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Can two adiabatic process paths intersect on a plot? This question is for both reversible and irreversible processes. There are some answers for this question on Quora, but they mostly address only for reversible adiabatic processes. Also I want a physical interpretation of the processes that is, if they don't intersect then why don't and what it would be like, physically for a system.

Here are answers on Quora-https://www.quora.com/Why-do-two-adiabatic-curves-never-cut-each-other

NOTE:- Please only answer if you are absolutely clear on your fundamentals of thermodynamics and can fully explain everything.

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    $\begingroup$ In the previous question you speak of, it is apparent that the person who answered you understands thermodynamics much better than you. Insulting a qualified person who was trying to help you doesn’t really motivate anybody to help you again. $\endgroup$ – knzhou Sep 7 '18 at 5:57
  • $\begingroup$ I am sorry for I didn't mean to insult someone. And I really respect their experience. But they said some things that were contradictory to concepts mentioned in some really credible sources. I also sent them links to these sources. Point is, these things leads discussion elsewhere & it takes far to longer to get to the answer. As you can see, we didn't get to one, in the question I mentioned. $\endgroup$ – shashank tyagi Sep 7 '18 at 6:28
  • $\begingroup$ A system undergoing an irreversible process generally aren't in thermal equilibrium (or arbitrarily close to it) and so aren't at any point on the $PV$ diagram. Different parts of the system will in general be a different pressures, so we can't talk about the pressure and even if it happens that the system does have a single pressure, away from equilibrium pressure and volume alone are not enough to determine the state of the system, so you cannot expect to say a great deal about what the system can and cannot do until the process is finished. $\endgroup$ – By Symmetry Sep 7 '18 at 9:41
  • $\begingroup$ If an adiabatic reversible process and an adiabatic irreversible process start out at the same state, doesn't that automatically mean they intersect? $\endgroup$ – Chet Miller Sep 7 '18 at 12:00
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    $\begingroup$ I don't see how you can possibly say that, when all I need to do is cite one single example in which an adiabatic reversible process and an adiabatic irreversible process start out in the same initial state. By the way, I don't know whether it's appropriate to ask this, but I'm going to do so anyway: Why are you spending so much of your valuable time speculating about esoteric things like this when you could be spending it so much more productively solving actual practice problems involving adiabatic reversible and irreversible processes? $\endgroup$ – Chet Miller Sep 7 '18 at 13:40
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To answer your first question, yes two (different) adiabatic paths can intersect in a plot. See State A in Fig 2. However, for this to be possible, at least one of them has to be irreversible. For different reversible adiabatic processes involving the same working substance, their plots cannot intersect. See Fig 1.

Now, if you could have two reversible adiabatic processes intersect and be connected by a reversible isothermal process, there would be a violation of the second law. In that context, the answer to the linked question where both adiabatic processes are shown as reversible, would be correct. That is because both the system and the surroundings would return to their original states with no increase in entropy.

However, since at least one adiabatic process must be irreversible, then the cycle A-B-C-A shown in Fig 2 below would not violate the Kelvin-Plank statement of the second law. Recall the law states:

“It is impossible to construct a device which operates on a cycle and produces no other effect (my emphasis) than the production of work and the transfer of heat from a single body”.

The irreversible adiabatic process (Process A-B) in Fig 2 produces an increase in entropy. Thus cycle A-B-C-A does constitute some other effect than the production of work and transfer of heat from a single body, and therefore does not violate the Kelvin-Plank statement of the second law.

Hope this helps enter image description here

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