24. The Second Law of Thermodynamics (cont.) and Entropy
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Fundamentals of Physics I24 / 24
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Transcript: English(auto-generated)
00:01
Okay, what I did last time was describe to you a certain heat engine called the Carnot engine. So, let me refresh your memory on what that was. The whole engine consisted of a gas at some pressure and
00:21
some volume, and the gas was subject to a cycle. And the cycle went like this. You start at some point, A, you expand at a certain temperature, T1, isothermally, till you come to the point B.
00:43
So, the way you should imagine that is that this gas is placed on a reservoir whose temperature is frozen, I mean, stuck at T1. Reservoirs never change the temperature. And what you do is you lift one grain after another of sand, so you let the gas expand, but you don't let it cool down.
01:03
The minute it tries to cool, some heat will go from below. So, you maintain the temperature and start at that volume and end at that volume. During this process, some amount of heat, Q1, enters the gas. Next thing you do, you let the gas expand some
01:21
more, but now you let it expand adiabatically. Adiabatic means you put the whole thing in a heat insulating material, so no heat can go in or out. Now, you can see if it expands, it's going to have to pay for it with its own internal energy, so the temperature will drop. So, you wait till it drops to
01:42
a lower temperature, T2, then you go backwards like this. That's when you start compressing the gas by putting some grains of sand back, But this time, the gas is kept in a lower temperature on top of the temperature reservoir,
02:01
T2, and you drop more sand on it to compress it. That is this part. During this portion, heat will actually be rejected by the gas. Why is heat rejected? Because normally when you compress a gas, you're doing work on it, the energy should go up, the energy goes up,
02:21
the temperature should go up, but you're not letting the temperature go up. You're forcing it to have the temperature T2, so it'll reject heat into this heat bath. And finally, you come back to the starting point by further compression, but adiabatic. And adiabatic means it's insulated from the outside world, no heat flow.
02:42
So, there are four parts. There are two things you have to notice about the Carnot engine. In fact, let me draw a schematic of the Carnot engine. This is the standard way people draw that. The engine takes some heat, Q1, from the furnace, a T1. It has some Q2 rejected at the
03:02
exhaust, which is a T2, and it delivers a certain amount of work, which by the law of conservation of energy, had to be Q1 minus Q2. Do you understand that? The work done by the gas is equal to that. And who paid for it? It's the reservoir,
03:23
upper reservoir furnished the heat, some heat was rejected downstairs, and the difference between the two is the work. Now, we call this a cyclic process because at the end of the day, the gas has come back to where it is.
03:41
That's very important. If you want to make an engine, it's not enough to convert. The whole purpose of heat engines was to burn something and get some work out of it. But you don't want it to be a one-shot thing. You want to be able to do it over and over and over again, and you can do that with this engine because after this cycle, it's back to where it started, and you can do this any number of times.
04:02
Now, we define an efficiency for this engine, which is what you get divided by what you pay for. What you get from the engine is, of course, the work W, and what you pay for is the coal that you burn, which is Q1. Now, writing W as Q1 minus Q2, you can write this as
04:23
1 minus Q2 over Q1. This is true for any engine. Any engine, even if it's not a Carnot engine, with the law of conservation of energy, you can write something like this. Anyway, let me say for the Carnot engine, I got this as the efficiency.
04:42
Now, why is the efficiency not 100%? That's because Q2 is not zero. You can say, why are we building a stupid engine that rejects some heat? Why not just use all of it in some other fashion? And that's what we're going to talk about. But assuming that the engine operates between two temperatures, the furnace temperature and the ambient
05:03
atmospheric temperature, you can calculate the efficiency for this engine by actually calculating Q2 and Q1. That's a simple problem of integrating the work done and going from here to here. You remember that? 1 minus nRT1 log VB over
05:26
VA, and downstairs you have nRT2 log VC over VB. Again, you've got to remember why that is true. When you go from A to B,
05:41
There is no change in internal energy because the temperature is the same. Therefore, the work done and the heat input are equal numerically, so Q2 is the work done and Q1 is the work done there, so that's what I got. Then, I did a little bit of calculation to show you, by using the fact that B and C are an adiabatic curve, A and D are an adiabatic
06:00
curve, that these logs actually cancel. These guys, of course, cancel to give the final result, 1 minus T2 over T1. That's the bottom line. I don't care what else you remember, but this is something I need for today's discussion. Efficiency of the Carnot engine is 1 minus T2 over T1.
06:27
Okay, now you can say, why are we interested in this very primitive engine containing a cylinder and gas and so on? Okay, this is the efficiency of this stupid engine. I'm sure people can build a better engine.
06:42
So, here is Carnot's claim. Mr. Carnot says, no engine can beat my engine. Okay, that's his claim. By beat my engine, he means no engine can be more efficient than my engine. And he's going to demonstrate
07:03
that. I'm going to demonstrate that for you. You don't get something for nothing. It's based on the following postulate, which I told you last time. I'm just going to write it, I will repeat it. We are going to postulate, as the second law of thermodynamics, that it's impossible to find a
07:22
process, the sole result of which is to transfer some heat from a cold body to a hot body. We all know you can transfer heat from a cold body to a hot body in a refrigerator. You take heat out of the freezer and dump it into the room. No one says that's wrong, but that's not the sole effect,
07:40
because you get an electric bill at the end of the month, because it's a compressor doing a lot of work. The claim is no other agency in the end should be affected in any fashion. If at the end of the day, all you had was heat flow uphill, that's not allowed. That's a postulate, and you have to grant them that postulate, which we accept to be phenomenologically valid. But taking that postulate,
08:02
Carnot will now show you that no engine can beat his engine. The key to the whole Carnot engine is that it is a reversible engine. By reversible engine, I mean at every step in the Carnot cycle, let's say going from here to here, I can also go backwards.
08:23
I can put a grain of sand and compress a piston, or I can take a grain of sand, it'll go back to where it was. You never far from equilibrium, and if you can go one way, you can go the other way. That also means the Carnot engine at the point A can go backwards to D, then to C, then to B, and then to A.
