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22. The Boltzmann Constant and First Law of Thermodynamics — Transcript

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0:01So, I had to leave you in the middle of something pretty
0:05exciting, so I'll come back and take it from there.
0:09So, what is it you have to remember from last time?
0:13You know, what are the main ideas I covered?
0:16One is, we took the notion of temperature, for which we have
0:19an intuitive feeling and turned it into something more
0:23quantitative, so you can not only say this is
0:25hotter than that, that's hotter than this,
0:27you can say by how much, by how many degrees.
0:29And in the end we agreed to use the absolute Kelvin scale for
0:34temperature. And the way to find the Kelvin
0:38scale, you take the gas, any gas that you like,
0:42like hydrogen or helium, at low concentration,
0:47and put that inside a piston and cylinder.
0:51That will occupy some volume and there's a certain pressure
0:55by putting weights on top, and you take the product of
0:59P times V, and the claim is for whatever
1:03gas you take, it'll be a straight line.
1:06Remember now, this is in Kelvin. Your centigrade scale is
1:10somewhere over here, but I've shifted the origin to
1:13the Kelvin scale. So, somewhere here will be the
1:16boiling point of water, somewhere the freezing point of
1:19water; that may be the boiling point
1:21of water. And if you took a different
1:24amount of a different gas, you'll get some other line.
1:28But they will always be straight lines if the
1:31concentration is sufficiently low.
1:33In other words, it appears that pressure times
1:37volume is some constant. I don't know what to call it.
1:41Say c, times this temperature.
1:48And you can use that to measure temperature because if you know
1:51two points on a straight line then you know that you can find
1:54the slope and then you can calibrate the thermometer,
1:57then for any other value of P times V that you
2:00get, you can come down and read your temperature.
2:03That's the preferred scale, and we prefer this scale
2:06because it doesn't seem to depend on the gas that you use.
2:10I can use one; you can use another one.
2:12People in another planet who have never heard of water--they
2:15can use a different gas. But all gases seem to have the
2:18property that pressure times volume is linearly proportional
2:22to this new temperature scale, measured with this new origin
2:26at absolute zero. There is really nothing to the
2:28left of this T = 0.
2:33The next thing I mentioned was, people used to think of the
2:37theory of heat as a new theory. You know, we got mechanics and
2:42all that stuff--levers and pulleys and all that.
2:45Then, you have this mysterious thing called heat,
2:47which has been around for many years but people started
2:50quantifying it by saying there's a fluid called the caloric fluid
2:53and hot things have a lot of it, and cold things have less of
2:56it, and when you mix them the caloric somehow flows from the
3:00hot to the cold. Then we defined specific heat,
3:03law of conservation of this caloric fluid that allows you to
3:07do some problems in calorimetry. You mix so much of this with so
3:11much of that, where will they end up?
3:13That kind of problem. So, that promoted heat to a new
3:17and independent entity, different from all other things
3:21we have studied. But something suggests that it
3:25is not completely alien or a new concept, because there seems to
3:29be a conservation of law for this heat,
3:31because the heat lost by the cold water was the heat gained
3:35by the hot water. I'm sorry.
3:37Heat gained by the cold water was the heat lost by the hot
3:40water. So, you have a conservation law.
3:42Secondly, we know another way to produce heat.
3:47Instead of saying put it on the stove, put it on the stove,
3:50in which case, there is something mysterious
3:52flowing from the stove into the water that heats it up,
3:55I told you there's a different thing you can do.
3:58Take two automobiles; slam them.
4:01This is not the most economical way to make your dinner but I'm
4:06just telling you as a matter of principle.
4:09Buy two Ferraris, slam them into each other and
4:12take this pot and put it on top and it'll heat up because
4:16Ferraris will heat up. The question is what happened
4:19to the kinetic energy of the two cars?
4:21That is really gone. So, in the old days,
4:23we would say, well, we don't apply the Law of
4:26Conservation of Energy because this was an inelastic
4:28relationship. That was our legal way out of
4:31the whole issue. But you realize now this
4:34caloric fluid can be produced from nowhere,
4:37because there was no caloric fluid before,
4:40but slamming the two cars produce this extra heat.
4:43So, that indicates that perhaps there's a relation between
4:47mechanical energy and heat energy--that when mechanical
4:50energy disappears, heat energy appears.
4:54So, how do you do the conversion ratio?
4:56You know, how many calories can you get if you sacrifice one
5:00joule of mechanical energy? So, Joule did the experiment.
5:03Not with cars. I mean, he didn't have cars at
5:06that time, so if he did he would've probably done it with
5:08cars. He had this gadget with him,
5:11which is a little shaft with some paddles and a pulley on the
5:15top, and you let the weight go down.
5:18And I told you guys the weight goes from here to here,
5:22the mgh loss will not be the gain in ½ mv^(2).
5:26Something will be missing. Keep track of the missing
5:29amount. So many joules--but meanwhile
5:31you find this water has become hot.
5:33You find then how many calories should have gone in,
5:36because we know the specific of water,
5:38we know the rise in temperature, we know how many
5:40calories were produced. And then, you compare the two
5:44and you find that 4.2 joules = 1 calorie.
5:52So, that is the conversion ratio of calories to joules.
5:55One joule, 4.2 joules of mechanical energy.
5:58So, in the example of the colliding cars,
6:01take the ½ mv^(2) for each car, turn it into joules,
6:04slam them together. If they come to rest,
6:07you've lost all of that, and then you take that and you
6:11write it as--divided by 4.2 and that's how much calories you
6:15have produced. If the car was made of just one
6:18material, it had a specific heat, then it would go up by a
6:21certain temperature you can actually predict.
6:23Okay. So today, I want to go a little
6:27deeper into the question of where is the energy actually
6:32stored in the car, and what is heat.
6:35We still don't know in detail what heat is.
6:37We just said car heats up and the loss of joules divided by
6:404.2 is the gain in calories. Now, we can answer in detail
6:46exactly what is heat. That's what we're going to talk
6:50about today. When we say something is
6:52hotter, what do we mean on a microscopic level?
6:55In the old days when people didn't know what anything was
6:58made of, they didn't have this understanding.
7:00And the understanding that I'm going to give you today is based
7:03on a simple fact that everything is made up of atoms.
7:06That was not known, and that's one of the greatest
7:09discoveries that, in the end,
7:11everything is made up of atoms, and atoms combine to form
7:14molecules and so on. So, how does that come into
7:17play? For that, I want you to take
7:19the simple example where the temperature enters.
7:23That is in the relation PV equal to some constant
7:27times temperature. Do you know what I'm talking
7:31about? Take some gas,
7:32make sure it's sufficiently dilute, put it into this piston,
7:36measure the weights on top of it,
7:39divide it by the area, to get the pressure,
7:41that's the pressure, that's the volume.
7:43The volume is the region here, multiply the product;
7:48then, if you heat up the gas by putting it on some hot plate,
7:52you'll find the product PV increases,
7:55and as the temperature increases, PV is
7:57proportional to T. We want to ask what is this
8:01proportionality constant. Suppose you were doing this.
8:05In the old days, this is what people did.
8:07What did we think should be on the right-hand side?
8:12What is going to control this particular constant for the
8:16given experiment? Do you know what it might be
8:20proportional to? Yes?
8:23Student: Amount of gas? Professor Ramamurti
8:25Shankar: Amount of gas. That's true,
8:26because if the amount of gas is zero, we think there's no
8:28pressure. When you say "amount of gas,"
8:30that's a very safe sentence because amount measured by what
8:35means? By what metric?
8:37Student: Probably number of particles?
8:41Professor Ramamurti Shankar: Right.
8:42Suppose you were not aware of particles.
8:44Then, what would you mean by "amount of gas?"
8:47Student: Mass. Professor Ramamurti
8:49Shankar: The mass. Now, if you guys ever said
8:51moles, I was going to shoot you down.
8:53You're not supposed to know those things.
8:55We are trying to deduce that. So, put yourself back in
8:58whatever stone ages we were in. We don't know anything else.
9:01Mass would be a reasonable argument, right?
