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# PV-diagrams and expansion work

Why work from expansion is the area under the curve of a PV-diagram. Why heat is not a state function and internal energy is a state function.  Created by Sal Khan.

## Want to join the conversation?

• I'm confused. Around the 2 minute mark you say we're assuming gradual incremental change, because otherwise the value of P for pressure would decrease. But won't it decrease anyway? Even if the change in volume were infinitesmally small, wouldn't that just create an equally small change in pressure? Thus when delta V goes up P always goes down? So it's less P by more dV. If the drop in P and the rise in delta V are proportional, then how can the value of W increase with increasing volume? AARGH!
(37 votes)
• Your question is 8+ years old, and my answer is most probably not relevant to you, but it may be relevant to future readers.
In the video, Sal didn't actually ignore the ∆P as the volume changed. He just took the ∆P in incremental steps instead of constantly taking its change, and we can clearly see in the PV diagram that Sal didn't ignore the ∆P because if he had done so, our area under the curve would've been a simple quadrangular.
Instead, Sal simply went down in the pressure for every new beginning of a new quasistatic process, kinda like how you go down in stairs, "abruptly", and not in a "smooth" way, like in a slide.
But why did Sal choose to"go down in stairs" instead of a "smooth slide"? Well, recall that Sal proved that W=Px∆V, and he used that fact to show the viewer that Work = area under the curve in a PV diagram.
So, he had to use stairs, because in each individual stair, the pressure is constant, in order to show us, the viewrs, that Work = the area under the curve.
Now, if we use smaller and smaller "stairs" to describe this process, i.e. infinitesimally small quasistatic processes, we practically get a smooth slide, just like how the actual process should be. Sal initially used "bigger" stairs to drive home the point that
Work = Area under the curve in a PV diagram, that's all there's to it.
(10 votes)
• Isn't the work being done by whoever took the rock off of the piston? Cuz rocks don't just disappear.
(8 votes)
• Yes there is work being done by the person removing the rock but that is not part of what is being calculated in this example.

The work being done by the gas in the container to lift the piston with the rest of the rocks is what the problem is about.
(6 votes)
• It doesn't make sense to me.
We're taking change in pressure(dP) to be negligible and thus take P in the formula as constant. At the same time, we're taking the (dV) change in volume as significant. When we're taking very small increments, how can we consider only dV and ignore dP? If dP is negligible, so is dV!
Also, when the graph is made, pressure is shown to change!
(7 votes)
• Robby's is a good answer (and the question is a good one, too). For each bar, we find the area by multiplying the TOTAL internal pressure by the CHANGE in volume, hence P and deltaV. But for each subsequent interval (bar) we do changge the pressure to a new value. We are not pretending that pressure is constant, we allow it to change from step to step, but the quantity we use is still total pressure (P), rather than deltaP.
(7 votes)
• After completing a full cycle, a certain amount of work is produced by the system. If the internal energy (U) of the system remains constant from the beginning of the cycle until it reaches the end, where does the produced work (energy) come from? Generating work without altering the energy of the system (U) seems like energy is generated from nothing.
(7 votes)
• The internal energy U stays the same because you have to input heat Q into the system when it delivers work W to the outside world.
The diagram is actually an example for an engine that converts heat into work. Have a look at the video about the Carnot Cycle!
(9 votes)
• its a fact that no perfect ideal gas exists..a gas may have similar characters..so if there is no ideal gas..what is the use of studying about it?
(3 votes)
• The ideal gas law turns a very complicated problem of solving a system with trillions of particles into a simple algebraic equation with the penalty of losing a small amount of accuracy. If you want to make the problem much more difficult to gain a small amount of additional accuracy, you are welcome to do so.
(8 votes)
• At the very end, Sal said that the change in U for ideal gases was 0. For the graph that had two different paths, both for fowards and backwards states, how does this apply since the amount of work done was different?
(3 votes)
• I hope I'm answering this question (wat too late)
But an ideal gas is defined as a gas where the molecules is so far apart that they dosn't react to eachother. This leads to that the internal energy for an ideal gas is only depending on Q (not pressure or volume). only heat.
Therefore U is still the same even if we change V or P
(I hope i got it right, maybe someone could fill in)
(2 votes)
• Why does Sal draw the PV-diagrams as linear functions (y=ax+b)? The equation behind the PV-diagram is P*V=n*R*T, so shouldn't the diagram by hyperbolic since n*R*T stays constant in these scenarios?
(2 votes)
• he has not drawn them as linear functions? they are clearly curves, that's why he has to use calculus and integration to find their areas.
(1 vote)
• Sal says PV=constant ..... hence P = constant / V ...so shouldn't the graph be like a curve ?
(2 votes)
• PV = constant if temperature is held constant. it doesn't have to be.
(3 votes)
• Isn't the inaccuracy in finding the work in each measured area coming from the fact that you are only measured the area of the rectangle instead of including the triangle above it as well? Assuming that the pressure changes form a straight line, in order to get an accurate value, couldn't you get the area of the rectangle, then use an equation to find the area of a right triangle to find the area above, then add both to find the correct area?
(1 vote)
• I suggest that you watch the calculus videos, it will help to make sense of finding the areas under curves. You are right, when dV is not infinitely small, then it is not exactly a perfect rectangle. But using calculus, we can make it so that we are adding up all the rectangles where dV is so infinitely small that the little triangle at the top doesn't exist. I'm sure Sal can explain this a lot better than me, in his Calculus videos :)
(4 votes)
• Also, i still dont get the idea why W=(deltaP)(deltaV) shouldnt be true.
and why W=P(deltaV) is true.
many say deltaP is negligible, but thats the same case for volume isnt it?
since its a quasi static process, volume and pressure both increase in infinitesimal increaments.
hence, if we account for deltaV why dont we account for the corresponding deltaP

