Calculating internal energy and work example | Chemistry | Khan Academy

In this video, we're going to do an example problem where we calculate internal energy and also calculate pressure-volume work. So we know the external pressure is 1.01 * 10^5 Pascals, and our system is some balloon. Let's say it's a balloon of argon gas.

Initially, our gas has a volume of 2.3 L, and then it transfers 485 joules of energy as heat to the surroundings. Once it does that, the final volume of our system is 2.05 L, and we're assuming here that the moles of gas didn't change. The question we're going to answer is: for this process, what is ΔU? So, what's 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, ΔU, is equal to the work done plus the heat transfer. Before we plug any numbers in here, the first thing I want to 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.

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 its internal energy should go down. Work, on the other hand, since V2 is less than V1, 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. 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 ΔU is equal to -485 joules. So that's our heat; we know it should have a negative sign because the heat was transferred to the surroundings, so -485 joules minus we should have a negative sign there, minus the external pressure, 1.01 * 10^5 Pascals.

So that's our external pressure times our change in volume. So that's our final volume, 2.05 L, minus our initial volume, which is 2.30 L. We could at this point be like, "Okay, we figured it all out! We just have to stick all of these numbers in the calculator, and we're done." 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 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 subtracting two things that don't have the same units, and that's bad. 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. The way I did this is by converting everything to the same units. If you take joules, which is already SI units, we can actually simplify it more in terms of other SI units. A joule is equal to 1 kg m²/s².

So joules is equal to one joule is equal to 1 kg m s over second squared. One Pascal, on the other hand, is 1 kilogram per m s². 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 cubic meters, and if we do that, everything is in terms of kilograms, meters, and seconds. This meter cancels out with one of these, and we end up with 1 kg m²/s² on both sides. So then, everything is in terms of joules.

That's not the only way you could have made sure that the units made sense. You could have converted them to something else. Basically, you just have to make sure that all 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 cubic meters so that everything works out in terms of joules. So if we do that, we get that -485 joules, 1.01 * 10^5 Pascals multiplied by -0.25 L, which is a 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 cubic meters. So, 1 liter is equal to 1 * 10^-3 m³, so now our liters cancel out, and Pascals times cubic meters gives us joules.

So that gives us that ΔU, our change in internal energy, is -485 joules. Then, if we plug this all into our calculator to calculate the work, we get positive 25.25 joules. If we add our heat and our work here, we get that the overall change in internal energy for this process is -40 joules.

So, the key things to remember here for this kind of problem are to double-check your signs for work and heat and also to make sure all of your units match.