9:04What's the argument? We know that if you have some
9:07amount of gas producing the pressure, and you put twice as
9:10much stuff, you would think it will produce
9:12twice as much pressure. Same reason why you think the
9:14expansion of a rod is proportionally change in
9:17temperature times the starting length.
9:19So, this mass is what's doing it.
9:21So, it's proportional to mass. It's a very reasonable guess.
9:24So, if you put more gas into your piston you think it'll
9:26produce more pressure. That's actually correct.
9:29So, let's go to that one particular sample in your
9:33laboratory that you did. So, put the mass that you had
9:36there. Then, you should put a constant
9:39still. I don't know what you want me
9:40to call this constant, say, c prime.
9:42This constant contains everything, but I pulled out the
9:46mass and the remaining constant I want to call c prime.
9:50This is actually correct. You can take a certain gas and
9:55you can find out what c prime is.
9:57But here is what people found. If you do it that way,
10:02the constant c prime depends on the gas you are
10:05considering. If you consider hydrogen gas,
10:08let's call that c prime for hydrogen.
10:12Somebody else puts in helium gas.
10:15Then you find the c prime for helium is one-fourth
10:20c prime for hydrogen.
10:28If you do carbon, it's another number.
10:31c prime for carbon is c prime for hydrogen
10:36divided by 12. So, each gas has a different
10:43constant. So, we conclude that yes,
10:46it's the mass that decides it but the mass has to be divided
10:50by different numbers for different gases to find the real
10:54effective mass in terms of pressure.
10:57In other words, one gram of hydrogen and one
11:01gram of helium do not have the same pressure.
11:05In fact, one gram of helium has to be divided by 4 to find its
11:12effect on pressure. So, you have to think about why
11:16is it that the mass directly is not involved.
11:18Mass has to be divided by a number, and the number is a
11:21nice, round number. 4 for this and 12 for that,
11:23and of course people figure out there's a long story I cannot go
11:27into, but I think you all know the answer.
11:29But now we are allowed to fast forward to the correct answer,
11:32because I really don't have the time to see how they worked it
11:36out, but from these integers and the
11:38way the gases reacted and formed complicated molecules,
11:41they figured out what's really going on is that you're dividing
11:45by a number that's proportional with the mass of the underlying
11:49fundamental entity, which would be an atom.
11:51In some case a molecule, but I'm just going to call
11:54everything as atom. So, if things,
11:56like, carbon, as atoms, weigh 12 times as
11:59much as things called hydrogen, then if you took some amount of
12:04carbon, you divide it by a number, like,
12:0712 to count the number of carbon atoms.
12:10Okay, so hydrogen you want to count the number of hydrogen
12:12atoms. So then, what really you want
12:15here is not the mass, but the number of atoms of a
12:20given kind. We are certainly free to write
12:23either a mass or the number of atoms, because the two are
12:27proportional. But the beauty of writing it
12:29this way, you write it in this fashion, by this new constant
12:33k, k is independent of the gas.
12:41So, you want to write it in a manner in which it doesn't
12:43depend on the gas. You can write it in terms of
12:46mass. If you did, for each mass
12:48you've got to divide by a certain number.
12:49TThen once you divide it by the number you can put a single
12:53constant in front. Or if you want a universal
12:56constant, what you should really be counting is the number of
13:00atoms or molecules.
13:04So, you couldn't have written it that way until you knew about
13:07atoms and molecules and people who are led to atoms and
13:09molecules by looking at the way gases interact,
13:11and it's a beautiful piece of chemistry to figure out really
13:16that there are entities which come in discreet units.
13:20Not at all obvious in the old days, that mass comes in
13:22discreet units called atoms, but that's what they deduced.
13:25So, this is called the Boltzmann Constant.
13:29The Boltzmann Constant has a value of 1.4 times 10^(-23),
13:38let's see, joules/Kelvin.
13:47That's it. Or joules/Kelvin or degrees
13:53centigrade.
14:01So, this is a universal constant.
14:08So, now what people like to do is they don't like to write the
14:11number [N], because if you write the number,
14:14in a typical situation, what's the number going to be?
14:18Take some random group gas. One gram, two grams,
14:21one kilogram, it doesn't matter.
14:23The number you will put in there is some number like
14:2610^(23) or 10^(25). That's a huge number.
14:30So, whenever a huge number is involved, what you try to do is
14:34to measure the huge number as a simple multiple off another huge
14:38number, which will be our units for
14:40measuring large numbers. For example,
14:42when you want to buy eggs, you measure in dozens.
14:45When you want to buy paper, you might want to measure it in
14:48thousands or five hundreds or whatever unit they sell them in.
14:51It's a natural unit. When you want to find
14:53intergalactic distances, you may use a light year.
14:56You use units so that in that unit, the quantity of interest
14:59to us is some number that you can count in your hands.
15:03When you count people's height, you use feet because it's
15:07something between 1 and 8, let's say.
15:09You don't want to use angstroms and you don't want to use
15:12millimeters. Likewise, when you want to
15:14simply count numbers, it turns out there's a very
15:17natural number called Avogadro's Number,
15:24and Avogadro's Number is 6 times 10^(23).
15:28There's no unit. It's simply a number,
15:30and that's called a mole. So, a mole is like a dozen.
15:34We wanted to buy 6 times 10^(23) eggs,
15:38you will say get me one mole of eggs.
15:43A mole is just a number. It's a huge number.
15:46You can ask yourself what's so great about this number?
15:49Why would someone think of this particular number?
15:51Why not some other number? Why not 10^(24)?
15:55Do you know what's special about this number?
16:00Yes? Student: [inaudible]
16:04Professor Ramamurti Shankar: Yes.
16:07If you like, a mole is such that one mole of
16:11hydrogen weighs one gram. And hydrogen is the simplest
16:16element with a nucleus of just a proton and the electron's mass
16:20is negligible. So, this, if you like,
16:22is the reciprocal of the mass of hydrogen.
16:26In other words, one over Avogadro's Number is
16:32the mass of hydrogen in grams, of a hydrogen atom in grams.
16:40So, you basically say, I want to count this large
16:43number so let me take one gram, which is my normal unit if
16:46you're thinking in grams. Then I ask, "How many hydrogen
16:49atoms does one gram of hydrogen contain?"
16:52That's the number. That's the mole.
16:54So, if you decide to measure the number of atoms you have in
16:58a given problem, in terms of this number,
17:01you write it as some other small number called moles,
17:04times the number in a mole, and you are free to write it
17:09this way. If you write it this way,
17:13then you write this nRT, R is the universal gas
17:20constant. What's n times the
17:23Boltzmann Constant. N_0 times the
17:25Boltzmann Constant. That happens to be 8.3 joules
17:29per degree centigrade or per Kelvin.
17:38Right? The units for R will be
17:40PV, which is units of energy divided by T.
17:45In terms of calories, I'd remember this as a nice,
17:47round number. Two calories per degree
17:50centigrade. Degrees centigrade and Kelvin
17:55are the same. The origins are shifted,
17:57but when you go up by one degree in centigrade or Kelvin,
18:00you go the same amount in temperature.
18:03So, this [R] is what they found out first,
18:06because they didn't know anything about atoms and so on.
18:10But later on when you go look under the hood of what the gas
18:15is made of, if you write it in terms of the number of actual
18:19atoms, you should use the little
18:21k, or you can write it in terms of number of moles,
18:23in which case use big R.
18:30And the relation between the two is simply this.
18:34If you're thinking of a gas and how many moles of gas do I have?
18:37For example, one gram of hydrogen would be
18:39one mole. Then you will use R.
18:41If you've gone right down to fundamentals and say,
18:43"How many atoms do I have?" and you put that here,
18:45you will multiply it by this very tiny number.
18:52Alright. Now, you start with this law
18:55and you ask the following question.
18:57On the left-hand side is the quantity P times
19:01V. On the right-hand side I have
19:04nRT, but let me write it now as NkT.
19:08You guys should be able to go back and forth between writing
19:11in terms of number of moles or the number of atoms.
19:15You'll like this because all numbers here will be small,
19:19of the order 1. R is a number like 8,
19:22in some units, and n would be 1 or 2
19:24moles. Here, this N will be a
19:26huge number, like 10^(23). k will be a tiny number
19:30like 10^(-23). Think in terms of atoms.