thankyou
(2 votes)
• Because since work is Force x Displacement, which is Pressure x Area x Displacement, the last two terms being deltaV, work is defined at a constant pressure. So we create an infinitesimal time clip wherein the change in pressure is so small, we don't care much about it. DeltaV is the change in the volume in that instant. Obviously that will be small, but P*deltaV won't be negligible. In case we try to incorporate changing pressure too, we end up with a deltaP*deltaV term, which, as you'd have learnt in calculus, is a product of two infinitesimal and hence can be discarded if it is added to terms with less than two infinitesimals. (P*deltaV + deltaP*deltaV is the same as P*deltaV)
(1 vote)

## Video transcript

In the last video, we saw that a system could do work by expanding. And in the situation we drew, we had a situation where the ceiling was movable. We had this piston and we, like in our process video, we had a bunch of pebbles. We removed a pebble, so the pressure in our system, if we assume that it was just so small that the pressure was constant, it pushed up on the piston with some force. We figured out that that force, since pressure is force per area, we just multiplied pressure times the area of our piston, and we got the amount of force we're applying. We apply that, and then we multiply that times the distance that we push the piston up, and then we get the amount of work that it did by expansion, or the expansion work. We said, well, you know we could have rewritten that. If you said pressure times our area, times our distance, we could instead write that as pressure, times the area, times the distance. And the area times the distance is the change in volume. And so we came up with a neat little formulation, that the work done by a system could be written as the pressure times the change in volume. So in this case, I wrote the internal energy formula, where it's the work done by the system. So I did a minus, right? Because when you do work, you are giving energy away to someone else. So in that situation, we did a minus. And so instead of writing work, we could say, minus the pressure, times the change in volume. And remember this is a quasi-static process. And we're doing it at very small increments. We're assuming that this change in volume is very small, and that the pressure is roughly constant while we're doing this. And of course that's not the case, right? If we did this, if this was a large change in volume, or if this happened all of a sudden, if these were really big pebbles, then our pressure will change as we expand. So it's hard to say what the pressure times the change in volume is. But if we assume things are really, really being done in very, very small increments, we could say, OK, let's say the pressure was constant over that small increment, and then we can multiply it by the change in volume. Now let's see how this can relate to some of what we've done before with the PV-diagram. And so far, all we've seen the PV-diagram, or what I used it for, is to kind of help explain the difference between quasi-static processes, or to say when macrostates are defined. But let me now do something more useful with it. And this will give you an idea, or start giving you an idea of why people who study thermodynamics love these so much. So before I did anything, when my canister was just here, I had all the pebbles on it. And we were in a state of equilibrium. I could describe all of its macrostates, its pressure, its volume, its temperature. I could describe its internal energy as well. So let me draw it here. So let's say I was at this state. This was state number 1. State number 1 was right there. And then, let's say I just start removing pebbles. Remember, if I just remove all the pebbles at once, the system's going to go into flux. We wouldn't be doing a quasi-static process, or a reversible process, which isn't always the same thing. But for our purposes, we wouldn't be in equilibrium the whole time. And we would have to wait to get to equilibrium. And at some point we'd have some pressure and volume that's down here. This is if we weren't doing it as a quasi-static process. Now we are, what I showed in the last video, we are doing it as a, or we're trying to get close to a quasi-static process, because we're doing it in small increments, with these little pebbles. And if these aren't small enough for you, you could do it in smaller pebbles. So we're moving incrementally. So, for example, in that last video, we maybe moved from there. We removed one pebble and we got right there. You remove another pebble and you go right there. You remove another pebble and you go right there. And the benefit of doing these quasi-static processes is you really get a path going from one state to the next. Let's say when you remove all but one of the pebbles, just, you know, this describes our path. So let's say we are in state 2 and we've removed all but one. Let me draw that. So state 2 will look something like this. I'll draw it really quick. So that's our container. That's our piston. We only have one