If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Quasistatic and reversible processes

Using theoretically quasi-static and/or reversible processes to stay pretty much at equilibrium. Created by Sal Khan.

Want to join the conversation?

  • spunky sam blue style avatar for user Chunmun
    What is the difference between reversible process and quasi static process according to sal in this video ?
    (10 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Reinhard Grünwald
      quasistatic the process is so slow, that the system is always in (or very close to) an equilibrium state
      reversible the process goes the other way as soon as you reverse the applied conditions.
      Quasistatic and reversible are not the same thing. For example you can mix two gases very slowly (quasistatic) but you can't reverse that easily. The reason is that the entropy between the starting point and the endpoint has increased and that's a main characteristic of an irreversible process.
      (43 votes)
  • female robot grace style avatar for user Strings to Eternity
    Are microstates also known as INTENSIVE properties ?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • mr pink red style avatar for user Yosharma999
    You say it is in equilibrium the whole time. That means an infinitesimally small particle is removed each and every time. So basically, the system remains where it was before even after removing all the particles one-by-one. But that doesn't happen when you remove all of them together. I don't get this.
    (0 votes)
    Default Khan Academy avatar avatar for user
    • leafers ultimate style avatar for user DAllenSchneider
      Think of this as "baby steps". If, instead of using infinitesimally small particles, let's use sand. After removing a single grain, the system would become chaotic, but only very mildly so. After a very short amount of time it would come back to equilibrium. We could take our measurements, and move on to the next grain. Reducing the size of those grains reduces the time it takes to reach equilibrium, such that an infinitesimally small particle will take an infinitesimally short amount of time.
      You might need to wrap your head around equilibrium in general to better understand this. The essential point is that the system becomes non-uniform temporarily. The temperature and pressure is not consistent all the way through. Theoretically the average temperature and pressure of all points within the volume will always fall along the line he plotted, but there is no way to measure the pressure at every point simultaneously. It is therefore necessary to wait for the system to come to equilibrium before gathering any meaningful information from it.
      (10 votes)
  • piceratops ultimate style avatar for user peter2310
    So, this is probably a definition issue, but why does a reversible process have to go through infinitesimal changes like a quasi-static process? Doesn't a process which does not lose energy to friction and heat dissipation and which is also able to fully reverse between two states already qualify as a reversible process? Why should it matter if I take one grain of sand away or a big amount of it if by doing the opposite I end up perfectly in the state I started in for both cases?

    Is it because the definition of a reversible process is that we need to have 'well-defined' system states to be measured and that the path from initial to final state, like that of a quasi-static process, must be shown? I'm wondering because making a big change, although you will experience a much more dynamic change (oscillations of states etc), you will still end up with equilibrium with time. So say you go from state 1 to state 2 by taking away 1 kg of sand grain by grain which takes a lot of time. Compare this with taing away 1 kg of sand instantaneously and waiting for equilibrium to occur. Both approaches will get from the same state 1 to the same state 2. Now reverse the process by doing the opposite and both will get back to the same initial state. Can the second approach of taking away a large instead of an infinitesimal amount still be called a 'reversible process'?