08:43
If the Carnot engine went backwards, it will look like this. It will take heat Q1, I'm sorry, it'll take heat Q2, somebody will give it work W, and it'll reject Q1 at this temperature. This is the Carnot refrigerator, and that's the
09:04
Carnot engine. The Carnot engine and refrigerator are the same machine, you just can make it run one way or the opposite way. That's what we are going to use. So, let me give you a demonstration that's for, as far as I can tell, is good enough for our present purpose, that you cannot beat the
09:22
Carnot engine. So, let us say a Carnot engine takes some heat, say 100 calories, delivers 20 calories and work, rejects 80 calories, and it's operating between some two temperatures, T1 and T2. This is the Carnot.
09:45
I'm taking a particular example where the efficiency is, what, one-fifth. But don't worry about the fact that, yeah, so it's one-fifth. So, 1 minus T2 over T1 should be one-fifth.
10:00
Now, if you say you have a better engine, you have a better efficiency, what you really mean is that your engine can take 100 calories and deliver more than 20 calories. Maybe it'll deliver 40 calories and reject only 60 calories, but it's also operating between the same two temperatures.
10:20
So, this is your engine, Y for your engine. Your engine is better than Carnot's engine. At least that's your claim. You're now going to get shot down. So, how am I going to shoot this down? I'm going to run the Carnot
10:40
engine backwards, first thing. Then, I'm going to get a Carnot engine which is twice as big as this Carnot engine. It's not more efficient. You've just got twice as much gas. If you take a Carnot engine, do the following things. Run it backward and make it twice as big. What will that Carnot engine do?
11:02
It will look like this. It will take up 160 calories from downstairs. It will want 40 calories input and it'll dump 200 calories upstairs. So, this is a two times Carnot. Let me put a star saying it's Carnot run backwards.
11:23
See, it doesn't take any special genius to make an engine bigger. It won't be more efficient. It'll have the same efficiency. But if you can build something with one piston, I can make the same design with a bigger piston or a smaller piston. I'm just saying I want it to be twice as big for the Carnot engine. So, the following simple reason.
11:40
Your engine's giving out 40 units of work. Mine refrigerator needs 40 to run. So, what we do is we directly take the output from your engine and feed it to my refrigerator. Okay? Your heat engine makes work. My refrigerator needs work and I've scaled mine so that its
12:03
appetite matches your output. That's all I've done. Now, let's draw a box around these guys. Don't look under the hood and see what you've got. At the end of a full cycle, when everything is done, all the gases, all the pistons, everything has come back to where it starts.
12:22
But no need to plug this gadget into the wall. You don't have to plug this into anything because this refrigerator is getting the power from this heat engine. So, it doesn't need external power. I look at the lower reservoir, I see 100 calories leaving. You see that? 60 coming down and 160 going up.
12:41
I look at the upper reservoir, I see 200 calories out, in and 100 out. So, basically, the combined gadgets, yours and mine, are equal to this gadget. It simply transfers heat from a cold body to a hot body with no other changes anywhere in
13:01
the universe, and that is not allowed. So, you cannot have an engine more efficient than the cardinal engine. The logic is pretty simple. These numbers I picked are representative, but you can take any number. As long as your engine does better than mine, instead of 40 calories, but made 30 calories needed. If it produced 30,
13:21
I'll get an engine which is one and a half times as bigger than the standard engine. I run it backwards, your 30 will feed my engine, and you will find once again heat is flowing from a cold body to a hot body with no other changes anywhere in the universe. That is forbidden by the postulate. So, this is how Carnot's engine, even though it's a very
13:42
primitive engine, is the standard for all engines. No engine can be better than the Carnot engine. And the key to the Carnot engine is that it's a reversible engine. If I had more time, I can show you that all reversible engines operating between two temperatures will have the same efficiency, namely that of the Carnot engine. Okay.
14:04
Now, what has the Carnot engine got to do with what I started saying earlier? You remember, I started saying earlier that there are certain things in our world that seem allowed but don't seem to happen. Like, you know, drop an egg, it doesn't come back in your hand.
14:22
You let the weight go down and turn the paddle. The paddle doesn't turn backwards and lift the weight. I said many things are forbidden, and there is a law that's going to forbid all of them, so I'm coming to that law. But it all started with a person asking a very practical question, how good can I make engines? And what Carnot is telling you
14:40
is the best efficiency any engine can have is this. Therefore, even today, hundreds of years after Carnot, no company in America or Japan, China, anywhere can build an internal combustion engine, for example, whose efficiency is better than this one. It's a theoretical maximum. In reality, efficiency will be less than
15:02
this because most engines have a loss. There is heat leaking out, there is friction that's good for nothing. So, in the end, efficiency will be always lower than this. Okay, now let me leave the practical domain and go to more theoretical issues that follow from this.
15:20
What I want you to notice is the following. For a Carnot engine, q1 over q2 turned out to be T1 over T2, right? I did that for you here. Therefore, I'm going to write it as follows. q1 over T1 minus q2 over T2 equal to 0.
15:46
That's just a simple rewrite. But I'm going to write this in another notation. My cycle had four parts, remember? I did this, then I did this, then I did this, then I did that. In this segment, I had q1 over T1. In this segment,
16:06
there was no q1 or q2, there was no q because this is the adiabatic process. In this segment, I had minus q2 over T2, and the last segment I had 0, and the whole thing is 0.
16:23
In other words, what I'm telling you is the following. At every stage, I'm telling you that if you looked at the heat absorbed by the system in stage i divided by the temperature of stage i,
16:41
and you added all of them, you get 0. My process had four stages. Stage 1 is here, stage 2 is adiabatic, stage 3 is this, and stage 4 is 0 again. In this summation, qi is defined to be delta qi is the heat input in stage i,
17:08
in the cycle or in the process. So, we defined q2 to be positive, even though it was rejected by the engine, but in this summation, the agreement I make is delta q is positive if heat comes into
17:23
the system, delta q is negative if heat leaves the system. So, why is that important? Now, this is the heart of the whole entropy concept. Remember I told you there is nothing called the heat in a system. You cannot look at a gas and
17:43
say that's the amount of heat in the system. Why? Because if you take a point there and say there is some amount of heat in the system, if you go through some kind of cycle and come back to the same point, then since you come back to the same point, internal energy doesn't change.