pebble left on top. And then of course we have the gas now. Let me write this down. This is state 2. And let me write state 1 was something like this. State 1, the ceiling was lower. We had a bunch of pebbles on top of it. And we had a smaller volume, and so the gas was bumping into the ceiling and the walls and the floor a lot more. I'll just draw the same number. So we had a higher pressure. So pressure was high and volume low. Now in state 2-- so this is pressure high. This pressure is this axis. This is volume. So we had high pressure, low volume. And we got to a situation after removing all but one pebble. And we're doing it slowly, so we're always in equilibrium. So we have a path. This is after removing each of the pebbles, so that our pressure and volume macro states are always well defined. But in state 2, we now have a pressure low and volume is high. The volume is high, you can just see that, because we kept pushing the piston up slowly, slowly, trying to maintain ourselves in equilibrium so our macrostates are always defined. And our pressure is lower just because we could have the same number of particles, but they're just going to bump into the walls a little bit less, because they have a little bit more room to move around. And that's all fair and dandy. So this describes the path of our system as it transitioned or as it experienced this process, which was a quasi-static process. Everything was defined at every point. Now we said that the work done at any given point by the system is the pressure times the change in volume. Now, how does that relate to here? Change in volume is just a certain distance along this x-axis. Along, more like I should call it the volume-axis. This is a change in volume. We started off at this volume, and let's say when we removed one pebble we got to this volume. Now, we want to multiply that times our pressure. Since we did it over such a small increment, and we're so close to equilibrium, we could assume that our pressure's is roughly constant over that period of time. So we could say that this is the pressure over that period of time. And so how much work we did, it's this pressure over here, times this volume, which is the area of this rectangle right there. And for any of you all who've seen my calculus videos, this should start looking a little bit familiar. And then what about when we could take our next pebble? Well now our pressure is a little bit lower. This is our new pressure. Our pressure is a little bit lower. And we multiply that times our new change in volume-- times this change in volume-- and we have that increment of work. Once again, this is the area of this rectangle. And if you keep doing that, the amount of work we do is essentially the area of all of these rectangles as we remove each pebble. And now you might say, especially those of you who haven't watched my calculus videos, gee, you know, this might be getting close, but the area of these rectangles isn't exactly the area of this curve. It's a little inexact. There's a little error here. And what I would say is, well if you're worried about that, what you should do is use smaller increments of volume. And if you want to have smaller changes in volume along each step, what you do is you remove even smaller pebbles. And this goes back to trying to get to that ideal quasi-static process. So if you did that of, eventually the delta V's would get smaller and smaller and smaller, and the rectangles would get thinner and thinner and thinner. You'd have to do it over more and more steps. But eventually you'll get to a point, if you assume really small changes in our delta V. In calculus world, that infinitely small changes, you write it as dV. So if you take a sum of all the pressures, times the dV's you get the area under this curve. So the way to think about it when you're looking at this PV-diagram, if someone says, you're going from this point to this pressure and this volume, to this pressure and this volume. And they say, how much work did you do? You say, oh, well I just had to figure out the area under this curve. If you want to know the real math behind it, if you could get your pressure as a function of volume-- and if you haven't watched the calculus videos you can ignore this little aside I'm going to do here. This is this curve right here. If you could write it this way, let's say you could write pressure as a function of volume. When you're in algebra, you learn a curve is, you know, y is a function of x. But here, y is the pressure and x is volume, so its pressure is a function of volume. So the area under this curve is the integral of the pressure as a function of volume, that's the height at any point, times our very small change in volume. So times our very small change in volume. And you take the sum from our starting volume, so volume initial to volume final. And we'll do this in the future, especially when we start touching on entropy. But this is a neat result. Even if you don't know the calculus, or if this confuses you, if you've never seen an integral before, you could ignore it. But you could