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Calculating internal energy and work example

Worked example calculating the change in internal energy for a gas using the first law of thermodynamics.

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  • piceratops tree style avatar for user andreilyskov
    Where is the 10^-3 coming from (multiplied by m^3?
    (11 votes)
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  • purple pi teal style avatar for user Gustavo Delazeri
    Why work is pressure multiplied by the change in volume?
    (13 votes)
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  • orange juice squid orange style avatar for user JL
    The drawing is somewhat confusing.
    It appears that 485J is being moved from V1 to V2. But if I am understanding this correctly, the 485J is leaving V1 and the result is V2 correct?
    (10 votes)
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  • hopper jumping style avatar for user Yuya Fujikawa
    I'm sorry but why is it that doing work on the system increases the internal energy in the first place? What exactly does work done to the system mean? like squeezing the baloon? but that just decreases the volume and increases the pressure and the energy shouldn't change..(?) Also, if Heat, Q, was transferred to the system, can I view that as like someone heated the system with fire or something? If so, that makes sense as molecules gain kinetic energy from heat. I really appreciate if you can answer this.
    (5 votes)
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    • piceratops seed style avatar for user RogerP
      When work is done on a system, energy is transferred to that system, which increases the internal energy of the system. Conversely, energy is lost from whatever is doing the work on the system.

      Heating a system with a fire is a classical way of transferring heat to the system. That was the basis of the steam engines that started the whole science of thermodynamics.
      (6 votes)
  • piceratops tree style avatar for user Jurgil Peronilla
    Isn't it 1L is equal to 1dm^3?
    (4 votes)
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  • blobby green style avatar for user Natrah Arifin
    At t=2.47, why do you add a minus sign there? It appears all of a sudden...
    (4 votes)
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    • mr pink red style avatar for user Anthony Scaletti
      That comes from the formula for "work" for a mechanically reversible system. She doesn't goes into the details, but for a mechanically reversible system:
      w= -Pext(delta V). You need the negative in front because when work is done "on the system" (which is positive), delta V (Vfinal-Vinitial) is negative (the ballon is compressed). So the two negatives make the whole term for work positive. Side note: remember that external pressure (Pext) is considered constant. Which makes the formula for work so simple. So we have not learned anything in this problem about the pressure inside. That would require more assumptions.
      (7 votes)
  • piceratops sapling style avatar for user Yasmin B
    Why is the sign for pressure is negative"
    (5 votes)
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    • boggle blue style avatar for user Andrew H.
      Go ahead and watch this part.
      We are considering the work that is done on the system. So if you compress the space, you are doing work to it.
      If the system (balloon) expands, it is doing work to the surrounding. In this case, the system loses energy by doing work, thus the work done is negative.
      (0 votes)
  • aqualine seedling style avatar for user kazimsyed9911
    In the formula the negative sign indicate that the pressure of gas that is in the system is working against the external pressure . the external pressure could be block or anything and in the video the balloon compress that means work is done by the surrounding leading to increase in internal energy and the work is positive
    (4 votes)
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  • male robot johnny style avatar for user thejosh210savage
    At about she defines joules in terms of other units(kgm2/s2) please could you tell me how she derived that?
    (1 vote)
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    • mr pants purple style avatar for user Ryan W
      It's one of the definitions of joules...

      The energy transferred to an object when a force of 1 Newton acts on that object a distance of 1 metre.

      J = N * m

      N = kg * m * s^-2

      Therefore: J = kg * m^2 * s^-2
      (5 votes)
  • leafers sapling style avatar for user V_GRNG
    Can anyone tell me that the answer should be -510.25J instead of -460J...Because the work is being done by the surroundings on the system the value of W should be considered positive i.e Delta U= -Q+W. Kindly let me know if I'm correct or incorrect
    (2 votes)
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    • male robot johnny style avatar for user yashasvee basotia
      Here we are considering the work done BY THE SYSTEM not the surroundings.In this case the balloon compresses so we know that work is being done on it,now this work is done by the atmosphere,so this work would be positive but when we see it in terms of the work done by system then the system did the same amount of of negative work and that's the reason for the -ve sign.
      (2 votes)

