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Work from expansion

How a system can do work by expanding. Created by Sal Khan.

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  • blobby green style avatar for user jan125djw
    When Volume increases, pressure must decrease, that was in the first lesson on thermodynamics.

    P1 x V1 = P2 x V2 (assuming there was no exchange of energy from outside).
    Here Volume increases by very small number and Pressure decreases also by very small number as seen in the pictures by Sal.

    Here we are assuming (as I have read some comments below), that changes in pressure are so infinitesimally small in quasistatic process that we can take pressure initial to the equation. But hey, I'm very confused now, because it seems illogical to me that volume isn't infinitesimly small also. Either both of them change or none of them changes(in my opinion).

    Please, is there any logical explanation?? Otherwise I can't accept this equation, for me this is Work = (change in Pressure)x(change in Volume) .

    But if Pressure x Volume haven't change has Internal Energy risen?

    Thank you for your kindness and please explain this to me :-)
    (2 votes)
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  • piceratops ultimate style avatar for user ∫∫ Greg Boyle  dG dB
    In these types of problems, you'll hear the term adiabatic used. What is an adiabatic process and how do you know if the system is adiabatic?
    (2 votes)
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    • duskpin ultimate style avatar for user John Linick
      In case a problem doesn't directly say "adiabatic" you can know a process (expansion or compression) is adiabatic if it occurs "rapidly". This is because if you compress something fast enough, the system doesn't have enough time to equibrilate with the environment and give off heat (Q=0). This means that a fast compression will result in the temperature increasing. You can use the equation Sal used in the video to see that in an adiabatic process ∆U = - W(by system)
      (6 votes)
  • blobby green style avatar for user patheriae
    At when Sal defines the pressure and force, why does he assume that the force applied on the ceiling of the piston is constant? As the volume increases by the force pushing the ceiling a distance "x", shouldn't the pressure decrease? If the pressure decreases in this process, why doesn't the force also decrease?
    (5 votes)
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  • blobby green style avatar for user anubiorvr
    while piston moving upwards by removing small weight over the piston, there is change in volume and pressure also. but why the change in pressure is not considered
    (2 votes)
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    • orange juice squid orange style avatar for user Ved Gupta
      Work done by a system is equal to the area enclosed between the graph and the V axis from V1 to V2 in a P - V graph. So, when P is not constant, the area enclosed is effected, and thus the work done. So, the change in pressure does makes a difference, just like a variable force.
      I don't know why it's neglected here (maybe due to calculus complexity)
      (4 votes)
  • leaf green style avatar for user jack
    Why change in volume is not infinitesimal but change in pressure is ?
    (2 votes)
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    • blobby green style avatar for user C Hart
      If you are still interested in this, I gave an answer below that addresses why we can ignore dP during this calculation, even though it may be on the same scale as dV. The short answer is that there are two factors in the multiplication, P and dV. dP has little effect when added to P, but dV has a large effect when multiplied.
      (4 votes)
  • leaf green style avatar for user Sammy
    why are heat and work not state functions?
    (2 votes)
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  • blobby green style avatar for user nishanthv
    One thing that really intrigues me is the gas applying equal pressure on every surface of the container (stated at ). Why is it then, that a pressurized box or bag explodes from the excessive pressure only from one side? I understand that it may be due to one part being weaker than the other, but what would happen in an ideal container that is perfectly uniform?
    (3 votes)
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  • leaf green style avatar for user aman bhardwaj
    How to calculate work when volume doesnot change
    (2 votes)
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  • blobby green style avatar for user Kay Flinshire
    If a container containing a specific amount of an ideal gas( no heat transmission) expand in volume, how will the internal pressure an tenperature change in respond to the change in volume?
    (2 votes)
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  • duskpin tree style avatar for user Neha
    In my physics textbook it says that work done by the system is positive and work done on the system is negative, but in my chemistry textbook its the opposite. Why are the conventions different?
    (2 votes)
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Video transcript