look at this intuitively and see the work I did is the area under this curve. Now, let me ask you one more thing. Let's say some work is being done to the system. So we start adding some marbles back. So let's say-- actually, let's say we're going from this direction. Let's say we start at state 2 and we go in that direction. So direction matters. So let's say we go in that direction right there. So I should put some arrows. And I'm overloading this picture so much. Actually let me just do a new picture, that's probably the best thing to do. So it's pressure, volume-- I'm actually going to do two. Let me just do pressure, volume. I'm going to do two graphs here. All right. So in the first one it's pressure, volume, pressure, volume. We started here at 1, and we went here to 2. So our system was essentially pushed up on the piston. And it could be a curve or a line, I'm not going to get too particular right now, but it was going in this direction. And so we can say that the work done was the pressure times the increase in volume at any moment. So the work done was the area under this curve. Now, if we started at position 2 and we go to position 1. 2 to 1. Now what's happening? Now we're compressing. So if we're going in that direction, you might say, oh OK, maybe the work done by the system is still the area under the curve. Well you'd be close. Because what's happening now? We're now compressing the system or adding the marbles back. We're putting energy into the system. So if we do that, remember, your work done by the system was pressure times an increase in volume. Now it's going to be your pressure times a decrease in volume. So when you go back in this direction, the area is not the work done by the system, it's the work done to the system. And maybe I'll do that in a different color, so green for work done to the system. Now let me throw you another little interesting idea. And this is actually a key idea. It's good to get the intuition here. So let me just draw a very simple PV-diagram again. So let's say we start at some state here. State 1. And I do something, you know, I'm in a quasi-static process and it, you know, it's doing something weird, and I get to state 2 here. And it's going in this direction. So my volume is increasing. So in this situation, what is the work done by the system? Easy enough, it's the area under this curve. Now let's say that I keep doing some type of quasi-static process, but it takes a different path. I'm doing something else, other than adding the marbles directly back. So my new path looks something like this to get back to state 1. So these arrows are going back. So now what is the work done to the system? well my volume is decreasing, so it's the area under the second curve. The area under the second curve is the work done to the system. So if I want to know what the net work the system did, going from state 1 to state 2, and then going back to state 1-- remember, this is a pressure and volume diagram-- what is it? Well the work that the system did was this whole area under this brown curve. And then it had some work done to it, which is the area under this magenta curve. So the net work it did is essentially the white, the whole area, minus this red area. So the net work it did would be essentially just the area inside this loop. And hopefully you don't have to know calculus to do this, although calculus you would actually use to compute these areas. But I just want to give you that intuition, that the area inside this closed loop is actually the amount of work that our system has done. And what's important is the direction that it's going. So it increased volume, then decreased volume, so it's kind of this clockwise motion. This is the work that our system has done, which, I don't know, to me is a pretty interesting thing. And later we can use this notion to come up with some other ideas behind our state variables I'll make one little aside here. Remember, our state variable pressure volume, we did stuff to it then we went back to that state. That stayed the same. And I want to say another thing. For our purposes, when we're dealing with ideal gases, where the internal energy is essentially the kinetic energy of the system, if we go and do all sorts of crazy stuff and come back, our internal energy hasn't changed. So the internal energy is always going to be the same at this point. So if I said, I did all of this stuff and came back here, what is my change in internal energy? It's 0. The change is 0. Now if I said I went from here to here, I would have a different internal energy and my change would be something real. But since this is a state function, it doesn't care how I got there. If I took all these loops and got back there, it just says, look, if I'm at this point in the PV-diagram, my internal energy is the same thing. So if I start at this point, and I finish again at this point, I have had no change in internal energy. And we'll talk more about that in the next video. But I just wanted to leave you there and get you this intuition behind the areas under the curves in the PV-diagram.