19:32That's what you do. Big numbers, small constants.
19:35When you think of moles, moderate numbers and moderate
19:39value of constants. We want to ask ourselves,
19:43"Is there a microscopic basis for this equation?"
19:47In other words, once we believe in atoms,
19:50do we understand why there is a pressure at all in a gas?
19:55That's what we're going to think about now.
19:57So, for this purpose, we will take a cube of gas.
20:06Here it is.
20:13This is a cube of side L by L by L.
20:20Inside this is gas and it's got some pressure,
20:23and I want to know what's the value of the pressure.
20:26You've got to ask yourself, why is there pressure?
20:28Remember, I told you what pressure means.
20:30If you take this face of the cube, for example,
20:33it's got to be nailed down to the other faces;
20:36otherwise, it'll just come flying out because the gas is
20:39pushing you out. The pressure is the force on
20:41this face divided by area. So, somebody inside is trying
20:45to get out. Those guys are the molecules or
20:47the atoms, and what they're doing is constantly bouncing off
20:52the wall, and every time this one bounces
20:55on a wall, its momentum changes from that to the other one.
20:59So, who's changing the momentum? Well, the wall is changing the
21:03momentum. It's reversing it.
21:04For example, if you bounce head-on and go
21:06back, your momentum is reversed. That means you push the wall
21:10with some force and the wall pushes you back with the
21:13opposite force. It's the force that you exert
21:15on the wall that I'm interested in.
21:17I want to find the force on the wall, say, this particular face.
21:21You can find the pressure on any face.
21:23It's going to be the same answer.
21:25I'm going to take the shaded face to find the pressure on it.
21:28Now, if you want to ask, what is the force exerted by me
21:32on any body, I know the force has a rate of change of
21:37momentum, because that is d/dt of
21:40mv, and m is a constant, and that's just
21:44dv/dt, which is ma.
21:47I'm just using old F = ma, but I'm writing it as a
21:50rate of change and momentum. Now, I have N molecules
21:54or N atoms, randomly moving inside the box.
21:58Each in its own direction, suffering collisions with the
22:02box, bouncing off like a billiard ball would at the end
22:05of the pool table and going to another wall and doing it.
22:09Now, that's a very complicated problem, so we're going to
22:12simplify the problem. The simplification is going to
22:15be, we are going to assume that one-third of the molecules are
22:18moving from left to right. One-third are moving up and
22:22down and one-third are moving in and out of the blackboard.
22:25If at all you make an assumption that the molecules
22:28are simply moving in the three primary directions,
22:32of course you will have to give equal numbers in these
22:34directions. Nothing in the gas that favors
22:36horizontal or vertical. In reality, of course,
22:38you must admit the fact they move in all directions,
22:41but the simplified derivation happens to give all the right
22:44physics, so I'm going to use that.
22:46So, N over three molecules are going back and
22:49forth between this wall, and this wall.
22:51I'm showing you a side view. The wall itself looks like this.
22:55The molecules go back and forth.
23:00Next assumption. All the molecules have the same
23:04speed, which I'm going to call v.
23:07That also is a gross and crude description of the problem,
23:12but I'm going to do that anyway and see what happens.
23:16So now, you ask yourself the following question.
23:19Take one particular molecule. When it hits the wall and it
23:24bounces back, its momentum changes from
23:28mv to -mv; therefore, the change in
23:31momentum is 2mv.
23:38How often does that change take place?
23:42You guys should think about that first.
23:45How often will that collision take place?
23:48Once you hit the wall here, you've got to go to the other
23:51wall and come back. So, you've got to go a distance
23:542L, and you're going at a speed v,
23:57the time it takes you is 2L over v.
24:00So, ΔP over ΔT is 2L divided by
24:04v. That gives me mv^(2)
24:07over L. That is the force due to one
24:10molecule. That's the average force.
24:14You realize it's not a continuous force.
24:17The molecule will hit the wall, there's a little force exchange
24:20between the two, then there's nothing,
24:22then you wait until it comes back and hits the wall again.
24:26If that were the only thing going on, what you would find is
24:29the wall most of the time, has no pressure and suddenly it
24:32has a lot of pressure and then suddenly nothing.
24:34But fortunately, this is not the only molecule.
24:37There are roughly 10^(23) guys pounding themselves against the
24:40wall. So, at any given instant,
24:42even if it's 10^(-5) seconds, there'd be a large number of
24:45molecules colliding. So, that's why the force will
24:48appear to be steady rather than a sharp noise.
24:51It looked very steady because somebody or other will be
24:54pushing against the wall.
24:59This is the force due to one molecule.
25:01The force due to all of them would be N over 3 times
25:05mv^(2) over L.
25:15N over 3 because of the N molecules,
25:17a third of them were moving in this direction.
25:19You realize the other two directions are parallel to the
25:23wall. They don't apply force on the
25:25wall. To apply force on the wall,
25:26you've got to be moving perpendicular to the wall.
25:29For example, if the planes that walls are
25:31coming out of the blackboard, moving in and out of the
25:34blackboard doesn't produce a force on this wall.
25:36That produces a force on the other two faces.
25:39So, as far as any one set of faces is concerned,
25:42in one plane, only the motion orthogonal to
25:45that is going to contribute. That's why you have N
25:48over 3. We're almost done.
25:50That's the average force. If you want,
25:52I can denote average by some F bar.
25:55Then what about the average pressure?
25:58The average pressure is the average force divided by the
26:01area of that face, which is F over
26:03L^(2), that gives me N over 3,
26:06mv^(2) over L^(3).
26:13Now, this is very nice because L^(3) is just the volume
26:18of my box.
26:23So, I take the L^(3), which is equal to the volume of
26:26my box, and I send it to the other side
26:29and write it as PV equals N over 3mv^(2).
26:44This is what the microscopic theory tells you.
26:46Microscopic theory says, if your molecules all have a
26:49single speed, they're moving randomly in
26:52space so that a third of them are moving back and forth
26:55against that wall and this wall, then this is the product
26:59PV. Experimentally,
27:00you find PV = NkT.
27:03So, you compare the two expressions and out comes one of
27:10the most beautiful results, which is that mv^(2)
27:16over 2 is 3 over 2kT. Now that guy deserves a box.
27:23Look what it's telling you. It's a really profound formula.
27:26It tells you for the first time a real microscopic meaning of
27:33temperature. What you and I call the
27:36temperature for gas is simply, up to these factors,
27:403/2 k, simply the kinetic energy of
27:43the molecules. That's what temperature is.
27:46If you've got a gas and you put your hand into the furnace and
27:49it feels hot, the temperature you're
27:50measuring is directly the kinetic energy of the molecules.
27:57That is a great insight into what temperature means.
28:01Remember, this is not true if T is measured in
28:06centigrade. If T were measured in
28:08centigrade, our freezing point of water mv^(2),
28:11would vanish. But that's not what's implied.
28:14T should be measured from absolute zero.
28:16It also tells you why absolute zero is absolute.
28:19As you cool your gas, the kinetic energy of molecules
28:22are decreasing and decreasing and decreasing,
28:24but you cannot go below not moving at all,
28:26right? That's the lowest possible
28:28kinetic energy. That's why it's absolute zero.
28:31At that point, everybody stops moving.
28:33That's why you have no pressure. Now, these results are modified
28:37by the laws of quantum mechanics, but we don't have to
28:41worry about that now. In the classical physics,
28:44it's actually correct to say that when the temperature goes
28:47to zero, all motion ceases. Now, this is the picture I want
28:52you to bear in mind when you say temperature.
28:55Absolute temperature is a measure of molecular agitation.
28:59More precisely, up to the constant k,
29:023/2 k, the kinetic energy of a
29:04molecule is the absolute temperature.
29:07That's for a gas. If you took a solid and you
29:09say, what happens when I heat the solid?
29:12You have a question? Yes?
29:15Student: [inaudible] Professor Ramamurti
29:19Shankar: You divide by 2 because ½ mv^(2) is a
29:23familiar quantity, namely, kinetic energy.
