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Understanding why enthalpy can be viewed as "heat content" in a constant pressure system. Created by Sal Khan.

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  • leaf green style avatar for user Wally
    I don't understand the claims at :( . There is an equation written :
    delta P * V = P * delta V
    and Sal says if the pressure is constant, these will cancel out.
    Pressure being constant means: delta P = 0. But P still has a value.
    So filling it in gives:
    0 * V = P * delta V
    0 = P * delta V (because P does have a value, even though it's constant)
    But P has a value, so the right term IS NOT 0 (unless delta V is 0, or P = 0).
    So i dont know how both terms cancel out if only the pressure is constant T_T, help?
    (33 votes)
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    • spunky sam blue style avatar for user Andyd
      Just to clarify further:
      delta(P.V)=delta(P).V + P.delta(V)

      Now, when Sal says that pressure is not changing, is constant, you can see that delta(P)=0 so that the first term is dropped, leaving...
      (16 votes)
  • piceratops ultimate style avatar for user Noah Major
    What is a state variable?
    (10 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      In thermodynamics, a state variable is a property of a system that depends only on the current equilibrium state of the system
      Examples of state variables are temperature, pressure, volume, internal energy, enthalpy, and entropy.
      A state variable does not depend on the path by which the system arrived at its present state.
      For example, if I have 1 L of a gas, I can heat it, cool it, compress it, expand it, etc. But if I end up with a volume of 1 L, the volume is the same as it was at the beginning.
      So, volume is a state variable.
      (27 votes)
  • blobby green style avatar for user Shamma Bhanvadia
    How does enthalpy relate to calorimetry?
    (11 votes)
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  • spunky sam blue style avatar for user Chunmun
    Can absolute enthalpy and absolute internal energy be determined ?
    (8 votes)
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    • piceratops ultimate style avatar for user elm.parkinson
      In theory - but it would take more than a lifetime. For instance, internal energy is the kinetic energy of all the particles (and all the other energy in a system, but lets assume this is an ideal gas). It would take forever to calculate the kinetic energy of every particle because they are probably all different, and there are millions of them. You would need a super computer and that would require a lot of energy, because the particles kinetic energy will change when they collide with each other so the computer would have to calculate it instantly.

      So in theory? Yes. But completely impossible in practice.
      (7 votes)
  • leafers ultimate style avatar for user ff142
    I don't get why ΔPV = PΔV at constant pressure. If pressure is constant, then wouldn't ΔP be 0 so ΔPV would be 0?
    (3 votes)
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  • mr pants teal style avatar for user SanFranGiants
    So is Work = Pressure * Volume, Change in Pressure * Volume, Pressure * Change in Volume or Change in Pressure * Change in Volume?
    (2 votes)
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    • aqualine seedling style avatar for user pkeranger121
      Work = Pressure * change in Volume, because by defnition Work = Force*distance. Well we know Pressure = Force/Area. Solving for Force, force= area*pressure. We also know that distance, lets say x is how far an area traveled. If we set it up in 3D like a cylinder, we see that the change in distance is the height of cylinder. Volume = Area*height, but since height is changing, so is volume.
      (6 votes)
  • blobby green style avatar for user mustafa_afg
    I think I understand what enthalpy 'is', what its used for. But I'm confused on one part: Lets just say dE (change in internal E) is 10 and Qp (=H) is 2. Will for any process with dE=10, Qp =2.
    And what if dE=20 , will Qp be 4? and if not what's the point really in comparing enthalpies.
    (4 votes)
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    • orange juice squid orange style avatar for user Maya Pawlikowski
      I love how you are rationalizing this! :D Great job!
      It also depends on other factors, but if you keep the increase of both sides of the equation the same, then the equation still works! :)

      (I'm not entirely sure, but that's what i think. Sorry to answer a year old question. If you found the accurate answer, could you please share it? Thanks!)

      Hope this helped! :D - Maya
      (1 vote)
  • blobby green style avatar for user pranayreddy037
    What is the exact difference between entropy and enthalpy?
    (3 votes)
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  • leaf blue style avatar for user Heatcliffe
    So if pressure isn’t constant can i Not define enthalpy ...... then is this equation wrong ; delH = delU + del(PV) = n.R.delT
    .... I seriously am confused plz help

    (3 votes)
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  • blobby green style avatar for user Raunaq Singh
    What is exactly enthalpy? Really confused! 😕
    (2 votes)
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Video transcript

