 In our conversation about work, one of the shorthand simplifications for work that we had come up with was called boundary work, which is the work associated with a moving boundary. And what I would like to do now is generate an even more specific list of shorthand simplifications for boundary work specifically. So let's start a list called simplifications of boundary work. And I will bring over our definition of boundary work just to remind us of what our boundary work actually is. You don't want to get too caught up in the simplifications. So for simplification number one, let's consider an isochoric process. In an isochoric process, that's one where the volume doesn't change. How does the integral of pressure with respect to volume behave? Well, if there is no change in volume, there is no boundary work. I mean, if dv here is zero, then work is zero. Doesn't matter what the pressure is. Neat. Boundary work is zero. So item number one, done. Cool. Let's make another property constant. What if we were considering an isobaric process instead of isochoric? Change in pressure is zero instead of change in volume is zero. Well, if our pressure is constant, it's going to come out of our integral. If it comes out of our integral, we're left with the pressure times the integral of dv, which is just delta v, which means that I would have pressure times delta v from one to two. Therefore, pressure times v2 minus v1. Okay, next property, let's consider an isothermal process. In an isothermal process, the temperature doesn't change. That's time job. The temperature doesn't change. Upper case t. Well, if we have the integral of pressure with respect to volume from v1 to v2, then in order to simplify this for an isothermal process, we have to have some way of relating pressure to temperature. And I don't have a convenient way for doing that in all cases. But I do have a convenient way of doing that for a specific case. That is the case of an ideal gas. So I will write this as isothermal asterisks of an ideal gas. And now we can recognize that for an ideal gas, pressure is related to temperature through the ideal gas law. Pressure times volume is equal to mass times specific gas constant times temperature. So I could say pressure is equal to mass times gas constant times temperature divided by volume. And if I plug that into my integral, I would have the integral of mass times specific gas constant times temperature divided by volume, d volume. And when I try to figure out how that integral is going to simplify, I should start with bringing out our constants. The mass is constant because in this process, I'm assuming that I have a closed system, then the specific gas constant will come out because it's a constant and the temperature will come out because it's an isothermal process. And then all I have to do is simplify the integral of one over volume, d volume. And remember that the integral of one over x dx is the natural log of x. So I'll write this as mass times specific gas constant times temperature times natural log of volume evaluated from V1 to V2. So this would be m times r times t times the natural log of V2 minus the natural log of V1. Then because of our logarithm rules, I can say mass times gas constant times temperature times the natural log of volume 2 over volume 1. And mass times gas constant times temperature doesn't have to be the beginning of this equation. Because remember from our ideal gas law, mass times gas constant times temperature is equal to pressure times volume. So I basically have three ways of writing this out. I could say mrt times the natural log of V2 over V1, or I could say p1v1 times the natural log of V2 over V1, or I could say p2v2 times the natural log of V2 over V1. All three forms are valid. I will also point out after I get rid of this, that because mrt can be written as p1 times V1, that is another way to think through this simplification step here. If you didn't want to just treat the masses being constant temporarily to bring it out, you could approach it by considering the pressure times the volume again. Point being the boundary work for an isothermal process of an ideal gas can be simplified into mrt times natural log of V2 over V1. Of course, if you wanted to use moles, you could say the number of moles times the universal gas constant times temperature times natural log of V2 over V1, or pressure times volume times the natural log of V2 over V1 provided that you use pressure and volume at the same state point. So isobaric, isochoric, isothermal, let's let's throw in one more. And for this one more, I want to simplify for a general type of problem. And that general type of problem is going to be a polytropic process, a polytropic process, if I open up a new page here, I think is easiest to get your head wrapped around if you consider what we've done so far. If you were to plot out pressure with respect to volume, and recognize that the work is the area under the curve, we started by considering an isochoric process, which would be a vertical line. And we said the area under that curve is zero. Great. Then we considered the next easiest case, which is a horizontal line. Great. That isobaric process is just going to simplify to a rectangle. You have