 Problem three, a closed piston-cylinder device filled with 10 kilograms of air undergoes a series of quasi-equilibrium processes shown here. Determine the specific boundary work in kilojoules per kilogram occurring between states 1 and 4. So, let's think about what's actually happening here. We have a gigantic apparently piston-cylinder device. I said it so it's now canon. I guess it would have to be pretty large in order to fit 10 10 kilograms of air, so that makes sense. Awesome. I have a large piston-cylinder device containing air and between states 1 and 4 here it is undergoing three processes. From 1 to 2, the specific volume is constant, the pressure is increasing. From 2 to 3, the specific volume and pressure are both increasing, and then from 3 to 4, the specific volume is increasing, but not the pressure. So, because the mass is constant, I know that because it says a closed piston-cylinder device, because the mass is constant, this specific volume increasing means that the actual volume is going to be increasing as well. So, between 1 and 2, the piston doesn't move. Between 2 and 3, the piston goes up quite a bit, and then from 3 to 4 it goes up a bit more. And I'm asked to calculate the specific boundary work. Let's step back to what boundary work actually means. We defined boundary work, which is that work associated with a moving boundary. In this case, the moving boundary is the raising piston. The total boundary work would be the integral of pressure with respect to volume. And what I want is actually the specific boundary work. So that would be the total boundary work divided by mass, which would be 1 over mass times the integral of pressure with respect to volume. And I could bring that mass inside the integral, and I could divide the differential of volume with mass, and end up with the differential of specific volume. So I could have the integral of pressure with respect to specific volume. And what I'm looking for is the specific boundary work between states 1 and 4. So this would be the integral of pressure with respect to specific volume from specific volume 1 to specific volume 4. And I could make this a little bit easier by recognizing that I could break that integral apart. So the specific boundary work from 1 to 4 would be the same as adding together the specific boundary works from 1 to 2 and then 2 to 3 and then 3 to 4. Which would be the integral of pressure with respect to specific volume from v1 to v2 plus the integral of pressure with respect to specific volume from v2 to v3 and then again from 3 to 4. I know that doesn't look like a P, but I swear it is. So at this point, I look back at how my pressure actually varies across the processes. Pressure is only constant from 3 to 4. So I could bring the P outside of the integral from 3 to 4 here and actually just take the pressure times the differential of specific volume. Pressure times delta specific volume. So pressure of 500 kilopascals times 0.2 meters cube per kilogram. However, from 1 to 2 and 2 to 3, this gets a little bit more complicated. Because I would have to actually calculate or determine the equation of this line, I have to know how pressure varies with specific volume in order to actually evaluate this integral. Except for one thing. I could take a shortcut by recognizing that the specific, the integral of pressure with respect to specific volume is just going to be the area under this curve because I'm given a graph of pressure and specific volume. I could actually just geometrically determine this area here and that would give me my answer. And I can do that by splitting it apart into some simpler geometric shapes. Add them together. That would give me my answer. So let's see. Let's do the area of this rectangle plus the area of this triangle plus the area of this rectangle. So the first rectangle would be 200 kilopascals times 0.4 cubic meters per kilogram. So I guess let's write that out. 200 kilopascals times 0.5 minus 0.1 cubic meters per kilogram. Plus the area of this triangle, which would be one half base times height. So that would be one half. The base is going to be 0.5 minus 0.1 cubic meters per kilogram. And then the height is 500 kilopascals minus 200 kilopascals. And plus the area of this rectangle over here. So this would be just to be consistent. Let's take 500 kilopascals times 0.7 minus 0.5 cubic meters per kilogram. So 500 kilopascals times 0.7 minus 0.5 cubic meters per kilogram. So I calculate these numbers and add them together. I'll get my total specific boundary work, if that makes any sense. So wake up calculator. We have work to do. 200 times 0.4 is 80. 0.5 times 0.4 times 300 is 60 and 500 times 0.2 is 100. So 80 plus 60 plus 100 would give me 240. I don't want to get into the habit of doing mental math because inevitably I'm going to screw something up. So let's do 80 plus 60 plus 100. That's 240. Awesome. Score one for mental math. Although I guess I did do mental math over here too. It's probably fine. Let's just ignore the problem and it'll go away surely. So I have 240. 240. Now the question becomes, what are the units on that? Well, I know that I have kilopascals times cubic meters per kilogram. So when I add all of these up, this is actually 240 kilopascals 240 kilopascals meters cubed per kilogram. Now let's think about what that means. A kilopascal is a thousand pascals and a pascal is defined as being a Newton per square meter. Then I know that I'm looking for an answer in kilojoules per kilogram. So I want to get this numerator into kilojoules. I'm going to end up with kilograms on the bottom here. This meter squared is going to cancel this cubed. So I'm left with Newton meters per kilogram. Well, that's convenient because I know that a joule is defined as being a Newton meter. So in order to get an answer in kilojoules per kilogram, I'm going to have to take this answer that I came up with, multiply it by a thousand and then divide it by a thousand again. Because there's a thousand joules in a kilojoule. So my answer of 240 is actually already in kilojoules per kilogram. So you'll notice that at no point in this geometric solution did I use the 10 kilograms? Or remember that we were supposed to or we were supposed to quote unquote have calculated this answer by figuring out the equations of these lines. So the equation of the line from one to two wouldn't have been a function of specific volume, it would have been a function of something else. This fails the vertical line test because one input could yield multiple outputs. So this integral here would just become zero. Then I could come up with the equation of the line from two to three and that would have given me let's see y plus mx plus or y equals mx plus b, in this case y is pressure and x is cubic meters per kilogram. I could extrapolate this line to come up with a y-intercept, that'd be b, m would be the slope which is rise of a run, which I could calculate, I could come up with this equation as well. And then three to four, this would be an easier integral because pressure would come out. But at no point in that solution either would I have used 10 kilograms. That means this 10 kilograms was more information than I needed, it was a red herring. It would have been useful if I were asked for the total boundary work, that just kilojoules occurring between one and four. But because I'm asked for kilojoules per kilogram, the kilograms, the number of kilograms, the mass doesn't actually matter. But that gives me my answer to question number three.