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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.

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