18:00
So, this is the work done. Therefore, the work done is equal to the heat and it's not zero. So, in this example, what would have happened is you would have added heat q to the system. So, if there is some notion of how much heat is there in the beginning, you've got that plus the q that you added, and yet you're back to where
18:22
you are. Therefore, you cannot define something and say that's the amount of q in the system, because you're able to add q to the system and bring it back to where it is. What happened, of course, is you added q and you did some work. But the point is, you cannot say the q here is so and so, but you can say the energy here is so and so,
18:40
because if you come back to the same point, p times v is equal to 3 is equal to RT, and the internal energy of a gas is proportional to T. So, the energy returns the whole value. But now I'm going to tell you, so listen very carefully. Energy is a state variable because it comes to the starting value if you come to the starting point.
19:01
It doesn't matter where you wander off in the PV diagram. Heat is not a state variable. There's nothing called the heat at that point, because I'm able to go on a loop, change the value of q or add some q to it, and I come to the same point. So, there's no unique q associated with that point. But there is a new, unique quantity, S, called entropy,
19:24
which has a fixed value at a given point. And if you go for a little loop in the PV diagram and you come to the same point, S will return to the starting value. So, who is this S? That S is defined by saying the change in the entropy is
19:45
the heat you add divided by the temperature. In a tiny little process, if you're at some temperature and you're at a little amount of q, you add a little amount of q, keep track of all the changes, and that change will
20:02
be zero as I showed you in this Carnot cycle. You can show more generally that if you take any path, not just the Carnot engine bounded by adiabatic and isothermals, but any path you take, you can show that the Δq added up is not zero,
20:20
but if you add the Δq over T, namely give a weighting factor 1 over T to the heat you add, then the positives and negatives cancel and give you exactly zero. In other words, q1 was not q2, q1 is bigger than q2. But q1 over T1 precisely balances q2 over T2. The heat absorbed at higher temperature, if divided by T1 to
20:44
compute the change in entropy, then in the upper part of the Carnot cycle here, the two cancel as far as the entropy change is concerned. So, entropy is a new variable. It's a mathematically discovered variable. What we find is that if I
21:02
postulate, there's a variable called entropy, the change in which in any process is the heat transfer divided by temperature, then that has the property that when you go around on a loop, you come back to net zero change. So, the change in energy can be associated with the number you
21:22
can call entropy. Entropy is like a height. Suppose you're walking around on some landscape, each point has a height. You can walk here and there and come back at every stage, you keep track of the change in height. If you come back to where you are, the change in height will add up to zero. That's what I told you in the case of a potential energy
21:40
function. It is just like a potential energy function. The internal energy and the entropy are now state variables. At every point, the gas has a certain internal energy and an entropy. They deserve to be called state variables because when you go on a loop and come back, they return to the starting values.
22:00
Now, you have no idea what this quantity stands for. You agree it's bizarre. Work done, we understand. Heat absorbed and heat rejected, we understand. What is ΔQ over T? Why does dividing by T make such a big difference? Why does it produce a new variable? We can see it is true, at least in the Carnot cycle. You can verify in detail
22:21
that the change in the total of all the ΔQs over T is in fact zero. So, we'll develop a feeling for what it means, but historically this is what happened. People realized, hey, that's another state variable. We introduce a new variable, we have no idea what it means, but it is a state variable so we'd better take it very seriously. So, I'm going to tell you
22:43
exactly what it means. But first, I want you to get some practice calculating the entropy change for a couple of processes. So, let us take, for example, one gram of water, or let's say m grams of some substance. It's got some specific heat, C, and I'll change the
23:04
temperature, say, from some T initial to some T final. I do that by keeping the temperature T initial. I'm going to start the system, starting the system at Ti, and I should never be far from equilibrium. That's one of the conditions
23:21
on this. In fact, you can say it should be near, always near equilibrium. So, the system starts at temperature T initial. I put it on a heat bath, which is infinitesimally warmer than this. Then, I let it heat up to that point. Then, I put it on another reservoir, slightly harder than this one.
23:42
At all stages, I want the system to be almost near equilibrium. In the sense of calculus, you can make the difference as small as you like, provided you do enough number of times. And slowly, I raise this guy from here to here. At every stage, the system has a well-defined temperature, there's a well-defined heat input, and I want to add it all up.
24:01
So, what's the change in entropy? S final minus S initial is equal to dQ over T, and dQ, summed over all the parts, I should write it as an integral, as MC ΔT over T. Do you see that? dQ is MC ΔT. But I've got to divide by T,
24:22
and that integral is done between some initial and final temperature, and we can do this integral rather trivially. It is MC log T final over T initial. That is the increase in energy, the change in energy, the change in energy, the change in grams of some substance, a specific heat,
24:40
C, that's heated from T initial to T final. We're just getting practice calculating this. We still don't know what this guy means, so don't worry about that right now. If you did not divide by T, what are you calculating? MC ΔT integrated is just MC times change in temperature. That's the old calorimetric problem you did.
25:01
How many calories does it take to raise the substance from some initial to final temperature? That you understand. When you divide by T is a thing you don't understand, but anyway, let's make sure we know how to calculate the increase in entropy when water or something is heated from a lower to a higher temperature. If you cooled it from a
25:21
higher to a lower temperature, you can use the same formula Tf over Ti, but Tf will now be smaller than Ti if you cooled it. So, log of a number less than one is negative, and the entropy change will be negative. So, let me do one more entropy calculation. That's going to be pretty important for us. That entropy calculation is
25:41
this. Take a gas and watch it expand isothermally from some starting point at some fixed temperature T. It goes from V1 to V2. What is the change in entropy when it goes from here to here? Again, the system must always
26:07
be in equilibrium or near equilibrium, so I can plot it as a point in the PV diagram. So, I slowly take grain after grain, do the whole thing I told you, and find S2 minus S1. This is 2 and this is 1.
26:21
That is equal to sum of all the heat transfers divided by temperature at every little step of the way. But remember, this is also the same as PΔD over T. Why? Because on an isothermal, dU is zero.
26:42
I've repeated it many times, but you've got to know this. ΔQ is PΔV. But P over T is nRT over VΔV. You also want to divide it by a T, you understand? P over T is nR over V.
