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I would like to ask a specific conceptual question which bothers me for quite some time! First of all i do know the difference in between reversible and irreversible processes. What is thought in Thermodynamics courses is the maximum work can be obtained via reversible processes. Now please consider the following example.

Air (assume as ideal gas) at 5 bar and 298.15K (25℃) is expanded to 1 bar and 298.15K by a mechanically reversible processes: Heating at constant pressure followed by cooling at constant volume.

When one considers the corresponding PV diagram , the work is calculated as the area under the curve which is obviously larger than the reversible isothermal expansion.

Here are my questions

1) How many different reversible paths can be drawn in between two different states at the same temperature (there can be infinite number of irreversible paths)?

2) How can heating an ideal gas at constant pressure and cooling at constant volume be a reversible process (these are not adiabatic or isothermal)?

3) Is it possible to say that : There can be many different reversible paths between two specified states, all of which will give larger work than corresponding irreversible paths but also vary in between themselves so that it is not possible to state which reversible path will give the highest work before specifying the path itself.

Thank you all in advance for your sincere help and answers.

UPDATE: From MIT thermodynamics course notes: the reversible one produces the maximum work of all possible processes between two states.

If so there should not be more than one reversible process between two states. Then how can the process path given in the question be reversible as we can already perform the same change with isothermal expansion.

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  • $\begingroup$ I think the definition of a reversible path is that it is fully defined by the endpoints. I think an irreversible path might not even be plottable on a PV-diagram unless it goes extremely slow, since the thermodynamic variables are only defined for equilibriums. Thermodynamics is not my profession so I might be wrong though. $\endgroup$ – Emil Feb 23 '16 at 23:58
  • $\begingroup$ "Maximum amount of work" simply relates to the definition of a reversible process. A reversible process is one that does not increase entropy. This implies a maximum amount of work because no work is lost to heat (via an increase in entropy). Every reversible process between two fixed states will give the same amount of work - any deviation would make the process irreversible . $\endgroup$ – Ben Bartlett Feb 24 '16 at 1:52
  • $\begingroup$ I'm having trouble understanding the process you described. In step 1, eating at constant pressure will lead to a higher temperature and a higher volume than the original. In step 2, cooling at constant volume can bring you back to the original temperature, but at a lower pressure. So you have not gotten back to the original state of 5 bars and 298. What am I missing? $\endgroup$ – Chet Miller Feb 24 '16 at 2:35
  • $\begingroup$ I have written the second state pressure as 5 bar which should be 1 bar. A mistake by me...Now it is corrected. $\endgroup$ – Altn Feb 24 '16 at 9:29
  • $\begingroup$ Related: http://physics.stackexchange.com/q/234648/59023 $\endgroup$ – honeste_vivere Aug 10 '16 at 12:09
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Answers to questions:

1) How many different reversible paths can be drawn in between two different states at the same temperature (there can be infinite number of irreversible paths)?

There are an infinite number of reversible paths between the two different states. They don't need to be isothermal and adibatic (or combinations of these), but it is easy to visualize a sequence of isothermal and adibatic steps to get from the intial end state to the final end state.

2) How can heating an ideal gas at constant pressure and cooling at constant volume be a reversible process (these are not adiabatic or isothermal)?

As I said, the reversible steps do not have to be adiabatic or isothermal. For example, consider polytropic steps.

3) Is it possible to say that : There can be many different reversible paths between two specified states, all of which will give larger work than corresponding irreversible paths but also vary in between themselves so that it is not possible to state which reversible path will give the highest work before specifying the path itself.

This is correct if you leave out the part about "corresponding irreversible paths." You can't associate a particular irreversible path with any particular reversible path, or vice versa. All the paths between the same two end states (both reversible and irreversible) will have the same $\Delta U$ and $\Delta S$

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  • $\begingroup$ I appreciate your answer. As Ben Bartlett replied reversible process yields maximum amount of work as no work is lost as heat. As you replied there are infinite number of reversible paths possible between two paths. My conclusion is :Possibility of infinite #of reversible and irreversible paths but in some cases irreversible paths may give larger work than any reversible ones. What can be said is any reversible path does the maximum work which can not be compared with other reversible and irreversible processes. Here maximum only implies in itself there is no work loss as heat! Correct? $\endgroup$ – Altn Feb 24 '16 at 9:41
  • $\begingroup$ Maybe this article I wrote will help with your understanding. It looks at a comparison between reversible and irreversible expansions and compressions that start out at the same state and result in the same volume change, comparing the reversible work to the irreversible work: physicsforums.com/insights/… $\endgroup$ – Chet Miller Feb 24 '16 at 11:59
  • $\begingroup$ I stand by what I said about the existence of an infinite number of reversible paths between the initial and final states, and that they don't all give the same amount of work. To understand this, all you need to do is think about reversible cycles that start and end at the same state, but with different amounts of area on the PV diagram. The only thing that is constant about this is the difference between the heat and the work (zero). $\endgroup$ – Chet Miller Feb 24 '16 at 12:36
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1) The constant temperature curves of Ideal gas are not closed, so this implies that given two points with same temperature they must be joined by a unique constant temperature path, the only reversible process at constant temperature. But if you want any reversible process joining the two points, then any curve on the surface V=NkT/P that joins them will do! (think the surface as embedded in R^3). It makes no sense to talk about irreversible path, since there is no actual path, I mean no curve in the surface V=NkT/P. A curve/path is by definition a reversible process.

2) May be I am missing your point, but again, as long as you can draw a curve on the surface defined by the state equation, the reversible process exists.

3) I do not see a simple answer to this. I will think about it.

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