Let me draw a good old PV diagram. That's my pressure axis, this is my volume axis. Just like that, I have pressure and volume. I showed several videos ago that if we start at some state here in the PV diagram, right there, and that I change the pressure and the volume to get to another state, and I do it in a quasistatic way, so essentially I'm always close to equilibrium, so my state variables are always defined, I could have some path that takes me to some other state right there. And this is my path. I'm going from this state to that state. And we showed that if I just did this, the work done by the system is the area under this curve. And then if I were to move back to the previous state, and then if I were, you know, by some path, just some random path that I happened to be drawing, the work done to the system would be the area under this light blue curve. So the net work done by the system ended up being the area inside of this path. So this is-- let me do it in different color. The net work done would be the area inside of this path when I go in this clockwise type of direction. So this is the net work done by the system. And now, we also know that if we're at some point on this PV diagram, that our state is the same as it was before. So if we go all the way here, and then go all the way back, all of our state variables will not have changed. Our pressure is the same as it was before. Our volume was the same as it was before because we went all the way back to that same point on the PV diagram. And our internal energy is also the same point as it was before, so our change in internal energy over this path, you're going to have a different internal energy here than you had here, but when you go around the circle and you get back, your change in internal energy is equal to zero. And we know that our change in internal energy is defined as, and this is from the first law of thermodynamics, the heat added to the system minus the work done by the system. Now, if we go on a closed loop on our PV diagram, then what's our change in internal energy? It's zero. So we get zero change in internal energy, because we're at the same state is equal to the heat applied to the system minus the work done or-- and I've done this little exercise multiple times. I think is probably the fourth or fifth time I'm doing it. We get that the heat added to the system, if we just add w to both sides, is equal to the work done by the system. So this area inside of this path, I already said, it's the work done by the system, and if you don't remember even where that came from, it was, remember, pressure times volume times change in volume is a little incremental change in work, and that's why it relates to the area. But we've done that multiple times. I won't go there just yet. So if you have any area here, some heat was added to the system, some net heat, right? Some heat was added here, and some heat was probably taken out here, but you have some net heat that's added to the system. And I use that argument to say why heat isn't a good state variable. Because-- and I had a whole video on this-- that if I define some state variable, let's just say, heat content. Let's say I want to define some state variable heat content. And I would say that the change in heat content would, of course, be equal to the change in heat. That's what I'm defining. If I'm adding heat to the system, my heat content should go up. But the problem with that heat content state variable was that, let's say over here, I say that the heat content is equal to 5. Now, I just showed you that if we go on some path here and we come back, and there's some area in this little path that I took, that some heat was added. So let's say that this area right here, so this is q is equal to the work done by the system, let's say it's equal to 2. So every time, if I start at heat content is equal to 5, that's just an arbitrary number, and I were to do this entire path, when I go back, the heat content would have to be 7. And then when I go back and do it again, my heat content would have to be 9. And it would have to increment by 2 every time I do this exact path. It would have to increment by the amount of area that this path goes around. So heat content can't be a state variable, because it's dependent on how you got there. A state variable-- and remember this. In order to be a state variable, if you're at this point, you have to have the same value. If your internal energy was 10 here, when you do the path and you come back, your internal energy will be 10 again. That's why internal energy is a valid state variable. It's dependent only on your state. If your entropy was 50 here, when you soon. go back and you do all sorts of crazy things, and you come back to this point, your entropy is once again 50. If your pressure here is 5 atmospheres, when you come back here, your pressure will be 5 atmospheres. Your state variable cannot change based on what path you took. If you're at a certain state, that's all that matters to the state variable. Now this heat content didn't work, and that's why we actually led into some videos where I divided by t and we got entropy, which was an interesting variation. But that's still not satisfying. What if we really wanted to develop something that could in some way be a state variable, but at the same time measure heat? So obviously, we're going to have to make some compromises, because if we just do a very arbitrary kind of heat content variable, then every time you go around this, it's going to change. That's not a valid state variable. So let's see if we can make up one. So let's just make up a definition. Let's call my new thing that I'm going to try to maybe approximate heat, let's call it h, and just as a little bit of a preview, we're going to call it enthalpy. And let's just define it. I'm just playing around. Let's just define it as the internal energy plus my pressure times my volume. So then what would my change in enthalpy be? So my change in enthalpy will be, of course, the change of these things. But I could just say, that's my change in my internal energy plus my change in pressure times volume. Now, this is interesting. And I want to make a point here. This, by definition, is a valid state variable. Why is it? Because it's the addition of other state variables, right? At any point in my PV diagram, and it's also true if I did diagrams that were entropy in temperature or anything that dealt with state variables, at any point on my diagram, u is going to be the same, no matter how I got there. p is by definition going to be the same. That's why it's at that point. v is definitely going to be the same point. So if I just add them up, this is a valid state variable because it's just the sum of a bunch of other valid state variables. So let's see if we can somehow relate this thing that we've already established as a valid state variable. From the get-go, from our