27:05
So, P is equal to nRT over V. I'm just saying PV is nRT. So, P over T is nR over V. Now, dV over V, when you sum or integrate, will give you nR log V2
27:21
over V1. That is the change in entropy of this gas when it went from volume V1 to volume V2 at a given temperature. You don't have to use calculus to do this. Does everybody understand why this could have been done a lot easier?
27:41
You don't need to do the integral here, because I showed you in studying the heat engine that the heat transfer Q is nRT log V2 over V1. Since the whole process takes place at fixed temperature,
28:00
instead of dividing by T at every step, you can just divide by T overall. Just bring the T here. That's why this is the change in entropy. In other words, during the whole process, if you're at one temperature, then integral of ΔQ over T is just 1 over T times integral of ΔQ, which is the total heat transfer.
28:22
Anyway, this is the change in entropy, S2 minus S1 is nR log V2 over V1. Okay, so what have I done
28:43
so far? Let's collect our thoughts here. I went to the Carnot engine today and reminded you what the efficiency of the Carnot engine was. Then I showed you no engine can be better than the Carnot engine. That's a separate story. That has to do with how
29:00
efficient things can be. Then, a byproduct of the Carnot engine was this great realization that if you go on a closed loop, and at every stage you add, you compute the heat absorbed, but divide by the temperature at that point, the sum of all those is zero. What that means is that there is a quantity S whose change,
29:24
if it is defined to be ΔQ over T, has a property the total change in S is zero when you go around a loop. That means at every point you can associate an S. S is the property of the point. So then, I said, let's get used to computing change in entropy.
29:42
I took one example of heating a substance of some mass and specific heat C by a temperature dT and the change in entropy was mc log Tf over Ti. You got the log because integral of dT over T was logarithm. Then I took an ideal gas, I let it expand isothermally, I found the entropy change,
30:03
I got this answer. By the way, here's an interesting exercise. I don't have time to do it, but you can ask the following question. If you tell me that at every point there's unique entropy, then the entropy difference between 2 and 1 should be independent of how I go from 1 to 2. You understand? If you're walking on a
30:21
mountain, you take two points, they have a height difference, and I can find the height difference by going on this path, keeping track of the change in height or any other path. In the end, the height difference between two points is a high difference. So, I could find the entropy another way. Let me show you another way that's easy for you guys to work out. If you come down like this and go to the right like this,
30:43
and find the entropy change, you will get the same answer as I got on this side. Okay, it's too tempting for me to just leave it here. Let me tell you how it works. Call this intermediate point, give a subscript 0 for all its parameters. Then, in this step,
31:02
when you come from here to here, the entropy change will be, let's take one mole of a gas. When I go from here to here, then for one mole of a gas, this heat transfer dQ is cv dt, and I divide by T, and I do the integral from T1 to this point T naught.
31:24
That's the entropy change here. Then, in the horizontal part, since I'm going at constant pressure, I go Cp dT over T from T naught up to back to T1, because this T1 and T2 are the same temperature.
31:43
I have to add all these to get the entropy change. Now, it's a two-step process. You come down because you're at constant volume and you're doing something to the gas, dQ is cv dt. Horizontally, you're at constant pressure, dQ is cp dt. That's the definition of specific heat at constant
32:01
volume, constant pressure. But notice the following. Cp is equal to cv plus R. So, put cv plus R here and look what you get. Then, you get cv times log T naught over T1 plus log T1 over T naught,
32:23
which we can write as the product here, cv times log T1 over T naught plus R log T1 over T naught. In other words, this log is log of T naught over T1, the next log is log of T1 over
32:40
T naught, and log A plus log B is log of AB. But when you do this, look what happens here. This all cancel. Log of 1 is 0. The total entropy change is cv times log T1 over T naught. But for a gas at constant
33:02
pressure, using PV equal to RT, this ratio of temperature is also the ratio of the volumes. So, this one, I'm sorry, I should write it more carefully. When I say T1,
33:21
I really meant the temperature at this point, second point. So, that temperature over this temperature is also that volume divided by the initial volume, which is V1. See, this is a confusing problem because T1 happens to be T naught. The correct way to do this for me would be to write T naught over T1 times T2 over T naught,
33:43
then realize the fact that a T2 and T1 are equal because I'm on an isotherm. That's what I should have done. So, I should cancel this factor because T2 is T1, but here T2 over T naught is V2 over V naught, which is V2 over V1. That's, of course, the entropy change I got in one
34:01
shot here. So, you can find entropy change any way you like. You usually pick the easiest path, all right? So, we have now learned how to find entropy change. We have no idea what it means. It seems as if when you heat
34:21
something, entropy goes up. When you cool something, entropy goes down. That's certainly correct because ΔQ is positive for heating, and you divide by T, you may get a logarithm, but we know at every infinitesimal portion, I'm adding positive numbers. And when I cool things, it goes down. Why not just call it temperature? Why do you need this new concept called entropy?
34:41
So, that's what the rest of the lecture is about. It's a very, very powerful and beautiful concept, so I want to explain it and make sure I get it right. Here is now the result for all this hard work. Remember I told you there are many, many, many phenomena that seem forbidden in our world, and we're not quite sure what
35:03
law to invoke to prevent all of them from happening. Do we want a law for each one that says if you drop a bottle and it shatters, it won't come back? If you drop an egg, it won't come back. If you let hot and cold mix, they're not going to unmix. Or if you let a gas trapped in half a room escape the whole room, it'll never untrap
35:23
and go back to half the room. A lot of things happen one way but not the other, and I said I'm looking for a mega law that will prevent all these things from happening. Now, I'm ready to state that law. So, this is the third law. I mean, the second law of thermodynamics, Carnot said it one way,
35:40
but I'm going to say it in a way that's very, very general. The second law of thermodynamics says Δs for the universe is either zero or positive. There you have it. That's the great law. The law says take any process.
