Uploaded by YaleCourses on 22.09.2008

Transcript:

So, I had to leave you in the middle of something pretty

exciting, so I'll come back and take it from there.

So, what is it you have to remember from last time?

You know, what are the main ideas I covered?

One is, we took the notion of temperature, for which we have

an intuitive feeling and turned it into something more

quantitative, so you can not only say this is

hotter than that, that's hotter than this,

you can say by how much, by how many degrees.

And in the end we agreed to use the absolute Kelvin scale for

temperature. And the way to find the Kelvin

scale, you take the gas, any gas that you like,

like hydrogen or helium, at low concentration,

and put that inside a piston and cylinder.

That will occupy some volume and there's a certain pressure

by putting weights on top, and you take the product of

P times V, and the claim is for whatever

gas you take, it'll be a straight line.

Remember now, this is in Kelvin. Your centigrade scale is

somewhere over here, but I've shifted the origin to

the Kelvin scale. So, somewhere here will be the

boiling point of water, somewhere the freezing point of

water; that may be the boiling point

of water. And if you took a different

amount of a different gas, you'll get some other line.

But they will always be straight lines if the

concentration is sufficiently low.

In other words, it appears that pressure times

volume is some constant. I don't know what to call it.

Say c, times this temperature.

And you can use that to measure temperature because if you know

two points on a straight line then you know that you can find

the slope and then you can calibrate the thermometer,

then for any other value of P times V that you

get, you can come down and read your temperature.

That's the preferred scale, and we prefer this scale

because it doesn't seem to depend on the gas that you use.

I can use one; you can use another one.

People in another planet who have never heard of water--they

can use a different gas. But all gases seem to have the

property that pressure times volume is linearly proportional

to this new temperature scale, measured with this new origin

at absolute zero. There is really nothing to the

left of this T = 0.

The next thing I mentioned was, people used to think of the

theory of heat as a new theory. You know, we got mechanics and

all that stuff--levers and pulleys and all that.

Then, you have this mysterious thing called heat,

which has been around for many years but people started

quantifying it by saying there's a fluid called the caloric fluid

and hot things have a lot of it, and cold things have less of

it, and when you mix them the caloric somehow flows from the

hot to the cold. Then we defined specific heat,

law of conservation of this caloric fluid that allows you to

do some problems in calorimetry. You mix so much of this with so

much of that, where will they end up?

That kind of problem. So, that promoted heat to a new

and independent entity, different from all other things

we have studied. But something suggests that it

is not completely alien or a new concept, because there seems to

be a conservation of law for this heat,

because the heat lost by the cold water was the heat gained

by the hot water. I'm sorry.

Heat gained by the cold water was the heat lost by the hot

water. So, you have a conservation law.

Secondly, we know another way to produce heat.

Instead of saying put it on the stove, put it on the stove,

in which case, there is something mysterious

flowing from the stove into the water that heats it up,

I told you there's a different thing you can do.

Take two automobiles; slam them.

This is not the most economical way to make your dinner but I'm

just telling you as a matter of principle.

Buy two Ferraris, slam them into each other and

take this pot and put it on top and it'll heat up because

Ferraris will heat up. The question is what happened

to the kinetic energy of the two cars?

That is really gone. So, in the old days,

we would say, well, we don't apply the Law of

Conservation of Energy because this was an inelastic

relationship. That was our legal way out of

the whole issue. But you realize now this

caloric fluid can be produced from nowhere,

because there was no caloric fluid before,

but slamming the two cars produce this extra heat.

So, that indicates that perhaps there's a relation between

mechanical energy and heat energy--that when mechanical

energy disappears, heat energy appears.

So, how do you do the conversion ratio?

You know, how many calories can you get if you sacrifice one

joule of mechanical energy? So, Joule did the experiment.

Not with cars. I mean, he didn't have cars at

that time, so if he did he would've probably done it with

cars. He had this gadget with him,

which is a little shaft with some paddles and a pulley on the

top, and you let the weight go down.

And I told you guys the weight goes from here to here,

the mgh loss will not be the gain in ½ mv^(2).

Something will be missing. Keep track of the missing

amount. So many joules--but meanwhile

you find this water has become hot.

You find then how many calories should have gone in,

because we know the specific of water,

we know the rise in temperature, we know how many

calories were produced. And then, you compare the two

and you find that 4.2 joules = 1 calorie.

So, that is the conversion ratio of calories to joules.

One joule, 4.2 joules of mechanical energy.

So, in the example of the colliding cars,

take the ½ mv^(2) for each car, turn it into joules,

slam them together. If they come to rest,

you've lost all of that, and then you take that and you

write it as--divided by 4.2 and that's how much calories you

have produced. If the car was made of just one

material, it had a specific heat, then it would go up by a

certain temperature you can actually predict.

Okay. So today, I want to go a little

deeper into the question of where is the energy actually

stored in the car, and what is heat.

We still don't know in detail what heat is.

We just said car heats up and the loss of joules divided by

4.2 is the gain in calories. Now, we can answer in detail

exactly what is heat. That's what we're going to talk

about today. When we say something is

hotter, what do we mean on a microscopic level?

In the old days when people didn't know what anything was

made of, they didn't have this understanding.

And the understanding that I'm going to give you today is based

on a simple fact that everything is made up of atoms.

That was not known, and that's one of the greatest

discoveries that, in the end,

everything is made up of atoms, and atoms combine to form

molecules and so on. So, how does that come into

play? For that, I want you to take

the simple example where the temperature enters.

That is in the relation PV equal to some constant

times temperature. Do you know what I'm talking

about? Take some gas,

make sure it's sufficiently dilute, put it into this piston,

measure the weights on top of it,

divide it by the area, to get the pressure,

that's the pressure, that's the volume.

The volume is the region here, multiply the product;

then, if you heat up the gas by putting it on some hot plate,

you'll find the product PV increases,

and as the temperature increases, PV is

proportional to T. We want to ask what is this

proportionality constant. Suppose you were doing this.

In the old days, this is what people did.

What did we think should be on the right-hand side?

What is going to control this particular constant for the

given experiment? Do you know what it might be

proportional to? Yes?

Student: Amount of gas? Professor Ramamurti

Shankar: Amount of gas. That's true,

because if the amount of gas is zero, we think there's no

pressure. When you say "amount of gas,"

that's a very safe sentence because amount measured by what

means? By what metric?

Student: Probably number of particles?

Professor Ramamurti Shankar: Right.

Suppose you were not aware of particles.

Then, what would you mean by "amount of gas?"

Student: Mass. Professor Ramamurti

Shankar: The mass. Now, if you guys ever said

moles, I was going to shoot you down.

You're not supposed to know those things.

We are trying to deduce that. So, put yourself back in

whatever stone ages we were in. We don't know anything else.

Mass would be a reasonable argument, right?

What's the argument? We know that if you have some

amount of gas producing the pressure, and you put twice as

much stuff, you would think it will produce

twice as much pressure. Same reason why you think the

expansion of a rod is proportionally change in

temperature times the starting length.

So, this mass is what's doing it.

So, it's proportional to mass. It's a very reasonable guess.

So, if you put more gas into your piston you think it'll

produce more pressure. That's actually correct.

So, let's go to that one particular sample in your

laboratory that you did. So, put the mass that you had

there. Then, you should put a constant

still. I don't know what you want me

to call this constant, say, c prime.

This constant contains everything, but I pulled out the

mass and the remaining constant I want to call c prime.

This is actually correct. You can take a certain gas and

you can find out what c prime is.

But here is what people found. If you do it that way,

the constant c prime depends on the gas you are

considering. If you consider hydrogen gas,

let's call that c prime for hydrogen.

Somebody else puts in helium gas.

Then you find the c prime for helium is one-fourth

c prime for hydrogen.

If you do carbon, it's another number.

c prime for carbon is c prime for hydrogen

divided by 12. So, each gas has a different

constant. So, we conclude that yes,

it's the mass that decides it but the mass has to be divided

by different numbers for different gases to find the real

effective mass in terms of pressure.

In other words, one gram of hydrogen and one

gram of helium do not have the same pressure.

In fact, one gram of helium has to be divided by 4 to find its

effect on pressure. So, you have to think about why

is it that the mass directly is not involved.

Mass has to be divided by a number, and the number is a

nice, round number. 4 for this and 12 for that,

and of course people figure out there's a long story I cannot go

into, but I think you all know the answer.

But now we are allowed to fast forward to the correct answer,

because I really don't have the time to see how they worked it

out, but from these integers and the

way the gases reacted and formed complicated molecules,

they figured out what's really going on is that you're dividing

by a number that's proportional with the mass of the underlying

fundamental entity, which would be an atom.

In some case a molecule, but I'm just going to call

everything as atom. So, if things,

like, carbon, as atoms, weigh 12 times as

much as things called hydrogen, then if you took some amount of

carbon, you divide it by a number, like,

12 to count the number of carbon atoms.

Okay, so hydrogen you want to count the number of hydrogen

atoms. So then, what really you want

here is not the mass, but the number of atoms of a

given kind. We are certainly free to write

either a mass or the number of atoms, because the two are

proportional. But the beauty of writing it

this way, you write it in this fashion, by this new constant

k, k is independent of the gas.

So, you want to write it in a manner in which it doesn't

depend on the gas. You can write it in terms of

mass. If you did, for each mass

you've got to divide by a certain number.

TThen once you divide it by the number you can put a single

constant in front. Or if you want a universal

constant, what you should really be counting is the number of

atoms or molecules.

So, you couldn't have written it that way until you knew about

atoms and molecules and people who are led to atoms and

molecules by looking at the way gases interact,

and it's a beautiful piece of chemistry to figure out really

that there are entities which come in discreet units.

Not at all obvious in the old days, that mass comes in

discreet units called atoms, but that's what they deduced.

So, this is called the Boltzmann Constant.

The Boltzmann Constant has a value of 1.4 times 10^(-23),

let's see, joules/Kelvin.

That's it. Or joules/Kelvin or degrees

centigrade.

So, this is a universal constant.

So, now what people like to do is they don't like to write the

number [N], because if you write the number,

in a typical situation, what's the number going to be?

Take some random group gas. One gram, two grams,

one kilogram, it doesn't matter.

The number you will put in there is some number like

10^(23) or 10^(25). That's a huge number.

So, whenever a huge number is involved, what you try to do is

to measure the huge number as a simple multiple off another huge

number, which will be our units for

measuring large numbers. For example,

when you want to buy eggs, you measure in dozens.

When you want to buy paper, you might want to measure it in

thousands or five hundreds or whatever unit they sell them in.

It's a natural unit. When you want to find

intergalactic distances, you may use a light year.

You use units so that in that unit, the quantity of interest

to us is some number that you can count in your hands.

When you count people's height, you use feet because it's

something between 1 and 8, let's say.

You don't want to use angstroms and you don't want to use

millimeters. Likewise, when you want to

simply count numbers, it turns out there's a very

natural number called Avogadro's Number,

and Avogadro's Number is 6 times 10^(23).

There's no unit. It's simply a number,

and that's called a mole. So, a mole is like a dozen.

We wanted to buy 6 times 10^(23) eggs,

you will say get me one mole of eggs.

A mole is just a number. It's a huge number.

You can ask yourself what's so great about this number?

Why would someone think of this particular number?

Why not some other number? Why not 10^(24)?

Do you know what's special about this number?

Yes? Student: [inaudible]

Professor Ramamurti Shankar: Yes.

If you like, a mole is such that one mole of

hydrogen weighs one gram. And hydrogen is the simplest

element with a nucleus of just a proton and the electron's mass

is negligible. So, this, if you like,

is the reciprocal of the mass of hydrogen.

In other words, one over Avogadro's Number is

the mass of hydrogen in grams, of a hydrogen atom in grams.

So, you basically say, I want to count this large

number so let me take one gram, which is my normal unit if

you're thinking in grams. Then I ask, "How many hydrogen

atoms does one gram of hydrogen contain?"

That's the number. That's the mole.

So, if you decide to measure the number of atoms you have in

a given problem, in terms of this number,

you write it as some other small number called moles,

times the number in a mole, and you are free to write it

this way. If you write it this way,

then you write this nRT, R is the universal gas

constant. What's n times the

Boltzmann Constant. N_0 times the

Boltzmann Constant. That happens to be 8.3 joules

per degree centigrade or per Kelvin.

Right? The units for R will be

PV, which is units of energy divided by T.

In terms of calories, I'd remember this as a nice,

round number. Two calories per degree

centigrade. Degrees centigrade and Kelvin

are the same. The origins are shifted,

but when you go up by one degree in centigrade or Kelvin,

you go the same amount in temperature.

So, this [R] is what they found out first,

because they didn't know anything about atoms and so on.

But later on when you go look under the hood of what the gas

is made of, if you write it in terms of the number of actual

atoms, you should use the little

k, or you can write it in terms of number of moles,

in which case use big R.

And the relation between the two is simply this.

If you're thinking of a gas and how many moles of gas do I have?

For example, one gram of hydrogen would be

one mole. Then you will use R.

If you've gone right down to fundamentals and say,

"How many atoms do I have?" and you put that here,

you will multiply it by this very tiny number.

Alright. Now, you start with this law

and you ask the following question.

On the left-hand side is the quantity P times

V. On the right-hand side I have

nRT, but let me write it now as NkT.

