Lecture 13. The Carnot Cycle.
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Transcript: English(auto-generated)
00:07
So before we get started, can I ask a question? How many people took the chemistry GRE?
00:22
Gee, and Mark raises his hand. Nobody took the chemistry GRE? How many people are going to graduate school? You guys didn't take the chemistry GRE?
00:41
Just the general GRE? Just, I'm just, okay.
01:02
So most of you know by now the midterm scores are posted and the key is posted. You can pick up your exam in your discussion section. Steven and Gee and Mark are going to have those for you. Next time I'll put the cover sheet on the exam properly and you'll get it back electronically. It takes longer actually.
01:21
It's not the greatest system right now, but that's the way we'll do it. If you have questions about the grading of your exam, don't be afraid to talk to Steven and Gee and Mark. Those are the two guys who did the grading and so ask them first. If you have a question or a conflict
01:40
about the way your exam was graded, then come and see me. We'll work it out. Quiz 4 is Friday and in a moment I'm going to show you that these quizzes are very important to us. I'm going to show you that in just a second. But let me just tell you I looked very carefully
02:02
about what's going to be on the quiz and what's going to be on it is the stuff at the beginning of the chapter and chapter 15, sort of the first two-thirds of the chapter which is all the way through page 496. If I look at the exercises, the stuff that you need to be able to do is sort of in this range of exercises right here,
02:21
15.20 is already stuff for next week. So these first 19 problems in the exercises any one of those would be a great quiz question. Okay? Now, this is what the histogram looked
02:44
like for Midterm 1. I'm not any happier with it than you are.
03:05
So if you did better than a 49 you actually did pretty well. I mean in this class the mean should be a B or of course that's not the case when you have a straight scale. These are A's, these are B's
03:23
and these are lower grades over here. Okay? So I don't know why it didn't go better than this. One concern I have is that you get in the groove
03:40
of taking the quizzes with the multiple guests and they're pretty straightforward problems. They're sort of plug and chug type problems on the quizzes. I mean let's face it because they're pretty short, these quizzes and then when you get to the exam and it's harder problems that have to be worked out, you're just not used to that?
04:02
Maybe? I don't know. Let me just say if you think I think this material is easy, you're wrong. I mean I'm not even a physical chemist by training. I'm an analytical chemist. I don't even know why I got assigned to teach this class. I mean I have enormous difficulty with this material.
04:24
If you think all the symbols look the same to you, they look the same to me too. U, S, Q, W. I mean I get them wrapped around my neck constantly. When I took that trip last week to Kansas, I was sitting
04:41
on the airplane with my calculator working out problems. When I got off the airplane I had about 10 pages of problems. I'm trying to remember how to do this stuff. So you guys realistically, you know if I've got 10 pages and I'm teaching the class I've had this material before
05:00
mind you, I mean it's just been a while. All right you guys should have 100 pages of notes, 100 pages of problems worked out in your notebooks at a minimum, wouldn't you agree? So I'm just saying that I don't know that you don't have 100 pages.
05:22
But if you don't, all right mid-term 2 is 3 weeks from Friday and it's going to be just like mid-term 1 in terms of length and difficulty. I mean I don't think mid-term 1 should have been too difficult.
05:45
All right maybe mid-term 2 will be a little bit less difficult but it's going to be pretty close to mid-term 1 and so I'm just telling you in these 3 weeks you need to really buckle down. If you haven't done this 100 pages of problems, I'm really suggesting that you focus
06:02
on doing the hard problems, working them out. I mean I think you realize that open book doesn't really bind you very much on this type of exam. I mean if you don't know, if you're not really familiar with the book because you haven't been paging back and through it, working problems, I mean if you own the book
06:20
because you've been doing all these problems then having the book there is pretty helpful. All right but if you're opening the book and you're like gosh, I'm trying to find a problem like this one that's been worked out here in the book, you just don't have time for that on this kind of test. I mean time is really an issue. Okay so I know this material is hard.
