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PHYS 200: Fundamentals of Physics I
Lecture 24
 The Second Law of Thermodynamics (cont.) and Entropy
Overview
The focus of the lecture is the concept of entropy. Specific examples are given to calculate the entropy change for a number of different processes. Boltzmann’s microscopic formula for entropy is introduced and used to explain irreversibility.
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htmlFundamentals of Physics IPHYS 200  Lecture 24  The Second Law of Thermodynamics (cont.) and EntropyChapter 1: Review of the Carnot Engine [00:00:00]Professor Ramamurti Shankar: 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 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 T_{1}, isothermally, until you come to point B. So, the way you should imagine that is that this gas is placed on a reservoir whose temperature is frozen, I mean stuck at T_{1}. 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. The minute it tries to cool some heat will go from below to maintain the temperature and start at that volume and end at that volume. During this process, some amount of heat Q_{1} enters the gas. Next thing you do, you let the gas expand some 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 a lower temperature T_{2}, 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 a temperature reservoir T_{2}, 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, the temperature should go up; but you’re not letting the temperature go up, you’re forcing it to have the temperature T_{2}. So, it will 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. 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 Q_{1} from the furnace at T_{1}, it has some Q_{2} rejected at the exhaust, which is at T_{2} and it delivers a certain amount of work which, by the Law of Conservation of Energy, has to be Q_{1}  Q_{2}. Do you understand that? The work done by the gas equal to that. And who paid for it? It is the reservoir, 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 gases come back to where it is. 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 the cycle, it’s back to where it started and you can do this any number of times. 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 Q_{1}. Now, writing W as Q_{1}  Q_{2} you can write this as 1 minus Q_{2} over Q_{1}. 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 with the Carnot engine I got this as the efficiency. Now, why is the efficiency not 100%? That’s because Q_{2} 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 atmospheric temperature, you can calculate the efficiency for this engine by actually calculating Q_{2} and Q_{1}. That’s a simple problem of integrating the work done and going from here to here. Do you remember that? 1  nRT_{1} log V_{B} over V_{A} and downstairs you have NR T_{2} log V_{C} over V_{D}. Again, you’ve got to remember why that is true. When you go from A to B there is no change in internal energy because temperature is the same. Therefore, the work done and the heat import are equal numerically, so Q_{2} is the work done and Q_{1} 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 on an adiabatic curve, A and D are on an adiabatic curve, that these logs actually cancel, these guys of course, cancel to give the final result 1 minus T_{2} over T_{1}. 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 T_{2} over T_{1}. Okay now, you can say why I’d be interested in this very primitive engine containing a cylinder and gas and so on. Okay, this efficiency of this stupid engine, I’m sure people can build a better engine. So, here is Carnot’s claim. Mr. Carnot say,s “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 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 now 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 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, because you get an electric bill at the end of the month, because there’s a compressor doing a lot of work. The claim is–not–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 up hill. That’s not allowed; that’s a postulate and you have to grant him that postulate which we accept to be phenomenologically valid. But taking that postulate, 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 on every step in the Carnot cycle, let’s say going from here to here, I can also go backwards. I can put a grain of sand and compress a piston, or I can take a grain it’ll go back to where it was. You’re never far from equilibrium and if you can go one way you can go the other way. That also means the Carnot engine, starting at the point A, can go backwards to D then C then to B and then to A. If the Carnot engine went backwards it would look like this: it will take heat Q_{1}–I’m sorry it will take heat Q_{2}, somebody will give it work W and it’ll reject Q_{1} at this temperature. This is the Carnot refrigerator and that’s the Carnot engine. The Carnot engine and the refrigerator are the same machine; you just can make it run oneway or the opposite way. That’s what we’re 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 Carnot engine. So, let us say a Carnot engine takes some heat, say 100 calories, delivers 20 calories in work, rejects 80 calories is operating between some two temperatures T_{1} and T_{2}. This is the Carnot. I am taking a particular example where the efficiency is what, onefifth. But don’t worry about the fact that, yeah, so it’s onefifth, so one minus T_{2} over