You guys should be able to go back and forth between writing

in terms of number of moles or the number of atoms.

You'll like this because all numbers here will be small,

of the order 1. R is a number like 8,

in some units, and n would be 1 or 2

moles. Here, this N will be a

huge number, like 10^(23). k will be a tiny number

like 10^(-23). Think in terms of atoms.

That's what you do. Big numbers, small constants.

When you think of moles, moderate numbers and moderate

value of constants. We want to ask ourselves,

"Is there a microscopic basis for this equation?"

In other words, once we believe in atoms,

do we understand why there is a pressure at all in a gas?

That's what we're going to think about now.

So, for this purpose, we will take a cube of gas.

Here it is.

This is a cube of side L by L by L.

Inside this is gas and it's got some pressure,

and I want to know what's the value of the pressure.

You've got to ask yourself, why is there pressure?

Remember, I told you what pressure means.

If you take this face of the cube, for example,

it's got to be nailed down to the other faces;

otherwise, it'll just come flying out because the gas is

pushing you out. The pressure is the force on

this face divided by area. So, somebody inside is trying

to get out. Those guys are the molecules or

the atoms, and what they're doing is constantly bouncing off

the wall, and every time this one bounces

on a wall, its momentum changes from that to the other one.

So, who's changing the momentum? Well, the wall is changing the

momentum. It's reversing it.

For example, if you bounce head-on and go

back, your momentum is reversed. That means you push the wall

with some force and the wall pushes you back with the

opposite force. It's the force that you exert

on the wall that I'm interested in.

I want to find the force on the wall, say, this particular face.

You can find the pressure on any face.

It's going to be the same answer.

I'm going to take the shaded face to find the pressure on it.

Now, if you want to ask, what is the force exerted by me

on any body, I know the force has a rate of change of

momentum, because that is d/dt of

mv, and m is a constant, and that's just

dv/dt, which is ma.

I'm just using old F = ma, but I'm writing it as a

rate of change and momentum. Now, I have N molecules

or N atoms, randomly moving inside the box.

Each in its own direction, suffering collisions with the

box, bouncing off like a billiard ball would at the end

of the pool table and going to another wall and doing it.

Now, that's a very complicated problem, so we're going to

simplify the problem. The simplification is going to

be, we are going to assume that one-third of the molecules are

moving from left to right. One-third are moving up and

down and one-third are moving in and out of the blackboard.

If at all you make an assumption that the molecules

are simply moving in the three primary directions,

of course you will have to give equal numbers in these

directions. Nothing in the gas that favors

horizontal or vertical. In reality, of course,

you must admit the fact they move in all directions,

but the simplified derivation happens to give all the right

physics, so I'm going to use that.

So, N over three molecules are going back and

forth between this wall, and this wall.

I'm showing you a side view. The wall itself looks like this.

The molecules go back and forth.

Next assumption. All the molecules have the same

speed, which I'm going to call v.

That also is a gross and crude description of the problem,

but I'm going to do that anyway and see what happens.

So now, you ask yourself the following question.

Take one particular molecule. When it hits the wall and it

bounces back, its momentum changes from

mv to -mv; therefore, the change in

momentum is 2mv.

How often does that change take place?

You guys should think about that first.

How often will that collision take place?

Once you hit the wall here, you've got to go to the other

wall and come back. So, you've got to go a distance

2L, and you're going at a speed v,

the time it takes you is 2L over v.

So, ΔP over ΔT is 2L divided by

v. That gives me mv^(2)

over L. That is the force due to one

molecule. That's the average force.

You realize it's not a continuous force.

The molecule will hit the wall, there's a little force exchange

between the two, then there's nothing,

then you wait until it comes back and hits the wall again.

If that were the only thing going on, what you would find is

the wall most of the time, has no pressure and suddenly it

has a lot of pressure and then suddenly nothing.

But fortunately, this is not the only molecule.

There are roughly 10^(23) guys pounding themselves against the

wall. So, at any given instant,

even if it's 10^(-5) seconds, there'd be a large number of

molecules colliding. So, that's why the force will

appear to be steady rather than a sharp noise.

It looked very steady because somebody or other will be

pushing against the wall.

This is the force due to one molecule.

The force due to all of them would be N over 3 times

mv^(2) over L.

N over 3 because of the N molecules,

a third of them were moving in this direction.

You realize the other two directions are parallel to the

wall. They don't apply force on the

wall. To apply force on the wall,

you've got to be moving perpendicular to the wall.

For example, if the planes that walls are

coming out of the blackboard, moving in and out of the

blackboard doesn't produce a force on this wall.

That produces a force on the other two faces.

So, as far as any one set of faces is concerned,

in one plane, only the motion orthogonal to

that is going to contribute. That's why you have N

over 3. We're almost done.

That's the average force. If you want,

I can denote average by some F bar.

Then what about the average pressure?

The average pressure is the average force divided by the

area of that face, which is F over

L^(2), that gives me N over 3,

mv^(2) over L^(3).

Now, this is very nice because L^(3) is just the volume

of my box.

So, I take the L^(3), which is equal to the volume of

my box, and I send it to the other side

and write it as PV equals N over 3mv^(2).

This is what the microscopic theory tells you.

Microscopic theory says, if your molecules all have a

single speed, they're moving randomly in

space so that a third of them are moving back and forth

against that wall and this wall, then this is the product

PV. Experimentally,

you find PV = NkT.

So, you compare the two expressions and out comes one of

the most beautiful results, which is that mv^(2)

over 2 is 3 over 2kT. Now that guy deserves a box.

Look what it's telling you. It's a really profound formula.

It tells you for the first time a real microscopic meaning of

temperature. What you and I call the

temperature for gas is simply, up to these factors,

3/2 k, simply the kinetic energy of

the molecules. That's what temperature is.

If you've got a gas and you put your hand into the furnace and

it feels hot, the temperature you're

measuring is directly the kinetic energy of the molecules.

That is a great insight into what temperature means.

Remember, this is not true if T is measured in

centigrade. If T were measured in

centigrade, our freezing point of water mv^(2),

would vanish. But that's not what's implied.

T should be measured from absolute zero.

It also tells you why absolute zero is absolute.

As you cool your gas, the kinetic energy of molecules

are decreasing and decreasing and decreasing,

but you cannot go below not moving at all,

right? That's the lowest possible

kinetic energy. That's why it's absolute zero.

At that point, everybody stops moving.

That's why you have no pressure. Now, these results are modified

by the laws of quantum mechanics, but we don't have to

worry about that now. In the classical physics,

it's actually correct to say that when the temperature goes

to zero, all motion ceases. Now, this is the picture I want

you to bear in mind when you say temperature.

Absolute temperature is a measure of molecular agitation.

More precisely, up to the constant k,

3/2 k, the kinetic energy of a

molecule is the absolute temperature.

That's for a gas. If you took a solid and you

say, what happens when I heat the solid?

You have a question? Yes?

Student: [inaudible] Professor Ramamurti

Shankar: You divide by 2 because ½ mv^(2) is a

familiar quantity, namely, kinetic energy.

That's why you divide by 2. Another thing to notice is that

every gas, whatever it's made of, at a given temperature has a

given kinetic energy because the kinetic energy per molecule on

the left-hand side is dependent on absolute temperature and

nothing else. So at certain degrees,

like 300 Kelvin, hydrogen kinetic energy would

be the same, carbon kinetic energy would also be the same.

The kinetic energy will be the same, not the velocity.

So, the carbon atom is heavier, it will be moving slower at

that temperature in order to have the same kinetic energy.

So, all molecules, all gases, have a given

temperature. All atoms, let me say,

at a given temperature in gaseous form will have the same

kinetic energy [per molecule]. Now, if you have a

solid--What's the difference between a gas and a solid?

In a gas, the atoms are moving anywhere they want in the box.

In a solid, every atom has a place.

If you take a two-dimensional solid, the atoms look like this.

They form a lattice or an array. That's because you will find

out that, this is more advanced stuff, that every atom finds

itself in a potential that looks like this.

Imagine on the ground you make these hollows.

Low points -- low potential; high points -- high potential.

Obviously, if you put a bunch of objects here they will sit at

the bottom of these little concave holes you've dug in the

ground. At zero degrees absolute all

atoms will sit at the bottom of their allotted positions;

that'll be a solid at zero temperature.

So in a solid, everybody has a location.

I've shown you a one-dimensional solid,

but you can imagine a three-dimensional solid where in

a lattice of three-dimensional points,

there's an assigned place for each atom and it sits there.

If you heat up that solid now, what happens is these guys

start vibrating. Now, here is where your

knowledge of simple harmonic motions will come into play.

When you take a system in equilibrium, it will execute

simple harmonic motion if you give it a real kick.

If you put it on top of a hotplate, the atoms in the hot

plate will bump into these guys and start them moving.

They will start vibrating. So, a hot solid is one in which

the atoms are making more and more violent oscillations around

their assigned positions. If you heat them more and more

and more, eventually you start doing this.

You go all the way from here to here;

there is nothing to prevent it from rolling over to the next

side. Once you jump the fence,

you know, think of a bunch of houses, okay?

Or a hole in the ground. You're living in a hole in the

ground, as you get agitated you're able to do more and more

oscillations so you can roll over to the next house.

Once that happens all hell breaks loose because you don't

have any reason to stay where you are.

You start going everywhere. What do you think that is?

Student: Melting. Professor Ramamurti

Shankar: Pardon me? Student: Melting.

Professor Ramamurti Shankar: That's melting.

That's the definition of melting.

Melting is when you can leap over this potential barrier,

potential energy barrier, and go to the next site.

The next side is just like this side.

If you can jump that fence, you can jump this one.

You go everywhere and you melt. That's the process of melting,

and once you have a liquid, atoms don't have a definite

location. Now, between a liquid and a

solid, there is this clear difference, but a liquid and a

vapor is more subtle. So, I don't want to go into

that. If you look at a liquid

locally, it will look very much like a solid in the sense that

inter-atomic spacing is very tightly constrained in liquid.

Whereas in a solid, if I know I am here,

I know if I go 100 times the basic lattice spacing,

there'll be another person sitting there.

That's called long-range order. In a liquid, I cannot say that.

In a liquid, I can say I am here.

Locally, the environment around me is known, but if you go a few

hundred miles, I cannot tell you a precise

location if some other atom will be there or not.

So, we say liquid is short-range positional order,

but not long-range order, and a gas has no order at all.

If I tell you there's a gas molecule here,

I cannot tell you where anybody else is because nobody has any

assigned location. Okay.

So, this is the picture you should have of temperature.

Temperature is agitated motion. Either motion in the vicinity

of where you are told to sit. If you're in a solid a motion

all over the box with more and more kinetic energy.

The next thing in this caricature is that it is

certainly not true that a third of the molecules are moving back

and forth. We know that's a joke, right?

Now, in this room there's no reason on earth a third of the

molecules are doing this than others are doing.

That's not approximation. They're moving in random

directions. So, if you really got the

stomach for it, you should do a pressure

calculation in which you assume the molecules of random

velocities sprinkled in all directions,

and after all the hard work, turns out you get exactly this

answer. So, that's one thing I didn't

want to do. But something I should point

out to you is the following. So, suppose I give you a gas at

300 Kelvin. You go and you take this

formula literally and you calculate from it a certain ½

mv^(2). If you knew the mass of the

atom, say, it's hydrogen, we know the mass of hydrogen.

Then, you find the velocity and you say okay,

this man tells me that anytime I catch a hydrogen atom,

it'll have this velocity at 300 Kelvin.

It may have random direction, but he tells me that's the

velocity square. Take the square root of that,

that's the velocity. It seems to us saying the

unique velocity to each temperature.

Well, that's not correct. Not only are the molecules

moving in random directions, they're also moving with

essentially all possible velocities.

In fact, there are many, many possible velocities and

this velocity I'm getting, in this formula,

is some kind of average velocity, or the most popular

one, or the most common one.

So, if you really go to a gas and you have the ability to look

into it and see for each velocity, what's the probability

that I get that velocity? The picture I've given you is

the probability of zero except at this one magical velocity

controlled by the temperature. But the real graph looks like

this.

It has a certain peak. It likes to have a certain

value. If you know enough about

statistics, you know there's a most probable value,

there's a median, there's a mean value.

There are different definitions. They will all vary by factors

of order 1, but the average kinetic energy will obey this

condition. Yes?

Student: [inaudible] Professor Ramamurti

Shankar: It's not really a Gaussian because if you draw the

nature of this curve, it looks like v^(2)e to

the -mv^(2) over 2kT.

That's the graph I'm trying to draw here.

So, it looks like a Gaussian in the vicinity of this,

but it's kind of skewed. It's forced to vanish at the

origin.

And it's not peaked at v = 0.

A real Gaussian peak at this point would be symmetric.

It's not symmetric; it vanishes here and it

vanishes infinity. So, this is called a

Maxwell-Boltzmann distribution. You don't have to remember any

names but that is the detailed property of what's happening in

a gas. So, a temperature does not pick

a unique velocity, but it picks this graph.

If you vary your temperature, look at what you have to do.

If you change the number T here,

if you double the value of T,

that means if you double the value of v^(2) here and

there, the graph will look the same.

So, at every temperature there is a certain shape.