06:43
It's really hard for me too and the only way around that and we do need to really learn this stuff, the only way to learn it is to do a lot of problems. That's the only way to learn this stuff. I try to make these good lectures but these lectures are not going to teach you this material.
07:01
You've got to work the problems. Now fortunately the quizzes are saving you to a large extent. This is how you're doing. I calculated this yesterday morning. No quizzes were dropped in the generation of this score.
07:22
I didn't drop any quizzes. So some of you missed quizzes. Some of you did poorly on one or more quizzes. Everybody, virtually everybody would move to the right on this diagram if we dropped two quizzes or even one. Okay so actually this histogram is a worst case scenario.
07:43
It would be further to the right if we had dropped quizzes. Okay so what this means to us is that the quizzes are in fact important. They are moving us from a mean in this range over here to a mean in the low B range and hopefully by the end
08:00
of the quarter this mean is going to move further to the right after we drop these quizzes and we do better on mid-term two. Right that's the goal. Okay so that is actually a low B and we're hoping that this moves over to the mid B range over here somewhere
08:22
and I think that's a realistic objective for us. Alright but it's not going to happen on its own. I mean mid-term two could be just as bad as mid-term one. I hope not. Alright help me out here.
08:40
Okay so the way I calculated this in case you're wondering is I just added your three quizzes together. Quizzes are out of five points. I multiplied by eight because each quiz is in reality worth 40 points. So that would be 120 plus 100 is 220 right? That's where I got this so this is scaled from zero to one.
09:01
So that would be 100 percent of the course points. If you got fives in all the quizzes and 100 on the mid-term the mid-term was actually worth 110 with the extra credit. Why? You'd be over here somewhere. Okay in case you're wondering how I calculated it.
09:20
So if you want to you can drop your own quiz, recalculate your how am I doing score and see where you compare on this diagram. Alright I understand most folks are going to move to the right a little bit. Okay so we have this guy Sadie Carnot he's interested
09:45
in steam engines and at the time that he's doing his research the steam engines are about five percent efficient. The Carnot limit for a steam engine turns out you can run a steam engine where the steam is at about 300 degrees C and the cold reservoir it's not room
10:08
temperature it's actually the boiling point of water. That's 100 degrees C. Okay and so in principle you could operate this thing at sort of 65 percent efficiency.
10:21
That's the Carnot limit for a steam engine where the steam is at 300 degrees C and the cold reservoir is the boiling point of water. Alright so that really is quite a bit lower than the Carnot limit. So he was asking two questions. Could he do better and would changing the fluid from water
10:45
to something else could that help? We haven't really addressed that question yet. Okay so the question is what limits does nature place on the amount of work that we can derive
11:00
from this temperature gradient right here? Right nature doesn't allow us to get an arbitrary amount of work out of this temperature gradient. There are well defined limits that's what we're talking about. That's why this is sort of an important subject. So Carnot derived this thing called the Carnot cycle.
11:20
Turns out he didn't really talk about entropy in his writings. He never used the word entropy and it actually took a second guy to sort through everything that he had done to reformulate it in a way that's useful to us in terms of what we now call the second law of thermodynamics.
11:43
All right but he derived this thing called the Carnot cycle which is the most efficient existing cycle capable of converting any given amount of thermal energy into work, he did that and there's another way to look at this if you inject a certain amount of work into the process
12:01
of transferring heat from a cold reservoir to a hotter reservoir why the temperature difference that you're able to generate using a certain amount of work, that's also it's the same thing. All right and so in his cycle we start here on this isotherm
12:25
and we do an isothermal expansion and then an adiabatic expansion and isothermal compression and then an adiabatic compression and the reason we care about this, this looks like a completely artificial thing to be focusing so much attention on.