T_{1} should be onefifth. Now, if you say you have a better engine, you have a better efficiency. What you really mean is that your engine can take a 100 calories and deliver more than 20 calories, maybe even deliver 40 calories and reject only 60 calories, but it’s also operating between the same two temperatures. So, this is your engine. U for–no, Y for your engine, okay; your engine is better than Carnot’s engine, okay that’s your claim. You’re now going to get shot down. So, how am I going to shoot this down? I am going to run the Carnot engine backwards first thing, then I am going to get a Carnot engine which is twice as big as this Carnot engine, it’s not more efficient, it has just got twice as much gas. If you take a Carnot engine and do the following things, run it backwards and make it twice as big. What will that Carnot engine do? 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 twotimes Carnot; let me put a star saying it’s Carnot run backwards. 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 smaller piston. I’m just saying I want it to be twice as big for the following simple reason. Your engine’s giving out 40 units of work; my 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 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. 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? 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 hot body with no other changes anywhere in the universe, and that is not allowed. Therefore, you cannot have an engine more efficient that the Carnot engine. The logic is pretty simple. These numbers picked are representative, but you can take any number as long as your engine does better than mine, instead of 40 calories, would have made 30 calories needed. If it produced 30, I’ll get an engine which is onehalf times as bigger than the standard engine; run it backwards, your 30 will feed my engine and you will find once again heat is flowing from a cold body to 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 primitive engine, is the standard for all engines. No engine can be better than the Carnot engine. Now, the key to the Carnot engine is that it’s a reversible engine. If I had more time I could show you that all reversible engines operating between two temperatures will have the same efficiency, namely that of the Carnot engine. Okay, now what has the Carnot engine got to do with what I started saying earlier? You remember I started saying earlier 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. 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 am 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 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’s better than this one. It’s a theoretical maximum. In reality, efficiency will be less than this because most engines have a loss; there is heat leaking, 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. What I want you to notice if the following: for a Carnot engine Q_{1} over Q_{2} turned out to be T_{1} over T_{2}. Right, I did that for you here. Therefore, I am going to write it as follows: Q_{1} over T_{1} minus Q_{2} over T_{2} equal to zero. That’s just a simple rewrite. But I am going to write this in another rotation. 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 Q_{1} over T_{1}. In this segment there was no Q_{1} or Q_{2}, there was no Q, because this is the adiabatic process. In this segment, I had minus Q_{2} over T_{2} and the last segment I had zero and whole thing is zero. In other words, what I’m telling you is the following. At every stage, if you looked at the heat absorbed by the system, in stage i, divided by the temperature of stage i and you added all of them you get zero. My process had four stages. Stage 1 is here, stage 2 is adiabatic, state 3 is this, and stage 4 is zero again. In this summation Q_{i} is defined to be ΔQ_{i} is the heat input in stage i in the cycle, or in the process. So, we define Q_{2} to be positive, even though it was rejected by the engine, but in this summation, the agreement I make is ΔQ is positive if the heat comes into the system, ΔQ 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 say that’s the amount of heat in the system. Why? Because if you take a point there and say there’s 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, this is the work done, therefore, the work done is equal to the heat and it is not zero. So, in this example what would’ve happened is you’d added heat Q to the system. So, if there is some notion of how much heat is there in the beginning, you got that plus the Q that you added and yet you are back to where 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. 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 the T, energy returns to the old 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. It doesn’t matter where you wander off in the PV diagram. Heat is not a state variable, there’s nothing called a heat at that point, because I am 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. Chapter 2: Calculating the Entropy Change [00:19:18]But there is a new unique quantity S, called entropy, 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 the heat you add divided by the temperature. In a tiny little process, if you are at some temperature T, you add a little amount of Q, keep track of all the changes, and that change will 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, but ΔQ/T, namely give a weighting factor of 1/T to the heat you add, then the positives and negatives cancel and give you exactly zero. In other words, Q_{1} was not Q_{2}, Q_{1} is bigger than Q_{2}, but Q_{1}/T_{1} precisely balances Q_{2}/T_{2}. The heat absorbed at higher temperature if divided by T_{1}, to compute the change in entropy, then in the upper part of the Carnot cycle here, and the lower part of the Carnot 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 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 net zero change. That means every point can be associated with a number you can call entropy. Entropy is like a height. Suppose you are walking around on some landscape, each point is 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 function. It is just like a potential energy function. 