If you go to your temperature, it will look more or less the

same, but it may be peaked at a different velocity if you go to

a higher temperature. Now, this is another thing I

want to tell you. If you took a box containing

not atoms but just radiation, in other words,

go inside a pizza oven. Take out all the air,

but the oven is still hot, and the walls of the oven are

radiating electromagnetic radiation.

Electromagnetic radiation comes in different frequencies,

and you can ask how much energy is contained in every possible

frequency range. You know, each frequency is a

color so you know that. So, how much energy is in the

red and how much is in the blue? That graph also looks like this.

That's a more complicated law called the Planck distribution.

That law also has a shape completely determined by

temperature. Whereas for atoms,

the shape is determined by temperature as well as the mass

of the molecules. In the case of radiation,

it's determined fully by temperature and the velocity of

light. You give me a temperature,

and I will draw you another one of these roughly bell-shaped

curves. As you heat up the furnace,

the shape will change.

So again, a temperature for radiation means a particular

distribution of energies at each frequency.

For a gas it means a distribution of velocities.

Has anybody seen that in the news lately, you know,

or heard about this? Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

Student: [inaudible] Professor Ramamurti

Shankar: About this particular graph for radiation.

The probability at each frequency of finding radiation

of the frequency in a furnace of some temperature T.

Yes? Student: [inaudible]

Professor Ramamurti Shankar: No,

but in current news. In the last few years,

what people did was the following.

It's one of the predictions of the Big Bang theory that the

universe was formed some 14 and a half billion years ago,

and in the earliest stages the temperature of the universe was

some incredibly high degrees, then as it expanded the

universe cooled, and today, at the current size,

it has got a certain average temperature,

which is a remnant of the Big Bang.

And that temperature means that we are sitting in furnace of the

Big Bang. But the furnace has cooled a

lot over the billions of years. The temperature of the universe

is around 3 degrees Kelvin. And the way you determine that

is you point your telescope in the sky.

Of course, you're going to get light from this star;

you're going to get light from that star.

Ignore all the pointy things and look at the smooth

background, and it should be the same in all directions.

And plot that radiation, and now they use satellites to

plot that, and you'll get a perfect fit to this kind of

furnace radiation, called Black Body Radiation.

And you read the temperature by taking that graph and fitting it

to a graph like this, but there'll be temperature.

In the case of light, this won't be velocity squared,

but it will be the frequency squared,

but read off the temperature that'll make this work and

that's what gives you 3.1 or something.

Near 3 degrees Kelvin. In fact, the data point for

that now if you got that in your lab then you will be definitely

busted for fudging your data because it's a perfect fit to

Black Body Radiation. One of the most perfect fits to

Black Body Radiation is the background radiation of the Big

Bang. And it's isotropic,

meaning it's the same in all directions, and this is one of

the predictions of the Big Bang is that that'll be the remnant

of the Black Body Radiation. Again, it tells you there's a

sense in which, if you go to intergalactic

space, that is your temperature. That's the temperature you get

for free. We're all living in that heat

bath at 3 degrees. You want more heat,

you've got to light up your furnace but this is everywhere

in the universe, that heat left over from

creation. Okay.

That's a very, very interesting subject.

You know, a lot of new physics is coming out by looking at just

the Black Body Radiation because the radiation that's coming to

your eye left those stars long ago.

So, what you see today is not what's happening today.

It's what happened long ago when the radiation left that

part of the universe. Therefore, we can actually tell

something about the universe not only now, but at earlier

periods. And that's the way in which we

can actually tell whether the universe is expanding or not

expanding or is it accelerating in its expansion,

or you can even say once it was decelerating and now it's

accelerating. All that information comes by

being able to look at the radiation from the Big Bang.

But for you guys, I think the most interesting

thing is that when you are in thermal equilibrium,

and you are living in a certain temperature, then the radiation

in your world and the molecules and atoms in your world,

will have a distribution of frequencies and velocities given

by that universal graph. Now in our class,

we will simplify life and replace this graph with a huge

peak at a certain velocity by pretending everybody's at that

velocity. We will treat the whole gas as

if it was represented by single average number.

So, when someone says find the velocity of molecules,

they're talking about the average velocity.

You know statistically that it's the distribution of answers

and an average answer. Because the average is what you

and I have to know. Namely, ½ mv^(2) is 3/2

kT, on average. Okay.

Now, I'm going to study in detail thermodynamics.

So, the system I'm going to study is the only one we all

study, which is an ideal gas sitting inside a piston.

It's got a temperature, it's got a pressure,

and it's got a volume. And I'm going to plot here

pressure and volume and I'm going to put a dot and that's my

gas. The state of my gas is

summarized by where you put the dot.

Every dot here is a possible state of equilibrium for the

gas. Remember, the gas,

if you look at it under the hood, is made up of 10^(23)

molecules. The real, real state of the gas

is obtained by saying, giving me 10^(23) locations and

10^(23) velocities. According to Newton,

that's the maximum information you can give me about the gas

right now, because with that and Newton's

laws I can predict the future. But when you study

thermodynamics, you don't really want to look

into the details. You want to look at gross

macroscopic properties and there are two that you need.

Pressure and volume. Now, you might say,

"What about temperature?" Why don't I have a third axis

for temperature? Why is there also not a

property? Yes?

Student: [inaudible] Professor Ramamurti

Shankar: Yeah. Because PV = NkT.

I don't have to give you T, if I know P and

V. There's not an independent

thing you can pick. You can pick P and

V independently. You cannot pick T.

Let me tell you, by the way, PV = NkT is

not a universal law. It's the law that you apply to

dilute gases. But we are going to just study

only dilute ideal gas. Ideal gas is one in which the

atoms and molecules are so far apart that they don't feel any

forces between each other unless they collide.

So, here is my gas. It's sitting here.

Now, what I do, I had a few weights on top of

it. Three weights.

I suddenly pull out one weight. Throw it out.

What do you think will happen? Well, I think this gas will now

shoot up, it'll bob up and down a few times.

Then after a few seconds, or a fraction of a second,

it'll settle down with a new location.

By "settle down," I mean after a while I will not see any

macroscopic motion. Then the gas has a new pressure

and a new volume. It's gone from being there to

being there.

What about in between? What happened in between the

starting and finishing points? You might say look,

if it was here in the beginning it was there later,

it must've followed some path. Not really.

Not in this process, because if you do it very

abruptly, suddenly throwing out one-third of the weights,

there's a period when the piston rushes up,

when the gas is not in equilibrium.

By that, I mean there is no single pressure you can

associate with the gas. The bottom of the gas doesn't

even know the top is flying off. It's at the old pressure.

At the top of the gas there's a low pressure.

So, different parts of the gas at different pressure,

we don't call that equilibrium. So, the dot,

representing this system, moves off the graph.

It's off. It's off the radar,

and only when it has finally settled down,

the entire gas can make up its mind on what its pressure wants

to be; you put it back here.

So, we have a little problem that we have these equilibrium

states, but when you try to go from one to another you fly off

the map. So, you want to find a device

by which you can stay on the PV diagram as you change

the state of the gas, and that brings us to the

notion of what you call a quasi-static process.

A quasi-static process is trying to have it both ways in

which you want to change the state of the gas,

and you don't want it to leave the PV diagram.

You want it to be always at equilibrium.

So, what you really want to do is not put in three big fat

blocks like this, but instead take a gas where

you have many, many grains of sand.

They can produce the pressure. Now, remove one grain of sand.

It moves a tiny bit and very quickly settles down.

It is again true during the tiny bit of settling down you

didn't know what it was doing, but you certainly nailed it at

the second location. You move one grain at a time,

then you get a picture like this and you can see where this

is going. You can make the grain smaller

and smaller and smaller and in a mathematical sense you can then

form a continuous line. That is to say,

you perform a process that leaves the system arbitrarily

close to equilibrium, meaning give it enough time to

readjust to the new pressure, settle down to the new volume,

take another grain and another grain.

And in the spirit of calculus, you can make these changes

vanishing so that you can really then say you did this.

Yes? Student: Are all of

these small processes reversible?

Professor Ramamurti Shankar: Pardon me?

Student: Are all of these small processes

reversible? Professor Ramamurti

Shankar: Yes. Such a process is also

called--you can call it quasi-static but one of the

features of that, it is reversible.

You've got to be a little careful when you say reversible.

What we mean by "reversible" is, if I took off a grain of

sand and it came from here to the next dot,

and I put the grain back, it'll climb back to where it

was. So, you can go back and forth

on this. But now, that's an idealized

process because if you had a friction, if you had any

friction between the piston and the walls,

then if you took out a grain and it went up,

you put the grain back it might not come back to quite where it

is. Because some of the frictional

losses you will never get back. You cannot put Humpty Dumpty

back. So, most of the time processes

are not reversible, even if you do them slowly,

if there is friction. So, assume it's a completely

frictionless system. Because if there is friction,

there is some heat that goes out somewhere and some energy is

lost somewhere and we cannot bring it back.

If we took a frictionless piston and on top of it moved it

very, very slowly, you can follow this graph.

That's the kind of thermodynamic process we're

talking about. In the old days,

when I studied a single particular of the xy

plane, I just said the guy goes from here to here to there.

That's very easy to study and there's no restriction on how

quickly or how fast it moved. Particles have trajectories no

matter how quickly they move. For a thermodynamic system,

you cannot move them too fast, because they are extended and

you are having a huge gas a single number called pressure,

so you cannot change one part of the gas without waiting for

all of them to communicate and readjust and achieve a global

value for the new pressure and you can move gradually.

That's why it takes time to drag along 10^(23) particles as

if they are the single number or two numbers characterizing them.

So, we'll be studying processes like this.

Now, this is called a state. Two is a state and one is a

state. Every dot here, that is a state.

Now, in every state of the system, I'm going to define a

new variable, which is called a quantity

called U, which stands for the internal

energy of the gas.

Internal energy is simply the kinetic energy of the gas

molecules. For solids and liquids,

there's a more complicated formula.

For the gas, internal energy is just the

kinetic energy. And what is that?

It is 3/2 kT per molecule times N.

I'm sorry, 3/2 Nk, yeah. 3/2 kT times that.

Or we can write it as 3/2 nRT.

But nRT is PV. You can also write it as 3/2

PV, so internal energy is just 3/2 PV.

That means at a given point on the PV diagram,

you have a certain internal energy.

If you are there, that's your internal energy.

Take there-halves of PV and that's the energy and that's

literally the kinetic energy of all the molecules in your box.

So, now I'm ready to write down what's called the First Law of

Thermodynamics that talks about what happens if you make a move

in the PV plane from one place to another place.

If you go from one place to another place,

your internal energy will change from U_1

to U_2. Let's call it ΔU.

We want to ask what causes the internal energy of the gas to

change. So, you guys think about it now.

Now that you know all about what's happening in the

cylinder, you can ask how I will change the energy?

Well, if you wanted to change the energy of a system,

there are two ways you can do it.

One is you can do work on the gas.

Another thing is you can put the gas on a hotplate.

If you put it on hotplate, we know it's going to get

hotter. If it gets hotter,

temperature goes up. If temperature goes up,

the internal energy goes up. So, there are two ways to

change the energy of a gas. The first one we call heat

input. That just means put it on

something hotter and let the thing heat it up.

Temperature will go up. Notice that the internal energy

of an ideal gas depends only on the temperature.

That's something very, very important.

I mention it every time I teach the subject and some people

forget and lose a lot of points needlessly.

So, I'll say it once more with feeling.

The energy of an ideal gas depends only on the temperature.

If the temperature is not changed;

energy has not changed. So, try to remember that for

what I do later. So, the change of the gas,

this cylinder full that I put some weights on top and I've got

gas inside, it can change either because I

did, I put in some heat, or the gas did some work.

By that, I mean if the gas expands by pushing out against

the atmosphere, then it was doing the work and

ΔW is the work done by the gas.

That's why it comes to the minus sign, because it's the

work done by the gas. If you do work, you lose energy.

So, what's the formula for work done?

Let's calculate that. If I've got a piston here,

it's the force times the distance.

But the force is the pressure times the area times the

distance. Now, you guys should know

enough geometry to know the area of the piston times the distance

it moves is the change in the volume.

So, we can write it as P times dV.

That leads to this great law. Let me write it on a new

blackboard because we're going to be playing around with that

law. This is law number one.

The change in the internal energy of a system is equal to

ΔQ - PΔV.

What does it express? It expresses the Law of

Conservation of Energy. It says the energy goes up,

either because you pushed the piston or the piston pushed you;

then you decide what the overall sign is,

or you put it on a hotplate. We are now equating putting it

on a hotplate as also equivalent to giving it energy,

because we identify heat as simply energy.

So, if you took the piston and you nailed the piston so it

cannot move, and you put it on a hotplate,

PdV part will vanish because there is no ΔV.

That's the way of heating it, it is called ΔQ.

Another thing you can do is thermally isolate your piston so

no heat can flow in and out of it,

and then you can either have the volume increase or decrease.

If the gas expanded, ΔV is positive and the

PΔ - PΔV is negative, and the ΔU would be

negative; the gas will lose energy.

That's because the molecules are beating up on the piston and

moving the piston. Remember, applying a force

doesn't cost you anything. But if the point of application

moves, you do work. And who's going to pay for it,

the gas? It'll pay for it through its

loss of internal energy. Conversely, if you push down on

the gas, ΔV will be negative and this will become

positive and the energy of the gas will go up.

