A thermodynamic process is reversible if the system remains in the equilibrium state throughout. That's what we are told in the book. My doubt is, though, can we judge whether a process is reversible or not by looking at the P-V diagram?

Technically, P-V diagrams should only be used for a reversible process, as irreversible processes involve inhomogeneous pressure and temperature. However, I have seen P-V diagrams for Diesel cycles and Otto cycles, which are not reversible because of the sudden explosion of fuels. I am very confused by this. What's wrong?

It appears that any path on a P-V diagram can be reversible if the process is performed very slowly. Is that true?

To summarise,

  1. Does reversibility have anything to do with the P-V diagram?
  2. If I do not understand the above correctly, what are some examples of irreversible cycles?
  • $\begingroup$ I have revised the first paragraph of my answer based on comments from Chet Miller. The essence has not changed, that is, you can't tell from a PV diagram alone if a process is reversible because it doesn't assure us that the system is in internal pressure equilibrium. Hope it helps clarify. $\endgroup$
    – Bob D
    Sep 1 '19 at 15:35

You can’t tell if a process is reversible or not just by looking at a PV diagram. The reason is it only tells you the external pressure and doesn't tell us if the gas in the system is in equilibrium internally. In the case of an ideal gas, if it is not in internal pressure equilibrium we can't apply the ideal gas law during the process because it only applies to equilibrium states. For an irreversible process we must wait until the end allowing internal equilibrium to be reached. In other words, we can only apply the law only at the beginning and end points of the process. For a reversible process, which also requires no friction, we know that the system is in equilibrium throughout the process

PV diagrams don’t have to be limited to reversible processes. But they need to be presented with a description of the process that allows you to determine whether or not the process is reversible.

Finally a process carried out very slowly (quasi-statically) can be reversible, but an additional requirement is there’s no friction involved.


  1. Not sure what you mean by “anything to do with “.

  2. Any process involving finite temperature or pressure differences or involving friction is irreversible. For example suppose you have a diagram involving a constant volume process (a vertical line on a PV diagram). It could involve a very slow (quasi static) extraction or addition of heat making the process reversible. Or it could be a sudden decrease or increase in external pressure making the process irreversible. The diagram doesn’t include the time it took for the pressure to drop or increase. You need to be given that information.

Hope this helps

  • $\begingroup$ The external pressure is exactly the same as the force per unit area exerted by the gas on the piston (assuming a massless, frictionless piston). The problem is that the force per unit area exerted by the gas on the piston includes viscous normal stress (which depends on the rate at which the gas is deforming), so that this force per unit area is not equal to the pressure calculated from the ideal gas law. $\endgroup$ Sep 1 '19 at 12:30
  • $\begingroup$ @ChetMiller Hi Chet. Thanks for that clarification. Although the pressure exerted at the surface of the piston equals the external pressure, if I said instead that the pressure in the gas is not uniform (not in internal equilibrium) due to viscous stresses would that essentially be the same as what your saying. Because if it is, I would be more comfortable revising my answer accordingly. $\endgroup$
    – Bob D
    Sep 1 '19 at 12:49
  • $\begingroup$ Not exactly. The compressive stress in the gas is a linear summation of the "pressure" $\rho RT$ and the viscous stress. Any non-uniformity of pressure (or total stress) is more the result of the gas having mass (i.e. inertia). So there are really two contributions to the total stress and stress variations in a gas during an irreversible process: viscous stresses and inertia. $\endgroup$ Sep 1 '19 at 14:44
  • $\begingroup$ @ChetMiller That's a little too complicated for me, and probably for the OP as well. In order to keep it simple, is it sufficient just to say that the gas is not in internal pressure equilibrium so we can't apply the ideal gas law during the irreversible process and must wait until the end allowing internal equilibrium to be reached. In other words, we can apply the law only at the end (equilibrium) points of the process? $\endgroup$
    – Bob D
    Sep 1 '19 at 15:19
  • $\begingroup$ Yes. I agree. Well said. $\endgroup$ Sep 1 '19 at 15:20

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