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Calculating Temperature Changes

Patrick Ford
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Alright folks, welcome back. So in this problem we have a mono atomic gas that's initially at some temperature to 93 Kelvin. And what happens here is that there's two processes that are going to happen to this gas. The first one is we're gonna add some amount of heat at constant pressure until it reaches point b. And then we're gonna remove the same amount of heat 2000 jewels at constant volume until we reach point c. So we're taking two different processes. And basically what we wanna do is we want to calculate the temperature when it reaches that final point over here. Now, the first thing we want to do actually is just draw the process is out on the PV diagram. In fact, that's what we're told to do. So let's go ahead and do that. So from A to B, we're going to add 2000 heat jewels of heat at constant pressure. So that's going to be an ice A barrick process. It's just going to look like a flat horizontal line. So it's going to look like this from A to B. We're going this way, and then from B to C, it's an ice. A volumetric process because we're removing heat at constant volume. So it's just gonna look like a flat or so it's gonna look like a straight vertical line. Except we're gonna go downwards like this because we're removing heat. Alright, so here's what's going on. I know the temperature at point a. I'm gonna call this T. A is equal to 293 Kelvin. But then I have to heat transfers and with heat transfers are going to come some temperature changes and I want to figure out ultimately, well, what's the temperature here? When you finally hit points see now what happens is remember according to our new equations that we've seen for Aisa barrick and ice, a volumetric processes. When you add or remove heats, there's going to be some temperature changes for both of them. So there's delta tease. So what's going on here? Is that when you're adding heat from A to B, there's gonna be a temperature change delta T. From A to B. Then when you remove heat from B to C, you're gonna have another temperature change, which is delta T from B to C. I'm just gonna call those delta T. Is, right, So here's what's going on. If I can sort of write an equation for this, my final temperature, this TC over here is gonna be my starting temperature. The 2 93 plus the two changes, it's going to be plus delta T. From A to B. Plus delta T. From B to C. So, really what happens is if I'm starting off at 2 93 I just have to add the two temperature changes from A to B and B. Two C. And then basically that's gonna be my final answer. Right? So when you add those two things, that's gonna be whatever the temperature is at sea. Okay, so that's kind of what's going on here, The rest is just figuring out the right equation and then just calculating a bunch of stuff. So let's start off with this delta T. From A to B. So we've got A to B. And remember this is an ice, so this is an ice, a barrack process. So which heat equation we're going to use, we're going to use Q equals N. C. P times delta T. So it's gonna be Q equals N. C. P times delta T. From A to B. In order to calculate this delta T. From A to B. You just have to move this stuff over to the other side and then just start plugging in some numbers here. So what's the Q. What's the heat transfer? Well we're going to add 2000 joules of heat, so this is going to be plus 2000 divided by and the number of moles which is just three. And then cp remember cp just depends on what type of gas it is and what kind of process we're dealing with a mono atomic gas here that's undergoing a constant pressure process. So we're going to use this value here. So we're gonna use three times five halves times 8.314. And that's going to give you your change in temperature When you work this out. What you should get here is you should get 32.1 Kelvin. Alright, so that's the first number over here. So I'm gonna write 32.1 now, we just have to do the same exact thing from B to C. So from B to C here, this is gonna be an ice a volumetric process. So now we're just gonna use this equation this Q equals and C V delta T from B to C. And we're gonna do the basically the same exact procedure, we just have to divide this N C. V over to the other side and start plugging in some numbers. So what's the queue here? The queue is we're removing 2000 joules of heat. So be very careful here because you're not adding 2000 joules of heat, you are subtracting it. So this queue here needs to be negative, this is minus 2000 divided by the moles is still three. And now we're going to use CV. So we're actually just gonna use three halves are instead of five halves. So this is gonna be three halves times 8.314. That's going to give me my delta T from B to C. And this is gonna be negative 53.5 kelvin. So what I want to point out something really quickly here, what you'll see is that for these two different processes we added or subtracted the same amount of heat but the temperature changes were different, it was 32.1 For the ice a barrick and negative 53. 3.5 for the ice of volumetric. And that's because remember this molder specific heat is a lower number for ice. A volumetric. So that means that for the same amount of heat transfer, there's actually more temperature change. All right, so basically I just have to plug in this, this is gonna be negative 53.5. And when you go ahead and work everything out to 93 plus 3 32.1 plus this number over here, what you're gonna end up with is to 71.6 kelvin, and that is your final answer. So basically, if I can sort of draw these sort of ice affirms here, you're gonna have these sort of icy Therms, it's kind of gonna look something like this, where this one here is the 2 93. This one here is going to be the um it's gonna be 2 plus 32.1. Um So that's just going to equal 325.1. And then you're gonna have um this is gonna be just to 71.6 over here. Right? So you end up sort of at a lower ice a therm than when you originally started. Alright, so that's it for this one guys, let me know if you have any questions