So, there are two ways to change the energy of these

molecules. In the end, all you want is you

want the molecules to move faster than before.

One is to put them on a hotplate where there are

fast-moving molecules. When they collide with the

slow-moving molecules, typically the slow one's a

little more faster and the fast one's a little more slower and

therefore will be a transfer of kinetic energy.

Or when you push the piston down, you can show when a

molecule collides with a moving piston.

It will actually gain energy. So, that's how you do work.

That's the first law.

So, let us now calculate the work done in a process where a

gas goes from here to here on an isotherm.

Isotherm is a graph of a given temperature.

So, this is a graph P times V equal to

constant, because PV = nRT.

If T is constant, PV is a constant,

it's the rectangular hyperbola. The product of the x and

y coordinates is constant, so when the x

coordinate vanished, the y will go to

infinity. y coordinate vanishes,

x will go to infinity. So, you want to take your gas

for a ride from here to here. Throughout it's at a certain

temperature T. What work is done by you?

That's a very nice interpretation.

The work done by you is the integral of PdV.

But what is integral of PdV?

That's P, and that's dV.

Pdv is that shaded region.

In other words, if you just write PdV it

makes absolutely no sense. If you go to a mathematician

and say, "Please do the integral for me!"

can the mathematician do this? What's coming in the way of the

mathematician actually doing the integral?

What do you have to know to really do an integral?

Student: You have to know the function.

Professor Ramamurti Shankar: You have to know

the function. If you just say P,

we'll say maybe P is a constant, in which case I'll

pull it out of the integral. But for this problem,

because PV is nRT, and T is a

constant, P is nRT divided

by V, and that's the function that you would need to

do the integral, and if you did that you will

find there's nRT. All of them are constants.

They come out of the integral, dv over V,

and integrate from the initial volume, the final volume.

And you guys know this is a logarithm, and the log of upper

minus log of lower is the log of the ratio.

And this gives me nRT ln (V_2

/V_1). So, we have done our first work

calculation. When the gas goes on an

isothermal trajectory from start to finish, from volume

V_1 to volume V_2,

the work done, this is the work done by the

gas. You can all see that gas is

expanding and that's equal to this shaded region.

By the way, I mention it now, I don't want to distract you,

but suppose later on I make it go backwards like this,

part of the way. The work done on the going

backwards part is this area, but with a minus sign.

I hope you will understand, if you go to the right the

area's considered positive. If you go to the left,

the area is considered negative.

If you do the integral and put the right limit,

you'll get the right answer. But geometrically,

the area under the graph in the PV diagram is the work

done if you're moving the direction of increasing volume.

If you would decrease the volume, for example,

if you just went back from here, the area looks the same

but the work done is considered negative.

You don't have to think very hard.

If you do the calculation going backwards, you will get a

ln of V_1 over

V_2. That'll automatically be the

negative of the log of V_2 over

V_1. But geometrically,

the area under the graph is the work, if you are going to the

right. Yes?

Student: What determines the shape of the curve that

links the first state to the second state?

Professor Ramamurti Shankar: Oh,

this one? Student: Mmm-hmm.

Professor Ramamurti Shankar: This'll be a graph,

PV equal to essentially a constant.

So, you take your gas, you see how many moles there

are. You know R,

you know the temperature, you promised not to change the

temperature. So, you'll move on a trajectory

so that the product PV never changes.

And in any xy plane, if you draw a graph where the

product xy doesn't change it'll have this shape called a

"rectangular hyperbola." It just means,

whenever one increases, the other should decrease,

keeping the product constant. That's why P is

proportional to the reciprocal of V when you do the

integral. Very good.

So, this is now the work done by the gas.

What is the heat input? The heat input is a change in

internal energy minus the work done.

Let me see. The law was ΔU = ΔQ -

ΔW. Yeah, let's go back to this law.

In this problem, ΔW is what I just

calculated, nRT, whatever the log,

V_2 over V_1.

What is ΔQ? How much heat has been put into

this gas? How do I find that?

Student: Take out the T?

Professor Ramamurti Shankar: Pardon me?

Student: You take out the T?

Professor Ramamurti Shankar: For the heat input

you mean? Yeah, you can use mc

ΔT, but you don't have to do anymore work.

By that, I mean you don't have to do any more cerebration.

What can you do with this equation to avoid doing further

calculations? Do you know anything else?

Yes? Student: [inaudible]

Professor Ramamurti Shankar: Yes.

This is what I told you is the fact that people do not

constantly remember, but you must.

This gas did not change its temperature.

Go back to equation number whatever I wrote down.

U = 3/2 nRT or something.

T doesn't change, U doesn't change.

That means the initial internal energy and final internal energy

are the same because initial temperature and final

temperature are the same. So, this guy has to be zero.

That means ΔQ is the same as ΔW in this

particular case. Yes?

Student: [inaudible] Professor Ramamurti

Shankar: When you say mc ΔT, you've got to be

careful of what formula you want to use.

I'll tell you why you cannot simply use mc ΔT.

If you've got a solid and you use mc ΔT,

that is correct, because when you heat the

solid, the heat you put in goes into heating up the solid.

Maybe let's ask the following question.

His question is the following. You're telling me you put heat

into a gas, right? And you say temperature doesn't

go up. How can that possibly be?

I always thought when I put heat into something,

temperature goes up. That's because you were

thinking about a solid, where if you put in heat it's

got to go somewhere and, of course, temperature goes up.

What do you think is happening to the gas here?

Think of the piston and weight combination.

When I want to go along this path from here to here,

you can ask yourself where is the heat input and where is the

change in energy, and why is there no change in

temperature? If you take a piston like this,

if you want to increase the volume, you can certainly take

off a grain of sand, right?

If you took the grain of sand and the piston will move up,

it will do work and it actually will cool down,

but that's not what you're doing.

You are keeping it on a hotplate at a certain

temperature so that if it tries to cool down,

heat flows from below to above maintaining the temperature.

So, what the gas is doing in this case is taking heat energy

from below and going up and working against the atmosphere

above. It takes in with one hand and

gives out to the other, without changing its energy.

So, when you study specific heat, which is my next topic,

you've got to be a little more careful when you talk about

specific heats of gases, and I will tell you why.

There is no single thing called specific heat for a gas.

There are many, many definitions depending on

the circumstances. But I hope you understand in

this case; you've got to visualize this.

It's not enough to draw diagrams and draw pictures.

What did I do to the cylinder to maintain the temperature and

yet let it expand? Expansion is going to demand

work on part of the gas. That's going to require a loss

of energy unless you pump in energy from below.

So, what I've done is that I take grain after grain,

so that the pressure drops and the volume increases,

but the slight expansion would have cooled it slightly but the

reservoir from below brings it back to the temperature of the

reservoir. So, you prop it up in

temperature. So, we draw the picture by

saying the gas went from here to here, and we usually draw a

picture like this and say heat flowed into the system during

that process. Alright, now I'll come to this

question that was raised about specific heat.

Now, specific heat, you always say is ΔQ

over ΔT or ΔT divided by the mass of the

substance. Now, it turns out that for a

gas, you've already seen that what you want to count is not

the actual mass, but the moles.

Because we have seen at the level of the ideal gas law,

the energy is controlled by not simply the mass,

but by the moles. Because every molecule gets a

certain amount of energy, namely 3/2 kT,

and you just want to count the number of molecules,

or the number of moles. Now, there are many,

many ways in which you can pump in heat into a gas and heat it

up and see how much heat it takes.

But let's agree that we will take one mole from now on and

not one kilogram. Not one kilogram.

We'll find out if you do it that way, the answer doesn't

seem to depend on the gas. That's the first thing.

Take a mole of some gas and call the specific heat as the

energy needed to raise the temperature of one mole by one

degree. So, this should not be m.

This should be the number of moles.

If you take one mole, you can say okay,

one mole of gas I was told has energy U = 3/2 RT.

Because it was there-halves nRT but n is one

mole. Now, you want to put in some

heat, and you really want the ΔQ over ΔT,

so I will remind you that ΔQ is ΔU +

PΔv.

The heat input into a gas is the change of energy plus P

Δv. And that's just from the first

law. So, if I'm going to divide

ΔQ by ΔT, there's a problem here.

Did you allow the volume to change or did you not allow the

volume to change? That's going to decide what the

specific heat is. In other words,

when a solid is heated, it expands such a tiny amount,

we don't worry about the work done by the expanding solid

against the atmosphere. But for a gas,

when you heat it, the volume changes so much that

the work it does against the external world is

non-negligible. Therefore, the specific heat is

dependent on what you allow the volume term to do.

So, there's one definition of specific heat called

C_V, and C_V is the

one at constant volume. You don't let the volume change.

In other words, you take the piston and you

clamp it. Now, you pump in heat from

below by putting it on a hotplate.

All the heat goes directly to internal energy.

None of that is lost in terms of expansion.

So, ΔV is zero. In that case,

ΔQ over ΔT at constant volume,

we denote that in this fashion, at constant volume,

this term is gone, and it just becomes ΔU

over ΔT. That's very easily done.

ΔU over ΔT is 3/2 R.

So, the specific heat of a gas at constant volume is 3 over

2R. When I studied solids,

I never bother about constant volume because a change in the

volume of a solid is so negligible when it's heated up,

it's not worth specifying that it was a constant volume

process. But for a gas,

it's going to matter whether it was constant volume or not.

Then, there's a second specific heat people like to define.

That's done as follows. You take this piston.

You have some gas at some pressure.

You pump in some heat but you don't clamp the piston.

You let the piston expand any way it wants at the same

pressure. For example,

if it's being pushed down by the atmosphere,

you let the piston move up if it wants to, maintaining the

same pressure. Well, if it moves up a little

bit, then the correct equation is the heat that you put in is

the change in internal energy plus P times ΔV,

where now P is some constant pressure,

say the atmospheric pressure, ΔV is the change in

volume. So, ΔQ needed now will

be more, because you're pumping in heat from below and you're

losing energy above because you're letting the gas expand.

Because you were letting the pressure be controlled from the

outside at some fixed value. So now, ΔQ--this

ΔU will be 3 over 2RΔT.

Now, what's the change in P times ΔV?

Here is where you should know your calculus.

The P times change in V is the same as the

change in PV, if P is a constant.

Right? Remember long back when I did

rate of change of momentum is d/dt of mv,

it's m times dv/dt,

because m doesn't change.

You can take it inside the change.

But now we use PV = RT. I'm talking about one mole.

PV = RT. That's a change in the quantity

RT, R is a constant, that's R times

ΔT so I put in here R times ΔT.

This is the ΔQ at constant pressure.

So, the specific heat of constant pressure is ΔQ

over ΔT, keeping the pressure constant.

You divide everything by ΔT you get 3 over

2R, plus another R, which is 5 over

2R.

So, the thing you have to remember, what I did in the end,

is that a gas doesn't have a single specific heat.

If we just say, put in some heat and tell me

how many calories I need to raise the temperature,

that's not enough. You have to tell me whether in

the interim, the gas was fixed in its volume,

or changed its volume, or obeyed some other condition.

The two most popular conditions people consider are either the

volume cannot change or the pressure cannot change.

If the volume cannot change, then the change in the heat you

put in goes directly to internal energy, from the First Law of

Thermodynamics. That gives you a specific heat

of 3 over 2R. If the pressure cannot change,

you get 5 over 2R. You can see

C_P is bigger than C_V

because when you let the piston expand,

then not all the heat is going to heat the gas.

Some of it is dissipated on top by working against the

atmosphere. Then, notice that I've not told

you what gas it is. That's why the specific heat

per mole is the right thing to think about because then the

answer does not depend on what particular gas you took.

Whether it's hydrogen or helium, they all have the same

specific heat per mole. They won't have the same

specific heat per gram, right?

Because one gram of helium and one gram of hydrogen don't have

the same number of moles. So, you have to remember that

we're talking about moles. The final thing I have to

caution you--very, very important.

This is for a monoatomic gas.

This is for a gas whose atom is the gas itself.

It's a point. Its only energy is kinetic

energy. There are diatomic gases,

by two of them [atoms] joined together,

they can form a dumbbell or something;

then the energy of the dumbbell has got two parts,

as you learned long ago. It can rotate around some axis

and it can also move in space. Then the internal energy has

also got two parts. Energy due to motion of the

center of mass and energy due to rotation.

Some molecules also vibrate. So, there are lots of

complicated things, but if you got only one guy,

or one atom, whatever its mass is,

it cannot rotate around itself and it cannot vibrate around

itself, so those energies all disappear.

So, we have taken the simplest one of a monoatomic gas,

a gas whose fundamental entity is a single atom rather than a

complicated molecule. And that's all you're

responsible for. I'll just say one thing.

C_P over C_V,

I want to mention it before you run off to do your homework.

I don't know if it comes up. It's called γ,

that's five-third for a monoatomic gas.

You can just take the ratio of the numbers.

If in some problem you find γ is not five-thirds,

do not panic. It just means it's a gas which

is not monoatomic. If it's not monoatomic,

these numbers don't have exactly those values.

We don't have to go beyond that. You just have to know there's a

ratio γ, which is five-thirds in the

simplest case, but in some problem,

somewhere in your life, you can get a γ which

is not five-thirds.

exciting, so I'll come back and take it from there.