29:27That's why you divide by 2. Another thing to notice is that
29:31every gas, whatever it's made of, at a given temperature has a
29:35given kinetic energy because the kinetic energy per molecule on
29:40the left-hand side is dependent on absolute temperature and
29:44nothing else. So at certain degrees,
29:47like 300 Kelvin, hydrogen kinetic energy would
29:50be the same, carbon kinetic energy would also be the same.
29:53The kinetic energy will be the same, not the velocity.
29:56So, the carbon atom is heavier, it will be moving slower at
30:00that temperature in order to have the same kinetic energy.
30:04So, all molecules, all gases, have a given
30:07temperature. All atoms, let me say,
30:09at a given temperature in gaseous form will have the same
30:12kinetic energy [per molecule]. Now, if you have a
30:16solid--What's the difference between a gas and a solid?
30:21In a gas, the atoms are moving anywhere they want in the box.
30:25In a solid, every atom has a place.
30:28If you take a two-dimensional solid, the atoms look like this.
30:36They form a lattice or an array. That's because you will find
30:40out that, this is more advanced stuff, that every atom finds
30:43itself in a potential that looks like this.
30:49Imagine on the ground you make these hollows.
30:52Low points -- low potential; high points -- high potential.
30:55Obviously, if you put a bunch of objects here they will sit at
30:59the bottom of these little concave holes you've dug in the
31:03ground. At zero degrees absolute all
31:05atoms will sit at the bottom of their allotted positions;
31:10that'll be a solid at zero temperature.
31:12So in a solid, everybody has a location.
31:16I've shown you a one-dimensional solid,
31:18but you can imagine a three-dimensional solid where in
31:21a lattice of three-dimensional points,
31:23there's an assigned place for each atom and it sits there.
31:27If you heat up that solid now, what happens is these guys
31:31start vibrating. Now, here is where your
31:34knowledge of simple harmonic motions will come into play.
31:37When you take a system in equilibrium, it will execute
31:41simple harmonic motion if you give it a real kick.
31:44If you put it on top of a hotplate, the atoms in the hot
31:47plate will bump into these guys and start them moving.
31:50They will start vibrating. So, a hot solid is one in which
31:54the atoms are making more and more violent oscillations around
31:59their assigned positions. If you heat them more and more
32:03and more, eventually you start doing this.
32:05You go all the way from here to here;
32:07there is nothing to prevent it from rolling over to the next
32:10side. Once you jump the fence,
32:12you know, think of a bunch of houses, okay?
32:15Or a hole in the ground. You're living in a hole in the
32:18ground, as you get agitated you're able to do more and more
32:21oscillations so you can roll over to the next house.
32:23Once that happens all hell breaks loose because you don't
32:26have any reason to stay where you are.
32:28You start going everywhere. What do you think that is?
32:32Student: Melting. Professor Ramamurti
32:33Shankar: Pardon me? Student: Melting.
32:35Professor Ramamurti Shankar: That's melting.
32:36That's the definition of melting.
32:38Melting is when you can leap over this potential barrier,
32:40potential energy barrier, and go to the next site.
32:43The next side is just like this side.
32:45If you can jump that fence, you can jump this one.
32:47You go everywhere and you melt. That's the process of melting,
32:50and once you have a liquid, atoms don't have a definite
32:53location. Now, between a liquid and a
32:55solid, there is this clear difference, but a liquid and a
32:58vapor is more subtle. So, I don't want to go into
33:01that. If you look at a liquid
33:02locally, it will look very much like a solid in the sense that
33:06inter-atomic spacing is very tightly constrained in liquid.
33:10Whereas in a solid, if I know I am here,
33:13I know if I go 100 times the basic lattice spacing,
33:16there'll be another person sitting there.
33:19That's called long-range order. In a liquid, I cannot say that.
33:22In a liquid, I can say I am here.
33:24Locally, the environment around me is known, but if you go a few
33:27hundred miles, I cannot tell you a precise
33:30location if some other atom will be there or not.
33:33So, we say liquid is short-range positional order,
33:36but not long-range order, and a gas has no order at all.
33:39If I tell you there's a gas molecule here,
33:41I cannot tell you where anybody else is because nobody has any
33:44assigned location. Okay.
33:47So, this is the picture you should have of temperature.
33:50Temperature is agitated motion. Either motion in the vicinity
33:54of where you are told to sit. If you're in a solid a motion
33:57all over the box with more and more kinetic energy.
34:01The next thing in this caricature is that it is
34:04certainly not true that a third of the molecules are moving back
34:08and forth. We know that's a joke, right?
34:10Now, in this room there's no reason on earth a third of the
34:13molecules are doing this than others are doing.
34:15That's not approximation. They're moving in random
34:17directions. So, if you really got the
34:18stomach for it, you should do a pressure
34:20calculation in which you assume the molecules of random
34:23velocities sprinkled in all directions,
34:25and after all the hard work, turns out you get exactly this
34:28answer. So, that's one thing I didn't
34:30want to do. But something I should point
34:32out to you is the following. So, suppose I give you a gas at
34:37300 Kelvin. You go and you take this
34:39formula literally and you calculate from it a certain ½
34:42mv^(2). If you knew the mass of the
34:44atom, say, it's hydrogen, we know the mass of hydrogen.
34:47Then, you find the velocity and you say okay,
34:50this man tells me that anytime I catch a hydrogen atom,
34:53it'll have this velocity at 300 Kelvin.
34:56It may have random direction, but he tells me that's the
34:58velocity square. Take the square root of that,
35:00that's the velocity. It seems to us saying the
35:03unique velocity to each temperature.
35:05Well, that's not correct. Not only are the molecules
35:08moving in random directions, they're also moving with
35:12essentially all possible velocities.
35:15In fact, there are many, many possible velocities and
35:18this velocity I'm getting, in this formula,
35:21is some kind of average velocity, or the most popular
35:25one, or the most common one.
35:27So, if you really go to a gas and you have the ability to look
35:32into it and see for each velocity, what's the probability
35:36that I get that velocity? The picture I've given you is
35:40the probability of zero except at this one magical velocity
35:44controlled by the temperature. But the real graph looks like
35:47this.
35:50It has a certain peak. It likes to have a certain
35:53value. If you know enough about
35:55statistics, you know there's a most probable value,
35:58there's a median, there's a mean value.
36:00There are different definitions. They will all vary by factors
36:03of order 1, but the average kinetic energy will obey this
36:07condition. Yes?
36:08Student: [inaudible] Professor Ramamurti
36:10Shankar: It's not really a Gaussian because if you draw the
36:13nature of this curve, it looks like v^(2)e to
36:18the -mv^(2) over 2kT.
36:23That's the graph I'm trying to draw here.
36:25So, it looks like a Gaussian in the vicinity of this,
36:27but it's kind of skewed. It's forced to vanish at the
36:30origin.
36:34And it's not peaked at v = 0.
36:36A real Gaussian peak at this point would be symmetric.
36:39It's not symmetric; it vanishes here and it
36:42vanishes infinity. So, this is called a
36:45Maxwell-Boltzmann distribution. You don't have to remember any
36:49names but that is the detailed property of what's happening in
36:53a gas. So, a temperature does not pick
36:56a unique velocity, but it picks this graph.
37:00If you vary your temperature, look at what you have to do.
37:03If you change the number T here,
37:06if you double the value of T,
37:08that means if you double the value of v^(2) here and
37:11there, the graph will look the same.
37:13So, at every temperature there is a certain shape.
37:15If you go to your temperature, it will look more or less the
37:19same, but it may be peaked at a different velocity if you go to
37:23a higher temperature. Now, this is another thing I
37:26want to tell you. If you took a box containing
37:30not atoms but just radiation, in other words,
37:35go inside a pizza oven. Take out all the air,
37:39but the oven is still hot, and the walls of the oven are
37:43radiating electromagnetic radiation.
37:46Electromagnetic radiation comes in different frequencies,
37:50and you can ask how much energy is contained in every possible
37:54frequency range. You know, each frequency is a
37:57color so you know that. So, how much energy is in the
37:59red and how much is in the blue? That graph also looks like this.
38:03That's a more complicated law called the Planck distribution.