12:42
All right I mean when are we actually in real life going to want to go from here to here or from here to here or from here to here? All right why are we going to care about this cycle? Well the answer turns out to be that any arbitrary cycle like this thing shown in red here, any arbitrary cycle
13:02
that you care about can be decomposed into a large number of Carnot cycles. All right this is a diagram from your book. Here's a Carnot cycle right here. All right and here's another one right next to it right here. All right I think if you look at these two Carnot cycles you can see that this one is going this direction,
13:20
this one is going this direction and so the two Carnot cycles that share that edge, what's going on at that edge is just going to cancel and this edge cancels between these two Carnot cycles and this edge cancels between these two Carnot cycles because what that edge represents is the two cycles going in along the same path in opposite directions.
13:45
Right and what you're left with after you cancel all of these paths here is you're left with the path on the outside of this thing that doesn't cancel. Right because there's nothing, there's no Carnot cycle next to it to cancel it out and so I think you can see
14:02
that here there's maybe 30 Carnot cycles and if you made this 3 million you could map out this red circle pretty accurately with little tiny Carnot cycles. Okay so the point is if we understand how one of these cycles works we can generalize what we learned
14:22
from one cycle to this infinite number of cycles or a large number of cycles. Okay so any cyclical process like this is constrained by these Carnot considerations, these Carnot efficiencies that we're talking about. This applies to this, that's why we spend
14:45
so much time studying it. Right we constantly bump into this Carnot limit in chemistry. Okay so this is what we said, blah, blah, blah, blah, blah.
15:00
How efficient is the heat engine? The efficiency is the work that we get out of it divided by the heat that's absorbed from the hot reservoir, that's how the efficiency is defined. Okay and so in this case it would be 5 over 20, we said this on Monday and we can derive then an expression
15:21
for this efficiency which is 1 minus the temperature of the cold reservoir divided by the temperature of the hot reservoir. Where did we get that equation? Can we derive it? So this is just going back to the stuff that we did on Monday, we can prove that this is right and the way that we do
15:42
that is we write out the work for each of the cycles, cycle should not, I don't mean cycle, I mean path, step. Step 1 is the isothermal expansion and step 2 is the adiabatic expansion, step 3 is the isothermal contraction,
16:02
and step 4 is the adiabatic contraction. Okay and we have equations for each one of these terms and it should be obvious to us that these two terms right here are going to cancel because my goodness they're identical except for the ordering of these limits.
16:21
TC is here, TC is here, and so if I reverse those limits I'm just going to put a minus sign in front of one of these integrals and they're going to cancel. So I'm only left with these two guys right here and these volume ratios here are related to one another and that's not obvious.
16:43
All right but the way that you can tell that they are is this equation right here which is on page 471 of your book. This describes the relationship between the volume at one point on an adiabat, V1 and T1, volume and temperature, and the volume and temperature, this is supposed to be 4,
17:11
that's supposed to be 4. So I'm talking about this .1 and this .4 so that's V4, that should be T4, V1, T1, those two points apply to this.
17:22
Okay so these, because these two points are both on the same adiabat this ratio has to be satisfied where gamma is CP over CV. Okay and the same thing is true here, did I miss this? Oh gosh, oh sorry, no this is, actually what I,
17:49
this is actually the way I wrote this is fine. Right this is T2, that's this temperature isotherm right here, this is T1, that's this isotherm right here and these are the two volumes right here. So this actually this is okay, sorry.
18:03
So is this, okay so if I write these two equations and rearrange them I get this equality right here and this equality right here, one from each one of those equations.
18:21
Then if I divide this guy by this guy the T1s go away, the T2s go away and I'm left with these volume ratios and so I can simplify this into this and once I've done
18:41
that it's obvious that these two logarithmic terms are going to cancel one another, they're the same. Okay so rather than canceling I can add them
19:01
and so I can express them both in terms of V2 and V1 or I could express them both in terms of V3 or V4 if I wanted to and so this is the expression that I get for the work. Integrated over the whole cycle now, this is the hot reservoir temperature, the cold reservoir temperature in these two volumes.