So, what is it you have to remember from last time?

You know, what are the main ideas I covered?

One is, we took the notion of temperature, for which we have

an intuitive feeling and turned it into something more

quantitative, so you can not only say this is

hotter than that, that's hotter than this,

you can say by how much, by how many degrees.

And in the end we agreed to use the absolute Kelvin scale for

temperature. And the way to find the Kelvin

scale, you take the gas, any gas that you like,

like hydrogen or helium, at low concentration,

and put that inside a piston and cylinder.

That will occupy some volume and there's a certain pressure

by putting weights on top, and you take the product of

P times V, and the claim is for whatever

gas you take, it'll be a straight line.

Remember now, this is in Kelvin. Your centigrade scale is

somewhere over here, but I've shifted the origin to

the Kelvin scale. So, somewhere here will be the

boiling point of water, somewhere the freezing point of

water; that may be the boiling point

of water. And if you took a different

amount of a different gas, you'll get some other line.

But they will always be straight lines if the

concentration is sufficiently low.

In other words, it appears that pressure times

volume is some constant. I don't know what to call it.

Say c, times this temperature.

And you can use that to measure temperature because if you know

two points on a straight line then you know that you can find

the slope and then you can calibrate the thermometer,

then for any other value of P times V that you

get, you can come down and read your temperature.

That's the preferred scale, and we prefer this scale

because it doesn't seem to depend on the gas that you use.

I can use one; you can use another one.

People in another planet who have never heard of water--they

can use a different gas. But all gases seem to have the

property that pressure times volume is linearly proportional

to this new temperature scale, measured with this new origin

at absolute zero. There is really nothing to the

left of this T = 0.

The next thing I mentioned was, people used to think of the

theory of heat as a new theory. You know, we got mechanics and

all that stuff--levers and pulleys and all that.

Then, you have this mysterious thing called heat,

which has been around for many years but people started

quantifying it by saying there's a fluid called the caloric fluid

and hot things have a lot of it, and cold things have less of

it, and when you mix them the caloric somehow flows from the

hot to the cold. Then we defined specific heat,

law of conservation of this caloric fluid that allows you to

do some problems in calorimetry. You mix so much of this with so

much of that, where will they end up?

That kind of problem. So, that promoted heat to a new

and independent entity, different from all other things

we have studied. But something suggests that it

is not completely alien or a new concept, because there seems to

be a conservation of law for this heat,

because the heat lost by the cold water was the heat gained

by the hot water. I'm sorry.

Heat gained by the cold water was the heat lost by the hot

water. So, you have a conservation law.

Secondly, we know another way to produce heat.

Instead of saying put it on the stove, put it on the stove,

in which case, there is something mysterious

flowing from the stove into the water that heats it up,

I told you there's a different thing you can do.

Take two automobiles; slam them.

This is not the most economical way to make your dinner but I'm

just telling you as a matter of principle.

Buy two Ferraris, slam them into each other and

take this pot and put it on top and it'll heat up because

Ferraris will heat up. The question is what happened

to the kinetic energy of the two cars?

That is really gone. So, in the old days,

we would say, well, we don't apply the Law of

Conservation of Energy because this was an inelastic

relationship. That was our legal way out of

the whole issue. But you realize now this

caloric fluid can be produced from nowhere,

because there was no caloric fluid before,

but slamming the two cars produce this extra heat.

So, that indicates that perhaps there's a relation between

mechanical energy and heat energy--that when mechanical

energy disappears, heat energy appears.

So, how do you do the conversion ratio?

You know, how many calories can you get if you sacrifice one

joule of mechanical energy? So, Joule did the experiment.

Not with cars. I mean, he didn't have cars at

that time, so if he did he would've probably done it with

cars. He had this gadget with him,

which is a little shaft with some paddles and a pulley on the

top, and you let the weight go down.

And I told you guys the weight goes from here to here,

the mgh loss will not be the gain in ½ mv^(2).

Something will be missing. Keep track of the missing

amount. So many joules--but meanwhile

you find this water has become hot.

You find then how many calories should have gone in,

because we know the specific of water,

we know the rise in temperature, we know how many

calories were produced. And then, you compare the two

and you find that 4.2 joules = 1 calorie.

So, that is the conversion ratio of calories to joules.

One joule, 4.2 joules of mechanical energy.

So, in the example of the colliding cars,

take the ½ mv^(2) for each car, turn it into joules,

slam them together. If they come to rest,

you've lost all of that, and then you take that and you

write it as--divided by 4.2 and that's how much calories you

have produced. If the car was made of just one

material, it had a specific heat, then it would go up by a

certain temperature you can actually predict.

Okay. So today, I want to go a little

deeper into the question of where is the energy actually

stored in the car, and what is heat.

We still don't know in detail what heat is.

We just said car heats up and the loss of joules divided by

4.2 is the gain in calories. Now, we can answer in detail

exactly what is heat. That's what we're going to talk

about today. When we say something is

hotter, what do we mean on a microscopic level?

In the old days when people didn't know what anything was

made of, they didn't have this understanding.

And the understanding that I'm going to give you today is based

on a simple fact that everything is made up of atoms.

That was not known, and that's one of the greatest

discoveries that, in the end,

everything is made up of atoms, and atoms combine to form

molecules and so on. So, how does that come into

play? For that, I want you to take

the simple example where the temperature enters.

That is in the relation PV equal to some constant

times temperature. Do you know what I'm talking

about? Take some gas,

make sure it's sufficiently dilute, put it into this piston,

measure the weights on top of it,

divide it by the area, to get the pressure,

that's the pressure, that's the volume.

The volume is the region here, multiply the product;

then, if you heat up the gas by putting it on some hot plate,

you'll find the product PV increases,

and as the temperature increases, PV is

proportional to T. We want to ask what is this

proportionality constant. Suppose you were doing this.

In the old days, this is what people did.

What did we think should be on the right-hand side?

What is going to control this particular constant for the

given experiment? Do you know what it might be

proportional to? Yes?

Student: Amount of gas? Professor Ramamurti

Shankar: Amount of gas. That's true,

because if the amount of gas is zero, we think there's no

pressure. When you say "amount of gas,"

that's a very safe sentence because amount measured by what

means? By what metric?

Student: Probably number of particles?

Professor Ramamurti Shankar: Right.

Suppose you were not aware of particles.

Then, what would you mean by "amount of gas?"

Student: Mass. Professor Ramamurti

Shankar: The mass. Now, if you guys ever said

moles, I was going to shoot you down.

You're not supposed to know those things.

We are trying to deduce that. So, put yourself back in

whatever stone ages we were in. We don't know anything else.

Mass would be a reasonable argument, right?

What's the argument? We know that if you have some

amount of gas producing the pressure, and you put twice as

much stuff, you would think it will produce

twice as much pressure. Same reason why you think the

expansion of a rod is proportionally change in

temperature times the starting length.

So, this mass is what's doing it.

So, it's proportional to mass. It's a very reasonable guess.

So, if you put more gas into your piston you think it'll

produce more pressure. That's actually correct.

So, let's go to that one particular sample in your

laboratory that you did. So, put the mass that you had

there. Then, you should put a constant

still. I don't know what you want me

to call this constant, say, c prime.

This constant contains everything, but I pulled out the

mass and the remaining constant I want to call c prime.

This is actually correct. You can take a certain gas and

you can find out what c prime is.

But here is what people found. If you do it that way,

the constant c prime depends on the gas you are

considering. If you consider hydrogen gas,

let's call that c prime for hydrogen.

Somebody else puts in helium gas.

Then you find the c prime for helium is one-fourth

c prime for hydrogen.

If you do carbon, it's another number.

c prime for carbon is c prime for hydrogen

divided by 12. So, each gas has a different

constant. So, we conclude that yes,

it's the mass that decides it but the mass has to be divided

by different numbers for different gases to find the real

effective mass in terms of pressure.

In other words, one gram of hydrogen and one

gram of helium do not have the same pressure.

In fact, one gram of helium has to be divided by 4 to find its

effect on pressure. So, you have to think about why

is it that the mass directly is not involved.

Mass has to be divided by a number, and the number is a

nice, round number. 4 for this and 12 for that,

and of course people figure out there's a long story I cannot go

into, but I think you all know the answer.

But now we are allowed to fast forward to the correct answer,

because I really don't have the time to see how they worked it

out, but from these integers and the

way the gases reacted and formed complicated molecules,

they figured out what's really going on is that you're dividing

by a number that's proportional with the mass of the underlying

fundamental entity, which would be an atom.

In some case a molecule, but I'm just going to call

everything as atom. So, if things,

like, carbon, as atoms, weigh 12 times as

much as things called hydrogen, then if you took some amount of

carbon, you divide it by a number, like,

12 to count the number of carbon atoms.

Okay, so hydrogen you want to count the number of hydrogen

atoms. So then, what really you want

here is not the mass, but the number of atoms of a

given kind. We are certainly free to write

either a mass or the number of atoms, because the two are

proportional. But the beauty of writing it

this way, you write it in this fashion, by this new constant

k, k is independent of the gas.

So, you want to write it in a manner in which it doesn't

depend on the gas. You can write it in terms of

mass. If you did, for each mass

you've got to divide by a certain number.

TThen once you divide it by the number you can put a single

constant in front. Or if you want a universal

constant, what you should really be counting is the number of

atoms or molecules.

So, you couldn't have written it that way until you knew about

atoms and molecules and people who are led to atoms and

molecules by looking at the way gases interact,

and it's a beautiful piece of chemistry to figure out really

that there are entities which come in discreet units.

Not at all obvious in the old days, that mass comes in

discreet units called atoms, but that's what they deduced.

So, this is called the Boltzmann Constant.

The Boltzmann Constant has a value of 1.4 times 10^(-23),

let's see, joules/Kelvin.

That's it. Or joules/Kelvin or degrees

centigrade.

So, this is a universal constant.

So, now what people like to do is they don't like to write the

number [N], because if you write the number,

in a typical situation, what's the number going to be?

Take some random group gas. One gram, two grams,

one kilogram, it doesn't matter.

The number you will put in there is some number like

10^(23) or 10^(25). That's a huge number.

So, whenever a huge number is involved, what you try to do is

to measure the huge number as a simple multiple off another huge

number, which will be our units for

measuring large numbers. For example,

when you want to buy eggs, you measure in dozens.

When you want to buy paper, you might want to measure it in

thousands or five hundreds or whatever unit they sell them in.

It's a natural unit. When you want to find

intergalactic distances, you may use a light year.

You use units so that in that unit, the quantity of interest

to us is some number that you can count in your hands.

When you count people's height, you use feet because it's

something between 1 and 8, let's say.

You don't want to use angstroms and you don't want to use

millimeters. Likewise, when you want to

simply count numbers, it turns out there's a very

natural number called Avogadro's Number,

and Avogadro's Number is 6 times 10^(23).

There's no unit. It's simply a number,

and that's called a mole. So, a mole is like a dozen.

We wanted to buy 6 times 10^(23) eggs,

you will say get me one mole of eggs.

A mole is just a number. It's a huge number.

You can ask yourself what's so great about this number?

Why would someone think of this particular number?

Why not some other number? Why not 10^(24)?

Do you know what's special about this number?

Yes? Student: [inaudible]

Professor Ramamurti Shankar: Yes.

If you like, a mole is such that one mole of

hydrogen weighs one gram. And hydrogen is the simplest

element with a nucleus of just a proton and the electron's mass

is negligible. So, this, if you like,

is the reciprocal of the mass of hydrogen.

In other words, one over Avogadro's Number is

the mass of hydrogen in grams, of a hydrogen atom in grams.

So, you basically say, I want to count this large

number so let me take one gram, which is my normal unit if

you're thinking in grams. Then I ask, "How many hydrogen

atoms does one gram of hydrogen contain?"

That's the number. That's the mole.

So, if you decide to measure the number of atoms you have in

a given problem, in terms of this number,

you write it as some other small number called moles,

times the number in a mole, and you are free to write it

this way. If you write it this way,

then you write this nRT, R is the universal gas

constant. What's n times the

Boltzmann Constant. N_0 times the

Boltzmann Constant. That happens to be 8.3 joules

per degree centigrade or per Kelvin.

Right? The units for R will be

PV, which is units of energy divided by T.

In terms of calories, I'd remember this as a nice,

round number. Two calories per degree

centigrade. Degrees centigrade and Kelvin

are the same. The origins are shifted,

but when you go up by one degree in centigrade or Kelvin,

you go the same amount in temperature.

So, this [R] is what they found out first,

because they didn't know anything about atoms and so on.

But later on when you go look under the hood of what the gas

is made of, if you write it in terms of the number of actual

atoms, you should use the little

k, or you can write it in terms of number of moles,

in which case use big R.

And the relation between the two is simply this.

If you're thinking of a gas and how many moles of gas do I have?

For example, one gram of hydrogen would be

one mole. Then you will use R.

If you've gone right down to fundamentals and say,

"How many atoms do I have?" and you put that here,

you will multiply it by this very tiny number.

Alright. Now, you start with this law

and you ask the following question.

On the left-hand side is the quantity P times

V. On the right-hand side I have

nRT, but let me write it now as NkT.