    **UPDATE
    Just realized that Sal explained what I wrote in my two comments to this question at the very end of the video (from ).
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user C Hart
      My take on it is that if you jumped from state A to B quickly (like by removing the large rock off the piston) you would lose a lot of energy to friction, air turbulence, heat, vibration, etc. There is probably also inefficiency in the inner gas having to quickly rearrange itself to equalize pressure and temperature. So I don't know if you can say "non-quasi-static" AND "no heat or energy loss" in this example. I don't think state 2 will be the same if you do this quickly and slowly, which was your assumption.
      (2 votes)
  • piceratops ultimate style avatar for user Gauri Jaswal
    Couldn't one measure the pressure and volume at every instant while the piston is oscillating? One could record the values for every instant and plot those numbers. Why go through all the trouble of reducing pressure grain by grain?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Faisal Islam
      watch the part of video from to and you will get some idea. What we want to do here is to show how we really get from state A to state B on the graph if we kept temperature constant. If we don't do this grain by grain, we wreak havoc on the system undergoing process. Every time you do the process, you will end up with a different pathway from state A to state B due to well... tons of possible macrostates thanks to microstates. If this is not enough, then let me state this bluntly: If you measure pressure and volume at every instant while piston is oscillating, then record and plot their graph, you will get one of the many possible graphs for the same process, even if you do everything exactly the same as before . Going through the trouble of reducing pressure grain by grain tremendously reduces the oscillations, and thus show how Pressure is truely related to volume if temperature is constant(inverse proportion, and well if you know how the graph looks like, you will surely understand why we don't want oscillations to ruin this graph)
      (5 votes)
  • blobby green style avatar for user Jamie Martin
    Would it be possible for a completely static process, similar to the ones shown in the video? For example, instead of removing a grain of sand at a time, what if you removed an atom or molecule at each time. Would there still be an extremely small period of the system being undefined or is it possible through removing an amount so small that there would be no time when the system is undefined or completely negligible?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Satwik Pasani
      Mathematically , if only infinitesimal quantities of pressure is reduced at a time, then the process is actually quasistatic (in the limit the quantity approaches zero). But in reality,, if any non-zero finite amount be removed at a time from "heap of molecules" exerting pressure, it would, (although for a super-smalll interval, be in inequilibrium.
      (2 votes)
  • purple pi teal style avatar for user Ramya Narayanan
    Why should a reversible reaction be quasi static? Using the rock analogy provided by Sal, if we replace the half rock we evaporated, it will oscillate again and reach the original state. Also, what has loss of energy go to do with the reaction being non-reversible? Does the air inside lose the energy? And even if it does, how does that change the PV graph?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user Matt B
      Great question! Let's pretend you have: A + heat <--> B
      So if you start with A and heat (energy) you can make B. And now that you have B, you can transform it to make A with heat. What happens now if the heat dissipates (you remove the heat)? Now you do not have one of the ingredients on the left side (heat) and so you cannot reform B. Therefore, the reaction does not become reversible anymore:
      A + heat <--- B
      (3 votes)
  • blobby green style avatar for user weirdmind1
    ,if we knew exactly how energy got dissipated in friction,how the energy got transferred in other atoms,and we knew all the other variables,couldn't we at least theoretically reverse the process?sorry for my energy and thanks in advance.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • spunky sam blue style avatar for user Ethan Dlugie
    At , Sal says that there is no loss of energy (in a frictionless world). But how is this possible? The gas in the container applies a force in the direction of the displacement of the piston (up), thus it must be doing positive work. Shouldn't the gas then have less energy?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Mohit Gupta
    why would the piston come down after going up when the weight of the block has been reduced to half??
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Icedlatte
      He gives the reason at .
      Once half is gone, remember what's going to happen to all the gas particles in the chamber.
      Volume is going to increase, the gas will instantly try to fill the empty gap that the lifted piston will leave behind as it goes up. They're going to make the piston go up and then down again to counterbalance, probably almost infinitely.
      I remember Einstein mentioned something about how a moving object never actually stops. Or it could be someone else. Conservation of Momentum.
      (1 vote)

