Cyclic Thermodynamic Processes - Video Tutorials & Practice Problems

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concept

Properties of Cyclic Thermodynamic Processes

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Hey guys, So we've solved lots of problems with PV diagrams, but one of the most important ones and common ones that you'll see is one where you start off with some position in the PV diagram, you'll do a sequence of steps and then you'll end right back where you started. And it basically just goes in a loop like this. So these are called cyclic thermodynamic processes and there's a couple of really important properties that you need to know about them. So in this video we're gonna dive right in, we're gonna solve this problem together and let's go ahead and check it out. So as a cyclic process is one where a system or a gas completes a sequence of steps, but it returns to its initial state. So that's what a cycle is. It returns back to the initial states. So there's two important properties here and rather than tell you want to show you. So let's just go ahead and take a look at our problem here. So we have this ideal gas and it goes through this process. We want to calculate the work that's done by the gas over the cycle. So the total amount of work done. So how do we do that? Well, let's see in part a if we want to calculate the total amount of work done, remember this is just a bunch of processes, sort of strung together. The way we calculate the network or the total work is you just have to add up all the work's done by all these individual processes, right, We've done that before. So basically if I have from A to B and B, two C and so on and so forth. The total work will just be the work done across all of these different steps. So the work done from A to B, B to C, C to D. And the work done from D. Back to A. Alright, so it might seem like there's a lot of terms we're gonna have to go calculate a lot of stuff, but let's just go ahead and refer to our table here just to make some of these things really easy. So, if you think about this, this sort of square here, a rectangle is really made up of two different types of steps. Aisa barrick and ice a volumetric. And what's easy about that is that we have the work equations for both of these. So it turns out what happens is remember in ice of violin metric processes, the work done is equal to zero. So that means that from A to B, there is no work done because there's no change in volume. Right? And then from C to D. This is also a vertical line. So there's no work done here. So, really, instead of four times, we really only have two. And these two are Sabbaruddin processes, which means we'll be able to use P delta v. Alright, so that just means that we're going to take this work equation and it simplifies to the work done from BBC, that's gonna be the pressure at B. It's the pressure here. So this is PB times the change in the volume. This is gonna be delta V. From B to C. Plus the work the pressure down the p. The pressure done here at D. Sorry, that's P. D. Here. And then times the change in the volume from D. Back to A. All right, so let's take a look at this. Basically, I'm just gonna start plugging in some numbers here at PBS 35. What's the change in the volume from B to C. We're going from 28 to 40 here. So what that means here is that this delta V. Is equal to 12 From here to here. Alright, so 28, you know the difference between 28 and 40? So this is gonna be 12. And then what about actually let's go ahead and plug it in, you know, 40 -28 just to do final minus initial. So then what about this PD here, this is gonna be 15 And then the change in the volume here. Now from D. back to a. Remember that this arrow here goes to the left, so my final minus initial is actually 28 -40. So it's gonna be 28 -40. Alright, so if you go ahead and work this out here, what you're gonna end up getting, is that the total amount of work done over this whole entire cycle is jewels. All right, So that's the answer. Let's move on to part B. Now, in part B. We want to calculate the area that is enclosed within the loop. The area that's enclosed is going to be this area that's sort of enclosed inside of this rectangle. So all I have to do is just find the area of this rectangle. How do I do that? Well, the area over here is just gonna be base times height. Right? That's the area of a rectangle. This is the area here. So base times height. What is my base? Well, my base here is 12 because that's the distance. B 2028 forties. This is 12. What about the heights? The height here is 15 minus 30 35 minus 15, which is 20. So the base here times the height is 20, sorry, 12 Times 20. And this is equal to 240 jewels as well. So, what happens here is you'll notice that these two numbers are the same. We calculated the work done. It's 2 40. The area is to 40 as well. And so that leads me to the first important property. You need to know which is that the total work done in a cyclic process is the area enclosed inside of the loop. The area enclosed inside here represents the total work done. And the reason for this is because if you think about what's happening for B. Two C, we have all of this area that's underneath the curve. But then from C to D. Right, well, sorry, from D. Back to a we're basically have this area underneath the curve here and this area here sort of just cancels itself out. Right. We have positive work done that, we have negative one work done. So this sort of effectively just cancels out and then what you end up with is just everything that's inside of the square or the rectangle or the loop in general, it actually works for any type of loop. So that's the sort of first process. The first property you need to know now what happens is in this process, this cycle here, we've gone in a clockwise loop, so that the top the work done at the top from BBC was positive and the work done at the bottom here was negative. So what that means here is that if the loop runs clockwise, the work done is positive. If we had run this loop in reverse, then what happens is that this work would have been negative and this one would have been positive. So your work done in the cycle would have been negative. So these are just some important properties you need to know if it's clockwise, it's going to be positive. If it's counterclockwise it's going to be negative. Alright. So now let's move on to part C. Now. So in parts that we want to calculate basically, we're giving all these heats from A to B, B to C, C to D and D. Today, we want to calculate the change in the internal energy for the cycle. So that just means we're going to calculate delta E. But now for the cycle, so how do we do that? Well, the only, the only equation we know for delta E is that the delta internal equals q minus W. But that was only for a single process. Well, it turns out what happens is instead of using delta equals q minus W. For each individual process, you're really over the cycle. Just gonna be adding a bunch of these, you know, q minus W. And then q minus W and q minus w. You're gonna be adding up all the delta ease for each one of these processes here. So you can actually sort of simplify this and you can actually just use delta E. For the cycle here is equal to Q. For the cycle minus the W. With the work done over the cycle, so you can do across multiple processes or cycles. It doesn't matter. Right? And so basically we can just use that delta E is equal to the total amount of heat transfer for the cycle, minus the total amount of work done over the cycle. And the reason this is helpful is because in a lot of problems you have to calculate the work done over the cycle first and then you'll have to use it in this equation here. So we already have with this delta, so that this w for the cycle is because we calculated that in part A. So in order to calculate the delta E. I just need to figure out what's the total amount of heat transfer over the cycle. Well, I'm actually told where all the heat transfers are for each of the different processes, the different legs or the steps here. So what I can do is just add up all of these things together here that will represent the total amount of heat transfer. So really this is just going to be 800 plus -600 -300 here, and then this is going to be minus. And then this is here. The work done is gonna be 240. Right? So this is this and this is this whole entire piece over here. When you go ahead and work this out, What you're gonna get is that the change in the internal energy over the cycle is equal to zero. When you plug in all these numbers you're going to get zero. And that's the second important property about cyclic processes that you need to understand because the internal energy depends only on basically where you are in the PV diagram and not how you got there and doesn't depend on the path that you take, the delta E equals zero over a cycle. There's no change in internal energy. As long as you start and end in the same place, it doesn't matter the path that you take, as long as you start and end in the same place, there's no change in internal energy. So what that means here is that there is no change in the internal energy. Then the total amount of heat transferred over the cycle is equal to the total amount of work done over the cycle. And that really just comes from this equation over here. If this is equal to zero, then these two things must equal each other. So that's another sort of important thing you need to know. Alright guys, that's it for this one. Let me know if you have any questions.