19:23
Okay that's the work for the whole cycle, the heat that's transferred in the hot part of the cycle is given by this equation right here, this is just the expression for the isothermal expansion that's occurring in the first step of the cycle. Okay and so the efficiency is just that whole thing divided
19:41
by that whole thing according to this equation right here, I just take the total work divide by the heat that's transferred in the hot part of the cycle and I should get the efficiency. And so when I do that look NR is going to cancel when I look at this ratio right?
20:00
Log V2 over V1 that's going to cancel so I'm going to be left with TH minus T over TH and that simplifies right away to that so we can prove this, we know enough to prove this, remembering these equations is the hard part but you don't have to remember them,
20:23
you can look them up. Okay now what's the entropy change for each of the four steps of the reversible Carnot cycle? We need a thermodynamic expression for S, let's say it's this, we'll come back to this later.
20:44
In other words the change in the entropy is the change in the heat provided the process is reversible divided by the T that the process is occurring at. Now this reversibility issue turns out to be important, that's the reason this is an equal sign,
21:00
turns out we'll come to that later. Okay and so if I want to know the delta S for this whole cycle I can just add up the delta S for these steps. The reason I can do that is because delta S is a state function okay and so if I want to know what it is for the whole cycle I can just add
21:21
up the delta S's for these individual steps and I will get delta S for the whole cycle. All right steps two and three, steps two and four rather, sorry, are adiabatic right?
21:41
Step three is reversible, isothermal compression. So Q is zero for those two steps, that's why that's zero. The entropy is zero for those adiabatic steps because Q is zero. All right for the isothermal steps, which are the first one
22:05
and the third one, all right these are the entropies, right Q is not zero otherwise they'd be adiabatic. And so what can we say about steps one and step three? Well we can look up Q all right for step one.
22:24
All right it's going to be equal to according to equations that we talked about on Monday and last Wednesday, a week ago, all right this is the equation that applies to an isothermal change in volume.
22:45
Hopefully that looks familiar, that was one that you needed for the midterm. All right what's not totally obvious to me is that Q is equal to minus W all right how do we know that this is true, well we know that DU equals DQ plus DW right?
23:04
We know U doesn't change during any isothermal process. Right? U doesn't change, in other words the change in U is equal to the constant volume heat capacity, in fact the constant volume heat capacity is defined as the change in U divided by change in T.
23:24
That's how the heat capacity is defined and so the change in U is just the heat capacity times delta T and it's an isothermal process, delta T is zero. All right so we know U is not changing that means Q has got to equal W rather DQ has got
23:44
to equal minus DW right if U is not changing so that's why that has to be true. Don't take anything for granted when you're looking at this stuff, I mean try to prove it to yourself that stuff like this as simple as that seems is true.
24:04
I'm constantly having to do that myself. Okay so what do we know, well we know for step one V2 is bigger than V1, it's an expansion. Right? We're going down that to bigger V, we know W is minus PDV that's true always all right
24:24
so we know that's going to be negative because P2 is bigger than V1 and there's a minus sign in front of this thing so delta V is positive but there's a minus sign. All right that means Q has got to be positive. If work is negative Q has got to be positive in fact it's got to be the opposite of work and remember we're talking
24:42
about QH, this is the Q at high temperature because we're talking about step one along the top of that isotherm okay and then if you look at this guy, this guy is exactly like this guy except he's happening now at TC isn't he and it's a compression not an expansion
25:01
and so okay so we've got this equation right here, we've got this equation right here, we're trying to conclude something about this delta S for the whole cycle, the sum of all the four steps
25:21
and what we have to remember is that we showed back like ten slides ago that V2 over V1 equals V3 over V4 and if you make that substitution these two terms are going to be equal and opposite because check this out, TC is in the numerator, it's going to cancel with TC in the denominator, same thing with TH, that cancels with that
25:43
and so these terms if that equals that these two terms are going to be equal to one another. Right? Well that wasn't obvious to me that that had to be true so that's why I went through and added these slides. So that means that this, I mean that's why I had to prove this, this really is a square box
26:03
in other words the entropy is not changing here or here and the temperature obviously is not changing here and here and so if you make a temperature entropy diagram out of a Carnot cycle it's a square box.