38:06That law also has a shape completely determined by
38:10temperature. Whereas for atoms,
38:12the shape is determined by temperature as well as the mass
38:17of the molecules. In the case of radiation,
38:20it's determined fully by temperature and the velocity of
38:22light. You give me a temperature,
38:24and I will draw you another one of these roughly bell-shaped
38:27curves. As you heat up the furnace,
38:29the shape will change.
38:34So again, a temperature for radiation means a particular
38:37distribution of energies at each frequency.
38:40For a gas it means a distribution of velocities.
38:48Has anybody seen that in the news lately, you know,
38:51or heard about this? Student: [inaudible]
38:53Professor Ramamurti Shankar: Pardon me?
38:56Student: [inaudible] Professor Ramamurti
38:58Shankar: About this particular graph for radiation.
39:00The probability at each frequency of finding radiation
39:04of the frequency in a furnace of some temperature T.
39:07Yes? Student: [inaudible]
39:11Professor Ramamurti Shankar: No,
39:14but in current news. In the last few years,
39:18what people did was the following.
39:20It's one of the predictions of the Big Bang theory that the
39:25universe was formed some 14 and a half billion years ago,
39:29and in the earliest stages the temperature of the universe was
39:32some incredibly high degrees, then as it expanded the
39:36universe cooled, and today, at the current size,
39:39it has got a certain average temperature,
39:42which is a remnant of the Big Bang.
39:44And that temperature means that we are sitting in furnace of the
39:49Big Bang. But the furnace has cooled a
39:52lot over the billions of years. The temperature of the universe
39:56is around 3 degrees Kelvin. And the way you determine that
40:01is you point your telescope in the sky.
40:03Of course, you're going to get light from this star;
40:06you're going to get light from that star.
40:07Ignore all the pointy things and look at the smooth
40:11background, and it should be the same in all directions.
40:15And plot that radiation, and now they use satellites to
40:18plot that, and you'll get a perfect fit to this kind of
40:22furnace radiation, called Black Body Radiation.
40:26And you read the temperature by taking that graph and fitting it
40:28to a graph like this, but there'll be temperature.
40:31In the case of light, this won't be velocity squared,
40:33but it will be the frequency squared,
40:35but read off the temperature that'll make this work and
40:38that's what gives you 3.1 or something.
40:40Near 3 degrees Kelvin. In fact, the data point for
40:44that now if you got that in your lab then you will be definitely
40:48busted for fudging your data because it's a perfect fit to
40:52Black Body Radiation. One of the most perfect fits to
40:55Black Body Radiation is the background radiation of the Big
40:58Bang. And it's isotropic,
41:00meaning it's the same in all directions, and this is one of
41:03the predictions of the Big Bang is that that'll be the remnant
41:07of the Black Body Radiation. Again, it tells you there's a
41:11sense in which, if you go to intergalactic
41:13space, that is your temperature. That's the temperature you get
41:17for free. We're all living in that heat
41:19bath at 3 degrees. You want more heat,
41:21you've got to light up your furnace but this is everywhere
41:24in the universe, that heat left over from
41:26creation. Okay.
41:29That's a very, very interesting subject.
41:32You know, a lot of new physics is coming out by looking at just
41:36the Black Body Radiation because the radiation that's coming to
41:41your eye left those stars long ago.
41:44So, what you see today is not what's happening today.
41:48It's what happened long ago when the radiation left that
41:52part of the universe. Therefore, we can actually tell
41:55something about the universe not only now, but at earlier
41:58periods. And that's the way in which we
42:00can actually tell whether the universe is expanding or not
42:04expanding or is it accelerating in its expansion,
42:07or you can even say once it was decelerating and now it's
42:10accelerating. All that information comes by
42:13being able to look at the radiation from the Big Bang.
42:16But for you guys, I think the most interesting
42:19thing is that when you are in thermal equilibrium,
42:22and you are living in a certain temperature, then the radiation
42:26in your world and the molecules and atoms in your world,
42:29will have a distribution of frequencies and velocities given
42:32by that universal graph. Now in our class,
42:35we will simplify life and replace this graph with a huge
42:39peak at a certain velocity by pretending everybody's at that
42:44velocity. We will treat the whole gas as
42:47if it was represented by single average number.
42:51So, when someone says find the velocity of molecules,
42:54they're talking about the average velocity.
42:56You know statistically that it's the distribution of answers
42:59and an average answer. Because the average is what you
43:03and I have to know. Namely, ½ mv^(2) is 3/2
43:06kT, on average. Okay.
43:11Now, I'm going to study in detail thermodynamics.
43:15So, the system I'm going to study is the only one we all
43:19study, which is an ideal gas sitting inside a piston.
43:27It's got a temperature, it's got a pressure,
43:30and it's got a volume. And I'm going to plot here
43:34pressure and volume and I'm going to put a dot and that's my
43:39gas. The state of my gas is
43:41summarized by where you put the dot.
43:44Every dot here is a possible state of equilibrium for the
43:50gas. Remember, the gas,
43:52if you look at it under the hood, is made up of 10^(23)
43:57molecules. The real, real state of the gas
44:00is obtained by saying, giving me 10^(23) locations and
44:0410^(23) velocities. According to Newton,
44:06that's the maximum information you can give me about the gas
44:09right now, because with that and Newton's
44:11laws I can predict the future. But when you study
44:13thermodynamics, you don't really want to look
44:15into the details. You want to look at gross
44:18macroscopic properties and there are two that you need.
44:21Pressure and volume. Now, you might say,
44:23"What about temperature?" Why don't I have a third axis
44:28for temperature? Why is there also not a
44:32property? Yes?
44:34Student: [inaudible] Professor Ramamurti
44:36Shankar: Yeah. Because PV = NkT.
44:42I don't have to give you T, if I know P and
44:45V. There's not an independent
44:47thing you can pick. You can pick P and
44:48V independently. You cannot pick T.
44:50Let me tell you, by the way, PV = NkT is
44:53not a universal law. It's the law that you apply to
44:57dilute gases. But we are going to just study
45:01only dilute ideal gas. Ideal gas is one in which the
45:05atoms and molecules are so far apart that they don't feel any
45:08forces between each other unless they collide.
45:11So, here is my gas. It's sitting here.
45:16Now, what I do, I had a few weights on top of
45:20it. Three weights.
45:21I suddenly pull out one weight. Throw it out.
45:24What do you think will happen? Well, I think this gas will now
45:29shoot up, it'll bob up and down a few times.
45:32Then after a few seconds, or a fraction of a second,
45:35it'll settle down with a new location.
45:37By "settle down," I mean after a while I will not see any
45:41macroscopic motion. Then the gas has a new pressure
45:45and a new volume. It's gone from being there to
45:49being there.
45:54What about in between? What happened in between the
45:58starting and finishing points? You might say look,
46:02if it was here in the beginning it was there later,
46:06it must've followed some path. Not really.
46:09Not in this process, because if you do it very
46:12abruptly, suddenly throwing out one-third of the weights,
46:16there's a period when the piston rushes up,
46:19when the gas is not in equilibrium.
46:21By that, I mean there is no single pressure you can
46:24associate with the gas. The bottom of the gas doesn't
46:27even know the top is flying off. It's at the old pressure.
46:30At the top of the gas there's a low pressure.
46:32So, different parts of the gas at different pressure,
46:35we don't call that equilibrium. So, the dot,
46:38representing this system, moves off the graph.
46:40It's off. It's off the radar,
46:43and only when it has finally settled down,
46:45the entire gas can make up its mind on what its pressure wants
46:48to be; you put it back here.
46:50So, we have a little problem that we have these equilibrium
46:54states, but when you try to go from one to another you fly off
46:58the map. So, you want to find a device
47:00by which you can stay on the PV diagram as you change
47:03the state of the gas, and that brings us to the
47:06notion of what you call a quasi-static process.
47:09A quasi-static process is trying to have it both ways in
47:12which you want to change the state of the gas,
47:15and you don't want it to leave the PV diagram.
47:18You want it to be always at equilibrium.
47:22So, what you really want to do is not put in three big fat
47:26blocks like this, but instead take a gas where
47:29you have many, many grains of sand.
47:33They can produce the pressure. Now, remove one grain of sand.
47:38It moves a tiny bit and very quickly settles down.