I've talked a lot about how our change in internal energy of a system can be due to some heat being added to the system, or some work being added to the system, or being done to the system. And I'm going to write it again the other way, just because you see it both ways. You could say that the change in internal energy could be the heat added to the system minus the work done by the system. So there's two questions that might naturally spring up in your head. One is, how is heat added to or taken away from a system? And how is work done, or done by, or done to a system? The heat, I think, is fairly intuitive. If I have a-- and we'll be a little bit more precise in the future of this, but I just want to give you the sense of what we're talking about-- if I have some system here, some particles in some type of a canister. And it's at temperature, I don't know, let's say it's a temperature T1. I'll even give it a-- say it's at 300 kelvin. If I want to add heat to this system, what I can do is I can place another system right next to it, maybe right next to it. Who knows what size it is. And it's got some particles there. But its temperature is much, much, much higher. So this system's temperature, I'll say temp T2 is equal to, I don't know, let's say it's 1000 kelvin. I'm just making up numbers. So what's going to happen in this situation, you're going to have heat transferred from this second system to the first system. So you're going to have heat going into the system. Now, heat and, work and even internal energy, this goes back to our conversation of macrostates versus microstates. Heat is changing the macrostate of our systems. This system is going to lose temperature. This system's going to gain temperature. But we know what's happening on a micro level. These molecules are going to lose kinetic energy. These molecules are going to gain kinetic energy. How is that actually happening? Well, we assume that there's some type of a container here. Maybe it's a solid wall. These molecules are going to bump into that wall, and are going to make the particles in that wall vibrate, and then they're going to make the particles in the green container's walls vibrate. And so when the green container's molecules touch the wall, they're going to bounce off with even more kinetic energy, with even more velocity, because of that vibration in the wall will push them back even further. So that's essentially how you get this transfer of kinetic energy, or this transfer of heat. I think that's fairly intuitive. If we put this next to a cooler, a system with lower temperature, we would lose kinetic energy, or would lose heat. And there's other ways that we can do it. We could compress the-- well, I don't want to talk about that just now, because that'll be touching on work. So, how can we add or subtract work to a system? And this one's a little bit more interesting. Let's go back to our piston example. Let me just draw some lines here. So I have my container. There you go. It's got a little movable ceiling to it. That's my piston. And go back to the example. Because what we're going to be dealing with-- especially once I go into the pressure volume diagram, the PV diagram that I'm about to go into-- we want to deal with quasi-static processes. Processes that are always close enough to equilibrium that we feel OK talking about macrostates like pressure and volume. Remember, that if we just did something crazy and the whole system is in flux, those macrostates aren't defined anymore. So we want to do a quasi-static process. So I'll have pebbles instead of one big rock. I'll draw the pebbles a little bigger this time. And I have some pressure. So that's my piston and it's being kept down by these rocks. It's being kept up by the pressure of the gas. The gas is bumping into this ceiling. It's bumping into everything. The pressure at every point in the container is the same. It's at equilibrium. Now, what happens in that example where I removed one rock from that? So let me copy and paste that. So if I remove one rock from this thing right here. Copy and paste. So that's the same thing. Now let me remove a rock. I'll remove this top one, was removed. What's going to happen? Well I now have less weight pushing down on the piston, and I have a certain amount of pressure pushing up. The system, it'll very temporarily go out of equilibrium, but it'll be a very small difference in how much we're pressing down on it, so hopefully it won't be a huge change in our equilibrium. We'll stay pretty close to it. But we know from the previous example, instead of this thing flying up, it's going to shift up a little bit. This is just going to shift up a little bit. Right when we do it it's going to be like that, right there. And let me fill in that part with black, because it's not like the space disappeared. So let me fill that in right there. So our little piston will move up a very small amount. And what I claim is, when this happened, when I removed this little pebble from here, the system did some work. And let's just think about that. So work, according to the definitions that you learned in first-year physics, and when you're using classical or dealing with classical mechanics, you learn that work is equal to force times distance. So if I'm claiming that when this piston moved up a little bit, when I removed that pebble, I'm claiming