pressure multiplied by the width, which would be V2 minus V1. The next simplest case is going to be a straight line between the two. But we can actually characterize more than just a straight line between one and two. What if we were to consider any process that has a curve that bends the same, we can describe all of these curves at once provided that the amount of change in slope is constant. What I like to think of as the curviness of the line, the bend of the line, that curviness of the line has to be constant for our process to be described with the expression that we get in a polytropic process. Polytropic, many curves, all of those are described by this. Pressure times volume raised to an exponent is constant. For our polytropic processes, if we can figure out n to represent a process, then our polytropic process simplification of boundary work will be much easier to apply than trying to integrate from scratch all over again. And this is also convenient because I can come up with some common n values. This n value is the polytropic exponent, and some common values for n include 0, 1, and infinity. When n is equal to 0, then volume raised to the 0th power is just 1, and I have pressure is equal to constant. So n is equal to 0 is equivalent to an isobaric process. For n is equal to 1, I have pressure times volume is constant. And that should also be a little bit familiar, because the process that we analyzed before this was a process where we had a constant temperature of an ideal gas. And in that situation, we had a constant mass, we had a constant gas constant, and we have a constant temperature, so pressure times volume is constant. Because constant times constant times constant is still just a constant. So a polytropic exponent value of 1 is representative of an isothermal process of an ideal gas. The next common value is what happens as n approaches infinity. Well, as n approaches infinity, the importance of pressure becomes less and less important. So we could say at infinity, I mean, if you think of infinity temporarily as a number, we could say we have an isochoric process. So these are kind of our touchstones. We can use the relative value of n to think through what sort of behavior is actually happening in this process. And I will actually add one to this list. You're not going to be familiar with it yet, but you will be in the future. And that way, if you come back to this, say hypothetically, when you're in thermo two, and you're trying to figure out the boundary work associated with a gas power cycle, and you've forgotten these simplifications for boundary work, so you're watching through this video again, when n is equal to k, which is our isentropic exponent Cp over Cv, that's equivalent to an isentropic process. So if we can simplify the boundary work for a situation where the pressure times volume raised to the n is constant, then I can plug in any situation where I know n, the n value into our boundary work, and come up with the amount of work associated. By the way, I will also point out this is pressure times the quantity volume to the n, not pressure times volume quantity to the n. That's in a very important distinction. It is pressure times quantity volume to the n. Okay. So I will jump back to our definition of boundary work, the integral of pressure with respect to volume, and I will plug in our relationship for a polytropic process. That's a situation where the pressure times the volume raised to the n value is constant. So I could say pressure is equal to constant over volume to the n value. I'm saying P can be written as the constant divided by volume to the n. And I really only remember the power rule from Calc 1, so I try to do everything with the power rule that I possibly can. So I'm just going to write this as constant times volume to the negative n. I mean remember that multiplying by one half is the same as dividing by two. I don't know if that helped, that reciprocal, multiplying by the reciprocal of a quantity is the same as dividing by that quantity. So then I have the integral of constant times volume to the negative n, d volume, integrated from volume one to volume two. And constant is constant, so constant comes out. And I have constant multiplied by the integral v1, excuse me, the integral volume to the negative n, d volume. Now what is the integral of volume to the negative n power with respect to volume? It's cool, I'll give you a second. It's kind of hard to wrap your brain around. The first time you're considering a situation like this, it's easiest to consider if you just step back to the integral of x to the a with respect to x is going to be one over a plus one times x to the a plus one dx. If a was two then you have one over three x to the third. Okay, so my integral is going to be constant multiplied by one over a is negative n, so I have negative n plus one times volume to the negative n plus one and I'm evaluating that from x1 to x2. And then I would have constant times one over negative n plus one times v2 to the negative n plus one. Isn't that fun by the way? The way that negative n plus one rolls off the tongue, negative n plus one, negative n plus one, negative n plus one rolls off way better than one minus n. Remember that, it'll