Video transcript

- [Voiceover] In this video we're gonna do an example problem, where we calculate in internal energy and also calculate pressure volume work. So, we know the external pressure is 1.01 times ten to the 5th pascal, and our system is some balloon, let's say it's a balloon of argon gas, and initially our gas has a volume of 2.3 liters, and then it transfers, the gas transfers 485 joules of energy as heat to the surroundings. Once it does that, the final volume of our system, is 2.05 liters. And we're assuming here, that the moles of gas didn't change. The question we're gonna answer is, for this process, what is delta U? So, what the change in internal energy for our system? We can use the first law of thermodynamics to answer this. The first law tells us that the change in internal energy, delta U, is equal to the work done, plus the heat transfer. Before we plug any numbers in here, the first thing I wanna do is make sure I have a good idea of what signs everything should be. I think that's one of the trickiest things in these kind of problems. So here, since our system transferred energy to the surroundings and not the other way around, Q should be negative, because when your system transfers energy to the surroundings, then it's internal energy should go down. Work on the other hand, since V two is less than V one, the volume of our system went down, which means the surroundings had to do work on the gas to get the volume to decrease. We would expect if the surroundings did work on our system, that would increase the internal energy. So that means the work done here is positive. We can also calculate work because we know the external pressure, we know it's constant, and work can be calculated as the external pressure times the change in volume, and we know both of those things, we know the external pressure and we know the initial and final volumes. So if we start plugging that in, we get that delta U is equal to negative 485 joules, so that's our heat, we know it should have a negative sign because the heat was transferred to the surroundings. So negative 485 joules, minus, we should have a negative sign there, minus the external pressure 1.01, times ten to the 5th pascals, so that's our external pressure times our change in volume. So that's our final pressure, 2.05 liters, minus our initial pressure, which is 2.30 liters. We could at this point be like," okay, "we figured it all out, "we just have to stick all of these numbers "in a calculator and we're done," and that's probably what my first instinct would often be, but there's one more thing that we should check before we actually plug in the numbers and have a party. And that's our units, we have our heat, in terms of joules, we probably want our change in internal energy in terms of joules too. On this other side, we're calculating our work here, and we have pascals, so pascals times, so we have joules, and joules, we have pascals times liters, so then the question is, okay, we're doing joules minus pascals, times liters, we need to make sure that whatever we calculate here, in terms of work, also has units of joules. Otherwise, we will be, we'll be subtracting two things that don't have the same unit, and that's bad. (giggles) We'll have to do some sort of unit conversion first. So, let's just double check that pascals times liters, will give us joules. So the way I did this, is by converting everything to the same units. So, if you take joules, which is already SI units, we can actually simplify it more, in terms of other SI units. So, a joule is equal to one kilogram meters squared per second squared. So, joules is equal to, one joule is equal to one kilogram meters squared over seconds squared. One pascal, pascals are also in terms of, are also SI units, and if we convert pascals to kilograms, meters, and seconds, we get that one pascal is one kilogram per meter seconds squared. What this tells us is that, we have to multiply this by units of volume, and whatever we multiply it by should give us units of joules. So, what we need to do here, is convert our liters to meters cubed. And if we do that, everything is in terms of kilograms, meters, and seconds, this meters cancels out with one of these, and we end up with one kilogram meters squared per second squared on both sides. So then, everything is in terms of joules. That's not the only way that you could've made sure that the units made sense. You could've converted them to something else, but basically you just have to make sure that all of the units you're using in your equation match each other if you're going to add them or subtract them. All of this is to say that we need to make sure we convert liters to meters cubed, so that everything works out in terms of joules. So if we do that, we get that minus 485 joules, 1.01 times ten to the 5th pascals, times negative .25 liters, which is the change in volume, which is negative, the volume went down, so the change in volume should be negative, and then we have to add one more thing here to convert our liters to meters cubed. So, one liter is equal to one, ten to the minus third, meters cubed. So now our liters cancel out, and pascals times meters cubed, gives us joules. So that gives us, that delta U, or change in internal energy, is negative 485 joules, then if we plug this all into our calculator to calculate the work, we get positive 25.25 joules. So if we add our heat and our work here, we get that the overall change in internal energy for this process is negative 460 joules. So, the key things to remember here, for this kind of problem, is to double check your signs for work and heat, and also to make sure all of your units match.