2

example

Signs in a Three-Step Cyclic Process

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5m

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Hey guys, So let's take a look at this problem here. This is a kind of conceptual question. There's no numbers involved here. We can sort of do all of this kind of just with logic. But basically we have is we have this process that runs a cycle in a PV diagram. So it goes in a loop and it starts back where it and it ends where it started from. We want to we want to complete this table, not by calculating a bunch of numbers, but just by inserting signs, whether it's positive, negative or zero for each one of these terms here, Q. W. And delta. Internal Alright, so let's get started with the first process from A to B. Now this A to B. Process is an ice, A barrick process. Now, if we're trying to relate these variables like heat work and internal energy, we're always going to start off with the first law of thermodynamics. Right? So what happens here is that we have delta E. Internal change in internal energy of the system equals heat added to the system minus the work done by the system. Now we know that this change in internal energy is a positive number one. We can kind of make sense of this. Is that if you look sort of like drawing a bunch of ice with terms you're sort of getting farther away from the origin. So the temperature should increase and the internal energy increases. Right? That's one way you can kind of make sense of that. So what about these other two variables Q and w. Well, remember that for whenever you are moving in a PV diagram to the right then the area that is underneath the curve here is going to be positive. So the area is positive for underneath the curve anytime you're moving to the right on a PV diagram, that's something that we've said before in a previous video. So therefore if this is gonna be positive then what about this heat? Well, you can kind of rearrange this equation and figure this out. If you rearrange this, you're gonna get the delta internal of the system. Plus the work done by the system equals the heat added to the system. Right? So this is Q two. So, if this is a positive number and this is a positive number, then no matter what the heat on the other side of the equation also has to be a positive number. So, basically these are both positives in our table. All right. So, it's kind of just using logic, you know, logical reasoning to figure out those numbers or those signs here. Let's move on to the second piece of this cyclic process. So, from B to C. Now, from B to C is an ice. A volumetric process going straight up on the PV diagram here. Okay, we're told is that the heat transfer is going to be positive. What about the work here? Well, remember that the key characteristic of an ice. A volumetric process, is that there is no work done. It's just zero. So, this is going to be zero over here. All right. So, what about the delta E. Internal? Remember you're just gonna write out your first law of thermodynamics? Delta internal of equals Q two minus W. By now, what we do is we just eliminated this variable here. There is no work done by the gas. So what happens here is that this is a positive number, then? The delta internal also has to be positive. Right? This also has to be positive one. We can think about this. Is that again? These are the ice affirms you're still getting farther away from the origin. So you're increasing an internal energy. Okay, so what about C. Back to a that's our last process. Now. This is not an ice, a barrick or an isil volumetric. So, we can't really use any of these equations that we have here. This wasn't one of our special thermodynamic processes. So, let's see how do we figure this out? Well, so, we have delta E. Internal of equals Q two minus W by All right. The problem is that we don't really know what any of these variables are. We're missing one key sort of characteristic about what type of thermodynamic process? This is this is a cycle. So, remember that the key characteristic of a cycle here is that delta E. For the cycle is equal to zero. So, what that means here is that delta E. From A to B plus delta E from B to C plus delta E from C to A should equal zero. You should have no change in internal energy. Now, what we just found here was that these two processes, the first two had positive changes in internal energy. So in other words this is positive and this is also positive. So what happens is in order for the total change in internal energy to zero over the entire cycle, then that means that this number here has to be negative, right? If this is two and three, this has to be negative. Five. So that it all cancels out to zero. It's just making up some numbers here. Right? So that's how you figure out this sign here. This has to be negative because of the properties of a cyclic process. Okay, now, what about the work and the heats? Well, again, we're gonna use the same rule that we did over here whenever you're moving to the left overall to the left on a PV diagram, the area. So the area that is underneath the graph like this is going to be negative. So this work done by from C back to a is gonna equal negative, right? So that has to be negative. And then finally, what about the Q? What what happens here is if this is a negative number and this is a negative number then when you rearrange this, what you're going to get here is the delta E. Internal of plus W. By equals Q two. So if this is negative And this is negative, then Q2 also has to be negative. So all these are negative here and on the bottom row. Alright, so that's kind of a conceptual example. You're using a lot of properties of cyclic processes and the first law of thermodynamics to sulfur them. So let me know guys, if you guys have any questions, that's it for this one.

