My book says 'Maximum work can be obtained only from thermodynamically reversible processes,' but why is it so? What is the cause?

Actually to me the definition of reversible process is confusing. It says that at each step during the process, equilibrium is maintained. But, let's say,we have to expand a gas in a cylinder headed by piston from pressure $P_i$ to $P_f$. Suppose,the process is done with innumerable infinitesimal steps. Then, say at each step, $dp$ is changed in the pressure.

Before the step, gas and the piston were exerting same pressure $P_i$ . Now, when the piston is raised, gas does expand to equalize the pressure,which is $P_i - dp$ . So, at the end of the step, the gas & the piston will exert same pressure $P_i - dp$ and hence will be in mechanical equilibrium.

During the step,the gas and the piston was not in equilibrium;it was at the end of the step the equilibrium is achieved. So, during each step of the process the gas would be at disequilibrium . So,will it really be a reversible process? It is contradictory with the definition. But it is true that during each step, the system is at disequilibrium though infinitesimally small by $dp$ . So, why does the definition tell that at each step,there is equilibrium?

So, I have three questions:

  1. Why is maximum work only obtained from reversible processes?

  2. According to the definition, at each step of the reversible process,there must be equilibrium. But during each step, there is infinitesimal disequilibrium. So, why does the definition say so in-spite of the disequilibrium during each step?

  3. How do infinitesimal steps make a process reversible? I will be very grateful if anyone answer these three questions.

Please help.

  • $\begingroup$ Check Carnot cycle. $\endgroup$
    – Xiaoge Su
    Sep 25, 2014 at 1:03
  • $\begingroup$ I haven't seen too many texts deal with the mathematical nuances of reversibility in the Carnot Cycle, in fact none!I know it's a really popular pedagogical tool in classical thermodynamics, but for the life of me I never found it at all helpful! $\endgroup$
    – user7232
    Sep 25, 2014 at 11:33
  • $\begingroup$ @user5764 Basically all irreversible paths by definition have to "generate" some new (previously non-existent) amount of entropy (i.e, always make $\Delta S_{univ} > 0$) hence will always have some more inefficiency relative to the reversible processes in terms of doing work (e.g, such as needing more input heat, and for the same amount of work, etc). The rest of the discussion on "infinitesimals/quasi-static/etc" just boils down to (listing) conditions for real processes under which said entropy "generation" can be minimized, well, ideally down to zero for the fully "reversible" case..... $\endgroup$
    – ManRow
    Jan 31 at 12:06

2 Answers 2


I'll give a mathematical answer, and an analogy also for ease of thought. These are my own intuitions, so they may work for you.

My mathematical reasoning:

A lot of thermodynamic processes are phrased very carefully in terms of reversibility. Now my intuition of this is that reversibility is a consequence of the maths we must use. Thermodynamics likes to look at continuous processes such that we can use integrals instead of summations in the calculations: this stands to reason since classical thermodynamics treats a collection of particles as moving, say, a piston continuously from one place to another, not in discrete steps.

Thus, one way to think about it is to think of reversibility as being necessary for an integral formulation. If we allow ourselves to try and think in infinitesimal steps, moving an infinitesimal step forward is just an infinitesimal away from going backwards and hence this is described as being reversible. In short, if a process really happens continuously then we must impose the condition of reversibility on it, otherwise the mathematics fails.

My analogy:

Why this gives maximum work can be seen by analogy with a real life classical system: the tug-of-war. If you're in a tug of war with some of your friends who are of equal size to you and of equal strength, you're going to put in maximum work in pulling the rope. If one side is of unequal size and strength it becomes much easier for the stronger side.

To be more serious, if your particles move your piston irreversibly (i.e. they never move backwards along the way) then no work has been expended in going backwards—only forwards. Reversible systems belie the work they've had to do since the physical outcome we actually see—the final outcome—shows the piston in the same place as the irreversible case, however what we haven't seen is the system battling backwards and forwards.

To conclude: if we imagine a set amount of work needs to be done to move a piston irreversibly from a point A to a point B, then the reversible case must provide maximum work since some of the time the system has battled going backwards and has still ended up in the same place as the irreversible case.

EDIT: How to Achieve Equilibrium and still have Movement

I think the source of your confusion lies in the statement 'it was at the end of the step of the equilibrium'. I know what you mean by this, but I don't think it's precise. Equilibrium doesn't mean to describe position as such, but the oscillation of position. Even if you've physically pushed your piston a distance ds, it can still be in equilibrium by the very fact that it can still move forwards and backwards in its next spot.

So in fact the periodicity of position is maintained at each step, which is what we mean by the equilibrium is maintained.

If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur.

  • $\begingroup$ Sir, what will you say about the disequilibrium during each step? How should I deal with it? Doesn't it contradict the definition? $\endgroup$
    – user5764
    Sep 25, 2014 at 8:26
  • $\begingroup$ I've edited my answer to your response, however the main point to note is: "If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur." $\endgroup$
    – user7232
    Sep 25, 2014 at 11:21
  • $\begingroup$ If we consider a chamber with a membrane in the middle separating air and vaccum on either sides,what is my system?If i take the whole chamber as my system,there is no work transfer or heat transfer(chamber is insulated) if i remove the membrane.Therefore this justifies that no work is done by gas in free expansion.But the air and vaccum vary in composition,that is they are not in equilibrium,then taking the whole chamber as my system incorrect.Then what should be my system?Just the air half chamber or just the vaccum half chamber? $\endgroup$ Nov 13, 2019 at 9:56

Reversibility is a statement that means, the system and the surroundings are perfectly balanced (in equilibrium) at ANY AND ALL instants in time. This allows us to define $T_{\mathrm{sys}} = T_{\mathrm{surr}}$ and $p_{\mathrm{sys}} = p_{\mathrm{ext}}$ ALWAYS. The latter statement on pressure allows then that we can calculate work in terms of system pressure rather than external pressure. The proof from this point is the mathematics to show, irreversible systems always do less work than reversible systems.

Otherwise, the approach you take with your analysis of pressure in steps is the same one that allowed the tortoise to beat Achilles in a race. It is called Zeno's paradox.

In specific answers to your numbered questions ...

1) Accepting the above definitions allows the mathematics to prove that reversible work is always greater than irreversible work. As a physical answer, we can define reversible work as $W_{\mathrm{rev}}$. Any irreversible process must have a point in time where the system and surroundings are not at equilibrium with each other. Let's use this to get work $W_{\mathrm{extra}}$, and call $W_{\mathrm{irr}} = W_{\mathrm{rev}} + W_{\mathrm{extra}}$. The final proof that $|W_{\mathrm{irr}}| < |W_{\mathrm{rev}}|$ is that $W_{\mathrm{extra}}$ is always work done ON the system over a full cycle (where the system returns to its starting point.

2) You mistake that a process taking (very small but) finite steps as a valid replacement for a process that takes infinitesimal steps. We would never go anywhere in life (literally!) when your approach would be valid. See Zeno's paradox.

3) Infinitesimal steps are not part of the definition of reversibility. They are part of the way to solve a reversible process properly. Reversibility is a definition about the state of system and surroundings during any point in time.


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