Calculating Works For Multiple Thermodynamic Processes

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everyone. So in the last couple videos we saw how to calculate the works done in thermodynamic processes, whether there were straight lines, like horizontal or slanted lines. Now, in some problems you're gonna have more than one thermodynamic process. So I'm gonna show you how to calculate the work's done when you have multiple processes that a system goes through. For example, let's say that a system goes not only just from from point A. To point B, but let's say it also goes from point B to point C. And you want to calculate the total work over the whole process. So let me show you how to do that. Remember that each process, like the one from A to B creates its own area under the curve. And so that's just an individual work, let's just say for simplicity this is 100 jewels. Now from B to C. That's another process that creates another area and therefore there is another work and has to be less than 100 because the area is smaller. So let's just call this 80. If I want to calculate the total work, it's just gonna be the sum of both of those things together. It's basically gonna be the total area under the curve. So it's just gonna be 100 plus 80 and that's just 180 jewels. Alright, so that's pretty straightforward. So in general what happens is that the total work done over multiple thermodynamic processes, it's just going to be the sum of all of the works done in each individual process. So as a general equation, this is going to be w total equals W one plus W two plus so on and so forth. However many individual works you have in that process. All right. And that problem. So in general what happens is that this was kind of simplistic because we just gave you the works uh and a lot of problems. You're not gonna be giving those works and you have to calculate them. So what happens is in your problems, there are four special thermodynamic processes that you're going to see over and over again. So what I want to do is sort of introduce you to those four. And what I want to do in this video is start building out a table and this this table here is gonna house all of the information and equations that you'll need will keep on building on top of this table and future videos. And eventually you'll have everything you need for these problems. So, let's go ahead and check this out here. The two I want to talk about in this video are Aisa barrick and isso volumetric. The other two will talk about later. Our ice a thermal and idiomatic. So it's a barrick. The word ice or the prefix isil just means constant and baric means pressure. You can think of a one bar as a unit of pressure, that's something you may remember, but also a barometer is something that measures pressure. So I so barrick just means constant pressure. All right. We actually saw these kinds of problems before this is whenever you had some constant pressure and you were changing from an initial to final volume. All right. It was a straight horizontal line and we saw that the work equation was P delta v. Next we're gonna talk about is a volumetric processes. As the name implies. This is when you have constant volume. You may also see this written as an ice a Coric process but I always just remember it as isil volumetric. This is basically kind of the opposite of this situation. Over here here you have constant volume but now you're gonna be changing from an initial to final pressure. So what this looks like on a PV diagram is it kind of just looks like a straight vertical line like this. So with an ice of volumetric that's gonna be a straight vertical line. So for this one you have constant pressure, changing volume for this one you have constant volume changing pressure. Alright, so now we know what these things look like. Let's go ahead and take a look at our example problem here. So in our example we have these ideal gasses some container here. And basically the idea is that we want to go from point A to point C. We're gonna take two different paths to get there A B C. And then a D C. In the second part. So let's just jump right into this first part here. We want to calculate the total amount of work done for the abc path, basically the top path. So this is a total amount of work done over a multi step process to the total work. It's just gonna be the work done for the two processes that make it up. We have one process from A to B, another one from B to C. Now this one here is gonna be a nice barbaric process because it's a straight horizontal line. This one's gonna be a nice a volumetric process because it's a straight vertical line. But the withholding work is just gonna be work done from A to B. Plus the work done from B to C. We just have to calculate each one of these things and then add them up at the end. So let's go ahead and do that. So the work done from A to B. How do we do that? We have to figure out what process it is first. So we take a look from A to B. We said it's an ice a barrick process. So we can actually just use this equation p delta V. We've seen that forces listed right there. Let's go ahead and use it. So this is p times delta V. The pressure for this process is 4000 and the change in the volume is just from 2-5. So this is gonna be three. Notice how we're going to the right here. So that work should be positive. So we don't have to insert a negative sign or anything like that. So this is 4000 times three. This is going to be 12,000 jewels. So now we just need to figure out what's the work done from B to C. So then we can just add it. Uh and then just figure out the total, right? So the work done from B to C. Here. So that's this ice, a volumetric process. So the question is, can we use this P delta v. Well, hopefully you guys realize that you cannot use this equation because the P is not going to remain constant. So you can't use this. And the only other way that we can calculate this area. So this work is by calculating the area that's under the curve. So let's try to do that. First. We have to draw the path of the process. We have to draw the shape that's underneath this curve. But if this is a straight vertical line, it's just a line like this. Then you can't draw a square or a rectangle or a triangle or anything like that. The only thing you can draw underneath this straight vertical line is another vertical line and the area of a vertical line is just zero. It's not a two dimensional shape, right? You can't draw a square or anything like that. So that's the big thing here. The area that's done, the area or the work that's done by a gas in a nice a volumetric process is just equal to zero because there is no change in volume. That's a super important thing that you need to know. So that means that this work is equal to zero, then that just means that the total work is just the 12,000 plus zero. And that just gives you 12,000 jewels for the whole entire process. So that's how to calculate and solve these kinds of problems here. And that's part of it. So let's go ahead and start part B. Now, you can actually just go ahead and pause the video and see if you can calculate this one yourself. Alright, so in this process here we're going from A to D and then D to see this is basically just the reverse of what we did in part a basically you're just switching the order of the processes. So the work done from a two D two C is gonna be the work done from A to D. And then the work done from D to C. So in this case this is the ice a volumetric in this case, this is the ice a barrack. Now, what that means here is remember isola volumetric processes do no work or have no work, that means delta v is equal to zero. So really what happens is that the the only work is due to the work done from D to C. This is an ice a barrick process that we can just use p times delta V. And so all you have to do is just look at the pressure, that the constant pressure of 2000 and then the change in the volume is the same, right? It's the same three. So this delta V is still equal to three and it's positive. And so if you calculate this, this is gonna be 6000 jewels. And so therefore this is the total amount of work done over the A. D. C. Process. All right. So that brings me to my very, very last point here, which is that we went from POINT A to POINT C. Our initial and final for the two paths were the same. We just took two different paths to get there. And what we saw was that the works done in each of these processes were not the same. So what happens here is that the work done between two states depends on the path that is taken. This is sometimes called path dependence. Alright guys, so that's it for this one. Let me know if you have any questions

2

Problem

Problem

How much work is done on a gas that expands from A to B along the path shown below?

A

9 × 10^{7} J

B

3.6 × 10^{7} J

C

3.3 × 10^{7} J

D

1.2 × 10^{7} J

3

Problem

Problem

A gas with an initial volume of 0.2 m^{3} is heated at constant volume, and the pressure increases from 2×10^{5} Pa to 5×10^{5}. Then, it compresses at constant pressure until it reaches a final volume of 0.12 m^{3}. Draw the two processes in the PV diagram below and find the total work done by the gas.

A

- 1.6 × 10^{4} J

B

4 × 10^{4} J

C

0 J

D

- 4 × 10^{4} J

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