36:03
If at the end of the process the entropy of the universe is bigger than it was before, that will happen. If the entropy of the universe is smaller than it was before, it will not happen. Now, we have to make sure that this law has anything whatsoever to do with all the
36:20
other things we have studied, and I will show you this will forbid everything that should be forbidden and allow everything that should be allowed. So, let me start with the second law, which is the most obvious formulation. Mr. Carnot's version of the second law was that you cannot have a process in which some heat Q goes from a hot body to a cold body. I'm sorry, this is allowed,
36:42
right? According to Carnot, this is not allowed, right? Heat can flow downhill, cannot flow uphill. Let's see the entropy change the universe for the two cases. entropy is the same as the change in entropy is. In this case, for this guy,
37:04
the change in entropy is the following. The upper reservoir lost some amount Q, so ΔQ is a negative number, at some temperature T1. The lower reservoir gained Q at temperature T2.
37:22
Now, what's the overall sign of this? Think about it. T1's bigger than T2, so this negative number is smaller than this positive number, so the whole thing is positive. That means it's allowed. It's okay. By the same token, if you take this process, when heat flows uphill, then this reservoir loses some
37:43
heat Q at temperature T2. The other one gains heat Q at temperature T1, but this is clearly less than zero because this positive number is smaller than this negative number. So, there you have a very simple example where if heat flows the wrong way, the entropy of the
38:01
universe goes up. And it's very simple. It's only because it is true the heat loss of this guy is the heat gain of that guy. That's the law of conservation of energy. So, energy doesn't change, but entropy changes because for entropy it's not the heat
38:20
transfer, it's heat transfer divided by temperature. Therefore, a heat loss at one temperature and a heat gain at a lower temperature don't cancel when it comes to entropy. In one case, the entropy goes up and it's allowed, in another case, entropy goes down and it's not allowed. Okay, so let's try. So, that's certainly one
38:41
process that you know should not be allowed, but it's forbidden by the third law by a computation of entropy. So, the reason when you put a hot and cold body together, heat flows from the cold to the hot, I'm sorry, from the hot to the cold is because that's the way the entropy will go up.
39:03
So, let's take one more example. Suppose I have some mass of water at some temperature T1 and an equal mass of water at temperature T2. One is hot and one is cold.
39:22
I put them together. What will happen? We believe you will then get this big mass, all at some common temperature T star with this T1 plus T2 over 2. This is just by symmetry. It's equal mass, equal specific heat. Where will they meet?
39:41
They will meet halfway. They will meet here. Energy is, of course, conserved and that's how we determine, in fact, where they will meet. But look at the entropy. What happens to the entropy change? For the entropy change, you should imagine this water being steadily heated by putting
40:00
it in contact with a lot of reservoirs so it's never far away from equilibrium and slowly bringing it to this point. Then, the Δs total will be m, specific heat is 1, then ΔT over T, starting from T1 to T star for one thing,
40:25
and dT over T from T2 to T star for the other one. You understand that? They both meet at T star. Upper limit is T star, lower one is T1 for one and T2 for the other. So, the change in entropy becomes m log T star squared over T1, T2.
40:51
Now, we have to ask, okay, you've got an entropy change from start to finish, but how do you know it's positive? Can anybody give some reason
41:00
why this has to be positive without looking into properties of logarithms and so on? Any reason why? Yes? Yeah, so that's one way to prove that, but I'm saying I'm going to prove it to you that way.
41:23
But can you think of a reason why at every step of the process, when I brought this down and when I pushed that up, the entropy change is always positive in each step. So, let me explain why. Imagine cooling this down a little bit by putting, taking some heat, ΔQ out of this guy. That's happening at some
41:42
temperature here. This one is gaining some and this one is gaining delta Q, but at a lower temperature, okay? Therefore, the gain has got the same ΔQ, but at a lower temperature. This is always at an upper temperature. Throughout the process, at every stage,
42:00
this guy is hotter than this guy. So, every calorie of heat it gains at a higher temperature. Every calorie it loses is divided by a higher temperature. Every calorie it gains is divided by a lower temperature. At every increment, at every infinitesimal step, you can see the total entropy change is positive. Now, to prove it, we will do what he just said.
42:21
If you want to prove this is positive, you want to show that the fellow inside the logarithm is bigger than 1. So, I'm asking, is T1 plus T2 squared over 4 bigger than T1T2? Or I'm saying, is T1 plus T2 squared bigger than 4T1T2?
42:40
If you rearrange this, you can show the left hand side becomes T1 minus T2 squared. That's, of course, a positive number, because this would be T1 squared plus T2 squared plus 2T1T2. When I bring this guy to the other side, it will become minus 2T1T2, and so it's clearly positive. So, when hot and cold meet and reach lukewarm,
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entropy of the universe has gone up. It follows, therefore, if lukewarm spontaneously separated into hot and cold, the entropy would go down, and that's why that doesn't happen. That's why if you leave a jar of water at one temperature, it doesn't spontaneously separate into a part on top,
43:21
which is cold, and a part on the bottom, which is hot. Such a separation would not violate anything. It would not violate the law of conservation of energy, but it will violate the law that entropy has to always go up. So, one way is allowed, another way is not allowed. So, you see over and over again that anything that's allowed is in the direction of increasing entropy.
43:41
So, you see, anything that's forbidden is in the direction of decreasing entropy. But who is this entropy? What does it mean? Well, that we have not understood at all from any of these calculations, because dQ over T doesn't speak to us the way du or dQ or pdv does. So, the last part of what I'm going to do is going to
44:03
explain to you what is the microscopic basis of entropy. So, why is entropy going up? Why is the entropy going up? Why do we understand the tendency of things to go in the direction of increasing entropy? All I've shown you now is that that is a mystical
44:21
quantity called entropy. Circumstantially, you find every time something is forbidden, it's because had it taken place, entropy would have gone down. You remember the examples? Heat flowing from hot to cold, entropy goes up, entropy goes down, allowed. Flowing from cold to hot, entropy goes down, not allowed. Hot and cold mix and become lukewarm, allowed,
44:43
entropy goes up. Lukewarm separates into hot and cold, not allowed, entropy goes down. So, there's definitely a correlation. And yet, we don't know what it means. So, we're going to talk about what it means. For that, I'm going to consider the following process. I take a gas and I put it in a room in a box where there's a
45:06
partition halfway. The molecules are stuck on the left side. The gas has reached equilibrium and it's done what it can, which is to spread out over this volume. Now, I suddenly remove the partition and rip it out. So, let's follow this gas.