47:40It is again true during the tiny bit of settling down you
47:43didn't know what it was doing, but you certainly nailed it at
47:46the second location. You move one grain at a time,
47:49then you get a picture like this and you can see where this
47:52is going. You can make the grain smaller
47:54and smaller and smaller and in a mathematical sense you can then
47:57form a continuous line. That is to say,
48:00you perform a process that leaves the system arbitrarily
48:03close to equilibrium, meaning give it enough time to
48:06readjust to the new pressure, settle down to the new volume,
48:10take another grain and another grain.
48:12And in the spirit of calculus, you can make these changes
48:15vanishing so that you can really then say you did this.
48:18Yes? Student: Are all of
48:21these small processes reversible?
48:23Professor Ramamurti Shankar: Pardon me?
48:24Student: Are all of these small processes
48:25reversible? Professor Ramamurti
48:25Shankar: Yes. Such a process is also
48:27called--you can call it quasi-static but one of the
48:30features of that, it is reversible.
48:32You've got to be a little careful when you say reversible.
48:35What we mean by "reversible" is, if I took off a grain of
48:38sand and it came from here to the next dot,
48:41and I put the grain back, it'll climb back to where it
48:45was. So, you can go back and forth
48:48on this. But now, that's an idealized
48:50process because if you had a friction, if you had any
48:53friction between the piston and the walls,
48:56then if you took out a grain and it went up,
48:59you put the grain back it might not come back to quite where it
49:03is. Because some of the frictional
49:06losses you will never get back. You cannot put Humpty Dumpty
49:11back. So, most of the time processes
49:13are not reversible, even if you do them slowly,
49:16if there is friction. So, assume it's a completely
49:19frictionless system. Because if there is friction,
49:21there is some heat that goes out somewhere and some energy is
49:25lost somewhere and we cannot bring it back.
49:28If we took a frictionless piston and on top of it moved it
49:31very, very slowly, you can follow this graph.
49:33That's the kind of thermodynamic process we're
49:36talking about. In the old days,
49:38when I studied a single particular of the xy
49:41plane, I just said the guy goes from here to here to there.
49:44That's very easy to study and there's no restriction on how
49:46quickly or how fast it moved. Particles have trajectories no
49:49matter how quickly they move. For a thermodynamic system,
49:52you cannot move them too fast, because they are extended and
49:56you are having a huge gas a single number called pressure,
50:00so you cannot change one part of the gas without waiting for
50:03all of them to communicate and readjust and achieve a global
50:07value for the new pressure and you can move gradually.
50:10That's why it takes time to drag along 10^(23) particles as
50:14if they are the single number or two numbers characterizing them.
50:18So, we'll be studying processes like this.
50:22Now, this is called a state. Two is a state and one is a
50:25state. Every dot here, that is a state.
50:28Now, in every state of the system, I'm going to define a
50:36new variable, which is called a quantity
50:41called U, which stands for the internal
50:47energy of the gas.
50:54Internal energy is simply the kinetic energy of the gas
50:57molecules. For solids and liquids,
50:59there's a more complicated formula.
51:01For the gas, internal energy is just the
51:03kinetic energy. And what is that?
51:06It is 3/2 kT per molecule times N.
51:12I'm sorry, 3/2 Nk, yeah. 3/2 kT times that.
51:19Or we can write it as 3/2 nRT.
51:29But nRT is PV. You can also write it as 3/2
51:33PV, so internal energy is just 3/2 PV.
51:36That means at a given point on the PV diagram,
51:39you have a certain internal energy.
51:41If you are there, that's your internal energy.
51:43Take there-halves of PV and that's the energy and that's
51:46literally the kinetic energy of all the molecules in your box.
51:50So, now I'm ready to write down what's called the First Law of
51:55Thermodynamics that talks about what happens if you make a move
51:59in the PV plane from one place to another place.
52:04If you go from one place to another place,
52:06your internal energy will change from U_1
52:09to U_2. Let's call it ΔU.
52:16We want to ask what causes the internal energy of the gas to
52:20change. So, you guys think about it now.
52:22Now that you know all about what's happening in the
52:24cylinder, you can ask how I will change the energy?
52:27Well, if you wanted to change the energy of a system,
52:29there are two ways you can do it.
52:31One is you can do work on the gas.
52:34Another thing is you can put the gas on a hotplate.
52:37If you put it on hotplate, we know it's going to get
52:39hotter. If it gets hotter,
52:41temperature goes up. If temperature goes up,
52:43the internal energy goes up. So, there are two ways to
52:47change the energy of a gas. The first one we call heat
52:51input. That just means put it on
52:54something hotter and let the thing heat it up.
52:58Temperature will go up. Notice that the internal energy
53:01of an ideal gas depends only on the temperature.
53:04That's something very, very important.
53:07I mention it every time I teach the subject and some people
53:10forget and lose a lot of points needlessly.
53:13So, I'll say it once more with feeling.
53:15The energy of an ideal gas depends only on the temperature.
53:19If the temperature is not changed;
53:20energy has not changed. So, try to remember that for
53:25what I do later. So, the change of the gas,
53:28this cylinder full that I put some weights on top and I've got
53:34gas inside, it can change either because I
53:38did, I put in some heat, or the gas did some work.
53:43By that, I mean if the gas expands by pushing out against
53:48the atmosphere, then it was doing the work and
53:52ΔW is the work done by the gas.
53:56That's why it comes to the minus sign, because it's the
53:59work done by the gas. If you do work, you lose energy.
54:03So, what's the formula for work done?
54:05Let's calculate that. If I've got a piston here,
54:09it's the force times the distance.
54:12But the force is the pressure times the area times the
54:18distance. Now, you guys should know
54:21enough geometry to know the area of the piston times the distance
54:24it moves is the change in the volume.
54:27So, we can write it as P times dV.
54:30That leads to this great law. Let me write it on a new
54:33blackboard because we're going to be playing around with that
54:37law. This is law number one.
54:41The change in the internal energy of a system is equal to
54:47ΔQ - PΔV.
54:58What does it express? It expresses the Law of
55:01Conservation of Energy. It says the energy goes up,
55:03either because you pushed the piston or the piston pushed you;
55:07then you decide what the overall sign is,
55:09or you put it on a hotplate. We are now equating putting it
55:13on a hotplate as also equivalent to giving it energy,
55:16because we identify heat as simply energy.
55:19So, if you took the piston and you nailed the piston so it
55:23cannot move, and you put it on a hotplate,
55:27PdV part will vanish because there is no ΔV.
55:31That's the way of heating it, it is called ΔQ.
55:34Another thing you can do is thermally isolate your piston so
55:37no heat can flow in and out of it,
55:39and then you can either have the volume increase or decrease.
55:42If the gas expanded, ΔV is positive and the
55:45PΔ - PΔV is negative, and the ΔU would be
55:49negative; the gas will lose energy.
55:51That's because the molecules are beating up on the piston and
55:55moving the piston. Remember, applying a force
55:57doesn't cost you anything. But if the point of application
56:00moves, you do work. And who's going to pay for it,
56:03the gas? It'll pay for it through its
56:05loss of internal energy. Conversely, if you push down on
56:09the gas, ΔV will be negative and this will become
56:12positive and the energy of the gas will go up.
56:15So, there are two ways to change the energy of these
56:18molecules. In the end, all you want is you
56:20want the molecules to move faster than before.
56:22One is to put them on a hotplate where there are
56:25fast-moving molecules. When they collide with the
56:28slow-moving molecules, typically the slow one's a
56:30little more faster and the fast one's a little more slower and
56:33therefore will be a transfer of kinetic energy.
56:35Or when you push the piston down, you can show when a
56:38molecule collides with a moving piston.
56:40It will actually gain energy. So, that's how you do work.
56:46That's the first law.
56:51So, let us now calculate the work done in a process where a
56:57gas goes from here to here on an isotherm.
57:05Isotherm is a graph of a given temperature.
57:08So, this is a graph P times V equal to
57:13constant, because PV = nRT.
57:17If T is constant, PV is a constant,
57:19it's the rectangular hyperbola. The product of the x and
57:22y coordinates is constant, so when the x
57:24coordinate vanished, the y will go to
57:26infinity. y coordinate vanishes,
57:27x will go to infinity. So, you want to take your gas
57:31for a ride from here to here. Throughout it's at a certain
57:35temperature T. What work is done by you?
