23. The Second Law of Thermodynamics

Entropy Equations for Special Processes

# Entropy Equations for Special Thermodynamic Processes

Patrick Ford

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Hey guys, so by now we've seen how to calculate the change in entropy by using this equation here, Delta S equals Q over T. Now remember this only works when you have ice, a thermal processes where the temperature remains constant. Now, unfortunately you may run across some problems in which the temperature will not be constant. So we're actually gonna need some other entropy equations for special thermodynamic processes. And that's what I want to show you in this video. The bad news is that unfortunately some of your textbooks may or may not show the derivations for these equations. The good news is the derivations don't really matter. And I've looked through every possible situation you might run across and have basically summarized it in this table here. So I'm just gonna show you a bunch of equations for some special processes and we'll do some examples together. Let's check it out. So the first process remember is just nicer thermal. We actually have seen this equation is just delta as equals Q over. T. Nothing new there. The second type of process we saw is where a substance changes phase. Like for instance if you have water that's going to ice or vice versa. We saw a couple of these and basically you're just still using Q over T. Remember because the temperature remains constant for a phase change, you're just basically substituting this equation for ml that's the kalorama tree equation. Now, one situation that we haven't seen is the next one where a substance changes temperature instead of phase, for example, if you have water that goes from 0 to 100 degrees but doesn't actually change into steam or something like that. Now in this situation this process you can't use delta S equals Q over T. Because the temperature is not gonna remain constant. So we're gonna need a new equation for this. Now again, some of your textbooks may or may not show this, I'm just gonna give it to you this Delta S equals M. C. Times Ln of two. The final over T initial. This may look a little bit familiar to you remember the equation for heat transfer and changing temperatures from kalorama tree was the Q equals M. Cat equation basically all that's happening here is instead of a delta T. You just have an L. N. Of T final over T initials. So that's one way you might remember that. Let's actually just go ahead and look at our first example here, we have to calculate the change in entropy when we have some amount of water that warms from 20 to 80°C. So we have a substance that is now changing temperature. So we're gonna use our new entropy change equation here. Alright, so we have dealt S. Is equal to M. C. Times this is Ln of T final over T initial. So we're going from 20 degrees to 80 degrees. That's my T. Initial. And the final and I have the amount of mass here which is the 0.25. All I have to do is convert these temperatures to kelvin. So this is 20 plus 2 73 which is 2 93. And this is gonna be 80 plus 2 73 which is gonna be 3 53. All right. So, we have everything we need here. We have the mass, we have the temperatures and we also have C which is the specific heat that's 41 86. So I'm just gonna start plugging some numbers here. Delta X is equal to 0.25 times 41 86. And now we have we have 350. Actually, sorry, I have the L N. Which I have to do the L N. So this is gonna be the Ln of 300 53 divided by 2 93. What you end up getting here is 100 and 95 jewels per kelvin. Notice how we got a positive number and that makes sense. We have to add some heat to the water to warm it from 20 to 80. So therefore you've spread out a little bit more energy and the change in entropy is positive. All right. So that's pretty much all there is to it. Let's go ahead and take a look at some other processes here. The next one you might see is an 80 aerobatic prop. Sometimes the problem will tell you that it's a idiomatic and this one's actually really straightforward. Remember that an idiomatic processes, there is no heat transfer which means Q equals zero. Now, what that means is in your Q over T. Equation, there is no Q. So if there's no heat transfer that means that there is no change in entropy delta is zero. This is one of the rare cases where you're gonna have no change in entropy for the system or the universe or something like that. Alright, the next one is called a free expansion. Now this is really unfortunate because some textbooks may refer to this as an A. D. A batic free expansion which kind of sounds like an idiomatic but these are actually different processes because in these processes the Q. Is not equal to zero. So what happens is in a free expansion. The difference is that you have a gas that suddenly expands to a larger volume. The example I always like to think of is kind of like a balloon, right? It's like a balloon like this. And all of a sudden the gas of the balloon pops and all the air just suddenly rushes out to a larger volume. There is a change in entropy because there's more randomness now in the system and the equation for this is gonna be end our times Ln of V final over V initial. You can kind of remember this because in an idiomatic free expansion you have a sudden rapid change in volume and this equation here is going to involve the new volume that rushes out to divided by the initial volume that it was originally in. Alright, so that's one way you can kind of remember that now. The last one is called is basically when you have the gas that's changing at any constant volume or pressure. Basically this is an ice, a barrick or ice, a volumetric process. I'm just gonna give you these equations there, N C V times Ln of T final over tea, initial. So that's one of them. The other one is N C P times N Ln of T final over T initial. Alright, so basically the only difference here is that you have a C B or C P. Remember those are just values that we can read off of this table every year. It just depends on what type of gas you're working with. Now we're gonna run, we're gonna go ahead and work out the second example here, but I have one final point to make. This is a lot of equations. So I kind of come up with one way to help you remember them. If you look at these last sort of four out of the five equations, you'll notice there's a pattern. So the first letter is either an N or an M. It's either moles or mass is basically how much stuff you have times a constant? Like see that's the specific heat or R, which is remember the universal gas constant. So it's some constant here, C or R. Or C. V or C. P. And then the last thing is always an L. N. And in this case you have an L. N. Of some final minus or over initial. So T final over T. Initial or V final over the initials. So that might be some way you might remember those equations. Alright, so let's take a look at our second example. Now we have three moles of a mono atomic gas and it's cooled from 3 50 to 300 at constant volume. So what's the change in entropy? Well, we're gonna start off with delta S. That's change in entropy. Which type of process is this? Let's just go to sort of go down our list. Is it changing temperature? Well, yes, it actually is. So do we use this equation here? Well, what happens is this is when you have a substance that's changing temperature. But our values here are M and C. We have mass. And the specific heat. And this problem all we're told is that we have three moles of a gas. So that's just N not M. So we're probably not gonna use this equation now. This is not an idiomatic process because it's not a dramatically it's not told were not told that it's idiomatic. We're told that it's cool that constant volume. So it's not this one either. It's not a free expansion. This thing isn't just suddenly expanding out to a larger volume. In fact, we're told that it's actually a constant volume, so it's definitely not a free expansion. So it's actually just gonna be this last one here. It's either one of an ice, a barrick or an ice, a volumetric process. We actually know which one it is because it says volumetric constant volume. So we're actually just gonna use this equation over here. Alright, so we're gonna use N. C. V times Ln of T. Final over T. Initial. Alright, that's the equation you want to use. So we know the number of moles. This is just gonna be to the cv here, depends on what type of gas it is. We're told in this problem that it's a mono atomic gas which means the C. V. Value we're gonna use is three halves are so it's gonna be three halves times are which is 8.314. Now we're going have to multiply the L. N. And what's the initial and final temperature. Well it's cooled from 350 to 300. So what happens is this is actually my final is my 300 here. My 3 50 is my initial. Don't get those confused. So you're gonna do 300 over 3 50. And when you work this out, what you're gonna get here is negative 5.8 jewels per kelvin. So in this case the entropy decreased and that makes sense because in order to cool this gas from 3 50 to 300 you have to extract or remove some heat and there's a decrease in entropy. Alright, so that's it for this one. Guys, let me know if you have any questions.

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