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.