23. The Second Law of Thermodynamics

Entropy Equations for Special Processes

# Entropy & Ideal Gas Processes

Patrick Ford

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Hey everybody. So in our example here we have an ideal mon atomic gas that's going through this process that's shown in our PV diagram. So in other words, we're taking a two step process here from A to B and then from B to C. And ultimately we want to calculate the change in entropy for this two step process. So in other words we want to calculate here is delta S total. Alright, now, when we've used delta s before, in the past we usually have two objects that are exchanging heat. But here we actually have one gas issues going through two processes. So we're just going to use the delta S from A to B plus the delta S from B to C. That's the total change in entropy. It's just the change from both of the processes. So we have this list of equations here that represents all of the different equations for different processes and to basically calculate this delta as total, we just have to figure out which one of these equations apply to both of these processes. That's the whole thing here. Okay, so let's get started, we're gonna look at the first process, the one from A to B. So what type of process is it? Um can we use the ice a thermal equation? Is it a phase change? Well, if you look at this process here, it's not gonna be ice a thermal, remember anything on a PV diagram? Uh a nice a thermal to PV diagram is just this curve over here and this represents constant pressure, but a to be actually goes steeper than that. It doesn't remain on the same line. It's not ice a thermal it's also not a phase change. We don't have a gas that's changed into a liquid or something like that. Um We could use this temperature change here. But usually what happens is we're not given, we don't have mass and the specific heat in this problem, we're only just given moles. So it's probably not going to be this equation. However, this process is a idiomatic because we're told that this curve is an idea that so because this process here is a diabetic, then we have a special value for delta S. It equals zero for idiomatic processes. Remember these are sort of the special cases or the special processes where there is no heat transfer Q equals zero. So in other words, what happens is this whole entire term just goes away and it becomes zero. So in other words, your delta s total just becomes the change in entropy of the second process. So we're just gonna jump down to the second process. What type is it? Well, if you look through your diagram, um you should be pretty familiar with this because this is just a flat horizontal line. So in other words, this is an ice, a barrick process. So which equation do we use for delta S. Again, it's not ice, a thermal, it's not a phase change, it's not a temperature change, it's not an automatic and it's also not a free expansion. This gas isn't suddenly exploding or, you know, into a new volume or something like that. It's basically just going along until it just hits another ice a therm. So, really, it has to be this equation over here. The equation of use for ideal gasses. So, this is gonna be n times big C. Now, which kind of big C. Do you remember? There's two of them, their CV and CP, That just depends on what type it is. We use cP for ideal gasses or sorry for ice a barracks. So, we're gonna use CP times the L. N. Of the temperatures. So, I'm going to call this TC over Tv that's final over initial. Right, So, this ice a therm here is T. C. This one over here is TB. Okay, so, that's basically the equation that we're gonna use, we have the number of moles and the cp here, we can figure out from this table, because, remember this is a mono atomic gas. So, really the only thing we need to figure out what this problem is what T C and T B. Are, what are the two temperatures that were going between in this process here. So, let's figure that out. So, let's look at T. C. First. Okay, so, T C. Well, the only thing we know about T. C is the pressure. We know the pressure is this value over here. But remember that's not enough to actually solve for this temperature. So we're gonna have to use that? We're gonna have to use the fact that TC is actually the same as T. A. Remember both of these uh these things here, these points are along the same ice a therm. So T C. Is actually equal to ta and this is really important because this ta here we have a bunch of information about, we know what the pressure is and the volume. So we can use PV equals NRT to find it. So in other words, P A V A equals N R T A. So ta is going to equal P P A V A divided by N R. All right. And we actually have all that stuff. So the pressure at point a is 1.5 Times 10 to the 5th vol is 0.022. And then we divide by the number of moles, which is 0.8 times the R, which is 8.314. When you work that out, what you're gonna get is 496 kelvin. So that's 4 96. That's what we're going to plug in over here. So now we just do the same thing to figure out tv. So how do we figure out tv? Well, this point over here. Oh sorry, at this point over here we also have the pressure and the volume. So we're basically just gonna do the exact same thing. So we're going to use P B V B equals N R T B. So what happens here is that T B is equal to? We've got PB VB over and our So in other words we've got 0.5 Times 10 to the 5th And we've got the fire volume which is 0.042 and then divided by 0.8 And then 8.314. When you work this out, what you're gonna get for TB is you are going to get 316. Now this kind of makes some sense here, right? So this is 316. This is 496. As you go away from the origin, the temperature should be increasing. So that makes total sense. So that's the number that we're gonna plug in for over here. Okay, so now we just have to go ahead and figure out and plug in our last equation. So this is gonna be n which is 0.8. The cp that we're gonna use is five halves times 8.314 and now we have times the Ln of T. C. Is 4 96 divided by T B, which is 3 16. And you work this out when you're gonna get, is that a delta s total. The total change in entropy For this process here is going to be 7.5 jewels per Kelvin. Alright, so that's it for this one, guys, let me know if you have any questions.

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