definition, this works because it's just the sum of completely valid state variables. So let's see if we can relate this somehow to heat. So we know what delta u is. If we're dealing with all of the internal energy or the change in internal energy, and I'm not going to deal with all the other chemical potentials and all of that, it's equal to the heat applied to the system minus the work done by the system, right? Let me put everything else there. The change in enthalpy is equal to the heat applied minus the work done-- that's just the change in internal energy-- plus delta PV. This is just from the definition of my enthalpy. Now this is starting to look interesting. What's the work done by a system? So I could write change in h, or enthalpy, is equal to the heat applied to the system minus-- what's the work done by a system? If I have some system here, it's got some piston on it, you know, if we're doing it in a quasistatic, I have those classic pebbles that I've talked about in multiple videos. When I apply heat or let's say I remove some of these pebbles, so I'm at a different equilibrium, but what's actually happening? When is the work being done? You have some pressure being applied up here, and this piston is going to be moving up, and your volume is going to increase. And we showed multiple videos ago that the work done by the system can be, and you can kind of view this as the volume expansion work, it's equal to pressure times change in volume. And let's add the other part. So this was our change in internal energy. And I had several videos where I show this. And now let me add the other part of the equation. So our enthalpy, our change in enthalpy, can be defined by this. Now. Something interesting is going on. I said that I wanted to define something, because I wanted to somehow measure heat content. My change in enthalpy will be equal to the heat added to the system, if these last two terms cancel out. If I can somehow get these last two terms to cancel out, then my change in enthalpy will be equal to this, if somehow these are equal to each other. So under what conditions are these equal to each other? Or another way, under what conditions is delta pressure times volume equal to pressure times delta volume? When does this happen? When can I make this statement? Because if I can make this statement, then these two terms are equivalent right here, and then you my change in enthalpy will be equal to the heat added. Well, the only way I can make this statement is if pressure is constant. Now why is that? Let's just think about it mathematically. If this is a constant, then if I just change-- you know, if this is just 5, 5 times a change in something is the same thing as the change in 5 times that thing, so it just mathematically works out. Or if you view it another way, if this is a constant, you can just factor it out, right? Well, if I said, the change in 5x, that would be equal to 5 times x final minus 5 times x initial. And you could say, well, that's just equal to 5 times x final minus x initial. Well, that's just equal to 5 times the change in x. It's kind of almost too obvious for me to explain. I think sometimes when you overexplain things, it might become more confusing. So this applies-- and the 5 I'm just doing as the analogy for a constant. So if pressure is constant, then this equation is true. So if pressure is constant-- so this is a key assumption-- then if heat is being applied in a constant pressure system-- so we could write it this way. I'll write it multiple times, because this is key. If pressure is constant, then our definition, our little thing we made up, this enthalpy thing, which we defined as internal energy plus pressure and volume, then in a constant pressure system, our change in enthalpy we just showed is equal to the heat added to the system because all of these two things become equivalent under constant pressure, so I should write that. This is only true when heat is added in a constant pressure system. So how does this gel with what we did up here on our PV diagram? What's happening in a constant pressure system? Let me draw our PV diagram. That's P, that's V. So what's happening in constant pressure? We're at some pressure right there. So if we're under constant pressure, that means we can only move along this line. So we could go from here to there and back to there, or we could go from there to there, back to there. So we could go there, all the way there, and then go back. But what do we see about this? Is there any area in this curve? I mean, there is no curve to speak of, because we're staying in a constant pressure. We've kind of squeezed out this diagram. We've made the forward path and the return path the same exact path. So because of this, you don't have that state problem because no net heat is being added to the system when you go from this point all the way to this point and then back to this point. So because of that, you can kind of see visually that enthalpy in a constant pressure, when you're not moving up and down in pressure, is the same thing as heat added. So you might say, hey Sal, this was a bit of a compromise, constant pressure, you know, that's a big assumption to make. Why is this useful at all? Well, it's useful, because most chemical reactions, especially ones that occur in an open beaker, or that might occur at sea level, and that should be a big clue, they occur at constant pressure. You know, if I'm sitting at the beach, and I have my chemistry set, and I have some beaker of something, and I'm throwing other stuff into it, and I'm looking for a reaction or something, it's a constant pressure system. This is going to be atmospheric pressure. 1 atmosphere. I'm sitting at sea level. So this is actually a very useful concept for everyday chemical expressions. It might not be so useful for engines, because engines always have pressure changing, but it's very useful for actual chemistry, for actually dealing with what's going to happen to a reaction at a constant pressure. So what we're going to see is that this enthalpy, you can kind of view it as the heat content when pressure is constant. In fact, it is the heat content when pressure is constant. So somehow-- well, not somehow, I showed you how-- we were able to make this definition, which by definition was a state variable, because it was the sum of other state variables, and if we just make that one assumption of constant pressure, it all of a sudden reduces to the heat content of that system. So we'll talk more in the future of measuring enthalpy, but you just have to say, if pressure is constant, enthalpy is the same thing as-- and it's really only useful when we're dealing with a constant pressure. But if we have a pressure constant, enthalpy can be imagined as heat content. And it's very useful for understanding whether chemical reactions need heat to occur or whether they release heat, so on and so forth. See