26:24
Now everything that we've said applies to only reversible processes. We've been talking about reversible processes so far and real life processes are virtually never reversible because to be reversible it's got to be infinitely slow.
26:43
So what if one or more of these steps of the process are irreversible? The efficiency of the process is not, in other words if the process is irreversible it's going to be less efficient than a reversible process or at best equal in efficiency. Limited, the efficiency
27:01
of the irreversible process is bounded by, if you will, the efficiency of the reversible process. Okay, so what does that mean? Well it means that if I write the sufficiency in terms of heats here's the efficiency of the irreversible process here, here's the efficiency of the reversible process here. If it's reversible now I'm going to write rev on it
27:22
to make sure that we understand and remember that it's reversible. Okay, so these ones are going to cancel. So we've got this guy right here. All right, and now what you need to remember is that these QCs here, those guys are negative.
27:44
QC and QC reversible are both negative. We showed that back on slide 36. I have no idea what slide this is, 45. How do you know that? Oh, it's down here.
28:01
Okay, so if I write these Qs as negative I've got to flip that equality sign around. Here it's greater than, sorry, here it's less than and here it's greater than, right? Greater than or equal to. All right, so if I make these negative I've got to flip that equality sign around and these reversible heats can be
28:24
equated with the temperature of the cold in the hot reservoir and so if I set this equal to this, I've got to use this greater than or equal to sign and in other words if I just cross multiply here I've got the
28:45
heat transferred from the cold reservoir divided by its temperature and the heat transferred from the hot reservoir over its temperature and there's a minus sign in front of this guy because it's negative. So for everything that, so this applies
29:01
to an irreversible process. If I move both of these guys over to the same side I can put a zero over here and say in general this has to be true. In other words if I evaluate these heats from the cold reservoir and the hot reservoir it's got to be less than or equal to zero for any irreversible process now or any process.
29:22
This applies to reversible or irreversible processes because there's an equal sign here. This is called the Clausius inequality. Who's Clausius?
29:41
So in other words if I integrate Q over T for a series, you'll recall this is the Q over T that applies for step one. This is Q over T that applies for step three. We've integrated for the whole cycle.
30:00
So if I integrate over a whole cycle Q over T that's got to be less than or equal to zero. This is really the most general statement of the Clausius inequality. He's the first German that we've talked about and what he did is he took Carnot's ideas
30:23
and translated them into what we now call the second law of thermodynamics. He named entropy and he tried to give it the symbol, you know, a good one but no sorry we're going to call it S. I don't know where S, where do you get S out of entropy?
30:45
Santa? I don't think so. Now you can memorialize Clausius with a T-shirt of your own and you can order this T-shirt online at a website that it will show you but look at this right here.
31:03
All right, no, I'm sorry, that was Joule. All right, the energy of the universe is constant, sorry, that wasn't Clausius. So this is totally wrong and I would ask that with me you join in the silent protest by not purchasing one of these $24 T-shirts.
31:25
Yeah, inaccurate and expensive. Okay, so this is the Clausius inequality but the second law of thermodynamics is this. All right, now is it obvious to you that this comes from this?
31:45
Well, it wasn't obvious to me. All right, so let's see if we can figure that out. Now, batteries are dying.
32:00
Here's how we can show that this is right. All right, state one, state two. We're going to go by some irreversible path between state one and state two and then we're going to go back on a reversible path. All right, irreversible, reversible, very simple.
32:20
Now, according to the Clausius inequality these two steps I can write two terms like this for these two steps. I'm going to go from state one to state two and I'm going to have a DQ over T for that and I'm going to go back from state two to state one except it's going to be reversible. And what the Clausius inequality says is hey, if there's any, if there are any irreversible steps that's
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equal sign goes away, see how that's less than or equals to? The equals only applies if everything's reversible. All right, so this is just going to be less than zero now because of this guy.