57:41That's a very nice interpretation.
57:43The work done by you is the integral of PdV.
57:49But what is integral of PdV?
57:51That's P, and that's dV.
57:54Pdv is that shaded region.
57:57In other words, if you just write PdV it
58:00makes absolutely no sense. If you go to a mathematician
58:03and say, "Please do the integral for me!"
58:04can the mathematician do this? What's coming in the way of the
58:10mathematician actually doing the integral?
58:14What do you have to know to really do an integral?
58:17Student: You have to know the function.
58:19Professor Ramamurti Shankar: You have to know
58:19the function. If you just say P,
58:21we'll say maybe P is a constant, in which case I'll
58:23pull it out of the integral. But for this problem,
58:25because PV is nRT, and T is a
58:29constant, P is nRT divided
58:32by V, and that's the function that you would need to
58:36do the integral, and if you did that you will
58:39find there's nRT. All of them are constants.
58:42They come out of the integral, dv over V,
58:45and integrate from the initial volume, the final volume.
58:48And you guys know this is a logarithm, and the log of upper
58:52minus log of lower is the log of the ratio.
58:55And this gives me nRT ln (V_2
59:00/V_1). So, we have done our first work
59:05calculation. When the gas goes on an
59:08isothermal trajectory from start to finish, from volume
59:11V_1 to volume V_2,
59:15the work done, this is the work done by the
59:18gas. You can all see that gas is
59:20expanding and that's equal to this shaded region.
59:29By the way, I mention it now, I don't want to distract you,
59:33but suppose later on I make it go backwards like this,
59:37part of the way. The work done on the going
59:40backwards part is this area, but with a minus sign.
59:43I hope you will understand, if you go to the right the
59:46area's considered positive. If you go to the left,
59:48the area is considered negative.
59:50If you do the integral and put the right limit,
59:52you'll get the right answer. But geometrically,
59:54the area under the graph in the PV diagram is the work
59:58done if you're moving the direction of increasing volume.
1:00:01If you would decrease the volume, for example,
1:00:04if you just went back from here, the area looks the same
1:00:07but the work done is considered negative.
1:00:10You don't have to think very hard.
1:00:13If you do the calculation going backwards, you will get a
1:00:15ln of V_1 over
1:00:16V_2. That'll automatically be the
1:00:18negative of the log of V_2 over
1:00:20V_1. But geometrically,
1:00:21the area under the graph is the work, if you are going to the
1:00:24right. Yes?
1:00:26Student: What determines the shape of the curve that
1:00:29links the first state to the second state?
1:00:32Professor Ramamurti Shankar: Oh,
1:00:33this one? Student: Mmm-hmm.
1:00:35Professor Ramamurti Shankar: This'll be a graph,
1:00:38PV equal to essentially a constant.
1:00:40So, you take your gas, you see how many moles there
1:00:42are. You know R,
1:00:43you know the temperature, you promised not to change the
1:00:46temperature. So, you'll move on a trajectory
1:00:48so that the product PV never changes.
1:00:51And in any xy plane, if you draw a graph where the
1:00:54product xy doesn't change it'll have this shape called a
1:00:58"rectangular hyperbola." It just means,
1:01:00whenever one increases, the other should decrease,
1:01:03keeping the product constant. That's why P is
1:01:06proportional to the reciprocal of V when you do the
1:01:10integral. Very good.
1:01:12So, this is now the work done by the gas.
1:01:16What is the heat input? The heat input is a change in
1:01:21internal energy minus the work done.
1:01:24Let me see. The law was ΔU = ΔQ -
1:01:29ΔW. Yeah, let's go back to this law.
1:01:35In this problem, ΔW is what I just
1:01:38calculated, nRT, whatever the log,
1:01:40V_2 over V_1.
1:01:43What is ΔQ? How much heat has been put into
1:01:49this gas? How do I find that?
1:01:52Student: Take out the T?
1:01:54Professor Ramamurti Shankar: Pardon me?
1:01:55Student: You take out the T?
1:01:58Professor Ramamurti Shankar: For the heat input
1:02:00you mean? Yeah, you can use mc
1:02:01ΔT, but you don't have to do anymore work.
1:02:03By that, I mean you don't have to do any more cerebration.
1:02:06What can you do with this equation to avoid doing further
1:02:10calculations? Do you know anything else?
1:02:13Yes? Student: [inaudible]
1:02:17Professor Ramamurti Shankar: Yes.
1:02:19This is what I told you is the fact that people do not
1:02:23constantly remember, but you must.
1:02:25This gas did not change its temperature.
1:02:27Go back to equation number whatever I wrote down.
1:02:31U = 3/2 nRT or something.
1:02:33T doesn't change, U doesn't change.
1:02:35That means the initial internal energy and final internal energy
1:02:38are the same because initial temperature and final
1:02:41temperature are the same. So, this guy has to be zero.
1:02:44That means ΔQ is the same as ΔW in this
1:02:50particular case. Yes?
1:02:53Student: [inaudible] Professor Ramamurti
1:02:57Shankar: When you say mc ΔT, you've got to be
1:03:01careful of what formula you want to use.
1:03:04I'll tell you why you cannot simply use mc ΔT.
1:03:10If you've got a solid and you use mc ΔT,
1:03:12that is correct, because when you heat the
1:03:14solid, the heat you put in goes into heating up the solid.
1:03:20Maybe let's ask the following question.
1:03:21His question is the following. You're telling me you put heat
1:03:25into a gas, right? And you say temperature doesn't
1:03:29go up. How can that possibly be?
1:03:31I always thought when I put heat into something,
1:03:33temperature goes up. That's because you were
1:03:36thinking about a solid, where if you put in heat it's
1:03:39got to go somewhere and, of course, temperature goes up.
1:03:42What do you think is happening to the gas here?
1:03:44Think of the piston and weight combination.
1:03:48When I want to go along this path from here to here,
1:03:54you can ask yourself where is the heat input and where is the
1:04:01change in energy, and why is there no change in
1:04:04temperature? If you take a piston like this,
1:04:07if you want to increase the volume, you can certainly take
1:04:12off a grain of sand, right?
1:04:14If you took the grain of sand and the piston will move up,
1:04:18it will do work and it actually will cool down,
1:04:21but that's not what you're doing.
1:04:23You are keeping it on a hotplate at a certain
1:04:26temperature so that if it tries to cool down,
1:04:29heat flows from below to above maintaining the temperature.
1:04:32So, what the gas is doing in this case is taking heat energy
1:04:36from below and going up and working against the atmosphere
1:04:40above. It takes in with one hand and
1:04:42gives out to the other, without changing its energy.
1:04:49So, when you study specific heat, which is my next topic,
1:04:51you've got to be a little more careful when you talk about
1:04:54specific heats of gases, and I will tell you why.
1:04:56There is no single thing called specific heat for a gas.
1:04:59There are many, many definitions depending on
1:05:02the circumstances. But I hope you understand in
1:05:04this case; you've got to visualize this.
1:05:07It's not enough to draw diagrams and draw pictures.
1:05:09What did I do to the cylinder to maintain the temperature and
1:05:13yet let it expand? Expansion is going to demand
1:05:16work on part of the gas. That's going to require a loss
1:05:20of energy unless you pump in energy from below.
1:05:22So, what I've done is that I take grain after grain,
1:05:26so that the pressure drops and the volume increases,
1:05:29but the slight expansion would have cooled it slightly but the
1:05:32reservoir from below brings it back to the temperature of the
1:05:35reservoir. So, you prop it up in
1:05:37temperature. So, we draw the picture by
1:05:40saying the gas went from here to here, and we usually draw a
1:05:43picture like this and say heat flowed into the system during
1:05:46that process. Alright, now I'll come to this
1:05:50question that was raised about specific heat.
1:05:54Now, specific heat, you always say is ΔQ
1:05:58over ΔT or ΔT divided by the mass of the
1:06:04substance. Now, it turns out that for a
1:06:07gas, you've already seen that what you want to count is not
1:06:11the actual mass, but the moles.