Video transcript

SAL: In the last video, where we talked about macrostates, we set up this situation where I had this canister, or the cylinder, and had this movable ceiling. I call that a piston. And the piston is being kept up by the pressure from the gas in the canister. And it's being kept down by, in the last example I had, a rock or a weight on top. And above that I had a vacuum. So essentially there's some force per area, or pressure, being applied by the bumps of the particles into this piston. And if this weight wasn't here-- let's assume that the piston itself or this movable ceiling itself, it has no mass-- if that weight wasn't there it would just be pushed indefinitely far, because there'd be no pressure from the vacuum. But this weight is applying some force on that same area downwards. So we're at some equilibrium point, some stability. And we plotted that on this PV diagram right here. I'll do it in magenta. So that's our state 1 that we were in right there. And then what I did in the last video, I just blew away half of this block. And as soon as I blew away half of this block, obviously the force that's being applied by the block will immediately go down by half, and so the gas will push up on it. And it happened so fast that, al of a sudden the gas is pushing up. Right when it happens, the gas near the top of the canister is going to have lower pressure, because it has less pushing up against it. The molecules that are down here don't even know that I blew away this block yet. It's going to take some time. And essentially the gas is going to push it up, and then maybe it'll oscillate down, and then push it up, and oscillate down a little bit. It'll take some time eventually until we get to another equilibrium state, where we have a new, probably, or definitely lower pressure. We definitely have a higher volume. I won't talk too much about it yet, but we probably have a lower temperature as well. And this is our new state. And our macrostate's pressure and volume are defined once we're at the new equilibrium, so we're right here. So my question in the last video was, how did we get here? Is there any way to have defined a path to get from our first state-- where pressure and volume were well defined, because the system wasn't thermodynamic equilibrium-- to get to our second state? And the answer was no. Because between this state and this state all hell broke loose. I had different temperatures at different points in the system. I could have had a different pressure here than I had up here. The volume might have been fluctuating from moment to moment. So when you're outside of equilibrium-- and I had written it down over there-- you cannot define, or you can't say that those macro variables are well defined. So there was no path that you could say how we got from-- erase this-- how we got from state 1 to state 2. You could just say, OK, we were in some type of equilibrium. So we were in state 1. Then I blew away half the rock. The pressure went down, the volume went up. The temperature also probably went down. And so I ended up in this other state once I reached equilibrium. And that's all fair and good, but wouldn't it have been nice if there was some way? If we could have said, look, you know, there's some way that we got from this point to this point? If we could perform my little rock experiment in a slightly different manner, so that all this hell didn't break loose, so that maybe at every point in between my macro variables are actually defined? So how could I do that? Remember, I said that the macro variables, the macrostates, whether it's pressure, temperature, volume, and there are others, but I said these are only defined when we are in a thermodynamic equilibrium. And that just means that things have reached a stability point. That, for example, the temperature is consistent throughout the system. If it's not consistent throughout the system, I shouldn't be talking about it. If the temperature is different here than it is up here, I shouldn't say that the temperature of the system is x. It's different at different points. I really can't make a well-defined statement about temperature, similar for pressure or for volume, because the volume is also fluctuating. But what if I perform that same experiment? That same process, I should call it. Let me draw it again. So I have my canister. And instead of starting with a rock, just one big rock-- let me draw, this is my piston right here, at the top of the movable ceiling of the cylinder. And I have some gas inside of it. Instead of having just one big rock like I had over here, how about I start with an equal weight of rock? But let's say I have a bunch of small pebbles that add up to that same rock. So just a bunch of, well, you know, just a pile of pebbles. You know, maybe they're sand. They're super duper small. Instead of just blowing away half of the sand all at once, like I did with that rock over there, and immediately jumping to that state and throwing the whole system into this undefined state of non-equilibrium. Instead of doing that, let me just do things very slowly and very gently. Let me just take out one grain of sand at a time. So if I just take out one grain of sand. And so I took out an infinitesimal amount of weight. So what's going to happen? Well this piston's going to move up a little bit. And let me draw that. So let me copy and paste it. So I just took out one little piece of sand. The force pushing down will be a little bit less. The pressure pushing down will be a little less. And so my piston-- let me see if I can draw this-- it will have moved up-- let me erase it-- it will have moved up a very infinitesimal-- infinitesimal means an infinitely small amount-- it would have moved an infinitely small amount of time. And so you wouldn't have thrown that system into this, you know, havoc that I did this last time. Of course, we haven't moved all the way here yet. But what we have done is, we would have moved from that point maybe to this other point right here that's just a little bit closer to there. I've just removed a little bit of the weight. So my pressure went down just a little bit. And my volume went up a just a little bit. Temperature probably went down. And the key here is I'm trying to do it in such small increments that as I do it, my system is pretty much super close to equilibrium. I'm just doing it just slow enough that at every step it achieves equilibrium almost immediately. Or it's almost in equilibrium the whole time I'm doing it. And then I do it again, and do it again. And I'll just draw my drawings a little less neat, just for the sake of time. Let's say I remove another little dot of sand that's infinitely small mass. And now my little piston will move just a little bit higher. And I have, remember I have one less sand up here than I had over here. And then my volume in my gas increases a little bit. My pressure goes down a little bit. And I've moved to this point here. What I'm doing here is I'm setting up what's called a quasi-static process. And the reason why it's called that is because it's almost static. It's almost in