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. 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 is 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 that the change in the total of all the ΔQs over T is in fact zero. So, we’ll develop up a feeling for what it means, but historically this is what happened. People realized, hey there’s another state variable. We introduce this 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 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, let’s say m grams of some substance. It’s got some specific heat C. And I’ll change the temperature say from some T initial for some T final. I do that by keeping the system–starting the system at T_{i} and I should never be far from equilibrium. That’s one of the conditions on this. In fact, you can say it should be near, always near equilibrium. So, the system starts at temperature T_{i}, I put it on a heat bath, which is infinitesimally warmer than this and I let it heat up at that point. Then, I put it on another reservoir, slightly hotter than this one. 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 welldefined temperature, has a welldefined heat input and I wanted to add it all up. So, what’s the change in entropy? S final minus S initial is equal to dQ over T, summed over all the parts, I should write this as an integral, mC ΔT over T. Do you see that, dQ is mC ΔT? But I got to divide by T and that integral is done between some initial and final temperature and we can do this integral rather trivially in an mC log T final over T initial. That is the increase in entropy of some m grams of some substance of specific heat 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 it 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. How many calories does it take to raise the substance from initial to final temperature? That you understand. When you divide by T, something 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 higher temperature. If you cooled it from a higher to lower temperature, you can use the same formula T_{f} over T_{i}, but T_{f} will now be smaller than T_{i}; if you cooled it, if the log of a number less than one is negative and the entropy changes it’ll be negative. So, let me do one more entropy calculation that’s going to be pretty important for us. That entropy calculation is this. Take a gas and watch it expand isothermally from some starting point, at some fixed temperature T that goes from V_{1} to V_{2}. What is the change in entropy when it goes from here to here? Again, the system must always 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 S_{2}  S_{1}, this is 2 and this is 1. That is equal to the sum of all the heat transfers divided by temperature at every little step of the way. But remember this is also the same as PΔV over T. Why? Because on an isothermal ΔU is zero, I repeated it many times but you’ve got to know this, ΔQ is PΔV. But P over T is nRT over V, ΔV. You also want to divide it by a T. Do you understand P over T is nR over V? So, P = 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 V_{2} over V_{1}. That is the change in entropy of this gas when it went from volume V and went to volume V_{2}, 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? 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 V_{2} over V_{1}. Since the whole process takes place at fixed temperature, 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 a change in entropy. In other words, during the whole process you feel at one temperature then integral of ΔQ over T is just 1 over T times integral of ΔQ, which is the total heat transfer. Anyway, this is the heat, this is the change in entropy, S_{2}  S_{1} is nR log V_{2} over V_{1}. Okay, so what have I done 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 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, if it is defined to be ΔQ over T, has a property the total change in S is zero and you go around a loop. That means at every point you can associate an S. S is a property of the point. Another example–so then, I said let’s get used to computing change in entropy. 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 of integral of dT over T was logarithm. Then, I took an ideal gas, I let it expand isothermally, I found the entropy change, 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 every point there is a unique entropy, then the entropy difference between 2 and 1 should be independent of how I go from 1 to 2. Do you understand? If you’re walking on a 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 the height difference. So, I could find the entropy another way. Let me show you another way that’s easy for you guys to work out. Come down like this and go to the right like this. And find the entropy change. You’ll get the same answer as I got on this thing. 