1:06:14Because we have seen at the level of the ideal gas law,
1:06:17the energy is controlled by not simply the mass,
1:06:21but by the moles. Because every molecule gets a
1:06:24certain amount of energy, namely 3/2 kT,
1:06:27and you just want to count the number of molecules,
1:06:29or the number of moles. Now, there are many,
1:06:31many ways in which you can pump in heat into a gas and heat it
1:06:36up and see how much heat it takes.
1:06:39But let's agree that we will take one mole from now on and
1:06:42not one kilogram. Not one kilogram.
1:06:48We'll find out if you do it that way, the answer doesn't
1:06:52seem to depend on the gas. That's the first thing.
1:06:55Take a mole of some gas and call the specific heat as the
1:06:59energy needed to raise the temperature of one mole by one
1:07:03degree. So, this should not be m.
1:07:05This should be the number of moles.
1:07:10If you take one mole, you can say okay,
1:07:13one mole of gas I was told has energy U = 3/2 RT.
1:07:22Because it was there-halves nRT but n is one
1:07:25mole. Now, you want to put in some
1:07:27heat, and you really want the ΔQ over ΔT,
1:07:32so I will remind you that ΔQ is ΔU +
1:07:35PΔv.
1:07:40The heat input into a gas is the change of energy plus P
1:07:43Δv. And that's just from the first
1:07:45law. So, if I'm going to divide
1:07:48ΔQ by ΔT, there's a problem here.
1:07:53Did you allow the volume to change or did you not allow the
1:07:56volume to change? That's going to decide what the
1:07:58specific heat is. In other words,
1:08:01when a solid is heated, it expands such a tiny amount,
1:08:04we don't worry about the work done by the expanding solid
1:08:08against the atmosphere. But for a gas,
1:08:11when you heat it, the volume changes so much that
1:08:14the work it does against the external world is
1:08:16non-negligible. Therefore, the specific heat is
1:08:20dependent on what you allow the volume term to do.
1:08:24So, there's one definition of specific heat called
1:08:27C_V, and C_V is the
1:08:31one at constant volume. You don't let the volume change.
1:08:34In other words, you take the piston and you
1:08:36clamp it. Now, you pump in heat from
1:08:39below by putting it on a hotplate.
1:08:41All the heat goes directly to internal energy.
1:08:44None of that is lost in terms of expansion.
1:08:47So, ΔV is zero. In that case,
1:08:50ΔQ over ΔT at constant volume,
1:08:53we denote that in this fashion, at constant volume,
1:08:57this term is gone, and it just becomes ΔU
1:09:00over ΔT. That's very easily done.
1:09:03ΔU over ΔT is 3/2 R.
1:09:11So, the specific heat of a gas at constant volume is 3 over
1:09:162R. When I studied solids,
1:09:17I never bother about constant volume because a change in the
1:09:20volume of a solid is so negligible when it's heated up,
1:09:23it's not worth specifying that it was a constant volume
1:09:26process. But for a gas,
1:09:27it's going to matter whether it was constant volume or not.
1:09:31Then, there's a second specific heat people like to define.
1:09:35That's done as follows. You take this piston.
1:09:39You have some gas at some pressure.
1:09:41You pump in some heat but you don't clamp the piston.
1:09:45You let the piston expand any way it wants at the same
1:09:49pressure. For example,
1:09:50if it's being pushed down by the atmosphere,
1:09:53you let the piston move up if it wants to, maintaining the
1:09:56same pressure. Well, if it moves up a little
1:09:59bit, then the correct equation is the heat that you put in is
1:10:03the change in internal energy plus P times ΔV,
1:10:07where now P is some constant pressure,
1:10:10say the atmospheric pressure, ΔV is the change in
1:10:14volume. So, ΔQ needed now will
1:10:16be more, because you're pumping in heat from below and you're
1:10:20losing energy above because you're letting the gas expand.
1:10:24Because you were letting the pressure be controlled from the
1:10:27outside at some fixed value. So now, ΔQ--this
1:10:32ΔU will be 3 over 2RΔT.
1:10:37Now, what's the change in P times ΔV?
1:10:40Here is where you should know your calculus.
1:10:42The P times change in V is the same as the
1:10:46change in PV, if P is a constant.
1:10:49Right? Remember long back when I did
1:10:51rate of change of momentum is d/dt of mv,
1:10:53it's m times dv/dt,
1:10:55because m doesn't change.
1:10:57You can take it inside the change.
1:10:58But now we use PV = RT. I'm talking about one mole.
1:11:03PV = RT. That's a change in the quantity
1:11:07RT, R is a constant, that's R times
1:11:11ΔT so I put in here R times ΔT.
1:11:15This is the ΔQ at constant pressure.
1:11:19So, the specific heat of constant pressure is ΔQ
1:11:23over ΔT, keeping the pressure constant.
1:11:27You divide everything by ΔT you get 3 over
1:11:302R, plus another R, which is 5 over
1:11:322R.
1:11:39So, the thing you have to remember, what I did in the end,
1:11:43is that a gas doesn't have a single specific heat.
1:11:46If we just say, put in some heat and tell me
1:11:48how many calories I need to raise the temperature,
1:11:51that's not enough. You have to tell me whether in
1:11:53the interim, the gas was fixed in its volume,
1:11:56or changed its volume, or obeyed some other condition.
1:11:59The two most popular conditions people consider are either the
1:12:03volume cannot change or the pressure cannot change.
1:12:06If the volume cannot change, then the change in the heat you
1:12:09put in goes directly to internal energy, from the First Law of
1:12:12Thermodynamics. That gives you a specific heat
1:12:14of 3 over 2R. If the pressure cannot change,
1:12:17you get 5 over 2R. You can see
1:12:20C_P is bigger than C_V
1:12:23because when you let the piston expand,
1:12:25then not all the heat is going to heat the gas.
1:12:29Some of it is dissipated on top by working against the
1:12:32atmosphere. Then, notice that I've not told
1:12:37you what gas it is. That's why the specific heat
1:12:42per mole is the right thing to think about because then the
1:12:46answer does not depend on what particular gas you took.
1:12:50Whether it's hydrogen or helium, they all have the same
1:12:54specific heat per mole. They won't have the same
1:12:57specific heat per gram, right?
1:12:59Because one gram of helium and one gram of hydrogen don't have
1:13:03the same number of moles. So, you have to remember that
1:13:06we're talking about moles. The final thing I have to
1:13:08caution you--very, very important.
1:13:10This is for a monoatomic gas.
1:13:16This is for a gas whose atom is the gas itself.
1:13:21It's a point. Its only energy is kinetic
1:13:24energy. There are diatomic gases,
1:13:26by two of them [atoms] joined together,
1:13:28they can form a dumbbell or something;
1:13:30then the energy of the dumbbell has got two parts,
1:13:33as you learned long ago. It can rotate around some axis
1:13:37and it can also move in space. Then the internal energy has
1:13:41also got two parts. Energy due to motion of the
1:13:43center of mass and energy due to rotation.
1:13:45Some molecules also vibrate. So, there are lots of
1:13:49complicated things, but if you got only one guy,
1:13:51or one atom, whatever its mass is,
1:13:53it cannot rotate around itself and it cannot vibrate around
1:13:57itself, so those energies all disappear.
1:13:59So, we have taken the simplest one of a monoatomic gas,
1:14:03a gas whose fundamental entity is a single atom rather than a
1:14:07complicated molecule. And that's all you're
1:14:09responsible for. I'll just say one thing.
1:14:11C_P over C_V,
1:14:13I want to mention it before you run off to do your homework.
1:14:15I don't know if it comes up. It's called γ,
1:14:17that's five-third for a monoatomic gas.
1:14:20You can just take the ratio of the numbers.
1:14:22If in some problem you find γ is not five-thirds,
1:14:25do not panic. It just means it's a gas which
1:14:29is not monoatomic. If it's not monoatomic,
1:14:31these numbers don't have exactly those values.
1:14:33We don't have to go beyond that. You just have to know there's a
1:14:37ratio γ, which is five-thirds in the
1:14:38simplest case, but in some problem,
1:14:40somewhere in your life, you can get a γ which
1:14:42is not five-thirds.

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