WEBVTT

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

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

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

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You know, what are the main
ideas I covered?

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

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an intuitive feeling and turned
it into something more

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quantitative,
so you can not only say this is

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hotter than that,
that's hotter than this,

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

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

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temperature.
And the way to find the Kelvin

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

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like hydrogen or helium,
at low concentration,

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and put that inside a piston
and cylinder.

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

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

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P times V,
and the claim is for whatever

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

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Remember now, this is in Kelvin.

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Your centigrade scale is
somewhere over here,

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but I've shifted the origin to
the Kelvin scale.

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So, somewhere here will be the
boiling point of water,

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somewhere the freezing point of
water;

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that may be the boiling point
of water.

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And if you took a different
amount of a different gas,

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you'll get some other line.

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But they will always be
straight lines if the

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concentration is sufficiently
low.

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In other words,
it appears that pressure times

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volume is some constant.

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I don't know what to call it.

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Say c,
times this temperature.

01:43.810 --> 01:48.420


01:48.420 --> 01:51.510
And you can use that to measure
temperature because if you know

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

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the slope and then you can
calibrate the thermometer,

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then for any other value of
P times V that you

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get, you can come down and read
your temperature.

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

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

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I can use one;
you can use another one.

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

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can use a different gas.

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But all gases seem to have the
property that pressure times

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volume is linearly proportional
to this new temperature scale,

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measured with this new origin
at absolute zero.

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There is really nothing to the
left of this T = 0.

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

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theory of heat as a new theory.

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You know, we got mechanics and
all that stuff--levers and

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pulleys and all that.

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Then, you have this mysterious
thing called heat,

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which has been around for many
years but people started

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

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

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

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hot to the cold.
Then we defined specific heat,

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law of conservation of this
caloric fluid that allows you to

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do some problems in calorimetry.

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You mix so much of this with so
much of that,

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where will they end up?

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That kind of problem.

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So, that promoted heat to a new
and independent entity,

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different from all other things
we have studied.

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But something suggests that it
is not completely alien or a new

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concept, because there seems to
be a conservation of law for

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this heat,
because the heat lost by the

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cold water was the heat gained
by the hot water.

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I'm sorry.
Heat gained by the cold water

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was the heat lost by the hot
water.

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So, you have a conservation law.

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Secondly, we know another way
to produce heat.

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

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in which case,
there is something mysterious

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

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

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Take two automobiles;
slam them.

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

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just telling you as a matter of
principle.

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Buy two Ferraris,
slam them into each other and

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

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Ferraris will heat up.

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The question is what happened
to the kinetic energy of the two

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cars?
That is really gone.

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So, in the old days,
we would say,

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well, we don't apply the Law of
Conservation of Energy because

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this was an inelastic
relationship.

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That was our legal way out of
the whole issue.

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But you realize now this
caloric fluid can be produced

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from nowhere,
because there was no caloric

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fluid before,
but slamming the two cars

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produce this extra heat.

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

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mechanical energy and heat
energy--that when mechanical

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energy disappears,
heat energy appears.

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So, how do you do the
conversion ratio?

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

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joule of mechanical energy?

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So, Joule did the experiment.

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

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

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cars.
He had this gadget with him,

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

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top, and you let the weight go
down.

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

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

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Something will be missing.

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Keep track of the missing
amount.

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So many joules--but meanwhile
you find this water has become

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hot.
You find then how many calories

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should have gone in,
because we know the specific of

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water,
we know the rise in

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temperature, we know how many
calories were produced.

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And then, you compare the two
and you find that 4.2 joules = 1

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calorie.

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

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One joule, 4.2 joules of
mechanical energy.

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So, in the example of the
colliding cars,

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

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slam them together.

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If they come to rest,
you've lost all of that,

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and then you take that and you
write it as--divided by 4.2 and

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that's how much calories you
have produced.

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If the car was made of just one
material, it had a specific

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heat, then it would go up by a
certain temperature you can

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actually predict.

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Okay.
So today, I want to go a little

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deeper into the question of
where is the energy actually

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stored in the car,
and what is heat.

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

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

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4.2 is the gain in calories.

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Now, we can answer in detail
exactly what is heat.

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That's what we're going to talk
about today.

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When we say something is
hotter, what do we mean on a

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microscopic level?

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

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made of, they didn't have this
understanding.

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

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

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

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discoveries that,
in the end,

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

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molecules and so on.

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So, how does that come into
play?

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For that, I want you to take
the simple example where the

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temperature enters.

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That is in the relation
PV equal to some constant

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times temperature.

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Do you know what I'm talking
about?

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Take some gas,
make sure it's sufficiently

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dilute, put it into this piston,
measure the weights on top of

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it,
divide it by the area,

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to get the pressure,
that's the pressure,

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that's the volume.

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The volume is the region here,
multiply the product;

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

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you'll find the product
PV increases,

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and as the temperature
increases, PV is

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proportional to T.

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We want to ask what is this
proportionality constant.

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Suppose you were doing this.

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In the old days,
this is what people did.

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

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What is going to control this
particular constant for the

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given experiment?

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Do you know what it might be
proportional to?

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Yes?
Student: Amount of gas?

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Professor Ramamurti
Shankar: Amount of gas.

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That's true,
because if the amount of gas is

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zero, we think there's no
pressure.

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When you say "amount of gas,"
that's a very safe sentence

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because amount measured by what
means?

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By what metric?
Student: Probably number

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of particles?
Professor Ramamurti

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Shankar: Right.

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Suppose you were not aware of
particles.

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

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Student: Mass.

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Professor Ramamurti
Shankar: The mass.

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Now, if you guys ever said
moles, I was going to shoot you

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down.
You're not supposed to know

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those things.
We are trying to deduce that.

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So, put yourself back in
whatever stone ages we were in.

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We don't know anything else.

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Mass would be a reasonable
argument, right?

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What's the argument?

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We know that if you have some
amount of gas producing the

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pressure, and you put twice as
much stuff,

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you would think it will produce
twice as much pressure.

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Same reason why you think the
expansion of a rod is

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proportionally change in
temperature times the starting

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length.
So, this mass is what's doing

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it.
So, it's proportional to mass.

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It's a very reasonable guess.

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

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produce more pressure.

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That's actually correct.

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

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laboratory that you did.

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So, put the mass that you had
there.

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Then, you should put a constant
still.

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I don't know what you want me
to call this constant,

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say, c prime.

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This constant contains
everything, but I pulled out the

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

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This is actually correct.

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You can take a certain gas and
you can find out what c

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prime is.
But here is what people found.

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If you do it that way,
the constant c prime

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depends on the gas you are
considering.

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If you consider hydrogen gas,
let's call that c prime

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for hydrogen.
Somebody else puts in helium

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gas.
Then you find the c

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prime for helium is one-fourth
c prime for hydrogen.

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If you do carbon,
it's another number.

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c prime for carbon is
c prime for hydrogen

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divided by 12.
So, each gas has a different

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constant.
So, we conclude that yes,

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

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by different numbers for
different gases to find the real

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effective mass in terms of
pressure.

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In other words,
one gram of hydrogen and one

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gram of helium do not have the
same pressure.

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

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effect on pressure.

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So, you have to think about why
is it that the mass directly is

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not involved.
Mass has to be divided by a

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number, and the number is a
nice, round number.

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4 for this and 12 for that,
and of course people figure out

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there's a long story I cannot go
into, but I think you all know

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the answer.
But now we are allowed to fast

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forward to the correct answer,
because I really don't have the

11:34.606 --> 11:36.656
time to see how they worked it
out,

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but from these integers and the
way the gases reacted and formed

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complicated molecules,
they figured out what's really

11:43.646 --> 11:47.526
going on is that you're dividing
by a number that's proportional

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with the mass of the underlying
fundamental entity,

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which would be an atom.

11:51.970 --> 11:54.190
In some case a molecule,
but I'm just going to call

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everything as atom.

11:55.120 --> 11:57.280
So, if things,
like, carbon,

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as atoms, weigh 12 times as
much as things called hydrogen,

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then if you took some amount of
carbon, you divide it by a

12:06.195 --> 12:08.875
number, like,
12 to count the number of

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carbon atoms.
Okay, so hydrogen you want to

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count the number of hydrogen
atoms.

12:13.370 --> 12:17.770
So then, what really you want
here is not the mass,

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but the number of atoms of a
given kind.

12:21.470 --> 12:25.380
We are certainly free to write
either a mass or the number of

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atoms, because the two are
proportional.

12:28.120 --> 12:31.810
But the beauty of writing it
this way, you write it in this

12:31.810 --> 12:34.990
fashion, by this new constant
k, k is

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independent of the gas.

12:36.710 --> 12:41.130


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

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depend on the gas.

12:44.529 --> 12:47.129
You can write it in terms of
mass.

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If you did, for each mass
you've got to divide by a

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certain number.
TThen once you divide it by the

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number you can put a single
constant in front.

12:54.500 --> 12:58.470
Or if you want a universal
constant, what you should really

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be counting is the number of
atoms or molecules.

13:01.900 --> 13:04.950


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

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atoms and molecules and people
who are led to atoms and

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molecules by looking at the way
gases interact,

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

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that there are entities which
come in discreet units.

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

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

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So, this is called the
Boltzmann Constant.

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

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let's see, joules/Kelvin.

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That's it.
Or joules/Kelvin or degrees

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centigrade.

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So, this is a universal
constant.

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

14:11.803 --> 14:14.343
number [N], because if you write
the number,

14:14.340 --> 14:18.770
in a typical situation,
what's the number going to be?

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Take some random group gas.

14:20.789 --> 14:22.349
One gram, two grams,
one kilogram,

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it doesn't matter.

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The number you will put in
there is some number like

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10^(23) or 10^(25).

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That's a huge number.

14:30.730 --> 14:34.540
So, whenever a huge number is
involved, what you try to do is

14:34.540 --> 14:38.540
to measure the huge number as a
simple multiple off another huge

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number,
which will be our units for

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measuring large numbers.

14:41.830 --> 14:44.320
For example,
when you want to buy eggs,

14:44.318 --> 14:45.888
you measure in dozens.

14:45.889 --> 14:48.359
When you want to buy paper,
you might want to measure it in

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

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It's a natural unit.

14:52.700 --> 14:55.140
When you want to find
intergalactic distances,

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you may use a light year.

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

14:59.990 --> 15:03.200
to us is some number that you
can count in your hands.

15:03.200 --> 15:07.160
When you count people's height,
you use feet because it's

15:07.161 --> 15:09.921
something between 1 and 8,
let's say.

15:09.919 --> 15:12.779
You don't want to use angstroms
and you don't want to use

15:12.783 --> 15:14.753
millimeters.
Likewise, when you want to

15:14.753 --> 15:17.073
simply count numbers,
it turns out there's a very

15:17.067 --> 15:19.137
natural number called Avogadro's
Number,

15:19.140 --> 15:24.520


15:24.519 --> 15:28.299
and Avogadro's Number is 6
times 10^(23).