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Okay, so since this is reversible we know it's equal to DS only because it's reversible. All right, and so I can flip these limits here. If I flip these limits I think you'll agree I've got
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to make this plus sign a minus sign, right? And now I can move this guy over to the other side and when I've done that I just derived this equation right here. All right, so this is the second law of thermodynamics.
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All right, it says that, well, okay, so what does it say? Look, it makes, it says there's three kinds of processes. First of all, if DS is greater than DQ over T integrated over the whole cycle, all right, that cycle can be spontaneous and if it's greater than DQ
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over T it's also irreversible. All right, if these two things are equal to one another it's reversible and if DS is less than DQ over T it doesn't happen spontaneously. You can make it happen,
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but it doesn't happen spontaneously. Did it just get yellow again? What's going on? Okay, if we're talking about an isolated system, an isolated system is one where there's no heat flow in or out, no mass flow either.
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Okay, and so if I want to understand what these inequalities translate into in an isolated system I just substitute zero for Q, right, because there's no heat that can go in and out of the system if it's isolated, okay? And so all of these inequalities just turn into I just substitute zero for this, boom, I get this.
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Okay, so the second law of thermodynamics allows us to classify a process as one that occurs spontaneously or one that doesn't. Very important to us as chemists. Okay, what are some examples?
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Here's some simple examples and this is the kind of problem that's right at the beginning of the exercises from exercise one to exercise 19. All right, if you're going to be looking at the reversible phase transition, vaporization, sublimation, melting, all occur at a particular temperature and with a defined enthalpy, all right, delta H.
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In this case the phase transition occurs if the phase transition occurs reversibly delta S is equal to delta H over T. So for example, here we have a plot of entropy on this axis and temperature on this axis.
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We are heating a frozen solid, all right? Entropy increases until we get to a phase transition where melting occurs and there's a stepwise increase in the entropy at the melting point.
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All right, what is that delta S? It's that. All right, if we know the delta H we just take this temperature, all right, we can calculate what that stepwise increase in entropy is and now we're heating the liquid. The entropy continues to go up. Now we get to the boiling point and by golly there's going to be a different delta H that applies there, delta H vaporization.
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All right, we divide by that temperature. We get that change in the entropy right there. Okay, so these stepwise increases in entropy happen as we're heating a solid through these phase transitions, all right, melting and boiling. There can also be reversible heating or cooling of a gas.
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Right? DS is DQ reversible over T. That means the heat capacity can be used to calculate the DQ, multiply that by DT or integrate, divide that by T and we're going to get DS. All right, we use the constant pressure heat capacity if it's a constant pressure process
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or constant volume heat capacity if it's a constant volume process. That part's pretty easy. Okay, so if I evaluate this integral, all right, it's an integral of 1 over T so I'm going to get log T final over T initial, all right, that's what the delta S is going to be equal to
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and so S goes up with the logarithm of this ratio. So here's what that looks like. Here's delta S plotted against that ratio right there, TF over TI, all right, this is a logarithmic increase. These are just different heat capacities.
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Okay? Entropy goes up, all right, that's why when you look at this diagram right here, here we're talking about a gas, all right, in principle that's what this is right here. Right, it's the entropy going up for this gas as we heat it. Right, it's happening in accordance
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with this equation right here, all right, which would predict a logarithmic increase in the entropy with temperature. All right, so we can calculate that very easily. We've got these, we've got this equation right here. We can also talk about the expansion or compression of a gas.
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Right, DQ is minus DW is PDV, so the change in the entropy is just Q over T, again DQ over T rather and in this case it's DQ is PDV and so if the gas is ideal, why?
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We can recast that in this form right here. We can just talk about the final and the initial volume and we can use that equation to calculate the change in the volume. So a constant TV is inversely proportional to P. All right, so I can express this in terms of volume or in terms of pressure, either one. All right, so I can calculate the delta S for that process too.
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That's another example from the beginning of Chapter 15, those exercises. Okay, so because S like use a state function you can add, yes, so any random process that you're thinking about you can break up into steps yourself.