You guys should be able to go back and forth between writing

in terms of number of moles or the number of atoms.

You'll like this because all numbers here will be small,

of the order 1. R is a number like 8,

in some units, and n would be 1 or 2

moles. Here, this N will be a

huge number, like 10^(23). k will be a tiny number

like 10^(-23). Think in terms of atoms.

That's what you do. Big numbers, small constants.

When you think of moles, moderate numbers and moderate

value of constants. We want to ask ourselves,

"Is there a microscopic basis for this equation?"

In other words, once we believe in atoms,

do we understand why there is a pressure at all in a gas?

That's what we're going to think about now.

So, for this purpose, we will take a cube of gas.

Here it is.

This is a cube of side L by L by L.

Inside this is gas and it's got some pressure,

and I want to know what's the value of the pressure.

You've got to ask yourself, why is there pressure?

Remember, I told you what pressure means.

If you take this face of the cube, for example,

it's got to be nailed down to the other faces;

otherwise, it'll just come flying out because the gas is

pushing you out. The pressure is the force on

this face divided by area. So, somebody inside is trying

to get out. Those guys are the molecules or

the atoms, and what they're doing is constantly bouncing off

the wall, and every time this one bounces

on a wall, its momentum changes from that to the other one.

So, who's changing the momentum? Well, the wall is changing the

momentum. It's reversing it.

For example, if you bounce head-on and go

back, your momentum is reversed. That means you push the wall

with some force and the wall pushes you back with the

opposite force. It's the force that you exert

on the wall that I'm interested in.

I want to find the force on the wall, say, this particular face.

You can find the pressure on any face.

It's going to be the same answer.

I'm going to take the shaded face to find the pressure on it.

Now, if you want to ask, what is the force exerted by me

on any body, I know the force has a rate of change of

momentum, because that is d/dt of

mv, and m is a constant, and that's just

dv/dt, which is ma.

I'm just using old F = ma, but I'm writing it as a

rate of change and momentum. Now, I have N molecules

or N atoms, randomly moving inside the box.

Each in its own direction, suffering collisions with the

box, bouncing off like a billiard ball would at the end

of the pool table and going to another wall and doing it.

Now, that's a very complicated problem, so we're going to

simplify the problem. The simplification is going to

be, we are going to assume that one-third of the molecules are

moving from left to right. One-third are moving up and

down and one-third are moving in and out of the blackboard.

If at all you make an assumption that the molecules

are simply moving in the three primary directions,

of course you will have to give equal numbers in these

directions. Nothing in the gas that favors

horizontal or vertical. In reality, of course,

you must admit the fact they move in all directions,

but the simplified derivation happens to give all the right

physics, so I'm going to use that.

So, N over three molecules are going back and

forth between this wall, and this wall.

I'm showing you a side view. The wall itself looks like this.

The molecules go back and forth.

Next assumption. All the molecules have the same

speed, which I'm going to call v.

That also is a gross and crude description of the problem,

but I'm going to do that anyway and see what happens.

So now, you ask yourself the following question.

Take one particular molecule. When it hits the wall and it

bounces back, its momentum changes from

mv to -mv; therefore, the change in

momentum is 2mv.

How often does that change take place?

You guys should think about that first.

How often will that collision take place?

Once you hit the wall here, you've got to go to the other

wall and come back. So, you've got to go a distance

2L, and you're going at a speed v,

the time it takes you is 2L over v.

So, ΔP over ΔT is 2L divided by

v. That gives me mv^(2)

over L. That is the force due to one

molecule. That's the average force.

You realize it's not a continuous force.

The molecule will hit the wall, there's a little force exchange

between the two, then there's nothing,

then you wait until it comes back and hits the wall again.

If that were the only thing going on, what you would find is

the wall most of the time, has no pressure and suddenly it

has a lot of pressure and then suddenly nothing.

But fortunately, this is not the only molecule.

There are roughly 10^(23) guys pounding themselves against the

wall. So, at any given instant,

even if it's 10^(-5) seconds, there'd be a large number of

molecules colliding. So, that's why the force will

appear to be steady rather than a sharp noise.

It looked very steady because somebody or other will be

pushing against the wall.

This is the force due to one molecule.

The force due to all of them would be N over 3 times

mv^(2) over L.

N over 3 because of the N molecules,

a third of them were moving in this direction.

You realize the other two directions are parallel to the

wall. They don't apply force on the

wall. To apply force on the wall,

you've got to be moving perpendicular to the wall.

For example, if the planes that walls are

coming out of the blackboard, moving in and out of the

blackboard doesn't produce a force on this wall.

That produces a force on the other two faces.

So, as far as any one set of faces is concerned,

in one plane, only the motion orthogonal to

that is going to contribute. That's why you have N

over 3. We're almost done.

That's the average force. If you want,

I can denote average by some F bar.

Then what about the average pressure?

The average pressure is the average force divided by the

area of that face, which is F over

L^(2), that gives me N over 3,

mv^(2) over L^(3).

Now, this is very nice because L^(3) is just the volume

of my box.

So, I take the L^(3), which is equal to the volume of

my box, and I send it to the other side

and write it as PV equals N over 3mv^(2).

This is what the microscopic theory tells you.

Microscopic theory says, if your molecules all have a

single speed, they're moving randomly in

space so that a third of them are moving back and forth

against that wall and this wall, then this is the product

PV. Experimentally,

you find PV = NkT.

So, you compare the two expressions and out comes one of

the most beautiful results, which is that mv^(2)

over 2 is 3 over 2kT. Now that guy deserves a box.

Look what it's telling you. It's a really profound formula.

It tells you for the first time a real microscopic meaning of

temperature. What you and I call the

temperature for gas is simply, up to these factors,

3/2 k, simply the kinetic energy of

the molecules. That's what temperature is.

If you've got a gas and you put your hand into the furnace and

it feels hot, the temperature you're

measuring is directly the kinetic energy of the molecules.

That is a great insight into what temperature means.

Remember, this is not true if T is measured in

centigrade. If T were measured in

centigrade, our freezing point of water mv^(2),

would vanish. But that's not what's implied.

T should be measured from absolute zero.

It also tells you why absolute zero is absolute.

As you cool your gas, the kinetic energy of molecules

are decreasing and decreasing and decreasing,

but you cannot go below not moving at all,

right? That's the lowest possible

kinetic energy. That's why it's absolute zero.

At that point, everybody stops moving.

That's why you have no pressure. Now, these results are modified

by the laws of quantum mechanics, but we don't have to

worry about that now. In the classical physics,

it's actually correct to say that when the temperature goes

to zero, all motion ceases. Now, this is the picture I want

you to bear in mind when you say temperature.

Absolute temperature is a measure of molecular agitation.

More precisely, up to the constant k,

3/2 k, the kinetic energy of a

molecule is the absolute temperature.

That's for a gas. If you took a solid and you

say, what happens when I heat the solid?

You have a question? Yes?

Student: [inaudible] Professor Ramamurti

Shankar: You divide by 2 because ½ mv^(2) is a

familiar quantity, namely, kinetic energy.

That's why you divide by 2. Another thing to notice is that

every gas, whatever it's made of, at a given temperature has a

given kinetic energy because the kinetic energy per molecule on

the left-hand side is dependent on absolute temperature and

nothing else. So at certain degrees,

like 300 Kelvin, hydrogen kinetic energy would

be the same, carbon kinetic energy would also be the same.

The kinetic energy will be the same, not the velocity.

So, the carbon atom is heavier, it will be moving slower at

that temperature in order to have the same kinetic energy.

So, all molecules, all gases, have a given

temperature. All atoms, let me say,

at a given temperature in gaseous form will have the same

kinetic energy [per molecule]. Now, if you have a

solid--What's the difference between a gas and a solid?

In a gas, the atoms are moving anywhere they want in the box.

In a solid, every atom has a place.

If you take a two-dimensional solid, the atoms look like this.

They form a lattice or an array. That's because you will find

out that, this is more advanced stuff, that every atom finds

itself in a potential that looks like this.

Imagine on the ground you make these hollows.

Low points -- low potential; high points -- high potential.

Obviously, if you put a bunch of objects here they will sit at

the bottom of these little concave holes you've dug in the

ground. At zero degrees absolute all

atoms will sit at the bottom of their allotted positions;

that'll be a solid at zero temperature.

So in a solid, everybody has a location.

I've shown you a one-dimensional solid,

but you can imagine a three-dimensional solid where in

a lattice of three-dimensional points,

there's an assigned place for each atom and it sits there.

If you heat up that solid now, what happens is these guys

start vibrating. Now, here is where your

knowledge of simple harmonic motions will come into play.

When you take a system in equilibrium, it will execute

simple harmonic motion if you give it a real kick.

If you put it on top of a hotplate, the atoms in the hot

plate will bump into these guys and start them moving.

They will start vibrating. So, a hot solid is one in which

the atoms are making more and more violent oscillations around

their assigned positions. If you heat them more and more

and more, eventually you start doing this.

You go all the way from here to here;

there is nothing to prevent it from rolling over to the next

side. Once you jump the fence,

you know, think of a bunch of houses, okay?

Or a hole in the ground. You're living in a hole in the

ground, as you get agitated you're able to do more and more

oscillations so you can roll over to the next house.

Once that happens all hell breaks loose because you don't

have any reason to stay where you are.

You start going everywhere. What do you think that is?

Student: Melting. Professor Ramamurti

Shankar: Pardon me? Student: Melting.

Professor Ramamurti Shankar: That's melting.

That's the definition of melting.

Melting is when you can leap over this potential barrier,

potential energy barrier, and go to the next site.

The next side is just like this side.

If you can jump that fence, you can jump this one.

You go everywhere and you melt. That's the process of melting,

and once you have a liquid, atoms don't have a definite

location. Now, between a liquid and a

solid, there is this clear difference, but a liquid and a

vapor is more subtle. So, I don't want to go into

that. If you look at a liquid

locally, it will look very much like a solid in the sense that

inter-atomic spacing is very tightly constrained in liquid.

Whereas in a solid, if I know I am here,

I know if I go 100 times the basic lattice spacing,

there'll be another person sitting there.

That's called long-range order. In a liquid, I cannot say that.

In a liquid, I can say I am here.

Locally, the environment around me is known, but if you go a few

hundred miles, I cannot tell you a precise

location if some other atom will be there or not.

So, we say liquid is short-range positional order,

but not long-range order, and a gas has no order at all.

If I tell you there's a gas molecule here,

I cannot tell you where anybody else is because nobody has any

assigned location. Okay.

So, this is the picture you should have of temperature.

Temperature is agitated motion. Either motion in the vicinity

of where you are told to sit. If you're in a solid a motion

all over the box with more and more kinetic energy.

The next thing in this caricature is that it is

certainly not true that a third of the molecules are moving back

and forth. We know that's a joke, right?

Now, in this room there's no reason on earth a third of the

molecules are doing this than others are doing.

That's not approximation. They're moving in random

directions. So, if you really got the

stomach for it, you should do a pressure

calculation in which you assume the molecules of random

velocities sprinkled in all directions,

and after all the hard work, turns out you get exactly this

answer. So, that's one thing I didn't

want to do. But something I should point

out to you is the following. So, suppose I give you a gas at

300 Kelvin. You go and you take this

formula literally and you calculate from it a certain ½

mv^(2). If you knew the mass of the

atom, say, it's hydrogen, we know the mass of hydrogen.

Then, you find the velocity and you say okay,

this man tells me that anytime I catch a hydrogen atom,

it'll have this velocity at 300 Kelvin.

It may have random direction, but he tells me that's the

velocity square. Take the square root of that,

that's the velocity. It seems to us saying the

unique velocity to each temperature.

Well, that's not correct. Not only are the molecules

moving in random directions, they're also moving with

essentially all possible velocities.

In fact, there are many, many possible velocities and

this velocity I'm getting, in this formula,

is some kind of average velocity, or the most popular

one, or the most common one.

So, if you really go to a gas and you have the ability to look

into it and see for each velocity, what's the probability

that I get that velocity? The picture I've given you is

the probability of zero except at this one magical velocity

controlled by the temperature. But the real graph looks like

this.

It has a certain peak. It likes to have a certain

value. If you know enough about

statistics, you know there's a most probable value,

there's a median, there's a mean value.

There are different definitions. They will all vary by factors

of order 1, but the average kinetic energy will obey this

condition. Yes?

Student: [inaudible] Professor Ramamurti

Shankar: It's not really a Gaussian because if you draw the

nature of this curve, it looks like v^(2)e to

the -mv^(2) over 2kT.

That's the graph I'm trying to draw here.

So, it looks like a Gaussian in the vicinity of this,

but it's kind of skewed. It's forced to vanish at the

origin.

And it's not peaked at v = 0.

A real Gaussian peak at this point would be symmetric.

It's not symmetric; it vanishes here and it

vanishes infinity. So, this is called a

Maxwell-Boltzmann distribution. You don't have to remember any

names but that is the detailed property of what's happening in

a gas. So, a temperature does not pick

a unique velocity, but it picks this graph.

If you vary your temperature, look at what you have to do.

If you change the number T here,

if you double the value of T,

that means if you double the value of v^(2) here and

there, the graph will look the same.

So, at every temperature there is a certain shape.