15:28.300 --> 15:30.470
There's no unit.
It's simply a number,

15:30.474 --> 15:32.304
and that's called a mole.

15:32.300 --> 15:34.890
So, a mole is like a dozen.

15:34.889 --> 15:38.879
We wanted to buy 6 times
10^(23) eggs,

15:38.877 --> 15:43.077
you will say get me one mole of
eggs.

15:43.080 --> 15:44.430
A mole is just a number.

15:44.430 --> 15:46.030
It's a huge number.

15:46.029 --> 15:49.489
You can ask yourself what's so
great about this number?

15:49.490 --> 15:51.690
Why would someone think of this
particular number?

15:51.690 --> 15:53.340
Why not some other number?

15:53.340 --> 15:58.470
Why not 10^(24)?
Do you know what's special

15:58.469 --> 16:00.309
about this number?

16:00.310 --> 16:04.070
Yes?
Student: [inaudible]

16:04.071 --> 16:07.811
Professor Ramamurti
Shankar: Yes.

16:07.809 --> 16:11.959
If you like,
a mole is such that one mole of

16:11.962 --> 16:14.572
hydrogen weighs one gram.

16:14.570 --> 16:18.200
And hydrogen is the simplest
element with a nucleus of just a

16:18.203 --> 16:21.053
proton and the electron's mass
is negligible.

16:21.049 --> 16:25.179
So, this, if you like,
is the reciprocal of the mass

16:25.177 --> 16:28.407
of hydrogen.
In other words,

16:28.405 --> 16:36.645
one over Avogadro's Number is
the mass of hydrogen in grams,

16:36.648 --> 16:40.978
of a hydrogen atom in grams.

16:40.980 --> 16:43.620
So, you basically say,
I want to count this large

16:43.623 --> 16:46.763
number so let me take one gram,
which is my normal unit if

16:46.762 --> 16:48.362
you're thinking in grams.

16:48.360 --> 16:51.640
Then I ask, "How many hydrogen
atoms does one gram of hydrogen

16:51.638 --> 16:53.038
contain?"
That's the number.

16:53.040 --> 16:56.170
That's the mole.
So, if you decide to measure

16:56.168 --> 16:59.698
the number of atoms you have in
a given problem,

16:59.700 --> 17:02.620
in terms of this number,
you write it as some other

17:02.623 --> 17:06.743
small number called moles,
times the number in a mole,

17:06.743 --> 17:10.333
and you are free to write it
this way.

17:10.329 --> 17:16.909
If you write it this way,
then you write this nRT,

17:16.911 --> 17:21.731
R is the universal gas
constant.

17:21.730 --> 17:24.070
What's n times the
Boltzmann Constant.

17:24.069 --> 17:25.959
N_0 times the
Boltzmann Constant.

17:25.960 --> 17:33.200
That happens to be 8.3 joules
per degree centigrade or per

17:33.197 --> 17:34.337
Kelvin.

17:34.340 --> 17:38.280


17:38.280 --> 17:40.070
Right?
The units for R will be

17:40.074 --> 17:42.444
PV, which is units of
energy divided by T.

17:42.440 --> 17:45.570


17:45.569 --> 17:47.929
In terms of calories,
I'd remember this as a nice,

17:47.928 --> 17:50.928
round number.
Two calories per degree

17:50.930 --> 17:55.050
centigrade.
Degrees centigrade and Kelvin

17:55.053 --> 17:57.813
are the same.
The origins are shifted,

17:57.810 --> 18:00.940
but when you go up by one
degree in centigrade or Kelvin,

18:00.939 --> 18:03.229
you go the same amount in
temperature.

18:03.230 --> 18:06.720
So, this [R]
is what they found out first,

18:06.716 --> 18:10.926
because they didn't know
anything about atoms and so on.

18:10.930 --> 18:15.170
But later on when you go look
under the hood of what the gas

18:15.166 --> 18:19.396
is made of, if you write it in
terms of the number of actual

18:19.402 --> 18:21.222
atoms,
you should use the little

18:21.224 --> 18:23.834
k, or you can write it in
terms of number of moles,

18:23.830 --> 18:25.160
in which case use big R.

18:25.160 --> 18:30.110


18:30.109 --> 18:34.019
And the relation between the
two is simply this.

18:34.019 --> 18:37.089
If you're thinking of a gas and
how many moles of gas do I have?

18:37.089 --> 18:39.569
For example,
one gram of hydrogen would be

18:39.573 --> 18:41.623
one mole.
Then you will use R.

18:41.619 --> 18:43.649
If you've gone right down to
fundamentals and say,

18:43.649 --> 18:44.849
"How many atoms do I have?"

18:44.849 --> 18:46.859
and you put that here,
you will multiply it by this

18:46.857 --> 18:47.617
very tiny number.

18:47.620 --> 18:52.110


18:52.110 --> 18:55.060
Alright.
Now, you start with this law

18:55.061 --> 18:57.791
and you ask the following
question.

18:57.789 --> 19:01.689
On the left-hand side is the
quantity P times

19:01.689 --> 19:04.529
V.
On the right-hand side I have

19:04.530 --> 19:08.050
nRT, but let me write it
now as NkT.

19:08.049 --> 19:11.909
You guys should be able to go
back and forth between writing

19:11.911 --> 19:15.381
in terms of number of moles or
the number of atoms.

19:15.380 --> 19:19.360
You'll like this because all
numbers here will be small,

19:19.359 --> 19:22.099
of the order 1.
R is a number like 8,

19:22.096 --> 19:24.356
in some units,
and n would be 1 or 2

19:24.355 --> 19:26.725
moles.
Here, this N will be a

19:26.730 --> 19:28.490
huge number, like 10^(23).

19:28.490 --> 19:31.600
k will be a tiny number
like 10^(-23).

19:31.600 --> 19:32.590
Think in terms of atoms.

19:32.590 --> 19:33.440
That's what you do.

19:33.440 --> 19:35.320
Big numbers, small constants.

19:35.319 --> 19:39.709
When you think of moles,
moderate numbers and moderate

19:39.710 --> 19:41.450
value of constants.

19:41.450 --> 19:45.960
We want to ask ourselves,
"Is there a microscopic basis

19:45.964 --> 19:47.724
for this equation?"

19:47.720 --> 19:50.980
In other words,
once we believe in atoms,

19:50.980 --> 19:55.790
do we understand why there is a
pressure at all in a gas?

19:55.789 --> 19:57.389
That's what we're going to
think about now.

19:57.390 --> 20:02.010
So, for this purpose,
we will take a cube of gas.

20:02.010 --> 20:06.810


20:06.810 --> 20:07.590
Here it is.

20:07.590 --> 20:13.510


20:13.509 --> 20:15.729
This is a cube of side L
by L by L.

20:15.730 --> 20:20.010


20:20.009 --> 20:22.999
Inside this is gas and it's got
some pressure,

20:22.996 --> 20:26.576
and I want to know what's the
value of the pressure.

20:26.579 --> 20:28.279
You've got to ask yourself,
why is there pressure?

20:28.279 --> 20:30.079
Remember, I told you what
pressure means.

20:30.079 --> 20:33.459
If you take this face of the
cube, for example,

20:33.461 --> 20:36.991
it's got to be nailed down to
the other faces;

20:36.990 --> 20:39.160
otherwise, it'll just come
flying out because the gas is

20:39.159 --> 20:41.649
pushing you out.
The pressure is the force on

20:41.651 --> 20:43.371
this face divided by area.

20:43.369 --> 20:45.849
So, somebody inside is trying
to get out.

20:45.849 --> 20:49.829
Those guys are the molecules or
the atoms, and what they're

20:49.830 --> 20:52.850
doing is constantly bouncing off
the wall,

20:52.849 --> 20:57.559
and every time this one bounces
on a wall, its momentum changes

20:57.564 --> 20:59.774
from that to the other one.

20:59.770 --> 21:02.240
So, who's changing the momentum?

21:02.240 --> 21:04.000
Well, the wall is changing the
momentum.

21:04.000 --> 21:04.900
It's reversing it.

21:04.900 --> 21:06.720
For example,
if you bounce head-on and go

21:06.721 --> 21:08.271
back, your momentum is reversed.

21:08.269 --> 21:11.889
That means you push the wall
with some force and the wall

21:11.887 --> 21:14.597
pushes you back with the
opposite force.

21:14.599 --> 21:17.139
It's the force that you exert
on the wall that I'm interested

21:17.138 --> 21:19.538
in.
I want to find the force on the

21:19.539 --> 21:21.869
wall, say, this particular face.

21:21.869 --> 21:23.609
You can find the pressure on
any face.

21:23.609 --> 21:25.039
It's going to be the same
answer.

21:25.039 --> 21:28.149
I'm going to take the shaded
face to find the pressure on it.

21:28.150 --> 21:32.780
Now, if you want to ask,
what is the force exerted by me

21:32.783 --> 21:37.163
on any body, I know the force
has a rate of change of

21:37.164 --> 21:40.684
momentum,
because that is d/dt of

21:40.675 --> 21:44.935
mv, and m is a
constant, and that's just

21:44.941 --> 21:47.251
dv/dt,
which is ma.

21:47.250 --> 21:50.580
I'm just using old F =
ma, but I'm writing it as a

21:50.583 --> 21:52.373
rate of change and momentum.

21:52.369 --> 21:56.259
Now, I have N molecules
or N atoms,

21:56.263 --> 21:58.943
randomly moving inside the box.

21:58.940 --> 22:02.440
Each in its own direction,
suffering collisions with the

22:02.440 --> 22:05.880
box, bouncing off like a
billiard ball would at the end

22:05.877 --> 22:09.757
of the pool table and going to
another wall and doing it.

22:09.759 --> 22:12.339
Now, that's a very complicated
problem, so we're going to

22:12.340 --> 22:13.400
simplify the problem.

22:13.400 --> 22:17.020
The simplification is going to
be, we are going to assume that

22:17.024 --> 22:20.534
one-third of the molecules are
moving from left to right.

22:20.529 --> 22:23.759
One-third are moving up and
down and one-third are moving in

22:23.764 --> 22:25.304
and out of the blackboard.

22:25.299 --> 22:28.769
If at all you make an
assumption that the molecules

22:28.765 --> 22:32.365
are simply moving in the three
primary directions,

22:32.369 --> 22:34.529
of course you will have to give
equal numbers in these

22:34.530 --> 22:36.630
directions.
Nothing in the gas that favors

22:36.626 --> 22:37.876
horizontal or vertical.

22:37.880 --> 22:40.270
In reality, of course,
you must admit the fact they

22:40.265 --> 22:42.745
move in all directions,
but the simplified derivation

22:42.746 --> 22:44.746
happens to give all the right
physics,

22:44.750 --> 22:46.460
so I'm going to use that.

22:46.460 --> 22:49.470
So, N over three
molecules are going back and

22:49.470 --> 22:51.890
forth between this wall,
and this wall.

22:51.890 --> 22:53.130
I'm showing you a side view.

