Alright, folks. Welcome back. So in this problem, we have a monoatomic gas that's initially at some temperature, 293 Kelvin. And what happens here is that there are two processes that are going to happen to this gas. The first one is we're going to add some amount of heat at constant pressure until it reaches point b, and then we're going to remove the same amount of heat, 2,000 joules, at constant volume until we reach point c. So we're taking two different processes and basically what we want to 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 processes 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 2,000 joules of heat at constant pressure. So that's going to be an isobaric 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 isovolumetric process because we're removing heat at a constant volume. So it's just going to look like a flat or so it's going to look like a straight vertical line, except we're going to 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 going to call this TA = 293 K. But then I have two 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 point C? Now, what happens is, remember, according to our new equations that we've seen for isobaric and isovolumetric processes, when you add or remove heats, there's going to be some temperature changes for both of them. So there's delta T's. So what's going on here is that when you're adding heat from A to B, there's going to be a temperature change delta T from A to B. Then when you remove heat from B to C, you're going to have another temperature change, which is delta T from B to C. I'm just going to call those delta T's. 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 going to be my starting temperature, the 293, 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 293, I just have to add the 2 temperature changes from A to B and B to C, and then, basically, that's going to be my final answer. Right? So when you add those two things, that's going to be whatever the temperature is at C. 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 isobaric process. So which heat equation are we going to use? We're going to use q = nCPΔT. So this is going to be q = n CPΔT 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 2,000 joules of heat. So this is going to be plus 2,000 divided by n the number of moles, which is just 3, and then CP. Remember, CP just depends on what type of gas it is and what kind of process. We're dealing with a monoatomic gas here that's undergoing a constant pressure process. So we're going to use this value here. So we're going to use 3·52·8.314, 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 K. Alright. So that's the first number over here. So I'm going to 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 going to be an isovolumetric process. So now we're just going to use this equation, this q = nCVΔT from B to C. And we're going to do basically the same exact procedure. We just have to divide this ncv over to the other side and start plugging in some numbers. So what's the q here? The q is we're removing 2,000 joules of heat. Be very careful here because you're not adding 2,000 joules of heat. You are subtracting it. So this q here needs to be negative. This is minus 2,000 divided by the moles is still 3, and now we're going to use CV. So we're actually just going to use 32R instead of 52. So this is going to be 32·8.314. That's going to give me my delta T from B to C, and this is going to be negative 53.5 K. So I want to point out something real
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Heat Equations for Special Processes & Molar Specific Heats
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