equilibrium the whole time. Every time I move a grain of sand I'm just moving a little bit closer. And obviously even a grain of sand, the reality is if I were to do this in real life, even a grain of sand on a small scale is going to reek a little bit of havoc on my system. This piston is going to go up a little bit. So say, let me just do even a smaller grain of sand, and do it even a little bit slower so that I'm always in equilibrium. So you can imagine this is kind of a theoretical thing. If I did an infinitely small grains of sand, and did it just slow enough so that it's just gently moved from this point to this point. But we like to think of it theoretically, because it allows us to describe a path. Because remember, why am I being so careful here? Why am I so careful to make sure that the state, the system is in equilibrium the whole time when I get from there to there? Because our macrostates, our macro variables like pressure, volume, and temperature, our only defined when we're in equilibrium. So if I do this process super slowly, in super small increments, it allows me to keep my pressure and volume and actually my temperature of macrostates at any point in time. So I could actually plot a path. So if I keep doing it small, small, small, I could actually plot a path to say, how did I get from state 1 to state 2 on this on this PV diagram. And you might say, hey, you know, Sal, this is all-- And I'll take a little step back here. I always found this really confusing. You know, you'll see a lot of talk in thermodynamic circles, or even in your book about-- it has to be a quasi-static process, and I always used to wonder, why are people going through these pains to describe this process where you're removing sand after sand? And the whole point is because you want to get as close to equilibrium the whole time you're doing it as possible so that your pressure and volume are defined the whole time. The reality is, in the real world you can never get something that's continuously defined, but you can just do really, really, really small increments. So that at each small increment you're at some equilibrium. And if you're not happy with that, you can do even smaller increments. So at some point, at some limiting point, you do have some type of continuous state change, while you're always in equilibrium. It's almost an oxymoron, because you're saying you're static, you're saying that you're in equilibrium the whole time, but clearly you're also changing the whole time. You keep removing little pieces of sand. But you're moving them just slowly enough that all that crazy up and down motion, and all of the flux, and all of the weird temperature changes don't happen. And it just, you know, just that it slowly, slowly, slowly creeps up. The reason why I'm even going through this exercise is because it's key when we start talking about thermodynamics and these PV diagrams, and we'll start talking about carnot engines and all of that, that we be able to at least theoretically describe the path that we take on this PV diagram. And we wouldn't have been able to do that if we can't assume that we're dealing with a quasi-static process. Now there's another term that you'll hear in thermodynamic circles that really, I mean, to me it really, I don't know, I had trouble comprehending it the first time I heard it, called reversible. And sometimes these terms quasi-static and reversible are used interchangeably, but there is a difference. Reversible processes are quasi-static, and most quasi-static processes are reversible, but there are a few special cases that aren't. But the idea of a reversible process is something that happens so slowly. So in this example I took off a grain of sand and I got to the state, but if I assume that no friction when, you know, when this piston moved up a little bit, in the real world, let's say if this piston was metal, when this rubs against the canister, there'd be a little bit of friction generated and a little bit of energy would be dissipated as friction or heat. But in a reversible process, we're assuming that, look, this is frictionless. When anything happens in the system, when we go from this state right here-- let's say this the state a, this is state b. So this is state a, this is state b. When we go from this state to this state, one, we're infinitesimally close to equilibrium the whole time, so all of our macrostates are well defined. And even more, when we move from one state to the other, there's no loss or dissipation of energy. So those are two important characteristics. One, infinitely close to equilibrium at all times, and no loss of energy. And the reason why that matters for a reversible process is because if we wanted, if we were sitting in state b, we could just add another grain of sand back in, push down this piston infinitely slowly, at an infinitely small increment, and get back to state a. So that's why it's called reversible. You could be at this point right here, and take out a little bit of sand, and get to this point right here. But if you want, since no energy was lost, you could add a little bit of sand, and get back to this point right here. Now the reality in the real world is, there is no such thing as a perfectly reversible process. There will always be, whenever you do anything, there will always be some energy or heat lost to the process. In the real world, if I moved down here, if I tried to put the sand back I would lose some energy and probably get to a little slightly different point. But you don't have to worry about that. The important takeaway from this video is that, in the situation I described there, there was no intermediate macrostate variables, because our system was in flux, it wasn't in equilibrium. So if we wanted to get intermediate states, we just have to essentially do this process slower. And so slow, I mean, it theoretically would take you forever, so we can only approximate it. But the sand gives you an idea of what we're talking about. And if we did it slowly with these infinitesimally small particles of sand, then we can define the state at every point along the process. And that's why we call it quasi-static, because at any point it's almost static. It's almost in equilibrium. So our pressures, volumes, and temperatures can be defined. And if we add to that the notion that we haven't lost any heat when we're going in one direction or another, we could say it's reversible, because if we took a piece of sand away, we can always add a little bit of sand next. Now, actually, with that said, let me give you the one example of maybe a quasi-static-- no, actually I'll save that for future video. Anyway, hopefully you understand that these are two concepts that used to really confuse me, and hopefully this clears it up a little bit. And I think more than what it is, I think the first time I read about them I'm like, OK, well what's the big deal? The big deal is, it allows you to define your macrostates for every state in between these two states that you care about. When you just did it as a regular kind of non-quasi-static process, in between you don't know what happened.