# Disequilibrium during infinitesimal steps of a thermodynamic reversible process & cause of maximum work to be achieved during reversible process?

In almost every sites and even in my book,it is written that "Reversible process is the process which is carried out infinitesimally slowly so that in each step thermodynamic equilibrium is maintained and any infinitesimal change in the condition can reverse the process to restore both system & the surroundings to their initial states" . Now 'a process is done infinitesimally slowly' means the process is done in infinitesimal steps . But during each step ,as beautifully examplified in the accepted answer of my another question by @André Neves, there is disequilibrium . At the end of the step,can only the equilibrium be reached . So, doesn't it bother the definition? But this is not the point. I want to know only by performing a process infinitesimally slowly using innumerable steps can make it reversible,while if I do the same process with a rapid change,it becomes irreversible. Is it so? Why? I know here comes one key that is relaxation time ie. during the irreversible process,the system does not get sufficient time during the process to attain equilibrium . But how can I avoid the fact that while doing a process infinitesimally slowly i.e. by performing innumerable steps to make the process reversible,disequilibrium exists during each step? Is not it contradictory to the definition which says that at all stage there is equilibrium? Do they don't care about the disequilibrium at each step? It is quite confusing to me; perform a process using infinitesimal steps during which disequilibrium exists and then write a definition which says at all the time equilibrium is maintained? Why? Another question which arises in my mind that how by performing reversible process ie. infinitesimally slowly to reach a certain state,maximum work is obtained and while doing the same process irreversibly ie. rapidly,least work is obtained. Why? Is work proportional to time? I will be very grateful if anyone answer these two questions. Plz help.

I think André Neves' answer to your other question says the opposite of what you think it says. But let me try to make it a little bit clearer for you, if I can.

In André Neves' answer, he talks about a gas in a piston that's weighed down by 1000 pieces of lead shot, each weighing a small amount $m$. After each lead shot is removed, there is indeed a period of disequilibrium before the system goes back to equilibrium at a different state. As a result of this disequilibrium, a little bit of heat $q$ is generated (mostly by friction as the piston oscillates up and down before it comes to rest) and transferred to the environment.

But these pieces of lead shot are not infinitesimal, they are merely quite small. Let's try to get a bit closer to the infinitesimal case, and consider 100,000 grains of sand, each weighing $m/100$. Now when you remove a grain of sand there's still a period of disequilibrium, but there's much less equilibrium at each step than there was with the lead shot, and (it turns out, once you do the math) that the amount of heat that's generated is much less than $q/100$. (If I'm not mistaken, it should be approximately $q/100^2$.) This means that after you've removed all those tiny grains of sand one at a time, the total amount of heat generated through friction will be much less than it was when you did it with lead shot.

Now, a quasi-static (or "reversible") process is an idealisation. You can't really remove an infinitesimal piece of weight from the piston, since the smallest thing you can remove is an atom, and in theory you'd have to wait an infinite time for the system to go to equilibrium after each step. (Well, ok, there are smaller things. But they're still bigger than infinitesimal no matter what you do you.) In reality there is no such thing as a quasi-static process, but because some things behave like them to a good approximation, people find it a useful thing to think about.

So what you have to imagine is that you keep doing this with smaller and smaller grains of sand, and that you keep going for ever, until they're basically zero mass, and there's an infinite number of them. It turns out, once you do the maths, that as the grains get smaller, the total heat that gets generated also gets smaller, and by the time you get to zero-size grains the total heat produced by friction will also be zero. This is what we call a reversible process, and as I said it doesn't really exist - but the point is, if you do this experiment with lead shot it will be a pretty good approximation, and the frictional heating will already be so small that you can basically ignore it.

Now, your second question was how the same process can be irreversible if done quickly but reversible if done infinitely slowly. I hope now the answer is clearer: when you do it quickly, it produces heat (often through friction as in the example above). But if you do it slowly it produces less heat, and if you're careful you can channel that same energy into useful work instead. So the system ends up in the same state after the irreversible and reversible versions of the same process, but its environment can be in a different state, containing more heat after the irreversible version than it does after the reversible one.

I will come up with an explanation that has to do with information.

One possible interpretation of the second principle of thermodynamics is that, during an irreversible process, there is a loss of information about the state of the system that can never be recovered (this interpretation is mostly due to Jaynes).

Now, when you perform (for instance) a transformation very very fast, the only thermodynamical states that are well defined are the initial and the final ones. The thermodynamic path in between these two states is unknown (and not even defined) because of the speed of the process and a lot of information is being lost then. Ultimately, you could even find a final state that has nothing to do with the initial state (because the final state is essentially set by the final constraints only).

A contrario, if you perform a very very slow transformation on a system, you can keep track of the state of the system at almost every time during the process and you minimize (in the infinitely slow process limit) the amount of information that is lost during the transformation; the process is therefore said to be reversible.

To make my point even clearer, in an infinitely slow process, you have a one-to-one correspondence between the initial state and any state along the thermodynamic path that leads to the final state; therefore you cannot have any loss of information. In a very fast irreversible process, that is not the case: you can have multiple initial states that will reach the same final state with the same protocol.

• Sir, I know nothing about information theory. But what you wrote is crystal-clear to me. Can you suggest me any book or site for studying information theory for beginner like me? – user36790 Sep 26 '14 at 10:34
• Sir,can you tell why is maximum work received only from reversible process? – user36790 Sep 26 '14 at 12:13
• You can have a look at the wiki page which is quite concise and simple and the references therein. Regarding the maximum work statement, I am not sure it is even true so won't try to discuss it here. – gatsu Sep 26 '14 at 12:36