22:53.130 --> 22:55.250
The wall itself looks like this.

22:55.250 --> 22:57.460
The molecules go back and forth.

22:57.460 --> 23:00.850


23:00.850 --> 23:04.900
Next assumption.
All the molecules have the same

23:04.902 --> 23:07.792
speed, which I'm going to call
v.

23:07.789 --> 23:12.109
That also is a gross and crude
description of the problem,

23:12.105 --> 23:16.265
but I'm going to do that anyway
and see what happens.

23:16.269 --> 23:19.669
So now, you ask yourself the
following question.

23:19.670 --> 23:21.200
Take one particular molecule.

23:21.200 --> 23:25.440
When it hits the wall and it
bounces back,

23:25.436 --> 23:30.806
its momentum changes from
mv to -mv;

23:30.809 --> 23:33.049
therefore, the change in
momentum is 2mv.

23:33.050 --> 23:38.760


23:38.759 --> 23:42.329
How often does that change take
place?

23:42.329 --> 23:45.929
You guys should think about
that first.

23:45.930 --> 23:48.480
How often will that collision
take place?

23:48.480 --> 23:51.780
Once you hit the wall here,
you've got to go to the other

23:51.781 --> 23:53.021
wall and come back.

23:53.019 --> 23:56.579
So, you've got to go a distance
2L, and you're going at a

23:56.577 --> 23:58.777
speed v,
the time it takes you is

23:58.779 --> 24:00.359
2L over v.

24:00.359 --> 24:04.239
So, ΔP over ΔT
is 2L divided by

24:04.243 --> 24:07.163
v.
That gives me mv^(2)

24:07.164 --> 24:10.854
over L.
That is the force due to one

24:10.845 --> 24:14.405
molecule.
That's the average force.

24:14.410 --> 24:17.760
You realize it's not a
continuous force.

24:17.759 --> 24:20.589
The molecule will hit the wall,
there's a little force exchange

24:20.588 --> 24:22.618
between the two,
then there's nothing,

24:22.617 --> 24:26.007
then you wait until it comes
back and hits the wall again.

24:26.009 --> 24:29.389
If that were the only thing
going on, what you would find is

24:29.387 --> 24:32.207
the wall most of the time,
has no pressure and suddenly it

24:32.214 --> 24:34.154
has a lot of pressure and then
suddenly nothing.

24:34.150 --> 24:37.090
But fortunately,
this is not the only molecule.

24:37.089 --> 24:40.849
There are roughly 10^(23) guys
pounding themselves against the

24:40.848 --> 24:42.598
wall.
So, at any given instant,

24:42.603 --> 24:45.623
even if it's 10^(-5) seconds,
there'd be a large number of

24:45.622 --> 24:46.842
molecules colliding.

24:46.839 --> 24:50.749
So, that's why the force will
appear to be steady rather than

24:50.747 --> 24:53.057
a sharp noise.
It looked very steady because

24:53.060 --> 24:55.340
somebody or other will be
pushing against the wall.

24:55.340 --> 24:59.370


24:59.369 --> 25:01.539
This is the force due to one
molecule.

25:01.539 --> 25:05.979
The force due to all of them
would be N over 3 times

25:05.982 --> 25:08.282
mv^(2) over L.

25:08.280 --> 25:15.470


25:15.470 --> 25:17.350
N over 3 because of the
N molecules,

25:17.354 --> 25:19.204
a third of them were moving in
this direction.

25:19.200 --> 25:23.010
You realize the other two
directions are parallel to the

25:23.005 --> 25:25.285
wall.
They don't apply force on the

25:25.288 --> 25:26.928
wall.
To apply force on the wall,

25:26.930 --> 25:29.330
you've got to be moving
perpendicular to the wall.

25:29.329 --> 25:31.339
For example,
if the planes that walls are

25:31.343 --> 25:34.013
coming out of the blackboard,
moving in and out of the

25:34.012 --> 25:36.632
blackboard doesn't produce a
force on this wall.

25:36.630 --> 25:39.490
That produces a force on the
other two faces.

25:39.490 --> 25:42.420
So, as far as any one set of
faces is concerned,

25:42.422 --> 25:45.042
in one plane,
only the motion orthogonal to

25:45.043 --> 25:47.043
that is going to contribute.

25:47.039 --> 25:48.629
That's why you have N
over 3.

25:48.630 --> 25:50.020
We're almost done.

25:50.020 --> 25:52.100
That's the average force.

25:52.099 --> 25:54.349
If you want,
I can denote average by some

25:54.354 --> 25:57.274
F bar.
Then what about the average

25:57.272 --> 25:59.672
pressure?
The average pressure is the

25:59.669 --> 26:02.449
average force divided by the
area of that face,

26:02.446 --> 26:04.796
which is F over
L^(2),

26:04.799 --> 26:07.619
that gives me N over 3,
mv^(2) over

26:07.619 --> 26:08.499
L^(3).

26:08.500 --> 26:13.020


26:13.019 --> 26:18.109
Now, this is very nice because
L^(3) is just the volume

26:18.108 --> 26:19.108
of my box.

26:19.110 --> 26:23.770


26:23.769 --> 26:26.179
So, I take the L^(3),
which is equal to the volume of

26:26.182 --> 26:29.532
my box,
and I send it to the other side

26:29.527 --> 26:35.577
and write it as PV equals
N over 3mv^(2).

26:35.580 --> 26:44.300


26:44.299 --> 26:46.809
This is what the microscopic
theory tells you.

26:46.809 --> 26:49.899
Microscopic theory says,
if your molecules all have a

26:49.904 --> 26:52.234
single speed,
they're moving randomly in

26:52.225 --> 26:55.435
space so that a third of them
are moving back and forth

26:55.438 --> 26:59.108
against that wall and this wall,
then this is the product

26:59.112 --> 27:00.942
PV.
Experimentally,

27:00.938 --> 27:03.528
you find PV =
NkT.

27:03.529 --> 27:10.139
So, you compare the two
expressions and out comes one of

27:10.137 --> 27:16.627
the most beautiful results,
which is that mv^(2)

27:16.625 --> 27:20.465
over 2 is 3 over 2kT.

27:20.470 --> 27:23.100
Now that guy deserves a box.

27:23.100 --> 27:25.130
Look what it's telling you.

27:25.130 --> 27:26.990
It's a really profound formula.

27:26.990 --> 27:33.010
It tells you for the first time
a real microscopic meaning of

27:33.014 --> 27:36.244
temperature.
What you and I call the

27:36.239 --> 27:40.189
temperature for gas is simply,
up to these factors,

27:40.194 --> 27:43.444
3/2 k,
simply the kinetic energy of

27:43.437 --> 27:46.537
the molecules.
That's what temperature is.

27:46.539 --> 27:49.019
If you've got a gas and you put
your hand into the furnace and

27:49.020 --> 27:50.500
it feels hot,
the temperature you're

27:50.496 --> 27:52.816
measuring is directly the
kinetic energy of the molecules.

27:52.820 --> 27:57.640


27:57.640 --> 28:01.820
That is a great insight into
what temperature means.

28:01.819 --> 28:06.149
Remember, this is not true if
T is measured in

28:06.147 --> 28:08.607
centigrade.
If T were measured in

28:08.609 --> 28:11.219
centigrade, our freezing point
of water mv^(2),

28:11.221 --> 28:14.511
would vanish.
But that's not what's implied.

28:14.509 --> 28:16.469
T should be measured
from absolute zero.

28:16.470 --> 28:19.970
It also tells you why absolute
zero is absolute.

28:19.970 --> 28:22.600
As you cool your gas,
the kinetic energy of molecules

28:22.597 --> 28:24.817
are decreasing and decreasing
and decreasing,

28:24.821 --> 28:26.891
but you cannot go below not
moving at all,

28:26.893 --> 28:28.943
right?
That's the lowest possible

28:28.936 --> 28:31.546
kinetic energy.
That's why it's absolute zero.

28:31.549 --> 28:33.669
At that point,
everybody stops moving.

28:33.670 --> 28:35.230
That's why you have no pressure.

28:35.230 --> 28:38.970
Now, these results are modified
by the laws of quantum

28:38.969 --> 28:42.849
mechanics, but we don't have to
worry about that now.

28:42.849 --> 28:46.099
In the classical physics,
it's actually correct to say

28:46.095 --> 28:49.765
that when the temperature goes
to zero, all motion ceases.

28:49.769 --> 28:54.859
Now, this is the picture I want
you to bear in mind when you say

28:54.859 --> 28:57.579
temperature.
Absolute temperature is a

28:57.582 --> 28:59.622
measure of molecular agitation.

28:59.619 --> 29:02.399
More precisely,
up to the constant k,

29:02.397 --> 29:04.717
3/2 k,
the kinetic energy of a

29:04.722 --> 29:07.372
molecule is the absolute
temperature.

29:07.370 --> 29:08.490
That's for a gas.

29:08.490 --> 29:11.530
If you took a solid and you
say, what happens when I heat

29:11.528 --> 29:14.458
the solid?
You have a question?

29:14.460 --> 29:17.470
Yes?
Student: [inaudible]

29:17.469 --> 29:21.579
Professor Ramamurti
Shankar: You divide by 2

29:21.578 --> 29:25.768
because ½ mv^(2) is a
familiar quantity,

29:25.770 --> 29:27.320
namely, kinetic energy.

29:27.320 --> 29:29.140
That's why you divide by 2.

29:29.140 --> 29:33.540
Another thing to notice is that
every gas, whatever it's made

29:33.541 --> 29:38.241
of, at a given temperature has a
given kinetic energy because the

29:38.236 --> 29:42.706
kinetic energy per molecule on
the left-hand side is dependent

29:42.711 --> 29:46.161
on absolute temperature and
nothing else.

29:46.160 --> 29:48.440
So at certain degrees,
like 300 Kelvin,

29:48.442 --> 29:51.812
hydrogen kinetic energy would
be the same, carbon kinetic

29:51.807 --> 29:53.847
energy would also be the same.

29:53.849 --> 29:56.669
The kinetic energy will be the
same, not the velocity.

29:56.670 --> 30:00.610
So, the carbon atom is heavier,
it will be moving slower at

30:00.607 --> 30:04.677
that temperature in order to
have the same kinetic energy.

30:04.680 --> 30:07.430
So, all molecules,
all gases, have a given

30:07.430 --> 30:09.570
temperature.
All atoms, let me say,

30:09.567 --> 30:12.607
at a given temperature in
gaseous form will have the same

30:12.614 --> 30:14.414
kinetic energy [per molecule].

30:14.410 --> 30:18.790
Now, if you have a
solid--What's the difference

30:18.786 --> 30:21.446
between a gas and a solid?

30:21.450 --> 30:25.130
In a gas, the atoms are moving
anywhere they want in the box.

30:25.130 --> 30:28.880
In a solid, every atom has a
place.

30:28.880 --> 30:32.030
If you take a two-dimensional
solid, the atoms look like this.