45:27
So, the gas initially is the point here. Then, there is a period when it goes off the radar because I told you, when you suddenly remove the wall,
45:41
the gas is not in a state of equilibrium. It doesn't even have a well-defined pressure. The minute you remove the wall, pressure is something here, pressure is zero in the vacuum here. So, you've got to wait a bit. So, gas has gone off the radar. We cannot talk about what's happening. Then, if you wait long enough, I get a gas that looks like this after some seconds or
46:01
whatever milliseconds. That is again gas in equilibrium. It's got a certain state. I can call it 2. What is the entropy change now? That's what I want to ask. What's the change in entropy here? Because I know that this is allowed and that is forbidden.
46:24
So, I want to ask, did the entropy go up? Now, how do we do the entropy calculation? Here's a wrong way to do the calculation. You go back and say ΔS is ΔQ over T. But this whole box, I forgot to mention, I'm sorry. This whole box is thermally isolated.
46:42
It's not in contact with anything. I just ripped out the partition in the middle. That's it. So, you might say, well, if it's thermally isolated, ΔQ is zero. So, you can divide by T. You can do what you want. So, this entropy change is zero. But that is a wrong argument.
47:03
Can you think about why that's not the way to do the entropy change and why that's the wrong analysis? First of all, could that be the right answer? Yes?
47:23
No, that is correct. But I'm saying this computation of entropy, the one-line calculation I did, namely there is no heat inflow, so ΔQ is zero, so ΔQ over T summed up is also zero. So, that's not how you do the entropy change. Is there any condition I made on computing?
47:42
Yes? Wait. In fact, here is the interesting thing. What's going to be the temperature difference between before and after? Which is going to be hotter or cooler? Any views on this?
48:10
Yep? Well, you can say it is cooler because it was insulated and expanded, but you cool down because you expand against an external pressure, right? Even when I remove this piston
48:22
here, there is no pressure pushing this gas back. You understand that? It doesn't do any work. It doesn't take any heat. So, what does that mean in the end when you settle down? What can you conclude then? No heat input, no work done. So, what does that mean in terms of temperature?
48:44
Same, because the internal energy cannot change. The ΔQ is zero, work done is zero. So, in fact, this is a surprise. This gas, when it expands, will be at the same temperature. You guys are thinking of the computer air, where if you let it expand, it cools down, because that is expanding
49:00
against the atmospheric pressure. This gas is expanding into a complete vacuum. It's got no piston to push again, nothing from outside pushing. So, these two points are at the same temperature. So, the correct answer, maybe you didn't catch on, but I will tell you the correct answer is there is an entropy increase in this problem, because the final state
49:23
definitely has a well-defined entropy you can calculate, because it's an equilibrium state. The initial one has a well-defined entropy. In between stages here are not on the PV diagram, because they were not equilibrium states. So, in this experiment, the way to calculate the entropy change is not to do ΔQ over T as it happened in
49:42
this experiment. What you really want to say is, my gas was here in the beginning, my gas is there in the end. Both are states of equilibrium, both have well-defined temperature, and I can find the entropy change in going from here to here for any process I want, as long as that process keeps the system near equilibrium, so I can follow dot by dot
50:02
where I'm moving and add the ΔQ over T. The correct rule for entropy change is the ΔQ over T computed on a path in which the system never strays from equilibrium. Now, your system strayed from equilibrium all the time, but we're not interested in how it got from here to here. We are saying, what's the entropy here, and what's the entropy here?
50:22
Since it's at the same temperature, I know one way to go from there to there is to follow the isotherm, because that'll connect the two points in this particular problem. Then, I know the entropy change is S2 minus S1 is nR log V2 over V1.
50:52
So, here is the subtle point I want to explain to you. This is worth understanding. A lot of people do all the problems and get everything right, but don't appreciate this particular question. The free expansion of a gas
51:02
into a vacuum takes it from one equilibrium state called one to another equilibrium state called two. In the actual expansion, the system went through states which cannot be even shown in the PV diagram. They were not equilibrium states. They didn't have well-defined pressure, didn't have well-defined temperature.
51:21
How do you define a temperature when half the box is empty? It doesn't have, you cannot call it the temperature of the gas. So, what do you mean by ΔQ over T? There is no T. Only when it settles down, there is a T. Every settled down equilibrium state is a well-defined entropy. What we're trying to do is to compute that difference. The way to compute the difference is to forget about
51:42
what actually happened and instead do the following. Take the gas like that and turn it this way if you like, just for convenience. Put a piston there at a halfway point. Put some weights on it so that
52:00
the pressure equals the pressure inside. And slowly remove little by little and keeping it on a reservoir at temperature T1, which is the temperature at which all of this happened, or at some temperature T. As you remove the grain of sand, what happens is the gas
52:22
gets some energy, that's a ΔQ over T, the reservoir loses some heat. In fact, the reservoir loses the same amount of heat, ΔQ, at the same temperature T. So, in fact, what happens in this process is the entropy of the universe does not change at all.
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The entropy gain of the gas is the entropy loss of the reservoir. But I use this as a device for finding the entropy gain of the gas in the other process. So, this is the part that's absolutely central to this whole
53:03
argument of how entropy is calculated. In a real spontaneous expansion of a gas where there was no reservoir, no nothing, we just broke the partition and expanded, we decided there is a change in entropy. That's an irreversible process. But to find the change in entropy in this problem, since 1 and 2 have well
53:23
defined entropies, we can find the change between 1 and 2 by going to a different experiment in which it was brought from 1 to 2, not in this crazy, irreversible way, but by putting it in contact with the heat bath at that temperature T, removing grain after grain, never straying far from equilibrium. That's when we find this
53:42
change. So, you replace the dotted line where you cannot compute anything by a solid line where you can, but since the change in entropy between 1 and 2 doesn't depend on how you got there, that's the answer. But this process, where you can actually see it, is the one in which ΔS of the universe is actually a zero.