be helpful later. Negative n plus one, just a fun thing to say, times v1 to negative n plus one. That's plus plus. That would be our programming language, form of our knowledge of work. So constant times one over negative n plus one times v2 to the power of negative n plus one minus one over negative n plus one times v1 to the negative n plus one. Okay, now I'm going to bring out one over negative n plus one at which point I have constant over negative n plus one times v2 to the negative n plus one minus v1 to the negative n plus one. And then I'm basically out of space, so I will start a new page. And this was saying boundary work of a polytropic process is constant over negative n plus one times the quantity v2 to the negative n plus one minus v1 to the negative n plus one. Well, constant is kind of unhelpful here because we didn't actually relate that to anything else other than describing what a polytropic process is. And the description was pressure times volume to the n is constant. So let's get rid of that constant by plugging in pressure times volume to the n. That would be pressure times volume to the n divided by negative n plus one times v2 to the negative n plus one minus v1 to the negative n plus one. Okay, now which pressure and volume do we actually use here? Is it one? Is it two? The answer is doesn't matter so long as they're evaluated at the same point. So I could say p1 times v1 or I could say p2 times v2, but I could not say p1v2 or p2v1. And just to make my life a little bit easier, I can actually distribute that back inside the parentheses at which point I would have one over negative n plus one multiplied by pressure times volume to the n times v2 to the negative n plus one minus pressure times volume to the n times v1 to the negative n plus one. And remember, I can plug in either state point so long as I do it the same for both. So when I'm bringing it inside the parentheses and I'm multiplying it by v2, I'm going to plug in my v2, excuse me, I'm going to plug in my state two properties. So v2 times v2 to the n multiplied by v2 to the negative n plus one. And then I'm going to subtract p1 times v1 to the n times v1 to the negative n plus one. And remember, when I'm multiplying a quantity to an exponent times the same quantity to a different exponent, I can combine the exponents together as an addition or subtraction process, like x squared times x to the third is equal to x to the two plus three. So I could write this as one over negative n plus one times the quantity p2 times v2 to the power of n minus n plus one minus p1 times v1 to the power of n minus n plus one. And how does that simplify? You're right, n minus n is zero. One is one. So I have p2 times v2 to the first power minus p1 v1 to the first power. Therefore, the boundary work can be written as p2 v2 minus p1 v1 divided by negative n plus one. So I write it as negative n plus one because I think that that phrase rolls off the tongue real well. It makes it easy for me to recall that when I'm working through a problem and I need to pull out the boundary work of a polytropic process, as opposed to flipping back through my notes, I can just remember it's p2 v2 minus p1 v1 and then something, what is it? Oh yes, it's negative n plus one. Neat. It is not negative one plus n because that's not nearly as fun to say as negative n plus one is also not one minus n because then I don't remember if it's n minus one and one minus n. It is easiest to remember if you can just remember negative n plus one. If it becomes muscle memory, you don't even have to remember it. So I will write this down back under my simplifications for boundary work. I had a polytropic process and that problem and that polytropic process is one where the pressure times the volume to the n is constant and the simplification is boundary work is v2 v2 minus p1 v1 to the negative n divided by negative n plus one. Okay. And while we're here, I will also point out that in some cases, instead of plugging in total boundary work, it might be more convenient to plug in specific boundary work. And remember that specific boundary work is total boundary work divided by mass. And if I were to bring that inside of my equation, I could write that as pressure times dv where v is specific volume. So I could say the specific boundary work is equal to the integral of pressure with respect to specific volume from specific volume one to specific volume two. And then the simplifications would become for an isochoric process, the boundary work, the specific boundary work would still be zero. For an isobaric process, we would have pressure times little v2 minus little v1. For the isothermal process of an ideal gas, we would just have r times t times the natural log of v2 over v1 or p1 times specific volume multiplied by natural log of v2 over v1 or p2 v2 specific volume two, that is pressure two times specific volume two times the natural log of v2 over v1. And in the polytropic process case, it would be p2 times specific volume two minus p1 times specific volume one. And those are our common simplifications of boundary work. That doesn't mean that these are the only simplifications of boundary work, but these are the ones that we're going to encounter the most.