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 zero for all its parameters. Then in this step, 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 the heat transfer dQ is C_{V}dT and I divide by T and I do the integral from T_{1} to this point T_{0}. That’s the entropy change here. Then in the horizontal part, since I’m going at constant pressure, I go C_{P}dT over T from T_{0} up to back to T_{1}, because this T_{1} and T_{2} are the same temperature. I have to add all these to get the entropy change. Now it’s a twostep process, you come down because you’re at constant volume and you’re doing something to the gas, dQ is C_{V}dT. Horizontally, you’re at constant pressure, dQ is C_{P}dT. That’s the definition of specific heat at constant volume, constant pressure. But notice the following: C_{P} is equal to C_{V} + R. So, put C_{V} + R here and look what you get, then you get C_{V} times log T_{0} over T_{1} plus log T_{1} over T_{0}, which we can write as the product here, plus R log T_{1} over T_{0}. In other words, this log is log of T_{0} over T_{1}, the next log is log of T_{1} over T_{0}, log A plus log B is log of AB. But when you do this look what happens here. This all cancels, log of 1 is zero; the total entropy change is R log T_{1} over T_{0}. But for a gas at constant pressure, using PV = 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 T_{1} I really meant the temperature at this point, second point. So, that temperature, or this temperature is also that volume divided by the initial volume, which is V_{1}. See, this is a confusing problem because T_{1} happens to be T_{0}. The correct way to do this for me would be to write T_{0} over T_{1} times T_{2} over T_{0}, then realize the fact that T_{2} and T_{1} are equal because I am on an isotherm. That’s what I should have done. So, I should cancel this factor because T_{2} is T_{1}, but here T_{2} over T_{0} is V_{2} over V_{0}, which is V_{2} over V_{1}. That’s, of course, the entropy change I got in one shot here. So, you can find entropy change anyway you like. You usually pick the easiest path. Alright, so we have now learned how to find entropy change. We have no idea what it means. It seems as if when you heat 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, and you make it a logarithm, but we know at every infinitesimal portion we’re adding positive numbers. And when I cooled things it goes down. Why not just call it temperature? Why do you need this new concept called entropy? So, that’s what the rest of the lecture is about. It’s a very, very powerful and beautiful concept so I wanted to explain it, 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 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 to the whole room, it’ll never untrap and go back to half the room. A lot of things happen one way but not the other and I said I am looking for a mega law that will prevent all these things from happening. Now, I’m ready to state that law. Chapter 3: The Second Law of Thermodynamics as a Function of Entropy [00:35:34]This is the third law–I mean, the Second Law of Thermodynamics. Carnot said it one way, 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, 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 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 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 allowed, right, according to Carnot, this is not allowed. Right, heat can flow downhill, cannot flow up hill. Let’s see the entropy change the universe with the two cases. In this case, with this guy, the change in entropy is the following: the upper reservoir lost some amount Q, so ΔQ is a negative number, add some temperature T_{1}, the lower reservoir gained Q at temperature T_{2}. Now, what’s the overall sign of this? Think about it. T_{1} is bigger than T_{2}, 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 up hill, then this reservoir loses some heat Q at temperature T_{2}, the other one gains heat Q at temperature T_{1}, 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 universe goes up. And it’s very simple, it’s only–it’s because, it’s true that the heat loss of this guy is the heat gain of that guy, that’s the Law of the Conservation of Energy. So, energy doesn’t change, but entropy changes, because for entropy it’s not the heat transferred, it’s heat transferred divided by temperature. Therefore, a heat loss at one temperature and a [equal] heat gain at a lower temperature don’t cancel when it comes to entropy. In one case the entropy goes up and is allowed; other case entropy goes down, that’s not allowed. Okay so let’s try, that’s certainly one 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. So, let’s take one more example. Suppose I have some mass of water at some temperature T_{1} and an equal mass of water at temperature T_{2}. One is hot and one is cold. I put them together, what will happen? We believe it will then get this big mass, all at some common temperature T star which is T_{1} + T_{2}/2; this is just by symmetry. It’s equal mass, equal specific heat, where will they meet? They will meet half way. They will meet here. Energies, of course, is conserved, and that’s how we determine in fact where they will meet. But look at the entropy. What happens with entropy change? With entropy change, you should imagine this water being steadily heated by putting it in contact with a lot of reservoir, so it’s never far 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 T_{1} to T star for one thing and dT over T from T_{2} to T star for the other one. Do you understand that? They both meet at T star, upper limit is T star, lower one is T_{1} for one and T_{2} for the other, so the change in entropy becomes M log T star squared over T_{1}T_{2}. Now, we have to ask, okay you got an entropy change from start to finish, but how do you know it’s positive? Can anybody give some reason why this has to be a positive without looking into properties of logarithms and so on? Any of these in, what–yes? Student: [inaudible] Professor Ramamurti Shankar: Yeah, that’s one way to prove that. But I am saying, I’m going to prove it to you that way. 