 Up next, three, two, one. A piston cylinder device initially contains 0.4 cubic meters of air at 100 kilopascals in 80 degrees Celsius. The air is now compressed, note, compressed, it's going to be a decrease of volume. In such a way that the temperature inside the device, a piston cylinder device initially contains 0.4 cubic meters of air at 100 kilopascals in 80 degrees Celsius. The air is now compressed, note, air is compressed. It's a work input. To 0.1 cubic meters, note that the volume is decreasing, that reinforces the whole compression, therefore work in thing. In such a way that the temperature inside the device remains constant, determine the work done during this process. So I will start to parse out what I know into state point properties. State point one has a volume, a temperature and a pressure. And then state point two has a volume and a temperature. So I'm going to say state point two has a volume of 0.1 cubic meters and T2 is equal to T1. Do you think it's reasonable to assume anything else is constant from state one to state two? You're right, probably reasonable to assume the mass is constant. Assumption, therefore, closed system. Which you could write as assuming the mass is constant as well. In that case, M2 is equal to M1. And we'll put an asterisk next to it to note that that's a result of our assumption. And now we'll pose the question, do we have both state points fixed? That is, do we have two independent intensive properties for both state points? Well, P1 and T1 give us our two independent intensive state points. Well P1 and T1 are our independent intensive properties as state point one from which we could theoretically come up with anything else since this is air that would be via the ideal gas law for air, including but not limited to specific volume one. What do I know about state point two? Well, temperature is an independent intensive property. And the other independent intensive property that I know about state point two is the volume. And the other independent intensive property, the thing that fixes state point two is the specific volume. Now I know what you're thinking. You're thinking, hold up, we don't have a specific volume drawn. You're right, we don't. But we can get to specific volume two. We know the volume goes from 0.4 to 0.1. And if I know the volume and the specific volume at state one, I could come up with mass. And if I know the mass of state one, that gives me the mass of state two. And if I were to use mass and specific volume at state two, I could come up with the total volume at state two. Or if I knew the mass and the total volume, I could come up with specific volume at state two. Going back a bit. You can think of this in one of two ways. One would be to say, well, if we can determine specific volume one, and we know the total volume of state of volume one, we could use those two facts to come up with mass one. And if we know the mass two is equal to mass one, and we know the total volume of two, we could use mass two and volume two to come up with specific volume two. But the much faster way to do it would be to recognize that the total volume goes from 0.4 to 0.1. It is quartered. So the specific volume must also go from whatever is at state one to one quarter of that value. Because specific volume is total volume divided by mass, if our total volume goes from 0.4 to 0.1, if it decreases by a factor of four, and the mass doesn't change, then the specific volume must also decrease by a factor of four. Therefore, the two independent intensive properties that I know about state one are pressure and temperature, and the two independent intensive properties that I know about state two are temperature and specific volume, and v2 is going to be v1 divided by four. Cool. Once we have our state point properties, we can determine anything else about them because they're fixed. Any independent intensive property we need, we got it. Now let's actually consider the rest of the problem. I want to know the work done. This is an isothermal process. So what I would do is I would refer to my simplifications of boundary work table that we had created painstakingly, and I will see that we had come up with a simplification for the isothermal process of an ideal gas. That happens to be the exact situation that we're considering. I guess I should add that let's assume air is ideal. We have no other way of proceeding because right now, since we don't know how to use the property lookup tables, we have no way of getting to actual properties for air using T1 to come up with anything else. So we're going to assume that air is ideal. General rule of thumb for this class. In thermal one, you can treat air as an ideal gas unless you're specifically told to do otherwise. Actually, you can treat any gas that appears on the ideal gas list in the appendix as an ideal gas, the exception to which being steam. Do not treat steam as an ideal gas ever in thermal one. Once we get to thermal two under very specific circumstances, we'll start to do that, but that's a story for another time. Broke. Note to future, John, I want to keep that rambling before this, but I want to stitch together the video here. Anyway, so now that the air is ideal, I can use the simplification of boundary work for an isothermal process of an ideal gas. That means I could use this form, I