If you go to your temperature, it will look more or less the

same, but it may be peaked at a different velocity if you go to

a higher temperature. Now, this is another thing I

want to tell you. If you took a box containing

not atoms but just radiation, in other words,

go inside a pizza oven. Take out all the air,

but the oven is still hot, and the walls of the oven are

radiating electromagnetic radiation.

Electromagnetic radiation comes in different frequencies,

and you can ask how much energy is contained in every possible

frequency range. You know, each frequency is a

color so you know that. So, how much energy is in the

red and how much is in the blue? That graph also looks like this.

That's a more complicated law called the Planck distribution.

That law also has a shape completely determined by

temperature. Whereas for atoms,

the shape is determined by temperature as well as the mass

of the molecules. In the case of radiation,

it's determined fully by temperature and the velocity of

light. You give me a temperature,

and I will draw you another one of these roughly bell-shaped

curves. As you heat up the furnace,

the shape will change.

So again, a temperature for radiation means a particular

distribution of energies at each frequency.

For a gas it means a distribution of velocities.

Has anybody seen that in the news lately, you know,

or heard about this? Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

Student: [inaudible] Professor Ramamurti

Shankar: About this particular graph for radiation.

The probability at each frequency of finding radiation

of the frequency in a furnace of some temperature T.

Yes? Student: [inaudible]

Professor Ramamurti Shankar: No,

but in current news. In the last few years,

what people did was the following.

It's one of the predictions of the Big Bang theory that the

universe was formed some 14 and a half billion years ago,

and in the earliest stages the temperature of the universe was

some incredibly high degrees, then as it expanded the

universe cooled, and today, at the current size,

it has got a certain average temperature,

which is a remnant of the Big Bang.

And that temperature means that we are sitting in furnace of the

Big Bang. But the furnace has cooled a

lot over the billions of years. The temperature of the universe

is around 3 degrees Kelvin. And the way you determine that

is you point your telescope in the sky.

Of course, you're going to get light from this star;

you're going to get light from that star.

Ignore all the pointy things and look at the smooth

background, and it should be the same in all directions.

And plot that radiation, and now they use satellites to

plot that, and you'll get a perfect fit to this kind of

furnace radiation, called Black Body Radiation.

And you read the temperature by taking that graph and fitting it

to a graph like this, but there'll be temperature.

In the case of light, this won't be velocity squared,

but it will be the frequency squared,

but read off the temperature that'll make this work and

that's what gives you 3.1 or something.

Near 3 degrees Kelvin. In fact, the data point for

that now if you got that in your lab then you will be definitely

busted for fudging your data because it's a perfect fit to

Black Body Radiation. One of the most perfect fits to

Black Body Radiation is the background radiation of the Big

Bang. And it's isotropic,

meaning it's the same in all directions, and this is one of

the predictions of the Big Bang is that that'll be the remnant

of the Black Body Radiation. Again, it tells you there's a

sense in which, if you go to intergalactic

space, that is your temperature. That's the temperature you get

for free. We're all living in that heat

bath at 3 degrees. You want more heat,

you've got to light up your furnace but this is everywhere

in the universe, that heat left over from

creation. Okay.

That's a very, very interesting subject.

You know, a lot of new physics is coming out by looking at just

the Black Body Radiation because the radiation that's coming to

your eye left those stars long ago.

So, what you see today is not what's happening today.

It's what happened long ago when the radiation left that

part of the universe. Therefore, we can actually tell

something about the universe not only now, but at earlier

periods. And that's the way in which we

can actually tell whether the universe is expanding or not

expanding or is it accelerating in its expansion,

or you can even say once it was decelerating and now it's

accelerating. All that information comes by

being able to look at the radiation from the Big Bang.

But for you guys, I think the most interesting

thing is that when you are in thermal equilibrium,

and you are living in a certain temperature, then the radiation

in your world and the molecules and atoms in your world,

will have a distribution of frequencies and velocities given

by that universal graph. Now in our class,

we will simplify life and replace this graph with a huge

peak at a certain velocity by pretending everybody's at that

velocity. We will treat the whole gas as

if it was represented by single average number.

So, when someone says find the velocity of molecules,

they're talking about the average velocity.

You know statistically that it's the distribution of answers

and an average answer. Because the average is what you

and I have to know. Namely, ½ mv^(2) is 3/2

kT, on average. Okay.

Now, I'm going to study in detail thermodynamics.

So, the system I'm going to study is the only one we all

study, which is an ideal gas sitting inside a piston.

It's got a temperature, it's got a pressure,

and it's got a volume. And I'm going to plot here

pressure and volume and I'm going to put a dot and that's my

gas. The state of my gas is

summarized by where you put the dot.

Every dot here is a possible state of equilibrium for the

gas. Remember, the gas,

if you look at it under the hood, is made up of 10^(23)

molecules. The real, real state of the gas

is obtained by saying, giving me 10^(23) locations and

10^(23) velocities. According to Newton,

that's the maximum information you can give me about the gas

right now, because with that and Newton's

laws I can predict the future. But when you study

thermodynamics, you don't really want to look

into the details. You want to look at gross

macroscopic properties and there are two that you need.

Pressure and volume. Now, you might say,

"What about temperature?" Why don't I have a third axis

for temperature? Why is there also not a

property? Yes?

Student: [inaudible] Professor Ramamurti

Shankar: Yeah. Because PV = NkT.

I don't have to give you T, if I know P and

V. There's not an independent

thing you can pick. You can pick P and

V independently. You cannot pick T.

Let me tell you, by the way, PV = NkT is

not a universal law. It's the law that you apply to

dilute gases. But we are going to just study

only dilute ideal gas. Ideal gas is one in which the

atoms and molecules are so far apart that they don't feel any

forces between each other unless they collide.

So, here is my gas. It's sitting here.

Now, what I do, I had a few weights on top of

it. Three weights.

I suddenly pull out one weight. Throw it out.

What do you think will happen? Well, I think this gas will now

shoot up, it'll bob up and down a few times.

Then after a few seconds, or a fraction of a second,

it'll settle down with a new location.

By "settle down," I mean after a while I will not see any

macroscopic motion. Then the gas has a new pressure

and a new volume. It's gone from being there to

being there.

What about in between? What happened in between the

starting and finishing points? You might say look,

if it was here in the beginning it was there later,

it must've followed some path. Not really.

Not in this process, because if you do it very

abruptly, suddenly throwing out one-third of the weights,

there's a period when the piston rushes up,

when the gas is not in equilibrium.

By that, I mean there is no single pressure you can

associate with the gas. The bottom of the gas doesn't

even know the top is flying off. It's at the old pressure.

At the top of the gas there's a low pressure.

So, different parts of the gas at different pressure,

we don't call that equilibrium. So, the dot,

representing this system, moves off the graph.

It's off. It's off the radar,

and only when it has finally settled down,

the entire gas can make up its mind on what its pressure wants

to be; you put it back here.

So, we have a little problem that we have these equilibrium

states, but when you try to go from one to another you fly off

the map. So, you want to find a device

by which you can stay on the PV diagram as you change

the state of the gas, and that brings us to the

notion of what you call a quasi-static process.

A quasi-static process is trying to have it both ways in

which you want to change the state of the gas,

and you don't want it to leave the PV diagram.

You want it to be always at equilibrium.

So, what you really want to do is not put in three big fat

blocks like this, but instead take a gas where

you have many, many grains of sand.

They can produce the pressure. Now, remove one grain of sand.

It moves a tiny bit and very quickly settles down.

It is again true during the tiny bit of settling down you

didn't know what it was doing, but you certainly nailed it at

the second location. You move one grain at a time,

then you get a picture like this and you can see where this

is going. You can make the grain smaller

and smaller and smaller and in a mathematical sense you can then

form a continuous line. That is to say,

you perform a process that leaves the system arbitrarily

close to equilibrium, meaning give it enough time to

readjust to the new pressure, settle down to the new volume,

take another grain and another grain.

And in the spirit of calculus, you can make these changes

vanishing so that you can really then say you did this.

Yes? Student: Are all of

these small processes reversible?

Professor Ramamurti Shankar: Pardon me?

Student: Are all of these small processes

reversible? Professor Ramamurti

Shankar: Yes. Such a process is also

called--you can call it quasi-static but one of the

features of that, it is reversible.

You've got to be a little careful when you say reversible.

What we mean by "reversible" is, if I took off a grain of

sand and it came from here to the next dot,

and I put the grain back, it'll climb back to where it

was. So, you can go back and forth

on this. But now, that's an idealized

process because if you had a friction, if you had any

friction between the piston and the walls,

then if you took out a grain and it went up,

you put the grain back it might not come back to quite where it

is. Because some of the frictional

losses you will never get back. You cannot put Humpty Dumpty

back. So, most of the time processes

are not reversible, even if you do them slowly,

if there is friction. So, assume it's a completely

frictionless system. Because if there is friction,

there is some heat that goes out somewhere and some energy is

lost somewhere and we cannot bring it back.

If we took a frictionless piston and on top of it moved it

very, very slowly, you can follow this graph.

That's the kind of thermodynamic process we're

talking about. In the old days,

when I studied a single particular of the xy

plane, I just said the guy goes from here to here to there.

That's very easy to study and there's no restriction on how

quickly or how fast it moved. Particles have trajectories no

matter how quickly they move. For a thermodynamic system,

you cannot move them too fast, because they are extended and

you are having a huge gas a single number called pressure,

so you cannot change one part of the gas without waiting for

all of them to communicate and readjust and achieve a global

value for the new pressure and you can move gradually.

That's why it takes time to drag along 10^(23) particles as

if they are the single number or two numbers characterizing them.

So, we'll be studying processes like this.

Now, this is called a state. Two is a state and one is a

state. Every dot here, that is a state.

Now, in every state of the system, I'm going to define a

new variable, which is called a quantity

called U, which stands for the internal

energy of the gas.

Internal energy is simply the kinetic energy of the gas

molecules. For solids and liquids,

there's a more complicated formula.

For the gas, internal energy is just the

kinetic energy. And what is that?

It is 3/2 kT per molecule times N.

I'm sorry, 3/2 Nk, yeah. 3/2 kT times that.

Or we can write it as 3/2 nRT.

But nRT is PV. You can also write it as 3/2

PV, so internal energy is just 3/2 PV.

That means at a given point on the PV diagram,

you have a certain internal energy.

If you are there, that's your internal energy.

Take there-halves of PV and that's the energy and that's

literally the kinetic energy of all the molecules in your box.

So, now I'm ready to write down what's called the First Law of

Thermodynamics that talks about what happens if you make a move

in the PV plane from one place to another place.

If you go from one place to another place,

your internal energy will change from U_1

to U_2. Let's call it ΔU.

We want to ask what causes the internal energy of the gas to

change. So, you guys think about it now.

Now that you know all about what's happening in the

cylinder, you can ask how I will change the energy?

Well, if you wanted to change the energy of a system,

there are two ways you can do it.

One is you can do work on the gas.

Another thing is you can put the gas on a hotplate.

If you put it on hotplate, we know it's going to get

hotter. If it gets hotter,

temperature goes up. If temperature goes up,

the internal energy goes up. So, there are two ways to

change the energy of a gas. The first one we call heat

input. That just means put it on

something hotter and let the thing heat it up.

Temperature will go up. Notice that the internal energy

of an ideal gas depends only on the temperature.

That's something very, very important.

I mention it every time I teach the subject and some people

forget and lose a lot of points needlessly.

So, I'll say it once more with feeling.

The energy of an ideal gas depends only on the temperature.

If the temperature is not changed;

energy has not changed. So, try to remember that for

what I do later. So, the change of the gas,

this cylinder full that I put some weights on top and I've got

gas inside, it can change either because I

did, I put in some heat, or the gas did some work.

By that, I mean if the gas expands by pushing out against

the atmosphere, then it was doing the work and

ΔW is the work done by the gas.

That's why it comes to the minus sign, because it's the

work done by the gas. If you do work, you lose energy.

So, what's the formula for work done?

Let's calculate that. If I've got a piston here,

it's the force times the distance.

But the force is the pressure times the area times the

distance. Now, you guys should know

enough geometry to know the area of the piston times the distance

it moves is the change in the volume.

So, we can write it as P times dV.

That leads to this great law. Let me write it on a new

blackboard because we're going to be playing around with that

law. This is law number one.

The change in the internal energy of a system is equal to

ΔQ - PΔV.

What does it express? It expresses the Law of

Conservation of Energy. It says the energy goes up,

either because you pushed the piston or the piston pushed you;

then you decide what the overall sign is,

or you put it on a hotplate. We are now equating putting it

on a hotplate as also equivalent to giving it energy,

because we identify heat as simply energy.

So, if you took the piston and you nailed the piston so it

cannot move, and you put it on a hotplate,

PdV part will vanish because there is no ΔV.

That's the way of heating it, it is called ΔQ.

Another thing you can do is thermally isolate your piston so

no heat can flow in and out of it,

and then you can either have the volume increase or decrease.

If the gas expanded, ΔV is positive and the

PΔ - PΔV is negative, and the ΔU would be

negative; the gas will lose energy.

That's because the molecules are beating up on the piston and

moving the piston. Remember, applying a force

doesn't cost you anything. But if the point of application

moves, you do work. And who's going to pay for it,

the gas? It'll pay for it through its

loss of internal energy. Conversely, if you push down on

the gas, ΔV will be negative and this will become

positive and the energy of the gas will go up.