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 temperature here, this one is gaining some ΔQ, 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, this guy’s hotter than this guy. So, every calorie of heat it gains at a higher temperature, every calorie it loses is divided by higher temperature, every calorie this gained is divided by 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. 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 T_{1} plus T_{2}^{2} over 4 bigger than T_{1}T_{2}? Or I’m saying is T_{1} plus T_{2}^{2} bigger than 4T_{1}T_{2}? If you rearrange this, you can show the lefthand side becomes T_{1}  T_{2}^{2}; that’s of course, a positive number because this will be T_{1}^{2} + T_{2}^{2} + 2T_{1}T_{2}. When I bring this guy to the other side, it’ll become minus 2T_{1}T_{2} and so it’s clearly positive. So, when hot and cold meet and create lukewarm, 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 which is cold and a part on 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; anything that is forbidden is in the direction of decreasing entropy. Chapter 4: The Microscopic Basis of Entropy [00:43:46]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, that last part of what I’m going to do is going to explain to you what is the microscopic basis of entropy. Why is the entropy going up? Why is–why do we understand the tendency of things to go the direction of increasing entropy? All I’ve shown you now is that it’s a mystical quantity called entropy, circumstantially you find every time something is forbidden, it’s because had it taken place entropy would have gone down. Do you remember the examples? Heat flowing from hot to cold entropy goes up, allowed. Flowing from cold to hot, entropy goes down, not allowed. Hot and cold mix and become lukewarm; allowed, entropy goes up. Luke warm 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 partition, half way. The molecules are stuck on the left side, 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, rip it out. So, let’s follow this gas. Initially, it is a point here. Then, there is a period when it goes off the radar because I told you when you suddenly remove the wall, the gas is not in a state of equilibrium. It doesn’t even have a welldefined pressure. The minute you remove the wall, pressure is something here, pressure is zero in the vacuum. So, you 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 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, that is forbidden, so I want to ask did the entropy go up? Now, how do we do the entropy calculation? Here’s the wrong way to do the calculation. You go back and say ΔS at ΔQ over T, but this whole box I forgot to mention–I’m sorry, this whole box is thermally isolated. 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 = 0 so you can divide by T, you can do what you want, so this entropy change is zero. But that is a wrong argument. Can you think about why that’s not the way to do the entropy change? Why that’s the wrong analysis? First of all, could that be the right answer? Yes. Student: It can’t because then that would mean that going backwards is not a valid process. Professor Ramamurti Shankar: No, that is correct, but I’m saying this computation of entropy. The one line calculation I did, namely there is no heat inflow, the ΔQ = 0, so ΔQ over T summed up is also zero. That’s not how you do the entropy change. Is there any condition I made on computing? Yes. Student: Well, uhm… there’s a gas [inaudible] Professor Ramamurti Shankar: Wait in fact, here’s 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? Yep. Student: [inaudible] Professor Ramamurti Shankar: Ah, 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? If, when–If I were to move this piston here, there is no pressure pushing this gas back. Do you understand that? Doesn’t do any work, doesn’t take any heat. So, what does that mean? In the end when you settle down, what can you conclude? No heat input, no work done, so what does that mean in terms of temperature? Student: Same. Professor Ramamurti Shankar: Same. Because the internal energy cannot change because ΔQ = 0, 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 against the atmospheric pressure. This gas is expanding into a complete vacuum. It’s got no piston to push against, nothing from outside pushing, so these two points are 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 definitely has a welldefined entropy you can calculate, because it’s an equilibrium state. Initial one has a welldefined 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 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 welldefined 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 where I’m moving and add the ΔQ over Ts. The correct rule for entropy change is ΔQ/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. Since this 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 S_{2}  S_{1} is nR log V_{2} over V_{1}. 