54:02
Why? Because you're putting in contact with the reservoir at the same temperature, so ΔQ over T for the gas and ΔQ over T for the reservoir exactly cancel because they're at the same temperature, and ΔQ for one is minus ΔQ for the other. So, don't confuse the two
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processes. One is a reversible process in which a gas is slowly allowed to expand from the initial state to the final state by removing grain after grain of some externally applied pressure in a process where the entropy of the universe doesn't change. Gas and reservoir change by equal amounts. But the punchline is, I don't care what happens to
54:42
the entropy of the universe and how much the gas gained, because that is the gain I need to find S2 minus S1 for this process. So, this is how we always find entropy change. In real life, processes happen in a way in which the entropy of the universe goes up, because take this box. In the real experiment, remember there was no
55:02
reservoir. It was only the box of gas. It expanded, its entropy went up, so entropy of the universe actually went up. That's why it's allowed. The entropy of the universe went up by an amount equal to the increase in entropy of the gas when it goes from one to two.
55:21
To find the increase in entropy of the gas when it goes from one to two, I create a second process in which I can compute the change by letting it go in a sequence of connected dots and finding the change. So, the way you used to find the change in entropy is not the way the entropy increase took place in the real process. So, I don't think I want to
55:42
solve this problem, but it's an important point and you should really think about that. All right, now the last point is, let's analyze this final formula I got for entropy change for free expansion of a gas. Let's take a simple case
56:05
where the gas expanded from some volume, v, to a volume 2 times v. What is S2 – S1? S2 – S1 is nR log 2. 2 happens to be
56:24
v2 over v1. The volume is doubled. Let me rewrite this as nk log 2, because the number of moles times R is the number of molecules times k. Let me rewrite that as
56:43
n log 2 to the n times k. Now, I will tell you what all these factors mean. And that comes from the
57:00
following. Thermodynamics only told you the formula for the change in the entropy of a gas. It didn't tell you what the entropy itself is. So, the formula for the entropy of a gas, without talking about the change, and also this formula, ΔS is ΔQ over T, doesn't even believe in atoms. You don't need atoms, you don't need molecules,
57:21
you don't need to know what anything is made up of. In fact, it was discovered long before atoms were proven to exist. Boltzmann is the guy who gave the formula for what the entropy of a gas is in terms of what the individual molecules are doing. That's the punchline of what's called statistical
57:42
mechanics, which gives you the microscopic origin of thermodynamics. Thermodynamics to this stage, talked about fluids and solids and gases, and nothing there requires that everything be made up of atoms. You don't have to know what it's made of. But if you tell me it's made up of atoms, you have a better idea of what entropy means. And here is Boltzmann's formula for entropy.
58:03
I'm going to write it down, Boltzmann constant times log ω. I'll tell you what ω is in a minute. This is such an important formula. It is written in Boltzmann's tombstone. So, when physicists go to Vienna, we skip the Arches, Reis, and everything else.
58:22
We go to Boltzmann's tomb, and we read this formula once more with some passion. That's the summary of a lifetime of work, but it's a great formula. And you will understand once you know the formula what all this means. So, let me tell you what ω means. I'm going to say in words, then I'm going to apply it to this problem.
58:41
If you took any gas, let it be under certain conditions that macroscopically have some pressure and some volume and so on. So, here is the gas at one instant. Now, we believe in atoms. Here are these atoms scattered all over the place.
59:02
Suppose I wait a little bit. What does the gas look like? It looks like that. If you can see every atom, the atomic configuration has changed. But macroscopically, nothing has changed. Instead of atom A being here, maybe A has gone there and B
59:21
has come here. In particular, if we divide the volume into some number of little cells that you can count, on the average, the gas has uniform density. There's some number of atoms per every little cube you can form. A little later, I move out and somebody else moves in. Macroscopically, it looks the same to me. Microscopically,
59:41
it looks the same to me. That's the way stuff is going on. ω is the number of different microscopic arrangements that agree with what is seen macroscopically. When I say microscopic arrangements, I really mean the following. Give every molecule a name.
01:00:00
So, it's going to be A, B, C, D, whatever. Then, put A here and B here, C here and D there. That's one arrangement. Then, permute them. Put D here and B here and C here and A there. That's another arrangement. To my naked eye, they all look the same, but they all produce the same macroscopic effect, but they're different microscopic arrangements. And you're supposed to take
01:00:22
the log of the number of those arrangements multiplied by this constant called Boltzmann's constant to get the entropy of the state. So, now let's apply it to this gas here. Now, the number of microscopic
01:00:40
arrangements, if you really want to get down to it, requires dividing the box into tiny little cells, maybe one cubic millimeter, and saying how many atoms are in each region. But we're going to do a crude calculation in which we'll only say the following. Either a molecule is on the left or it's on the right. We are saying these are the only two positions open to the
01:01:00
gas, left or right. Of course, in the left there is a left corner, top and left corner, bottom, but don't worry about it. Just say there are only two things it can do. It can be on the left or it can be on the right. Now, let's ask ourselves, if the gas looks like this,
01:01:21
how many ways can it be in this arrangement? As compared to how many ways can it be in this arrangement? So, it is like the following. Molecules are moving randomly.