So, there are two ways to change the energy of these

molecules. In the end, all you want is you

want the molecules to move faster than before.

One is to put them on a hotplate where there are

fast-moving molecules. When they collide with the

slow-moving molecules, typically the slow one's a

little more faster and the fast one's a little more slower and

therefore will be a transfer of kinetic energy.

Or when you push the piston down, you can show when a

molecule collides with a moving piston.

It will actually gain energy. So, that's how you do work.

That's the first law.

So, let us now calculate the work done in a process where a

gas goes from here to here on an isotherm.

Isotherm is a graph of a given temperature.

So, this is a graph P times V equal to

constant, because PV = nRT.

If T is constant, PV is a constant,

it's the rectangular hyperbola. The product of the x and

y coordinates is constant, so when the x

coordinate vanished, the y will go to

infinity. y coordinate vanishes,

x will go to infinity. So, you want to take your gas

for a ride from here to here. Throughout it's at a certain

temperature T. What work is done by you?

That's a very nice interpretation.

The work done by you is the integral of PdV.

But what is integral of PdV?

That's P, and that's dV.

Pdv is that shaded region.

In other words, if you just write PdV it

makes absolutely no sense. If you go to a mathematician

and say, "Please do the integral for me!"

can the mathematician do this? What's coming in the way of the

mathematician actually doing the integral?

What do you have to know to really do an integral?

Student: You have to know the function.

Professor Ramamurti Shankar: You have to know

the function. If you just say P,

we'll say maybe P is a constant, in which case I'll

pull it out of the integral. But for this problem,

because PV is nRT, and T is a

constant, P is nRT divided

by V, and that's the function that you would need to

do the integral, and if you did that you will

find there's nRT. All of them are constants.

They come out of the integral, dv over V,

and integrate from the initial volume, the final volume.

And you guys know this is a logarithm, and the log of upper

minus log of lower is the log of the ratio.

And this gives me nRT ln (V_2

/V_1). So, we have done our first work

calculation. When the gas goes on an

isothermal trajectory from start to finish, from volume

V_1 to volume V_2,

the work done, this is the work done by the

gas. You can all see that gas is

expanding and that's equal to this shaded region.

By the way, I mention it now, I don't want to distract you,

but suppose later on I make it go backwards like this,

part of the way. The work done on the going

backwards part is this area, but with a minus sign.

I hope you will understand, if you go to the right the

area's considered positive. If you go to the left,

the area is considered negative.

If you do the integral and put the right limit,

you'll get the right answer. But geometrically,

the area under the graph in the PV diagram is the work

done if you're moving the direction of increasing volume.

If you would decrease the volume, for example,

if you just went back from here, the area looks the same

but the work done is considered negative.

You don't have to think very hard.

If you do the calculation going backwards, you will get a

ln of V_1 over

V_2. That'll automatically be the

negative of the log of V_2 over

V_1. But geometrically,

the area under the graph is the work, if you are going to the

right. Yes?

Student: What determines the shape of the curve that

links the first state to the second state?

Professor Ramamurti Shankar: Oh,

this one? Student: Mmm-hmm.

Professor Ramamurti Shankar: This'll be a graph,

PV equal to essentially a constant.

So, you take your gas, you see how many moles there

are. You know R,

you know the temperature, you promised not to change the

temperature. So, you'll move on a trajectory

so that the product PV never changes.

And in any xy plane, if you draw a graph where the

product xy doesn't change it'll have this shape called a

"rectangular hyperbola." It just means,

whenever one increases, the other should decrease,

keeping the product constant. That's why P is

proportional to the reciprocal of V when you do the

integral. Very good.

So, this is now the work done by the gas.

What is the heat input? The heat input is a change in

internal energy minus the work done.

Let me see. The law was ΔU = ΔQ -

ΔW. Yeah, let's go back to this law.

In this problem, ΔW is what I just

calculated, nRT, whatever the log,

V_2 over V_1.

What is ΔQ? How much heat has been put into

this gas? How do I find that?

Student: Take out the T?

Professor Ramamurti Shankar: Pardon me?

Student: You take out the T?

Professor Ramamurti Shankar: For the heat input

you mean? Yeah, you can use mc

ΔT, but you don't have to do anymore work.

By that, I mean you don't have to do any more cerebration.

What can you do with this equation to avoid doing further

calculations? Do you know anything else?

Yes? Student: [inaudible]

Professor Ramamurti Shankar: Yes.

This is what I told you is the fact that people do not

constantly remember, but you must.

This gas did not change its temperature.

Go back to equation number whatever I wrote down.

U = 3/2 nRT or something.

T doesn't change, U doesn't change.

That means the initial internal energy and final internal energy

are the same because initial temperature and final

temperature are the same. So, this guy has to be zero.

That means ΔQ is the same as ΔW in this

particular case. Yes?

Student: [inaudible] Professor Ramamurti

Shankar: When you say mc ΔT, you've got to be

careful of what formula you want to use.

I'll tell you why you cannot simply use mc ΔT.

If you've got a solid and you use mc ΔT,

that is correct, because when you heat the

solid, the heat you put in goes into heating up the solid.

Maybe let's ask the following question.

His question is the following. You're telling me you put heat

into a gas, right? And you say temperature doesn't

go up. How can that possibly be?

I always thought when I put heat into something,

temperature goes up. That's because you were

thinking about a solid, where if you put in heat it's

got to go somewhere and, of course, temperature goes up.

What do you think is happening to the gas here?

Think of the piston and weight combination.

When I want to go along this path from here to here,

you can ask yourself where is the heat input and where is the

change in energy, and why is there no change in

temperature? If you take a piston like this,

if you want to increase the volume, you can certainly take

off a grain of sand, right?

If you took the grain of sand and the piston will move up,

it will do work and it actually will cool down,

but that's not what you're doing.

You are keeping it on a hotplate at a certain

temperature so that if it tries to cool down,

heat flows from below to above maintaining the temperature.

So, what the gas is doing in this case is taking heat energy

from below and going up and working against the atmosphere

above. It takes in with one hand and

gives out to the other, without changing its energy.

So, when you study specific heat, which is my next topic,

you've got to be a little more careful when you talk about

specific heats of gases, and I will tell you why.

There is no single thing called specific heat for a gas.

There are many, many definitions depending on

the circumstances. But I hope you understand in

this case; you've got to visualize this.

It's not enough to draw diagrams and draw pictures.

What did I do to the cylinder to maintain the temperature and

yet let it expand? Expansion is going to demand

work on part of the gas. That's going to require a loss

of energy unless you pump in energy from below.

So, what I've done is that I take grain after grain,

so that the pressure drops and the volume increases,

but the slight expansion would have cooled it slightly but the

reservoir from below brings it back to the temperature of the

reservoir. So, you prop it up in

temperature. So, we draw the picture by

saying the gas went from here to here, and we usually draw a

picture like this and say heat flowed into the system during

that process. Alright, now I'll come to this

question that was raised about specific heat.

Now, specific heat, you always say is ΔQ

over ΔT or ΔT divided by the mass of the

substance. Now, it turns out that for a

gas, you've already seen that what you want to count is not

the actual mass, but the moles.

Because we have seen at the level of the ideal gas law,

the energy is controlled by not simply the mass,

but by the moles. Because every molecule gets a

certain amount of energy, namely 3/2 kT,

and you just want to count the number of molecules,

or the number of moles. Now, there are many,

many ways in which you can pump in heat into a gas and heat it

up and see how much heat it takes.

But let's agree that we will take one mole from now on and

not one kilogram. Not one kilogram.

We'll find out if you do it that way, the answer doesn't

seem to depend on the gas. That's the first thing.

Take a mole of some gas and call the specific heat as the

energy needed to raise the temperature of one mole by one

degree. So, this should not be m.

This should be the number of moles.

If you take one mole, you can say okay,

one mole of gas I was told has energy U = 3/2 RT.

Because it was there-halves nRT but n is one

mole. Now, you want to put in some

heat, and you really want the ΔQ over ΔT,

so I will remind you that ΔQ is ΔU +

PΔv.

The heat input into a gas is the change of energy plus P

Δv. And that's just from the first

law. So, if I'm going to divide

ΔQ by ΔT, there's a problem here.

Did you allow the volume to change or did you not allow the

volume to change? That's going to decide what the

specific heat is. In other words,

when a solid is heated, it expands such a tiny amount,

we don't worry about the work done by the expanding solid

against the atmosphere. But for a gas,

when you heat it, the volume changes so much that

the work it does against the external world is

non-negligible. Therefore, the specific heat is

dependent on what you allow the volume term to do.

So, there's one definition of specific heat called

C_V, and C_V is the

one at constant volume. You don't let the volume change.

In other words, you take the piston and you

clamp it. Now, you pump in heat from

below by putting it on a hotplate.

All the heat goes directly to internal energy.

None of that is lost in terms of expansion.

So, ΔV is zero. In that case,

ΔQ over ΔT at constant volume,

we denote that in this fashion, at constant volume,

this term is gone, and it just becomes ΔU

over ΔT. That's very easily done.

ΔU over ΔT is 3/2 R.

So, the specific heat of a gas at constant volume is 3 over

2R. When I studied solids,

I never bother about constant volume because a change in the

volume of a solid is so negligible when it's heated up,

it's not worth specifying that it was a constant volume

process. But for a gas,

it's going to matter whether it was constant volume or not.

Then, there's a second specific heat people like to define.

That's done as follows. You take this piston.

You have some gas at some pressure.

You pump in some heat but you don't clamp the piston.

You let the piston expand any way it wants at the same

pressure. For example,

if it's being pushed down by the atmosphere,

you let the piston move up if it wants to, maintaining the

same pressure. Well, if it moves up a little

bit, then the correct equation is the heat that you put in is

the change in internal energy plus P times ΔV,

where now P is some constant pressure,

say the atmospheric pressure, ΔV is the change in

volume. So, ΔQ needed now will

be more, because you're pumping in heat from below and you're

losing energy above because you're letting the gas expand.

Because you were letting the pressure be controlled from the

outside at some fixed value. So now, ΔQ--this

ΔU will be 3 over 2RΔT.

Now, what's the change in P times ΔV?

Here is where you should know your calculus.

The P times change in V is the same as the

change in PV, if P is a constant.

Right? Remember long back when I did

rate of change of momentum is d/dt of mv,

it's m times dv/dt,

because m doesn't change.

You can take it inside the change.

But now we use PV = RT. I'm talking about one mole.

PV = RT. That's a change in the quantity

RT, R is a constant, that's R times

ΔT so I put in here R times ΔT.

This is the ΔQ at constant pressure.

So, the specific heat of constant pressure is ΔQ

over ΔT, keeping the pressure constant.

You divide everything by ΔT you get 3 over

2R, plus another R, which is 5 over

2R.

So, the thing you have to remember, what I did in the end,

is that a gas doesn't have a single specific heat.

If we just say, put in some heat and tell me

how many calories I need to raise the temperature,

that's not enough. You have to tell me whether in

the interim, the gas was fixed in its volume,

or changed its volume, or obeyed some other condition.

The two most popular conditions people consider are either the

volume cannot change or the pressure cannot change.

If the volume cannot change, then the change in the heat you

put in goes directly to internal energy, from the First Law of

Thermodynamics. That gives you a specific heat

of 3 over 2R. If the pressure cannot change,

you get 5 over 2R. You can see

C_P is bigger than C_V

because when you let the piston expand,

then not all the heat is going to heat the gas.

Some of it is dissipated on top by working against the

atmosphere. Then, notice that I've not told

you what gas it is. That's why the specific heat

per mole is the right thing to think about because then the

answer does not depend on what particular gas you took.

Whether it's hydrogen or helium, they all have the same

specific heat per mole. They won't have the same

specific heat per gram, right?

Because one gram of helium and one gram of hydrogen don't have

the same number of moles. So, you have to remember that

we're talking about moles. The final thing I have to

caution you--very, very important.

This is for a monoatomic gas.

This is for a gas whose atom is the gas itself.

It's a point. Its only energy is kinetic

energy. There are diatomic gases,

by two of them [atoms] joined together,

they can form a dumbbell or something;

then the energy of the dumbbell has got two parts,

as you learned long ago. It can rotate around some axis

and it can also move in space. Then the internal energy has

also got two parts. Energy due to motion of the

center of mass and energy due to rotation.

Some molecules also vibrate. So, there are lots of

complicated things, but if you got only one guy,

or one atom, whatever its mass is,

it cannot rotate around itself and it cannot vibrate around

itself, so those energies all disappear.

So, we have taken the simplest one of a monoatomic gas,

a gas whose fundamental entity is a single atom rather than a

complicated molecule. And that's all you're

responsible for. I'll just say one thing.

C_P over C_V,

I want to mention it before you run off to do your homework.

I don't know if it comes up. It's called γ,

that's five-third for a monoatomic gas.

You can just take the ratio of the numbers.

If in some problem you find γ is not five-thirds,

do not panic. It just means it's a gas which

is not monoatomic. If it's not monoatomic,

these numbers don't have exactly those values.

We don't have to go beyond that. You just have to know there's a

ratio γ, which is five-thirds in the

simplest case, but in some problem,

somewhere in your life, you can get a γ which

is not five-thirds.