Hey, everybody. So in our example here, we have an ideal monoatomic gas that's going through this process that's shown in our PV diagram. In other words, we're taking a 2 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 2 step process. So, what we want to calculate here is Δs total. Alright? Now when we've used Δs before in the past, we usually have 2 objects that are exchanging heats. But here, we actually have 1 gas that's just going through 2 processes. So we're just gonna use the Δsab plus the Δsbc. 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 Δs 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? Can we use the isothermal equation? Is it a phase change? Well, if you look at this process here, it's not gonna be isothermal. Remember, anything on a PV diagram is just this curve over here, and this represents constant pressure. But a to b actually goes steeper than that. It doesn't remain on the same line. It's not isothermal. It's also not a phase change. We don't have a gas that's changing to a liquid or something like that. 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 gonna be this equation. However, this process is adiabatic, because we're told that this curve is an adiabatic. So because this process here is adiabatic, then we have a special value for Δs. It equals 0 for adiabatic processes. Remember these are sort of the special cases or the special processes where there is no heat transfer, q=0. So in other words, what happens is this whole entire term just goes away and it becomes 0. So in other words, your Δs 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, you should be pretty familiar with this because this is just a flat horizontal line. So in other words, this is an isobaric process. So which equation do we use for Δs? Again, it's not isothermal. It's not a phase change. It's not a temperature change. It's not adiabatic, 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 isotherm. So, really, it has to be this equation over here, the equation we use for ideal gases. So this is gonna be n*C_P *ln(TC/TB). That's final over initial. Right? So this isotherm here is TC. 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 monoatomic gas. So really, the only thing we need to figure out with this problem is what TC and TB are. What are the two temperatures that we're going between in this process here? So let's figure that out. So let's look at TC first. Okay? So TC. Well, the only thing we know about TC 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 what we're gonna have to use is we're gonna have to use the fact that TC is actually the same as TA. Remember both of these, these points are along the same isotherm. So TC 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=nRT to find it. So in other words, PAVA=nRTA. So TA is gonna equal PAVA divided by nR. Alright? And we actually have all that stuff. So the pressure at point a is 1.5 times 105. Volume 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 496. That's what we're gonna plug in over here. So now we just do the same thing to figure out TB. So how do we figure out TB? Well, this point over here Or sorry. 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 gonna use PBVB=nRTB. So what happens here is that TB is equal to, we've got PBVB over nR. So in other words, we've got 0.5 times 105 and we've got the final 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 gonna 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 or plug in our last equation. So this is gonna be n which is 0.8. The CP that we're gonna use is 5/2 times 8.314, and now we have times the ln of TC is 496 divided by TB, which is 316, and you work this out what you're gonna get is that a Δs total. The total change in entropy for this process here is gonna be 7.5 joules per Kelvin. Alright. So that's it for this one guys. Let me know if you have any questions.
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Entropy Equations for Special Processes
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