could use this form, I could use this form. The world is my oyster. Now, which one do you want to use? I mean, they're all functionally equivalent here. It's just a matter of which one is going to be the least amount of effort to actually apply. And I don't want to do any more math than I have to. So since I have P1 and V1 already, I might as well use this form here. Does that make sense? I mean, I could go calculate the specific gas constant for air. We can look up the universal gas constant. We can look up the molar mass of air. We can divide the two, come up with a specific gas constant. I could use the volume and determine the specific volume and come up with a mass. Then I could take that mass times the specific gas constant, multiply by 80 degrees Celsius, and convert that to Kelvin. Or I could just take 100 kilopascals times 0.4 cubic meters, right? Work harder, work smarter, not harder. By thinking through it a little bit, we just cut, like, five calculations off of our work. And we're decreasing the probability that a rounding error is going to contribute to. And we're decreasing the relative effect of rounding error in our calculations by hand by eliminating those calculation steps. I will also point out that theoretically, we could use our volume at state two and the pressure at state two just as well, because we know 0.1 cubic meters and 100 kilopascals. But since they're both the same, I'm gonna use P1 and V1 with, go back a bit. Now, theoretically, we could also use P2 V2 because we could use the fact that our specific volume at state two is divided by four and the temperature to come up with pressure from the adiogas law and then apply that pressure in our specific volume at state two. But even that, now, theoretically, we could use P2 and V2 as well because we know V2 and we could determine P2 with the adiogas law. But again, we should eliminate as many steps as we possibly can. So using P1 V1 is the easiest method. So I'm going to apply bodywork is equal to P1 V1 times the natural log of V2 over V1, because I've already come up with the simplification or the bodywork of an isothermal process of an adiogas. We had a whole video dedicated to those simplifications. And now we are harvesting the fruits of our efforts by not having to derive this again. So bodywork for this process is going to be P1 times V1 times the natural log of V2 over V1. So I have a 100 kilopascal pressure at state one. He said, hoping that he actually had that pressure. And scrolling up just to be extra sure. Yep, 100 kilopascals and V1 is 0.4 cubic meters. So 0.4 cubic meters. And then I'm multiplying by the natural log of V2 over V1. That was 0.1 over 0.4, which is the natural log of one over four. And then I want kilojoules, probably, unless I'm doing a weird unit conversion, kilojoules. So I will start at my destination and work backwards, a kilojoules, 1,000 joules. A joule is a Newton meter and a Newton is a kilogram meter per second squared. And then I can also do a kilopascal is 1,000 pascals. Any pascal is a Newton per square meter and a Newton is a kilogram meter per second squared. But I will point out that but like we've talked about before, if we recognize that kilojoules and kilopascals both have the 1,000 magnitude in them, I don't have to go all the way down into our base dimensions. I could just say the following. A kilojoule is a kilonewton times a meter and a kilopascal is a kilonewton per square meter. A kilonewton cancels kilonewton, kilopascal, cancels kilopascals, kilodjoules, cancels kilodjoules, kilodjoules is what we want, cubic meters cancels meters and square meters. Therefore, our result is in kilodjoules. By thinking through what units you're going to be deploying in your conversions as you're breaking them down, you can sometimes cut off steps from your conversion. You can sometimes cut off steps in your conversion process and eliminate how many steps you have to go through. So 100 times 0.4 times the natural log of a quarter won't give us our answer in kilodjoules. So calculator, if you would please, 100, not C100, 100 times 0.4 times the natural log, which is over here of 0.25. Or if you wanna be just a little bit more accurate here, I could say 0.1 divided by 0.4, which is actually what it is. And I get negative 55.45. If you recall our conversation about what positive versus negative boundary work implies, you should know what that means. Expansion will be a positive boundary work, compression will be a negative boundary work. We got a negative number, which means we have a compression process. That makes sense because we were told that the air is compressed. So it makes sense that we got a compression boundary work. Furthermore, remember that expansion is work out and compression is work in. So I would characterize this as a work in of 55.45 kilodjoules, which does that make sense? Because we define our boundary work with a positive differential for volume, positive boundary work is a work out. Negative boundary work is still a work out. It's just in the opposite direction as a result of the negative. So a work in of 55.45 or a work out of negative 55.45, those are the ways that we could write the boundary work in terms of what's actually happening in our process here. A third way of writing this would be work is done on the air as opposed to work is done by the air.