that this system here did some work. So I'm claiming that it applied a force to this piston, and it applied that force to the piston for some distance. So let's figure out what that is, and if we can somehow relate it to other macro properties that we know reasonably well. Well we know the pressure and the volume, right? We know the pressure that's being exerted on the piston, at least at this point in time. And what's pressure? Pressure is equal to force per area. Remember, this piston, you're just seeing it from the side, but it's a kind of a flat plate or a flat ceiling on top of this thing. And at what distance did it move it? You know I could blow it up a little bit. It moved it some-- I didn't draw it too big here-- some x, some distance x. So this change, it moved it up some distance x there, right? So what is the force that it pushed it up? Well, the force, we know its pressure, the pressure's force per area. So if we want to know the force, we have to multiply pressure times area. If we multiply both sides of this times area, we get force. So we're essentially saying the area of this little ceiling to this container right there, you know, it could be, I could draw with some depth, but I think you know what I'm talking about. It has some area. It's probably the same area as the base of the container. So we could say that the force being applied by our system-- let me do it in a new color-- the force is equal to our pressure of the system, times the area of the ceiling of our container of the piston. Now that's the force. Now what's the distance? The distance is this x over here. The distance is-- I'll do it in blue-- it's this change right here. I didn't draw it too big, but that's that x. Now let's see if we can relate this somehow. Let me draw it a little bit bigger. And I'll try to draw in three dimensions. So let me draw the piston. What color did I do it in? I did it in that brown color. So our piston looks something-- I'll draw it as a elipse-- the piston looks like that. And it got pushed up. So it got pushed up some distance x. Let me see how good I can-- whoops. Let me copy and paste that same-- So the piston gets pushed up some distance x. Let me draw that. It got pushed up some distance x. And we're claiming that our-- oh sorry, this is the force. Sorry, let me be clear. This is the force, and this is the distance. So work is equal to our force, which is our pressure times our area, times the distance. I want to be very clear with that. Because when I wrote this I said, OK, the force that we're applying is the pressure we're applying, times the area of our cylinder. This is the area of our cylinder right here. That's the area of our cylinder right there. So if you do the pressure times this area, you get the force. And then we moved it some distance x. Now, we could rearrange this. We could say that the work is equal to our pressure times our area, times x. What's this? What's this area, this area right here, times x? Well that's going to be our change in volume, right? This area times some height is some volume. And that's essentially how much our container has changed in volume. When we pushed this piston up, the volume of our container has increased. You can see that, even looking from the side. Our rectangle got a little bit taller. When you look at it with a little bit of depth, you see the rectangle also didn't get taller. We have some surface area. Surface area times height is volume. So this right here, this term right here, is a change in volume. So we can write work now in terms of things that we know. We can write work done by our system. Work done is equal to pressure times our change in volume. Now this has a very interesting repercussion here. So we could-- actually many-- we can rewrite our internal energy formulas. So, for example, we can write internal change and internal energy is now equal to heat added to the system, plus the work-- let me say minus the work done by the system. Well what is the work done by the system? Well it's the pressure of the system times how much the system expanded. In this case, the system is pushing these marbles, or these pieces of sand up. It's doing work. If we were doing it the other way, if we were adding the sand, and we were pushing down on our little canister, we would be doing work to the system. So this is the situation where I'm doing here, where I'm removing the sand and the piston goes up, essentially the gas is pushing up on the piston, the system is doing the work. So if we go back to our little formula, that internal energy is heat minus the work done by a system, so done by, then we can write this as, this is equal to the heat added to the system minus this quantity, the pressure of the system, times the change in volume. And it's interesting, if the volume is increasing, then the system is doing work. And this applies-- we're going to talk a lot more about engines in the future-- but that's how engines do work. They have a little explosion that goes on inside of a cylinder that pushes up on the piston, and then that piston moves a bunch of other stuff that eventually turns wheels. So the volume increases, you're actually doing work. So I'm going to leave you there in this video. In the next video, we're going to relate this, this new way of writing our internal energy formula, and we're going to relate it to the PV diagram.