30:32.030 --> 30:36.630


30:36.630 --> 30:39.410
They form a lattice or an array.

30:39.410 --> 30:42.150
That's because you will find
out that, this is more advanced

30:42.151 --> 30:44.941
stuff, that every atom finds
itself in a potential that looks

30:44.939 --> 30:45.589
like this.

30:45.590 --> 30:49.350


30:49.349 --> 30:52.869
Imagine on the ground you make
these hollows.

30:52.870 --> 30:55.530
Low points -- low potential;
high points -- high potential.

30:55.529 --> 30:59.439
Obviously, if you put a bunch
of objects here they will sit at

30:59.435 --> 31:03.085
the bottom of these little
concave holes you've dug in the

31:03.085 --> 31:05.875
ground.
At zero degrees absolute all

31:05.883 --> 31:10.223
atoms will sit at the bottom of
their allotted positions;

31:10.220 --> 31:12.280
that'll be a solid at zero
temperature.

31:12.279 --> 31:16.619
So in a solid,
everybody has a location.

31:16.619 --> 31:18.619
I've shown you a
one-dimensional solid,

31:18.619 --> 31:21.409
but you can imagine a
three-dimensional solid where in

31:21.409 --> 31:23.619
a lattice of three-dimensional
points,

31:23.619 --> 31:27.429
there's an assigned place for
each atom and it sits there.

31:27.430 --> 31:31.610
If you heat up that solid now,
what happens is these guys

31:31.614 --> 31:34.344
start vibrating.
Now, here is where your

31:34.342 --> 31:37.832
knowledge of simple harmonic
motions will come into play.

31:37.829 --> 31:41.129
When you take a system in
equilibrium, it will execute

31:41.126 --> 31:44.356
simple harmonic motion if you
give it a real kick.

31:44.359 --> 31:47.439
If you put it on top of a
hotplate, the atoms in the hot

31:47.437 --> 31:50.567
plate will bump into these guys
and start them moving.

31:50.570 --> 31:51.750
They will start vibrating.

31:51.750 --> 31:56.550
So, a hot solid is one in which
the atoms are making more and

31:56.545 --> 32:01.335
more violent oscillations around
their assigned positions.

32:01.339 --> 32:05.089
If you heat them more and more
and more, eventually you start

32:05.087 --> 32:07.167
doing this.
You go all the way from here to

32:07.173 --> 32:08.853
here;
there is nothing to prevent it

32:08.851 --> 32:10.571
from rolling over to the next
side.

32:10.569 --> 32:14.439
Once you jump the fence,
you know, think of a bunch of

32:14.435 --> 32:17.365
houses, okay?
Or a hole in the ground.

32:17.369 --> 32:19.799
You're living in a hole in the
ground, as you get agitated

32:19.801 --> 32:22.321
you're able to do more and more
oscillations so you can roll

32:22.318 --> 32:23.468
over to the next house.

32:23.470 --> 32:26.490
Once that happens all hell
breaks loose because you don't

32:26.485 --> 32:28.635
have any reason to stay where
you are.

32:28.640 --> 32:30.560
You start going everywhere.

32:30.560 --> 32:32.030
What do you think that is?

32:32.030 --> 32:33.020
Student: Melting.

32:33.019 --> 32:34.719
Professor Ramamurti
Shankar: Pardon me?

32:34.720 --> 32:35.340
Student: Melting.

32:35.339 --> 32:36.259
Professor Ramamurti
Shankar: That's melting.

32:36.259 --> 32:38.119
That's the definition of
melting.

32:38.119 --> 32:40.869
Melting is when you can leap
over this potential barrier,

32:40.868 --> 32:43.418
potential energy barrier,
and go to the next site.

32:43.420 --> 32:45.410
The next side is just like this
side.

32:45.410 --> 32:47.160
If you can jump that fence,
you can jump this one.

32:47.160 --> 32:49.230
You go everywhere and you melt.

32:49.230 --> 32:52.010
That's the process of melting,
and once you have a liquid,

32:52.008 --> 32:53.908
atoms don't have a definite
location.

32:53.910 --> 32:57.060
Now, between a liquid and a
solid, there is this clear

32:57.060 --> 33:00.270
difference, but a liquid and a
vapor is more subtle.

33:00.269 --> 33:01.379
So, I don't want to go into
that.

33:01.380 --> 33:04.910
If you look at a liquid
locally, it will look very much

33:04.912 --> 33:08.712
like a solid in the sense that
inter-atomic spacing is very

33:08.706 --> 33:10.926
tightly constrained in liquid.

33:10.930 --> 33:13.320
Whereas in a solid,
if I know I am here,

33:13.316 --> 33:16.376
I know if I go 100 times the
basic lattice spacing,

33:16.376 --> 33:19.126
there'll be another person
sitting there.

33:19.130 --> 33:20.950
That's called long-range order.

33:20.950 --> 33:22.710
In a liquid, I cannot say that.

33:22.710 --> 33:24.330
In a liquid,
I can say I am here.

33:24.329 --> 33:27.659
Locally, the environment around
me is known, but if you go a few

33:27.663 --> 33:30.273
hundred miles,
I cannot tell you a precise

33:30.266 --> 33:33.516
location if some other atom will
be there or not.

33:33.519 --> 33:36.409
So, we say liquid is
short-range positional order,

33:36.409 --> 33:39.829
but not long-range order,
and a gas has no order at all.

33:39.829 --> 33:41.669
If I tell you there's a gas
molecule here,

33:41.668 --> 33:44.398
I cannot tell you where anybody
else is because nobody has any

33:44.403 --> 33:45.393
assigned location.

33:45.390 --> 33:49.300
Okay.
So, this is the picture you

33:49.297 --> 33:50.857
should have of temperature.

33:50.860 --> 33:52.970
Temperature is agitated motion.

33:52.970 --> 33:56.010
Either motion in the vicinity
of where you are told to sit.

33:56.009 --> 33:59.889
If you're in a solid a motion
all over the box with more and

33:59.892 --> 34:01.342
more kinetic energy.

34:01.339 --> 34:04.399
The next thing in this
caricature is that it is

34:04.395 --> 34:08.575
certainly not true that a third
of the molecules are moving back

34:08.580 --> 34:10.810
and forth.
We know that's a joke, right?

34:10.809 --> 34:13.399
Now, in this room there's no
reason on earth a third of the

34:13.395 --> 34:15.575
molecules are doing this than
others are doing.

34:15.580 --> 34:16.400
That's not approximation.

34:16.400 --> 34:17.810
They're moving in random
directions.

34:17.809 --> 34:19.689
So, if you really got the
stomach for it,

34:19.689 --> 34:22.269
you should do a pressure
calculation in which you assume

34:22.272 --> 34:24.622
the molecules of random
velocities sprinkled in all

34:24.621 --> 34:26.941
directions,
and after all the hard work,

34:26.941 --> 34:29.061
turns out you get exactly this
answer.

34:29.059 --> 34:30.979
So, that's one thing I didn't
want to do.

34:30.980 --> 34:34.950
But something I should point
out to you is the following.

34:34.949 --> 34:37.969
So, suppose I give you a gas at
300 Kelvin.

34:37.969 --> 34:40.959
You go and you take this
formula literally and you

34:40.961 --> 34:43.831
calculate from it a certain ½
mv^(2).

34:43.829 --> 34:46.019
If you knew the mass of the
atom, say, it's hydrogen,

34:46.016 --> 34:47.316
we know the mass of hydrogen.

34:47.320 --> 34:50.040
Then, you find the velocity and
you say okay,

34:50.044 --> 34:53.394
this man tells me that anytime
I catch a hydrogen atom,

34:53.387 --> 34:56.047
it'll have this velocity at 300
Kelvin.

34:56.050 --> 34:58.570
It may have random direction,
but he tells me that's the

34:58.573 --> 35:00.683
velocity square.
Take the square root of that,

35:00.675 --> 35:01.595
that's the velocity.

35:01.599 --> 35:04.889
It seems to us saying the
unique velocity to each

35:04.890 --> 35:07.110
temperature.
Well, that's not correct.

35:07.110 --> 35:10.920
Not only are the molecules
moving in random directions,

35:10.917 --> 35:14.297
they're also moving with
essentially all possible

35:14.302 --> 35:16.982
velocities.
In fact, there are many,

35:16.980 --> 35:20.760
many possible velocities and
this velocity I'm getting,

35:20.760 --> 35:23.610
in this formula,
is some kind of average

35:23.614 --> 35:26.254
velocity, or the most popular
one,

35:26.250 --> 35:27.800
or the most common one.

35:27.800 --> 35:32.310
So, if you really go to a gas
and you have the ability to look

35:32.312 --> 35:36.452
into it and see for each
velocity, what's the probability

35:36.454 --> 35:38.604
that I get that velocity?

35:38.599 --> 35:42.269
The picture I've given you is
the probability of zero except

35:42.272 --> 35:46.072
at this one magical velocity
controlled by the temperature.

35:46.070 --> 35:47.620
But the real graph looks like
this.

35:47.620 --> 35:50.890


35:50.890 --> 35:52.180
It has a certain peak.

35:52.179 --> 35:54.019
It likes to have a certain
value.

35:54.019 --> 35:57.129
If you know enough about
statistics, you know there's a

35:57.134 --> 35:59.274
most probable value,
there's a median,

35:59.268 --> 36:00.708
there's a mean value.

36:00.710 --> 36:02.170
There are different definitions.

36:02.170 --> 36:05.440
They will all vary by factors
of order 1, but the average

36:05.438 --> 36:07.888
kinetic energy will obey this
condition.

36:07.890 --> 36:09.920
Yes?
Student: [inaudible]

36:09.920 --> 36:12.370
Professor Ramamurti
Shankar: It's not really a

36:12.367 --> 36:15.047
Gaussian because if you draw the
nature of this curve,

36:15.050 --> 36:21.880
it looks like v^(2)e to
the -mv^(2) over

36:21.875 --> 36:24.895
2kT.
That's the graph I'm trying to

36:24.896 --> 36:26.866
draw here.
So, it looks like a Gaussian in

36:26.868 --> 36:29.108
the vicinity of this,
but it's kind of skewed.

36:29.110 --> 36:30.820
It's forced to vanish at the
origin.

36:30.820 --> 36:34.150


36:34.150 --> 36:36.680
And it's not peaked at v
= 0.

36:36.679 --> 36:39.969
A real Gaussian peak at this
point would be symmetric.

36:39.970 --> 36:42.320
It's not symmetric;
it vanishes here and it

36:42.315 --> 36:43.695
vanishes infinity.

36:43.699 --> 36:48.139
So, this is called a
Maxwell-Boltzmann distribution.

36:48.139 --> 36:51.639
You don't have to remember any
names but that is the detailed

36:51.643 --> 36:53.983
property of what's happening in
a gas.

36:53.980 --> 36:58.240
So, a temperature does not pick
a unique velocity,

36:58.239 --> 37:00.499
but it picks this graph.