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 into a vacuum takes it from one equilibrium state called 1 to another equilibrium state called 2. In the actual expansion, the system went through a stage 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. How do you find a temperature when half the box is empty? Doesn’t have it, you cannot it T 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 welldefined entropy. What we’re trying to do is to compute that difference. The way to compute the difference is to forget about what actually happened and instead do the following. Take the gas like that and turn it this way, [90 degree rotation] if you like. Just for convenience, put a piston there at the halfway point, put some weights on it so that the pressure equals the pressure inside and slowly remove little by little, and keeping it on a reservoir a temperature T_{1}. Which is the temperature at which all of this happens, or at some temperature T. As you remove the grain of sand, what happens is the gas gets some energy, that’s the 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 entropy of the universe does not change at all. The entropy gain of the gas; this is an entropy loss of the reservoir. But I used this as a device for finding the entropy gain of the gas in the other process. So, this is the part that is absolutely central to this whole argument 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, 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 welldefined 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 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 at zero. Why? Because you are 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 processes. One is a reversible process in which a gas is slowly allowed to expand from the initial state of the final stage 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 punch line is I don’t care what happens in reservoir, I want to know how much the gas gained, because that is the gain I need to find S_{2}  S_{1} for this process. So, this is how we always find entropy change. In real life processes happen in a way in which entropy of the universe goes up. Because take this box, in the real experiment, remember there is no reservoir, there’s 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 1 to 2. To find the increase in entropy of the gas when it goes from 1 to 2, 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 use 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 repeat it anymore, but it’s an important point and you should really think about that. Alright, now the last point is, what does–Let’s analyze this final formula I got for entropy change for free expansion of a gas. Let’s take a simple case where the gas expanded from some volume V to volume 2 times V. What is S_{2}  S_{1}? S_{2}  S_{1} = nR log 2. 2 happens to be V_{2}/V_{1}; 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 log of 2 to the N times k. Now, I will tell you what all these factors mean and that comes from the following. Thermodynamics only told you the formula for the change in the entropy of a gas, 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, 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 punch line of what’s called statistical mechanics, which gives you the microscopic origin of thermodynamics. Thermodynamics to this stage talk 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. I am going to write it down. Boltzmann constant times log ω. I’ll tell you what omega 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 orchestras and everything else, we go to Boltzmann’s tomb and we read this formula once more with some passion. That’s a 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 omega means. I am going to say it in words and then I am going to apply it to this problem. If you took any gas, let it be under certain conditions that macroscopically has 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. Suppose I wait a little bit, what did 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 has come here. In particular, if you divide the volume into some number of little cells that you can count, on the average the gas’ uniform density, there are some number of atoms, per every little cube you can form, a little later I move on and somebody else moves in. Macroscopically, it looks the same to me. Microscopically, stuff is going on. Omega 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, 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 the log of the number of these arrangements multiply 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 arrangements, if you really want to get down to it, requires dividing the box into tiny little cells, maybe 1 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: that either a molecule is on the left or is on the right. We are saying these are the only two positions open to the gas; left or right. Of course, in the left, there is left corner top and left corner bottom, but don’t worry about it. Just say that 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. How many ways can it be in this arrangement? As compared to how many ways it can be in this arrangement? It is like the following: molecules are moving randomly, a given molecule has got 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 lefthand side. You can see it’s like tossing a coin N times and wanting heads every time. Because you want every guy in a random process to end up picking 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, 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 left side. That looks like left, left, left, right, left, left, left. But that’s one arrangement. But have I done my complete homework here? Are there more arrangements? I want you to think about it. Student: [inaudible] arrangements Professor Ramamurti Shankar: Pardon me. Student: [inaudible] arrangements. 