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You can decompose any process you want into steps. Calculate the delta S for those steps. You will always get the right final delta S because it's a state function. It doesn't matter what path you take. All right, you will get the same right delta S if you calculate the delta S of each path correctly and add them.
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Calculate the entropy change when argon gas at 25 degrees C and 1 atm in a container and volume 100 cubic centimeters is compressed to 5 cubic centimeters, simultaneous to be heated to 45. So, we compress the gas, we compress argon and we heat it.
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Okay, we don't have any equation for heating and compressing but we have equations for each one by itself and so we can just heat and then compress or compress and then heat. Add them up. So, the total S for this process is just going to be the delta S for compressing plus the delta S for heating.
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This is process one. We'll call it isothermal compression. We'll do that first. We'll compress first and then we'll heat without changing the volume. Isothermal compression, we expect delta S to be negative for that based on the equations that I just showed you
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and delta S for the isovolumetric heating, that's going to be positive. Remember, that's that logarithmic increase. Entropy goes up logarithmically with temperature. All right, so in this case we don't know if this is going to be a spontaneous transformation or not. We don't know if delta S is going to be positive
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or negative because we've got one term that's positive, one term that's negative. We don't know which one's bigger. Okay, so it's useful to actually think that through and just say, you know, if these were both positive or both negative you could say, all right, delta S is definitely going to be overall positive or overall negative.
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Okay, so process one is compression to constant temperature. That's this equation right here. All right, and all I've done is substituted for the number of moles, I've just substituted PV over RT because I don't know. It doesn't tell me how many moles I've got, but I can calculate that from the information I've got here.
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All right, I can just cancel these Rs. Oops, better not. All right, they don't cancel for gosh sakes because I need this to be moles and that is joules per kelvin per mole.
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Right, it's different Rs. This one's .08206, this one is 8.31451. Okay, and if I do the dimensional analysis by golly, yeah, that's the way I need it to be because I want the units to be joules per kelvin. All right, be a little careful.
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Okay, so if I plug the numbers in, 1 atm, .1 liters, .08206, 298.15, 8.31451, volumes, boom, boom, boom, minus .102 joules per kelvin, good. We expected it to be minus.
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All right, do I have any intuition about the size of this number? Zero. I have absolutely no clue if that makes sense. I mean the fact that it's not 10 to the 20 reassures me a little. All right, so I don't know.
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Now we're going to do step two, all right, that's the equation that we need for step two because we're only changing the temperature at constant volume. That's the heat capacity right there, all right, for an ideal gas which we can expect argon to behave pretty darn ideally, I think you'd agree. We know what that heat capacity is going to be for constant volume, it's 3R over 2.
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That's the constant volume heat capacity of any ideal gas. That's the number of moles, all right, 3R over 2, yes, units will be yes, we need to multiply by mole. All right, so always think about the units, all right, the units of this heat capacity if they're 3R over 2 are joules per kelvin per mole so I've got
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to multiply by moles to get rid of that because I want the units to be joules per kelvin, okay, and so that's the same number of moles that we used before. That's the constant volume heat capacity. We know the initial and final temperatures are, yes, yes,
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yes, yes, yes, yes, kelvin temperatures and this is the delta S that we calculated. It's a lot smaller, all right, it's positive. We expected it to be positive, okay, and so then the total delta S is just the sum.
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That's the, what did we do first? Heated and then we changed the volume, right? No, we, what did we, oh, sorry, changed the volume for then we, sorry, yes, all right, okay,
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and so that's the sum of those two things, minus .0987 joules per kelvin. Do I have any clue whether that number makes sense? No. I really don't have any intuition
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about these entropy changes, all right, but I can say because that's a minus sign that this is not a spontaneous process, all right, that's what the minus sign tells me. It's not a spontaneous process. And we care about the magnitude, too, but the most important thing is to understand that.
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Okay, oh my goodness, it's exhausting, isn't it? Okay, so study for the quiz on Friday, we'll see you then.