37:00.500 --> 37:03.460
If you vary your temperature,
look at what you have to do.

37:03.460 --> 37:06.150
If you change the number
T here,

37:06.150 --> 37:08.840
if you double the value of
T,

37:08.840 --> 37:11.640
that means if you double the
value of v^(2) here and

37:11.637 --> 37:13.467
there, the graph will look the
same.

37:13.469 --> 37:15.929
So, at every temperature there
is a certain shape.

37:15.929 --> 37:19.429
If you go to your temperature,
it will look more or less the

37:19.433 --> 37:23.123
same, but it may be peaked at a
different velocity if you go to

37:23.115 --> 37:24.595
a higher temperature.

37:24.599 --> 37:27.999
Now, this is another thing I
want to tell you.

37:28.000 --> 37:33.710
If you took a box containing
not atoms but just radiation,

37:33.711 --> 37:37.821
in other words,
go inside a pizza oven.

37:37.820 --> 37:41.310
Take out all the air,
but the oven is still hot,

37:41.306 --> 37:45.306
and the walls of the oven are
radiating electromagnetic

37:45.311 --> 37:48.521
radiation.
Electromagnetic radiation comes

37:48.515 --> 37:52.425
in different frequencies,
and you can ask how much energy

37:52.427 --> 37:55.917
is contained in every possible
frequency range.

37:55.920 --> 37:58.750
You know, each frequency is a
color so you know that.

37:58.750 --> 38:01.140
So, how much energy is in the
red and how much is in the blue?

38:01.140 --> 38:03.110
That graph also looks like this.

38:03.110 --> 38:06.750
That's a more complicated law
called the Planck distribution.

38:06.750 --> 38:10.500
That law also has a shape
completely determined by

38:10.498 --> 38:12.958
temperature.
Whereas for atoms,

38:12.960 --> 38:17.360
the shape is determined by
temperature as well as the mass

38:17.364 --> 38:18.914
of the molecules.

38:18.909 --> 38:21.109
In the case of radiation,
it's determined fully by

38:21.106 --> 38:22.896
temperature and the velocity of
light.

38:22.900 --> 38:25.840
You give me a temperature,
and I will draw you another one

38:25.840 --> 38:27.800
of these roughly bell-shaped
curves.

38:27.800 --> 38:30.690
As you heat up the furnace,
the shape will change.

38:30.690 --> 38:34.190


38:34.190 --> 38:37.780
So again, a temperature for
radiation means a particular

38:37.781 --> 38:40.721
distribution of energies at each
frequency.

38:40.719 --> 38:43.179
For a gas it means a
distribution of velocities.

38:43.180 --> 38:48.890


38:48.889 --> 38:51.319
Has anybody seen that in the
news lately, you know,

38:51.320 --> 38:52.390
or heard about this?

38:52.389 --> 38:55.239
Student: [inaudible]
Professor Ramamurti

38:55.237 --> 38:56.687
Shankar: Pardon me?

38:56.690 --> 38:58.590
Student: [inaudible]
Professor Ramamurti

38:58.585 --> 39:00.785
Shankar: About this
particular graph for radiation.

39:00.789 --> 39:03.999
The probability at each
frequency of finding radiation

39:04.002 --> 39:07.702
of the frequency in a furnace of
some temperature T.

39:07.700 --> 39:11.890
Yes?
Student: [inaudible]

39:11.887 --> 39:14.867
Professor Ramamurti
Shankar: No,

39:14.866 --> 39:16.666
but in current news.

39:16.670 --> 39:19.970
In the last few years,
what people did was the

39:19.969 --> 39:23.149
following.
It's one of the predictions of

39:23.150 --> 39:27.670
the Big Bang theory that the
universe was formed some 14 and

39:27.673 --> 39:31.353
a half billion years ago,
and in the earliest stages the

39:31.349 --> 39:34.509
temperature of the universe was
some incredibly high degrees,

39:34.510 --> 39:37.150
then as it expanded the
universe cooled,

39:37.154 --> 39:41.164
and today, at the current size,
it has got a certain average

39:41.155 --> 39:43.965
temperature,
which is a remnant of the Big

39:43.971 --> 39:47.031
Bang.
And that temperature means that

39:47.027 --> 39:50.767
we are sitting in furnace of the
Big Bang.

39:50.769 --> 39:54.049
But the furnace has cooled a
lot over the billions of years.

39:54.050 --> 39:59.040
The temperature of the universe
is around 3 degrees Kelvin.

39:59.039 --> 40:03.149
And the way you determine that
is you point your telescope in

40:03.154 --> 40:05.294
the sky.
Of course, you're going to get

40:05.293 --> 40:07.343
light from this star;
you're going to get light from

40:07.340 --> 40:09.630
that star.
Ignore all the pointy things

40:09.626 --> 40:13.336
and look at the smooth
background, and it should be the

40:13.341 --> 40:15.131
same in all directions.

40:15.130 --> 40:18.980
And plot that radiation,
and now they use satellites to

40:18.980 --> 40:22.830
plot that, and you'll get a
perfect fit to this kind of

40:22.831 --> 40:26.061
furnace radiation,
called Black Body Radiation.

40:26.059 --> 40:28.959
And you read the temperature by
taking that graph and fitting it

40:28.963 --> 40:31.363
to a graph like this,
but there'll be temperature.

40:31.360 --> 40:33.650
In the case of light,
this won't be velocity squared,

40:33.645 --> 40:35.355
but it will be the frequency
squared,

40:35.360 --> 40:38.330
but read off the temperature
that'll make this work and

40:38.332 --> 40:40.592
that's what gives you 3.1 or
something.

40:40.590 --> 40:42.540
Near 3 degrees Kelvin.

40:42.539 --> 40:46.359
In fact, the data point for
that now if you got that in your

40:46.359 --> 40:50.179
lab then you will be definitely
busted for fudging your data

40:50.179 --> 40:53.739
because it's a perfect fit to
Black Body Radiation.

40:53.739 --> 40:56.989
One of the most perfect fits to
Black Body Radiation is the

40:56.986 --> 40:59.166
background radiation of the Big
Bang.

40:59.170 --> 41:01.950
And it's isotropic,
meaning it's the same in all

41:01.946 --> 41:05.546
directions, and this is one of
the predictions of the Big Bang

41:05.548 --> 41:09.268
is that that'll be the remnant
of the Black Body Radiation.

41:09.269 --> 41:11.979
Again, it tells you there's a
sense in which,

41:11.981 --> 41:15.741
if you go to intergalactic
space, that is your temperature.

41:15.739 --> 41:17.729
That's the temperature you get
for free.

41:17.730 --> 41:20.320
We're all living in that heat
bath at 3 degrees.

41:20.320 --> 41:22.850
You want more heat,
you've got to light up your

41:22.853 --> 41:25.393
furnace but this is everywhere
in the universe,

41:25.387 --> 41:27.477
that heat left over from
creation.

41:27.480 --> 41:30.340
Okay.
That's a very,

41:30.342 --> 41:32.262
very interesting subject.

41:32.260 --> 41:36.850
You know, a lot of new physics
is coming out by looking at just

41:36.847 --> 41:41.427
the Black Body Radiation because
the radiation that's coming to

41:41.434 --> 41:44.324
your eye left those stars long
ago.

41:44.320 --> 41:48.470
So, what you see today is not
what's happening today.

41:48.469 --> 41:52.079
It's what happened long ago
when the radiation left that

41:52.080 --> 41:53.590
part of the universe.

41:53.590 --> 41:56.950
Therefore, we can actually tell
something about the universe not

41:56.952 --> 41:58.822
only now, but at earlier
periods.

41:58.820 --> 42:02.650
And that's the way in which we
can actually tell whether the

42:02.648 --> 42:06.478
universe is expanding or not
expanding or is it accelerating

42:06.477 --> 42:09.367
in its expansion,
or you can even say once it was

42:09.372 --> 42:11.352
decelerating and now it's
accelerating.

42:11.349 --> 42:14.899
All that information comes by
being able to look at the

42:14.898 --> 42:16.868
radiation from the Big Bang.

42:16.869 --> 42:19.569
But for you guys,
I think the most interesting

42:19.570 --> 42:22.690
thing is that when you are in
thermal equilibrium,

42:22.690 --> 42:26.150
and you are living in a certain
temperature, then the radiation

42:26.146 --> 42:29.376
in your world and the molecules
and atoms in your world,

42:29.380 --> 42:32.990
will have a distribution of
frequencies and velocities given

42:32.989 --> 42:34.579
by that universal graph.

42:34.579 --> 42:37.729
Now in our class,
we will simplify life and

42:37.725 --> 42:42.135
replace this graph with a huge
peak at a certain velocity by

42:42.144 --> 42:45.444
pretending everybody's at that
velocity.

42:45.440 --> 42:50.120
We will treat the whole gas as
if it was represented by single

42:50.116 --> 42:52.936
average number.
So, when someone says find the

42:52.941 --> 42:55.331
velocity of molecules,
they're talking about the

42:55.325 --> 42:56.335
average velocity.

42:56.340 --> 42:59.600
You know statistically that
it's the distribution of answers

42:59.595 --> 43:00.915
and an average answer.

43:00.920 --> 43:04.470
Because the average is what you
and I have to know.

43:04.469 --> 43:07.749
Namely, ½ mv^(2) is 3/2
kT, on average.

43:07.750 --> 43:13.500
Okay.
Now, I'm going to study in

43:13.496 --> 43:15.666
detail thermodynamics.

43:15.670 --> 43:19.550
So, the system I'm going to
study is the only one we all

43:19.554 --> 43:23.444
study, which is an ideal gas
sitting inside a piston.

43:23.440 --> 43:27.220


43:27.219 --> 43:30.519
It's got a temperature,
it's got a pressure,

43:30.518 --> 43:32.358
and it's got a volume.

43:32.360 --> 43:36.590
And I'm going to plot here
pressure and volume and I'm

43:36.588 --> 43:39.698
going to put a dot and that's my
gas.

43:39.699 --> 43:44.379
The state of my gas is
summarized by where you put the

43:44.380 --> 43:47.840
dot.
Every dot here is a possible

43:47.836 --> 43:51.386
state of equilibrium for the
gas.

43:51.389 --> 43:54.889
Remember, the gas,
if you look at it under the

43:54.894 --> 43:58.014
hood, is made up of 10^(23)
molecules.

43:58.010 --> 44:01.850
The real, real state of the gas
is obtained by saying,

44:01.849 --> 44:05.689
giving me 10^(23) locations and
10^(23) velocities.

44:05.690 --> 44:08.170
According to Newton,
that's the maximum information

44:08.168 --> 44:10.248
you can give me about the gas
right now,

44:10.250 --> 44:12.860
because with that and Newton's
laws I can predict the future.

44:12.860 --> 44:14.410
But when you study
thermodynamics,

44:14.405 --> 44:16.695
you don't really want to look
into the details.

44:16.699 --> 44:19.969
You want to look at gross
macroscopic properties and there

44:19.965 --> 44:21.335
are two that you need.