3

Problem

Problem

An ideal gas is taken through the four processes shown below. The changes in internal energy for three of these processes are as follows: ΔE_{AB} = +82 J; ΔE_{BC} = +15 J; ΔE_{DA} = –56 J. Find the change in internal energy for the process from C to D.

A

- 153 J

B

41 J

C

- 41 J

D

Cannot determine

4

example

Heat Added Over Complete Cycle

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3m

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Alright guys, so let's take a look at our example problems. So we have this gas that's undergoing this cyclic thermodynamic process in our PV diagram. And we want to do is calculate the total amount of heat that's added over a complete cycle. So basically we just want to figure out well what is Q total? Now? There's a couple of ways you might think about this. You might think well the total amount of heat is just gonna be if I add together the heat from all of the processes. So, for example, this is gonna be cue from A to B plus que from B to C plus que from C back to a. Now you might be thinking this because well, let's see, B two C is an ice. A barrack process, that's S O. P. And so therefore we have an equation for that. Right? So we could probably figure that out. This is the C2 a process is going to be a nicer volumetric and we also have an equation for that. The trouble is we don't really have an equation when it comes to this A to B process. This isn't one of our special processes. So we don't have an equation to calculate just the heat for that. Alright, so this actually is not gonna work because there's no way to to figure out what QA B is. So this actually isn't going to work. And instead we're going to have to think about another property of cyclic processes. That's going to help us. So remember that for cyclic process, the change in the internal energy is always equal to zero. If you start and end in the same place on a PV diagram, there's no change in internal energy. So remember what that means. Is that the Q. Over the cycle is equal to W over the cycle, right? By the first law of thermodynamics. So really this Q over the cycle here is actually just gonna be equal to the total amount of work that's done in this in this cyclic process here. Remember that is actually pretty easy to calculate because the work done over the cycle is actually just the area that is inside of the loop that the sort of cycle runs around. Right? So really this is equal to the work done over the cycle and it's also equal to the heat that's added over the cycle. Those mean the same exact thing. So really this is just gonna be the area that's inside of the loop. And we have a couple of different ways. We can figure this out right? Since we have all the values for pressure and volume. We can kind of just use the area of a triangle like this, right? It's just a triangle. So we can use the area which is 1/2 base Times height. So let's just say that this is the heights, let's just say that this is the base like this. So really what happens is that Q. For the cycle is going to equal one half. Now the base here goes from four back to one. So the base here is going to be three and then the height of this thing is gonna be the difference between 40 and 10. So this base here is three and the height here is 30. So I'm just gonna do one half of three times 30. And what I'm gonna get here is 45 jewels. All right. Now we're missing one thing here. We're missing the rules for whether the work is positive or negative, remember that? This work cycle here. If it's clockwise, the work done over, the cycle is going to be Positive. If it's counterclockwise, the work done by the cycle is going to be negative. What do we have here? We have a loop that's running counterclockwise, it's this one here. So therefore our answer has to have a negative sign. It's going to be negative. 45 jewels for the work. And also the heat added over the cycle. So basically what this means here, is that more heat is removed from this cycle here than it is added over. And so the overall heat is going to be negative 45. All right, So that's this one. Let me know if you have any questions

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