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 molecule, molecule Moe or Joe or Poe, whoever’s here. You can pick them in N ways. Already when one is on the left and N  1 on the right, there are more ways in which that can happen. So, your eye sees one on the right, N  1 on the left. There are N ways in which that can happen. It’s like saying if I toss coins, I want 1 head and 99 tails. Well, it turns out there are a 100 ways in which it can happen, because the one head I got could have occurred in one of the hundred different turns. Now, suppose I want 2 heads and 98 tails, then as you know the way to do that is 100 times 99 divided by 1 times 2. These are some combinatorics you learn in high school. So, you find that if you plot number heads over the total number, and ask for how many ways in which that can happen, all heads can occur in only way, every coin has to be head, 1 head less, 1 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 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 in one way and which you can get all tails. But as you start scrambling them there are more and more ways. That means for the molecule, if you have a 100 molecules, the number of states omega is largest when they’re equally distributed. In fact, as the number molecules goes from 100 to 1000 to million to 10^{23}, this function is so sharply peaked that the approximate formula for omega, 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 the Boltzmann formula and, wait let me keep the formula here, and look at S = K log ω, then you find S_{1} is when everybody is on the left, that is k log 1, and S_{2} is 50/50 mix, half on the left and half on the right, that entropy is k log 2^{N},. Therefore, you can see that’s what I wrote down here, K log 2^{N}. So, the way to understand entropy is the following. If I give you a gas and ask, “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 the head, they have to stay on the head, these molecules can move around. The question is if they were all left to begin with, how long can they 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, 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 cold gas–This is hot and this is cold, with a thermally insulating partition, then all the hots are on the left and all the colds are on the right. If you remove the thermal insulation, but not even remove it, just replace it by a heatconducting thing, eventually the temperatures would equalize. So, what you find every time is, there’s many, many ways in which you can pack your atoms–if you put all the hot on the left and all the cold on the right, that can be achieved in fewer ways than in which you let them go wherever they like. [He should have said, “If you put all hot ones on one side and cold on the other, there are fewer ways of doing that, than if you let them go wherever they liked”] So, what you find is entropy is the direct measure of how disordered your system is. One technical measure of disorder is to ask how many microscopic arrangements can lead to what I see, and take the log of that number. And what you find is this has 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, there is a slight chance it will. That chance is one part and 2^{N}, where N is 10^{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, for suddenly all the gas in the whole room to come to where I am. 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’ll 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. Because if you took hot water and cold water, you are 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 bunch of coins which are all left and all right, but they’re allowed to flip as a function time, but they’ll never remain that way. Okay, so what’ll 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, they’ll be initially separated 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’s more ways to remain pink than to remain separated. Likewise, if pink spontaneously separated into red and colorless, that is very, very improbable, because entropy for that state would 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 arrangement in more ways. And since microscopic motion is random, it’s like tossing coins to ask for N heads many thousand times; it’s like asking all the 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. It’s just the entropy of the whole universe will not go down. So, all of life is an example of lowering entropy because, you know, 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, you will find there’s some corresponding increase of entropy 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 of the refrigerator. If you took care of all of that, entropy of the universe will go up or remain same; it really doesn’t go down. [end of transcript] Back to Top 
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