44:21.340 --> 44:22.820
Pressure and volume.

44:22.820 --> 44:25.450
Now, you might say,
"What about temperature?"

44:25.449 --> 44:30.609
Why don't I have a third axis
for temperature?

44:30.610 --> 44:33.950
Why is there also not a
property?

44:33.950 --> 44:35.960
Yes?
Student: [inaudible]

44:35.956 --> 44:37.736
Professor Ramamurti
Shankar: Yeah.

44:37.740 --> 44:42.970
Because PV = NkT.

44:42.969 --> 44:45.739
I don't have to give you
T, if I know P and

44:45.744 --> 44:47.124
V.
There's not an independent

44:47.120 --> 44:47.760
thing you can pick.

44:47.760 --> 44:49.300
You can pick P and
V independently.

44:49.300 --> 44:50.910
You cannot pick T.

44:50.909 --> 44:53.489
Let me tell you,
by the way, PV = NkT is

44:53.487 --> 44:54.717
not a universal law.

44:54.719 --> 44:58.209
It's the law that you apply to
dilute gases.

44:58.210 --> 45:03.410
But we are going to just study
only dilute ideal gas.

45:03.409 --> 45:06.929
Ideal gas is one in which the
atoms and molecules are so far

45:06.932 --> 45:10.632
apart that they don't feel any
forces between each other unless

45:10.634 --> 45:14.914
they collide.
So, here is my gas.

45:14.910 --> 45:16.060
It's sitting here.

45:16.059 --> 45:20.099
Now, what I do,
I had a few weights on top of

45:20.100 --> 45:21.310
it.
Three weights.

45:21.310 --> 45:23.000
I suddenly pull out one weight.

45:23.000 --> 45:26.750
Throw it out.
What do you think will happen?

45:26.750 --> 45:31.550
Well, I think this gas will now
shoot up, it'll bob up and down

45:31.554 --> 45:34.054
a few times.
Then after a few seconds,

45:34.046 --> 45:37.076
or a fraction of a second,
it'll settle down with a new

45:37.084 --> 45:39.744
location.
By "settle down," I mean after

45:39.739 --> 45:42.839
a while I will not see any
macroscopic motion.

45:42.840 --> 45:47.700
Then the gas has a new pressure
and a new volume.

45:47.699 --> 45:50.899
It's gone from being there to
being there.

45:50.900 --> 45:54.420


45:54.420 --> 45:56.790
What about in between?

45:56.789 --> 46:01.239
What happened in between the
starting and finishing points?

46:01.239 --> 46:04.929
You might say look,
if it was here in the beginning

46:04.926 --> 46:08.756
it was there later,
it must've followed some path.

46:08.760 --> 46:11.160
Not really.
Not in this process,

46:11.163 --> 46:14.533
because if you do it very
abruptly, suddenly throwing out

46:14.534 --> 46:17.944
one-third of the weights,
there's a period when the

46:17.938 --> 46:20.618
piston rushes up,
when the gas is not in

46:20.618 --> 46:23.138
equilibrium.
By that, I mean there is no

46:23.140 --> 46:25.790
single pressure you can
associate with the gas.

46:25.789 --> 46:29.279
The bottom of the gas doesn't
even know the top is flying off.

46:29.280 --> 46:30.520
It's at the old pressure.

46:30.519 --> 46:32.509
At the top of the gas there's a
low pressure.

46:32.510 --> 46:35.440
So, different parts of the gas
at different pressure,

46:35.440 --> 46:37.300
we don't call that equilibrium.

46:37.300 --> 46:39.560
So, the dot,
representing this system,

46:39.563 --> 46:40.913
moves off the graph.

46:40.910 --> 46:43.230
It's off.
It's off the radar,

46:43.228 --> 46:45.528
and only when it has finally
settled down,

46:45.532 --> 46:48.962
the entire gas can make up its
mind on what its pressure wants

46:48.961 --> 46:50.861
to be;
you put it back here.

46:50.860 --> 46:54.510
So, we have a little problem
that we have these equilibrium

46:54.512 --> 46:58.352
states, but when you try to go
from one to another you fly off

46:58.354 --> 47:00.674
the map.
So, you want to find a device

47:00.671 --> 47:03.851
by which you can stay on the
PV diagram as you change

47:03.847 --> 47:06.477
the state of the gas,
and that brings us to the

47:06.475 --> 47:09.035
notion of what you call a
quasi-static process.

47:09.039 --> 47:12.259
A quasi-static process is
trying to have it both ways in

47:12.259 --> 47:15.069
which you want to change the
state of the gas,

47:15.070 --> 47:18.450
and you don't want it to leave
the PV diagram.

47:18.449 --> 47:22.049
You want it to be always at
equilibrium.

47:22.050 --> 47:26.490
So, what you really want to do
is not put in three big fat

47:26.488 --> 47:29.988
blocks like this,
but instead take a gas where

47:29.992 --> 47:33.032
you have many,
many grains of sand.

47:33.030 --> 47:34.440
They can produce the pressure.

47:34.440 --> 47:38.280
Now, remove one grain of sand.

47:38.280 --> 47:40.520
It moves a tiny bit and very
quickly settles down.

47:40.519 --> 47:43.299
It is again true during the
tiny bit of settling down you

47:43.304 --> 47:46.294
didn't know what it was doing,
but you certainly nailed it at

47:46.286 --> 47:47.476
the second location.

47:47.480 --> 47:50.500
You move one grain at a time,
then you get a picture like

47:50.500 --> 47:52.820
this and you can see where this
is going.

47:52.820 --> 47:56.190
You can make the grain smaller
and smaller and smaller and in a

47:56.190 --> 47:59.290
mathematical sense you can then
form a continuous line.

47:59.289 --> 48:01.939
That is to say,
you perform a process that

48:01.941 --> 48:05.371
leaves the system arbitrarily
close to equilibrium,

48:05.369 --> 48:08.379
meaning give it enough time to
readjust to the new pressure,

48:08.384 --> 48:11.934
settle down to the new volume,
take another grain and another

48:11.925 --> 48:14.085
grain.
And in the spirit of calculus,

48:14.090 --> 48:17.420
you can make these changes
vanishing so that you can really

48:17.416 --> 48:18.846
then say you did this.

48:18.850 --> 48:21.710
Yes?
Student: Are all of

48:21.710 --> 48:23.360
these small processes
reversible?

48:23.360 --> 48:24.030
Professor Ramamurti
Shankar: Pardon me?

48:24.030 --> 48:25.170
Student: Are all of
these small processes

48:25.173 --> 48:25.863
reversible?
Professor Ramamurti

48:25.864 --> 48:26.174
Shankar: Yes.

48:26.170 --> 48:28.930
Such a process is also
called--you can call it

48:28.931 --> 48:31.631
quasi-static but one of the
features of that,

48:31.631 --> 48:32.921
it is reversible.

48:32.920 --> 48:35.550
You've got to be a little
careful when you say reversible.

48:35.550 --> 48:38.780
What we mean by "reversible"
is, if I took off a grain of

48:38.776 --> 48:41.366
sand and it came from here to
the next dot,

48:41.369 --> 48:45.539
and I put the grain back,
it'll climb back to where it

48:45.538 --> 48:47.998
was.
So, you can go back and forth

48:48.001 --> 48:50.341
on this.
But now, that's an idealized

48:50.342 --> 48:53.542
process because if you had a
friction, if you had any

48:53.541 --> 48:56.311
friction between the piston and
the walls,

48:56.309 --> 48:59.389
then if you took out a grain
and it went up,

48:59.385 --> 49:03.815
you put the grain back it might
not come back to quite where it

49:03.819 --> 49:06.779
is.
Because some of the frictional

49:06.782 --> 49:09.412
losses you will never get back.

49:09.409 --> 49:11.909
You cannot put Humpty Dumpty
back.

49:11.909 --> 49:14.689
So, most of the time processes
are not reversible,

49:14.693 --> 49:17.593
even if you do them slowly,
if there is friction.

49:17.590 --> 49:20.180
So, assume it's a completely
frictionless system.

49:20.179 --> 49:23.499
Because if there is friction,
there is some heat that goes

49:23.502 --> 49:27.002
out somewhere and some energy is
lost somewhere and we cannot

49:27.000 --> 49:29.470
bring it back.
If we took a frictionless

49:29.473 --> 49:32.323
piston and on top of it moved it
very, very slowly,

49:32.319 --> 49:33.969
you can follow this graph.

49:33.969 --> 49:36.729
That's the kind of
thermodynamic process we're

49:36.729 --> 49:38.549
talking about.
In the old days,

49:38.550 --> 49:41.180
when I studied a single
particular of the xy

49:41.175 --> 49:44.425
plane, I just said the guy goes
from here to here to there.

49:44.429 --> 49:46.969
That's very easy to study and
there's no restriction on how

49:46.965 --> 49:48.315
quickly or how fast it moved.

49:48.320 --> 49:50.840
Particles have trajectories no
matter how quickly they move.

49:50.840 --> 49:54.400
For a thermodynamic system,
you cannot move them too fast,

49:54.395 --> 49:57.885
because they are extended and
you are having a huge gas a

49:57.889 --> 50:01.779
single number called pressure,
so you cannot change one part

50:01.776 --> 50:05.426
of the gas without waiting for
all of them to communicate and

50:05.430 --> 50:09.020
readjust and achieve a global
value for the new pressure and

50:09.024 --> 50:10.734
you can move gradually.

50:10.730 --> 50:14.410
That's why it takes time to
drag along 10^(23) particles as

50:14.414 --> 50:18.674
if they are the single number or
two numbers characterizing them.

50:18.670 --> 50:22.100
So, we'll be studying processes
like this.

50:22.100 --> 50:24.570
Now, this is called a state.

50:24.570 --> 50:26.070
Two is a state and one is a
state.

50:26.070 --> 50:28.740
Every dot here, that is a state.

50:28.739 --> 50:36.269
Now, in every state of the
system, I'm going to define a

50:36.266 --> 50:41.596
new variable,
which is called a quantity

50:41.603 --> 50:47.013
called U,
which stands for the internal

50:47.010 --> 50:48.760
energy of the gas.

50:48.760 --> 50:54.210


50:54.210 --> 50:57.760
Internal energy is simply the
kinetic energy of the gas

50:57.760 --> 50:59.720
molecules.
For solids and liquids,

50:59.723 --> 51:01.563
there's a more complicated
formula.

51:01.559 --> 51:03.769
For the gas,
internal energy is just the

51:03.767 --> 51:06.367
kinetic energy.
And what is that?

51:06.369 --> 51:12.179
It is 3/2 kT per
molecule times N.

51:12.180 --> 51:17.790
I'm sorry, 3/2 Nk, yeah.

51:17.790 --> 51:19.610
3/2 kT times that.

51:19.610 --> 51:25.780
Or we can write it as 3/2
nRT.

51:25.780 --> 51:29.850


51:29.850 --> 51:32.430
But nRT is PV.

