entropy argument, that a reversible (quasi static) and irreversible adiabatic process starting at the same initial equilibrium state cannot wind up in the same final equilibrium state. If the final equilibrium pressures are the same (which is the case in your example) you will find that the final temperatures and volumes will not be the same. See my answer to the following:
Reversible vs. Irreversible Expansion
Hope this helps.