01:01:41
A given molecule has got a 50-50 chance of being on the left or being on the right. You are demanding that all the N molecules on their own, there's no partition now, on their own be on the left-hand side. You can see it's like tossing a coin N times and wanting heads every time, because you want every guy in a
01:02:05
machine to be left. So, if you want them all to be left, there's only one way to do that, and that's like saying, I tossed a million coins, I wanted all heads. There's only one way to do that. So, that arrangement, if you like, we will say looks like this. Left, left,
01:02:23
left, left, left, left for everybody. That's the arrangement here. Take another one where one molecule is over here and all the others are here on the other side. That looks like left, left, left, right, left,
01:02:41
left, left, but that's one arrangement. But have I done my complete homework here, or are there more arrangements? I want you to think about it. Pardon me? Student 2, N arrangement. Professor Ramamurti Shankar This is only one of the arrangements for the molecule on the right. There's another arrangement in which I can have the second
01:03:02
molecule, molecule Mo or Joe or Po, whoever is here, you can pick them in N ways. So, what I'm saying is, already when one is on the left and N minus one on the right, there are more ways in which that can happen. So, your eye sees one on the right, N minus one on the left, but there are N ways in which that can happen. It's like saying if I toss
01:03:20
coins, I want one head and 99 tails. Well, it turns out there are 100 ways in which it can happen because the one head I got could have occurred in one of the two heads. So, there are 100 different terms. Now, suppose I want two heads and 98 tails. Then, as you know,
01:03:41
the way to do that is 100 times 99 divided by 1 times 2. These are some combinatorics you learned in high school. So, you find that if you plot number of heads over the total number and ask for how many ways in which that can
01:04:02
happen, all heads can occur in only one way. Every coin has to be a head. One head less, and one tail and 99 heads can occur in N ways, then it'll become more and more and more, and you will find the 50-50 chance of half the number of heads and half the number of tails has the biggest
01:04:21
occurrence rate because that's the most number of ways to get 50-50 than anything else. And 100 heads is as bad as 100 tails. Again, there's only one way in which you can get all heads and one way in which you can get all tails, but as you start scrambling them, there are more and more ways. That means for the
01:04:41
molecule, if you have 100 molecules, the number of states ω is the largest when they're equally distributed. In fact, as the number of molecules goes from 100 to 1000 to a million to 10 to the 23, this function is so sharply peaked that the approximate
01:05:00
formula for ω when it's at the 50-50 thing is in fact 2 to the N. Namely, all the arrangements more or less belong to this configuration. Therefore, if you go back to Boltzmann formula and let me keep the formula here and look at S equals k log ω, then you find S one is when everybody is on the left.
01:05:26
So, the entropy is k log 2 to the N. That is k log 1. And S two is 50-50 mix, half on the left and half on the right. That entropy is k log 2 to the N.
01:05:44
Therefore, you can see that's what I wrote down here, k log 2 to the N. So, the way you understand entropy is the following. If I give you a guess and ask,
01:06:02
will they ever like to be in the left half of the box on their own? Of course, they won't. But if you start them out that way, you force them to be on the left, they'll be on the left. The question is, if you remove the partition, what would they likely do? Unlike coins, which once they land on their head, have to stay on
01:06:21
their head, these molecules can move around. The question is, if they were all left to begin with, how long can that possibly last? If they start moving randomly, they will always end up doing this because there are many, many more ways to do this than to do this. So, things go from one arrangement,
01:06:43
macroscopic to another one, left to their own device, because the final state can be realized in many, many more ways than the initial state. Similarly, if you combine a hot gas and a cold gas, this is hot and this is cold, with a thermally
01:07:00
insulating partition, then all the hearts are on the left, all the coals are on the right. If you remove the thermal insulation, or not even remove it, just replace it by a heat conducting thing, eventually the temperatures will equalize. So, what you find every time is, there's many, many ways in which you can pack your atoms.
01:07:20
If you put all the hot on the left, all the cold on the right, that can be achieved in fewer ways than in which you let them all go wherever they like. So, what you find is entropy is a direct measure of how disordered your system is. One technical measure of disorder is to ask, how many microscopic
01:07:41
arrangements can lead to what I see, and take the log of that number. And what you find is, this is a very low entropy, this is a very high entropy, because this configuration, in other words, if you took a gas in a whole room and asked, will it ever go to this configuration? That is a slight chance it will.
01:08:01
So, that chance is 1 part and 2 to the n, where n is 10 to the 23. So, it's not something you should wait for, but it can happen. So, the second law of thermodynamics is a statistical law. Microscopically, it's perfectly allowed for a gas to suddenly, all the gas in the whole room can come to where I am.
01:08:21
It's allowed, but you don't hold your breath, because that's not likely to happen. The odds for that is, again, one part in 1 over 2 to some huge number. On the other hand, if in this part of the room we release some gas, it will very quickly spread out, because there are more ways to do it. So, we understand completely now why certain things occur and why they don't.
01:08:42
Because if you took hot water and cold water, you're separating them to fast molecules and slow molecules. As long as they're in different insulated containers, that's the best they can do. But if you put them in contact, so that they no longer have to be separated, the question is, will they remain separated? It's like saying, I have a bunch of coins which are all left and all right,
01:09:02
but they're allowed to flip as a function of time, but they'll never remain that way. Okay, so what will happen is the cold molecules and hot will mingle and soon the container will have cold everywhere and hot everywhere and you will get some intermediate temperature. Or if you took two different dyes, you know, water here, on top of it you pour some red paint,
01:09:21
they'll be initially separated, but after a while they will mix. Because there's no reason for the hot, for the red molecules to stay on the top forever. They would like to occupy the whole box and so would the colorless ones, and eventually it will become pink. Because there are more ways to remain pink than to remain separated. Likewise, if it's pink spontaneously separated into red and colorless,
01:09:43
that is very, very improbable because entropy for that state will be lower. So, you have to understand the second law of thermodynamics is saying certain things occur because if you go in that direction you can realize that there is no real arrangement in more ways. And since microscopic motion is random, it's like tossing coins,
01:10:03
to ask for n heads when you toss n times is like asking all molecules to be on the left side of the room. To ask for 50-50 head and tails is like asking for molecules to populate the room equally. Okay, so that's called the arrow of time. But you've got to be careful about one thing. The entropy of a part of the universe can go down.
01:10:23
Just the entropy of the whole universe will not go down. So, all of life is an example of lowering entropy because the creation of life or the creation of tomatoes out of mud is a highly organizing process. So, the entropy there is really going down. But if you kept track of the rest of the world,
01:10:41
you will find that some corresponding increase of entropy is somewhere else. Or take your freezer. You know, your refrigerator sucks heat out of your freezer. ΔQ over T for that is negative, but somewhere there's bigger heat emitted outside from the exhaust in your refrigerator. If you took care of all of that, entropy of the universe will go up, or remain the same, but it doesn't go down.
01:11:02
All right, I have given you guys three problems. You should go on the website today. Two of them should take about five minutes, other one maybe ten minutes. I want you to think about calculating entropy changes. That's part of this syllabus, that's part of our course. It'll come online today. So, I have asked for them to be due either on Friday or
01:11:21
Monday. I don't know which one, whichever one, matter what. But it's a 30-minute exercise. I would suggest you guys attack it when you go back. Okay, so this is it. This is the end of part one. I'll see probably many of you when we do part two to look at the ENM and quantum mechanics.
01:11:42
Thanks a lot.