51:32.429 --> 51:35.469
You can also write it as 3/2
PV, so internal energy is

51:35.467 --> 51:36.527
just 3/2 PV.

51:36.530 --> 51:39.290
That means at a given point on
the PV diagram,

51:39.293 --> 51:41.263
you have a certain internal
energy.

51:41.260 --> 51:43.810
If you are there,
that's your internal energy.

51:43.809 --> 51:46.959
Take there-halves of PV
and that's the energy and that's

51:46.959 --> 51:50.209
literally the kinetic energy of
all the molecules in your box.

51:50.210 --> 51:55.050
So, now I'm ready to write down
what's called the First Law of

51:55.049 --> 51:59.969
Thermodynamics that talks about
what happens if you make a move

51:59.968 --> 52:04.648
in the PV plane from one
place to another place.

52:04.650 --> 52:06.800
If you go from one place to
another place,

52:06.804 --> 52:09.804
your internal energy will
change from U_1

52:09.798 --> 52:11.268
to U_2.

52:11.270 --> 52:12.470
Let's call it ΔU.

52:12.470 --> 52:16.960


52:16.960 --> 52:20.000
We want to ask what causes the
internal energy of the gas to

52:19.996 --> 52:22.306
change.
So, you guys think about it now.

52:22.309 --> 52:24.669
Now that you know all about
what's happening in the

52:24.669 --> 52:27.169
cylinder, you can ask how I will
change the energy?

52:27.170 --> 52:29.860
Well, if you wanted to change
the energy of a system,

52:29.859 --> 52:31.669
there are two ways you can do
it.

52:31.670 --> 52:34.060
One is you can do work on the
gas.

52:34.059 --> 52:37.669
Another thing is you can put
the gas on a hotplate.

52:37.670 --> 52:39.860
If you put it on hotplate,
we know it's going to get

52:39.862 --> 52:41.242
hotter.
If it gets hotter,

52:41.237 --> 52:42.387
temperature goes up.

52:42.389 --> 52:45.389
If temperature goes up,
the internal energy goes up.

52:45.389 --> 52:49.049
So, there are two ways to
change the energy of a gas.

52:49.050 --> 52:52.450
The first one we call heat
input.

52:52.449 --> 52:56.559
That just means put it on
something hotter and let the

52:56.564 --> 52:58.044
thing heat it up.

52:58.040 --> 52:59.400
Temperature will go up.

52:59.400 --> 53:03.540
Notice that the internal energy
of an ideal gas depends only on

53:03.538 --> 53:06.528
the temperature.
That's something very,

53:06.534 --> 53:09.524
very important.
I mention it every time I teach

53:09.522 --> 53:12.672
the subject and some people
forget and lose a lot of points

53:12.670 --> 53:14.940
needlessly.
So, I'll say it once more with

53:14.937 --> 53:17.117
feeling.
The energy of an ideal gas

53:17.117 --> 53:19.307
depends only on the temperature.

53:19.309 --> 53:20.869
If the temperature is not
changed;

53:20.870 --> 53:22.640
energy has not changed.

53:22.639 --> 53:26.589
So, try to remember that for
what I do later.

53:26.590 --> 53:31.500
So, the change of the gas,
this cylinder full that I put

53:31.496 --> 53:35.596
some weights on top and I've got
gas inside,

53:35.599 --> 53:40.539
it can change either because I
did, I put in some heat,

53:40.539 --> 53:43.009
or the gas did some work.

53:43.010 --> 53:48.090
By that, I mean if the gas
expands by pushing out against

53:48.092 --> 53:52.182
the atmosphere,
then it was doing the work and

53:52.177 --> 53:56.077
ΔW is the work done by
the gas.

53:56.079 --> 53:59.989
That's why it comes to the
minus sign, because it's the

53:59.986 --> 54:01.646
work done by the gas.

54:01.650 --> 54:03.560
If you do work, you lose energy.

54:03.559 --> 54:05.659
So, what's the formula for work
done?

54:05.660 --> 54:08.080
Let's calculate that.

54:08.079 --> 54:11.739
If I've got a piston here,
it's the force times the

54:11.735 --> 54:15.835
distance.
But the force is the pressure

54:15.838 --> 54:19.818
times the area times the
distance.

54:19.820 --> 54:23.070
Now, you guys should know
enough geometry to know the area

54:23.072 --> 54:26.552
of the piston times the distance
it moves is the change in the

54:26.552 --> 54:29.122
volume.
So, we can write it as P

54:29.117 --> 54:31.927
times dV.
That leads to this great law.

54:31.929 --> 54:35.309
Let me write it on a new
blackboard because we're going

54:35.312 --> 54:37.632
to be playing around with that
law.

54:37.630 --> 54:38.510
This is law number one.

54:38.510 --> 54:41.660


54:41.659 --> 54:47.689
The change in the internal
energy of a system is equal to

54:47.687 --> 54:49.837
ΔQ - PΔV.

54:49.840 --> 54:58.320


54:58.320 --> 55:00.360
What does it express?

55:00.360 --> 55:02.360
It expresses the Law of
Conservation of Energy.

55:02.360 --> 55:05.280
It says the energy goes up,
either because you pushed the

55:05.278 --> 55:08.588
piston or the piston pushed you;
then you decide what the

55:08.589 --> 55:11.539
overall sign is,
or you put it on a hotplate.

55:11.539 --> 55:15.239
We are now equating putting it
on a hotplate as also equivalent

55:15.240 --> 55:18.050
to giving it energy,
because we identify heat as

55:18.045 --> 55:21.625
simply energy.
So, if you took the piston and

55:21.629 --> 55:26.569
you nailed the piston so it
cannot move, and you put it on a

55:26.566 --> 55:29.136
hotplate,
PdV part will vanish

55:29.139 --> 55:30.999
because there is no ΔV.

55:31.000 --> 55:34.660
That's the way of heating it,
it is called ΔQ.

55:34.659 --> 55:37.609
Another thing you can do is
thermally isolate your piston so

55:37.610 --> 55:39.410
no heat can flow in and out of
it,

55:39.409 --> 55:42.439
and then you can either have
the volume increase or decrease.

55:42.440 --> 55:45.700
If the gas expanded,
ΔV is positive and the

55:45.702 --> 55:49.422
PΔ - PΔV is negative,
and the ΔU would be

55:49.421 --> 55:51.941
negative;
the gas will lose energy.

55:51.940 --> 55:55.580
That's because the molecules
are beating up on the piston and

55:55.577 --> 55:56.787
moving the piston.

55:56.789 --> 55:58.839
Remember, applying a force
doesn't cost you anything.

55:58.840 --> 56:01.910
But if the point of application
moves, you do work.

56:01.909 --> 56:04.049
And who's going to pay for it,
the gas?

56:04.050 --> 56:07.540
It'll pay for it through its
loss of internal energy.

56:07.539 --> 56:10.789
Conversely, if you push down on
the gas, ΔV will be

56:10.789 --> 56:14.099
negative and this will become
positive and the energy of the

56:14.095 --> 56:16.755
gas will go up.
So, there are two ways to

56:16.763 --> 56:19.203
change the energy of these
molecules.

56:19.199 --> 56:21.899
In the end, all you want is you
want the molecules to move

56:21.903 --> 56:22.903
faster than before.

56:22.900 --> 56:25.790
One is to put them on a
hotplate where there are

56:25.786 --> 56:27.256
fast-moving molecules.

56:27.260 --> 56:29.320
When they collide with the
slow-moving molecules,

56:29.318 --> 56:31.758
typically the slow one's a
little more faster and the fast

56:31.763 --> 56:34.383
one's a little more slower and
therefore will be a transfer of

56:34.379 --> 56:36.919
kinetic energy.
Or when you push the piston

56:36.920 --> 56:40.320
down, you can show when a
molecule collides with a moving

56:40.322 --> 56:42.902
piston.
It will actually gain energy.

56:42.900 --> 56:46.210
So, that's how you do work.

56:46.210 --> 56:47.090
That's the first law.

56:47.090 --> 56:51.590


56:51.590 --> 56:57.630
So, let us now calculate the
work done in a process where a

56:57.628 --> 57:02.208
gas goes from here to here on an
isotherm.

57:02.210 --> 57:05.740


57:05.739 --> 57:08.369
Isotherm is a graph of a given
temperature.

57:08.369 --> 57:13.119
So, this is a graph P
times V equal to

57:13.122 --> 57:17.132
constant, because PV =
nRT.

57:17.130 --> 57:19.380
If T is constant,
PV is a constant,

57:19.381 --> 57:20.931
it's the rectangular hyperbola.

57:20.929 --> 57:23.139
The product of the x and
y coordinates is

57:23.135 --> 57:25.175
constant, so when the x
coordinate vanished,

57:25.178 --> 57:26.688
the y will go to
infinity.

57:26.690 --> 57:29.230
y coordinate vanishes,
x will go to infinity.

57:29.230 --> 57:34.520
So, you want to take your gas
for a ride from here to here.

57:34.519 --> 57:37.029
Throughout it's at a certain
temperature T.

57:37.030 --> 57:41.270
What work is done by you?

57:41.269 --> 57:43.159
That's a very nice
interpretation.

57:43.159 --> 57:49.389
The work done by you is the
integral of PdV.

57:49.389 --> 57:51.189
But what is integral of
PdV?

57:51.190 --> 57:54.610
That's P,
and that's dV.

57:54.610 --> 57:57.420
Pdv is that shaded
region.

57:57.420 --> 58:00.160
In other words,
if you just write PdV it

58:00.161 --> 58:01.831
makes absolutely no sense.

58:01.829 --> 58:04.349
If you go to a mathematician
and say, "Please do the integral

58:04.349 --> 58:08.159
for me!"
can the mathematician do this?

58:08.159 --> 58:13.169
What's coming in the way of the
mathematician actually doing the

58:13.174 --> 58:15.684
integral?
What do you have to know to

58:15.680 --> 58:17.070
really do an integral?

58:17.070 --> 58:19.160
Student: You have to
know the function.

58:19.159 --> 58:19.979
Professor Ramamurti
Shankar: You have to know

58:19.979 --> 58:21.089
the function.
If you just say P,

58:21.087 --> 58:23.057
we'll say maybe P is a
constant, in which case I'll

58:23.064 --> 58:24.144
pull it out of the integral.

58:24.139 --> 58:27.529
But for this problem,
because PV is

58:27.525 --> 58:30.905
nRT, and T is a
constant,

58:30.909 --> 58:34.709
P is nRT divided
by V, and that's the

58:34.713 --> 58:38.053
function that you would need to
do the integral,

58:38.050 --> 58:41.390
and if you did that you will
find there's nRT.

58:41.390 --> 58:42.530
All of them are constants.

58:42.530 --> 58:45.600
They come out of the integral,
dv over V,

58:45.596 --> 58:48.886
and integrate from the initial
volume, the final volume.

58:48.889 --> 58:52.439
And you guys know this is a
logarithm, and the log of upper

58:52.437 --> 58:55.187
minus log of lower is the log of
the ratio.

