Entropy Equations for Special Processes - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Entropy Equations for Special Thermodynamic Processes

Video duration:

8m

Play a video:

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.

2

Problem

Problem

3 moles of an ideal gas are in the left side of an hourglass-shaped container, separated by a thin barrier. The right side is completely empty, but the volume of the left and right sides are equal. The barrier is suddenly removed, and the gas freely expands into the vacuum. What is the change in entropy?

A

17.3 J/K

B

0 J/K

C

–17.3 J/K

D

49.9 J/K

3

example

Entropy & Calorimetry

Video duration:

6m

Play a video:

everybody. So in this problem here, we have a mixing of two different temperatures of water. So, we have 100 and 95 kg at 30 Celsius and five kg of boiling. And then the first part, we're gonna calculate the final temperature of the mixture of water. So, remember, this is gonna be like a classic kalorama tree problem. And in the second part, we're gonna calculate the total change in the entropy. So, let's go ahead and get started here before I started with an equation, I just want to go ahead and draw a diagram. So we've got this hot tub like this and I filled it with some massive water. So this is gonna be 1 95 and this temperature here, this tea is gonna be 30 Celsius. I'm gonna convert this to Calvin real quick, which is gonna be 303. Then what I'm gonna do is I'm gonna add Some hot water to it. This hot water here is gonna be five kg at initial temperature of boiling. Remember boiling point for water is just th equals 100. So, in other words, it's gonna be 100 which is going to be 373 kelvin. So, what I'm gonna do is here, I'm gonna call the hot water, mh and th the colder water is going to be C and T C. All right, so, you mix these two waters together and then ultimately what you end up with here, As you just end up with a total amount of water of 200 kg, right, The five plus the 195. And it said in a new equilibrium temperature, we've got these two waters that mix different temperatures, there's going to be an exchange of heat from the hot to the cold and they're going to end up with some equilibrium temperature. And that's actually the first thing that I want to calculate here in part a I want to calculate the equilibrium temperature. Now, we've done this in an earlier video on kalorama tree, but we've basically come up with an equation for the equilibrium temperature. It's this really long one that it's like M C. T. S, M C. T. S and M. C. S and M. C. S. You're basically just gonna go ahead and plug and chug this. This is kind of a tedious part of this problem. But basically this is just gonna be the MCC for water and T. C. We're gonna use C. For water for everything. Plus this is going to be M H. C. For water th and then you divide this by M. C. C. W. Plus M H C. W. So if you plug this in, where you're gonna get is 1 95 41 times the initial temperature. So this is going to be the 303. Alright, this equation you can use Celsius or kelvin. I'm just gonna use kelvin. So this is gonna be plus five times 41 Pull Times 3, 73. And then you're going to divide by uh 1 41 86 plus five and then 41 86. Alright, so it's kind of tedious but when you plug all that stuff and just make sure you enter it carefully in your calculator with parentheses and all that stuff. What you're gonna get here is 304.8 Kelvin. Now if you look at this number this should make perfect sense to you. Remember this colder water here is at 303. There's much less of the hotter water which is at 3 73. So when you mix them together the final temperature should be pretty close to the initial temperature of the colder water. Right? So we've got 304.8 and that makes total sense. Alright, so now let's move on to the second part here. What is the change in entropy? So in this problem here, what we have is we have multiple objects that are exchanging heats and therefore changing entropy is right. So the hotter water gives off some heat to the colder water. So in other words, there's a delta s total here, that's the second step, which is right, or delta, delta is total equation and it's gonna be made up of two things. The water doesn't exchange heat with the hot tub or the air or anything like that. The only thing that's sort of happening in this problem is that we just have the two waters that are exchanging. In other words, the delta S total is gonna be delta S. H plus delta S. C. So now we have to do here is to solve for this variable here, we're gonna have to figure out which equation we use for delta S. So remember that we've actually come up with a couple of different equations for different processes. So the first thing we should check one, the first equation should check is if you can even use Q over T. Is it a nice a thermal process? And the thing is, it's not remember, it's not a nicer thermal process because both of these waters here start off after their initial temperatures and then the final one is gonna be different. So this is not a constant temperature process. So is it a phase change? Well, we don't have water, it's changing to ice or gas or anything like that. It's not a phase change. What about the temperature change? And actually this is the equation we're going to use. So it's going to be M C L N A T final over T initial. That's that. And that's actually gonna be the case for both of these terms here. So the first one is going to be NH cw and then times Ln of t final, that's our equilibrium temperature over T initial, which is just the temperature of the hot water plus the M. C. C. W. Times the Ln of final temperature over the temperature of the cold water. So you just add those M. C. L. N. Terms together you plug everything in and when you add them together that should be your delta S total. So when you plug everything which you should get is you should get five times 86 times the Ln of T. Final. Which is the 304.8 divided by the 73 Plus. Or sorry? Yeah. Plus this is going to be 195 41 times Ln Of 3 04.8 divided by 303. And then when you work this out you should get here, is that this number here just becomes negative. 4222. And this number here works out to 4835. So when you add them together, your delta s total Should be jules per Kelvin. Alright, so this is basically just confirming the second law of thermodynamics. We had these two processes and basically one of them resulted in a decrease in entropy. But there was another part of the problem that increased more entropy. So the overall change here in entropy of the universe is positive and that's exactly what we should expect. Alright folks, so that's it for this one

4

example

Entropy & Ideal Gas Processes

Video duration:

6m

Play a video:

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.

Do you want more practice?

We have more practice problems on Entropy Equations for Special Processes