I would like to ask the following questions:

  1. What is the significance of reversible and irreversible nature of thermodynamic process? (I understand that reversible processes are quasi-static, happening infinitesimally, but what is the direct outcome of being reversible and irreversible?)
  2. What does the "reversible" work and heat flow (Wrev and Qrev) signify? It is hard to understand why there are things like "lost work" due to irreversibility of a process.

Searching in the textbook and online learning sites do not seem to help - it would be great if it can be explained in simpler terms and with examples/analogies.


  • $\begingroup$ What do you mean by "signify" here? $\endgroup$
    – ACuriousMind
    Commented Apr 2, 2015 at 12:23
  • 1
    $\begingroup$ By "signify", I mean the "results of the gas being rev./irrev., in practical terms". For example, why the need to specify "reversible" work/heat flow in certain situations? (e.g. dS = dQrev/T instead of just dS = dQ/T) $\endgroup$
    – CHECK009
    Commented Apr 2, 2015 at 13:39
  • $\begingroup$ Related physics.stackexchange.com/q/488619/226902 $\endgroup$
    – Quillo
    Commented Jan 27, 2023 at 2:48

1 Answer 1


Reversible processes are important because they are related to the efficiency of a process. Take for examples a pair hotplates, one at 100C and one at 0C. In a theoretically ideal setting you could extract some work $W_0$ from this system until the two hotplates reached equilibrium. Then we would say the process is reversible, because in the same ideal world you could input $W_0$ into the system and once again create the same original temperature gradient in the hotplates.

In the real world, because of friction, poor design, and other factors, you might only extract $.7W_0$. We would say that some work in the system is lost to irreversible entropy. This is entropy which can't be reversed unless you add in extra work from the outside. You can partially reverse this system, but because you weren't able to extract all the possible work, you won't be able to reach the original temperature gradient. You might only get one hotplate to 80C and the other to 20C.

In the worst case, you simply touch the two hotplates together. In this case you can't extract any work from the system, and all potential work is lost to irreversible entropy. There is nothing you can do to restore the system to it's original state without doing extra work on the system added from the outside. So to more specifically answer your questions:

2) Reversible work and heat flow are just the parts of a system that make sense when you run it in reverse. So for example, compressing a gas with a piston make sense in reverse. But touching two hotplates together does not make sense in reverse.

1) The significance of reversibility versus irreversibility is related to efficiency. We can't extract work from an irreversible process. So systems like engines are designed in such a way to minimize the irreversible entropy and heat flow.

  • $\begingroup$ Thanks for the detailed answer! Can I clarify what you mean by "extracting work", since I can't quite picture what form of work you're referring to. (specifically, the hotplates - "work" isn't visible like compressing/expanding gas in piston) $\endgroup$
    – CHECK009
    Commented Apr 2, 2015 at 18:09
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    $\begingroup$ With the hotplates I was specifically imagining a basic carnot cycle type of engine that uses a piston to do some mechanical work. The hotplates would be your heatsinks. But I'm sure there are other possible engines you could come up with. Maybe some kind of fuel cell that could extract the work chemically. $\endgroup$ Commented Apr 2, 2015 at 21:00

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