58:55.190 --> 59:00.750
And this gives me nRT
ln (V_2

59:00.751 --> 59:03.301
/V_1).

59:03.300 --> 59:06.630
So, we have done our first work
calculation.

59:06.630 --> 59:10.240
When the gas goes on an
isothermal trajectory from start

59:10.238 --> 59:13.648
to finish, from volume
V_1 to volume

59:13.649 --> 59:16.249
V_2,
the work done,

59:16.252 --> 59:18.592
this is the work done by the
gas.

59:18.590 --> 59:22.130
You can all see that gas is
expanding and that's equal to

59:22.131 --> 59:23.461
this shaded region.

59:23.460 --> 59:29.020


59:29.019 --> 59:33.279
By the way, I mention it now,
I don't want to distract you,

59:33.279 --> 59:37.169
but suppose later on I make it
go backwards like this,

59:37.171 --> 59:40.341
part of the way.
The work done on the going

59:40.340 --> 59:43.740
backwards part is this area,
but with a minus sign.

59:43.739 --> 59:46.329
I hope you will understand,
if you go to the right the

59:46.325 --> 59:47.735
area's considered positive.

59:47.739 --> 59:49.679
If you go to the left,
the area is considered

59:49.675 --> 59:51.505
negative.
If you do the integral and put

59:51.506 --> 59:53.566
the right limit,
you'll get the right answer.

59:53.570 --> 59:56.550
But geometrically,
the area under the graph in the

59:56.549 --> 59:59.829
PV diagram is the work
done if you're moving the

59:59.832 --> 1:00:01.962
direction of increasing volume.

1:00:01.960 --> 1:00:04.580
If you would decrease the
volume, for example,

1:00:04.576 --> 1:00:07.766
if you just went back from
here, the area looks the same

1:00:07.773 --> 1:00:10.393
but the work done is considered
negative.

1:00:10.389 --> 1:00:13.069
You don't have to think very
hard.

1:00:13.070 --> 1:00:15.340
If you do the calculation going
backwards, you will get a

1:00:15.338 --> 1:00:16.838
ln of
V_1 over

1:00:16.837 --> 1:00:17.847
V_2.

1:00:17.849 --> 1:00:19.439
That'll automatically be the
negative of the log of

1:00:19.440 --> 1:00:21.000
V_2 over
V_1.

1:00:21.000 --> 1:00:23.400
But geometrically,
the area under the graph is the

1:00:23.399 --> 1:00:25.259
work, if you are going to the
right.

1:00:25.260 --> 1:00:28.070
Yes?
Student: What determines

1:00:28.068 --> 1:00:31.648
the shape of the curve that
links the first state to the

1:00:31.648 --> 1:00:33.408
second state?
Professor Ramamurti

1:00:33.412 --> 1:00:34.302
Shankar: Oh,
this one?

1:00:34.300 --> 1:00:35.430
Student: Mmm-hmm.

1:00:35.429 --> 1:00:38.309
Professor Ramamurti
Shankar: This'll be a graph,

1:00:38.311 --> 1:00:40.661
PV equal to essentially
a constant.

1:00:40.659 --> 1:00:42.939
So, you take your gas,
you see how many moles there

1:00:42.936 --> 1:00:43.986
are.
You know R,

1:00:43.989 --> 1:00:46.369
you know the temperature,
you promised not to change the

1:00:46.370 --> 1:00:48.690
temperature.
So, you'll move on a trajectory

1:00:48.692 --> 1:00:51.122
so that the product PV
never changes.

1:00:51.119 --> 1:00:54.529
And in any xy plane,
if you draw a graph where the

1:00:54.529 --> 1:00:58.309
product xy doesn't change
it'll have this shape called a

1:00:58.305 --> 1:01:00.005
"rectangular hyperbola."

1:01:00.010 --> 1:01:02.130
It just means,
whenever one increases,

1:01:02.132 --> 1:01:05.462
the other should decrease,
keeping the product constant.

1:01:05.460 --> 1:01:08.890
That's why P is
proportional to the reciprocal

1:01:08.894 --> 1:01:11.474
of V when you do the
integral.

1:01:11.470 --> 1:01:15.640
Very good.
So, this is now the work done

1:01:15.635 --> 1:01:18.315
by the gas.
What is the heat input?

1:01:18.320 --> 1:01:24.090
The heat input is a change in
internal energy minus the work

1:01:24.094 --> 1:01:25.784
done.
Let me see.

1:01:25.780 --> 1:01:31.260
The law was ΔU = ΔQ -
ΔW.

1:01:31.260 --> 1:01:32.260
Yeah, let's go back to this law.

1:01:32.260 --> 1:01:35.490


1:01:35.489 --> 1:01:38.019
In this problem,
ΔW is what I just

1:01:38.018 --> 1:01:40.488
calculated, nRT,
whatever the log,

1:01:40.485 --> 1:01:43.625
V_2 over
V_1.

1:01:43.630 --> 1:01:46.910
What is ΔQ?

1:01:46.909 --> 1:01:50.719
How much heat has been put into
this gas?

1:01:50.720 --> 1:01:52.410
How do I find that?

1:01:52.409 --> 1:01:54.549
Student: Take out the
T?

1:01:54.550 --> 1:01:55.550
Professor Ramamurti
Shankar: Pardon me?

1:01:55.550 --> 1:01:58.070
Student: You take out
the T?

1:01:58.070 --> 1:02:00.190
Professor Ramamurti
Shankar: For the heat input

1:02:00.190 --> 1:02:01.680
you mean?
Yeah, you can use mc

1:02:01.679 --> 1:02:03.869
ΔT, but you don't have to
do anymore work.

1:02:03.869 --> 1:02:06.879
By that, I mean you don't have
to do any more cerebration.

1:02:06.880 --> 1:02:10.740
What can you do with this
equation to avoid doing further

1:02:10.744 --> 1:02:13.434
calculations?
Do you know anything else?

1:02:13.430 --> 1:02:17.640
Yes?
Student: [inaudible]

1:02:17.638 --> 1:02:19.928
Professor Ramamurti
Shankar: Yes.

1:02:19.929 --> 1:02:23.179
This is what I told you is the
fact that people do not

1:02:23.175 --> 1:02:25.375
constantly remember,
but you must.

1:02:25.380 --> 1:02:27.820
This gas did not change its
temperature.

1:02:27.820 --> 1:02:31.000
Go back to equation number
whatever I wrote down.

1:02:31.000 --> 1:02:33.400
U = 3/2 nRT or
something.

1:02:33.400 --> 1:02:35.250
T doesn't change,
U doesn't change.

1:02:35.250 --> 1:02:38.500
That means the initial internal
energy and final internal energy

1:02:38.502 --> 1:02:41.032
are the same because initial
temperature and final

1:02:41.032 --> 1:02:42.532
temperature are the same.

1:02:42.530 --> 1:02:44.100
So, this guy has to be zero.

1:02:44.099 --> 1:02:50.489
That means ΔQ is the
same as ΔW in this

1:02:50.489 --> 1:02:53.699
particular case.
Yes?

1:02:53.699 --> 1:02:57.119
Student: [inaudible]
Professor Ramamurti

1:02:57.120 --> 1:03:01.030
Shankar: When you say mc
ΔT, you've got to be

1:03:01.028 --> 1:03:04.098
careful of what formula you want
to use.

1:03:04.099 --> 1:03:10.279
I'll tell you why you cannot
simply use mc ΔT.

1:03:10.280 --> 1:03:12.330
If you've got a solid and you
use mc ΔT,

1:03:12.332 --> 1:03:14.122
that is correct,
because when you heat the

1:03:14.122 --> 1:03:16.832
solid, the heat you put in goes
into heating up the solid.

1:03:16.830 --> 1:03:20.060


1:03:20.059 --> 1:03:21.689
Maybe let's ask the following
question.

1:03:21.690 --> 1:03:23.180
His question is the following.

1:03:23.179 --> 1:03:26.569
You're telling me you put heat
into a gas, right?

1:03:26.570 --> 1:03:29.970
And you say temperature doesn't
go up.

1:03:29.970 --> 1:03:31.260
How can that possibly be?

1:03:31.260 --> 1:03:33.720
I always thought when I put
heat into something,

1:03:33.718 --> 1:03:34.868
temperature goes up.

1:03:34.869 --> 1:03:37.559
That's because you were
thinking about a solid,

1:03:37.557 --> 1:03:40.647
where if you put in heat it's
got to go somewhere and,

1:03:40.654 --> 1:03:42.704
of course, temperature goes up.

1:03:42.699 --> 1:03:44.859
What do you think is happening
to the gas here?

1:03:44.860 --> 1:03:48.080
Think of the piston and weight
combination.

1:03:48.079 --> 1:03:54.059
When I want to go along this
path from here to here,

1:03:54.064 --> 1:04:01.114
you can ask yourself where is
the heat input and where is the

1:04:01.105 --> 1:04:04.825
change in energy,
and why is there no change in

1:04:04.826 --> 1:04:07.856
temperature?
If you take a piston like this,

1:04:07.862 --> 1:04:12.412
if you want to increase the
volume, you can certainly take

1:04:12.407 --> 1:04:14.797
off a grain of sand,
right?

1:04:14.800 --> 1:04:18.330
If you took the grain of sand
and the piston will move up,

1:04:18.325 --> 1:04:21.165
it will do work and it actually
will cool down,

1:04:21.171 --> 1:04:23.461
but that's not what you're
doing.

1:04:23.460 --> 1:04:26.470
You are keeping it on a
hotplate at a certain

1:04:26.467 --> 1:04:29.677
temperature so that if it tries
to cool down,

1:04:29.679 --> 1:04:32.959
heat flows from below to above
maintaining the temperature.

1:04:32.960 --> 1:04:36.900
So, what the gas is doing in
this case is taking heat energy

1:04:36.898 --> 1:04:40.698
from below and going up and
working against the atmosphere

1:04:40.702 --> 1:04:42.492
above.
It takes in with one hand and

1:04:42.487 --> 1:04:44.527
gives out to the other,
without changing its energy.

1:04:44.530 --> 1:04:49.140


1:04:49.139 --> 1:04:51.819
So, when you study specific
heat, which is my next topic,

1:04:51.816 --> 1:04:54.536
you've got to be a little more
careful when you talk about

1:04:54.541 --> 1:04:56.911
specific heats of gases,
and I will tell you why.

1:04:56.909 --> 1:04:59.459
There is no single thing called
specific heat for a gas.

1:04:59.460 --> 1:05:02.140
There are many,
many definitions depending on

1:05:02.141 --> 1:05:03.361
the circumstances.

1:05:03.360 --> 1:05:05.260
But I hope you understand in
this case;

1:05:05.260 --> 1:05:07.410
you've got to visualize this.

1:05:07.409 --> 1:05:09.659
It's not enough to draw
diagrams and draw pictures.

1:05:09.659 --> 1:05:13.109
What did I do to the cylinder
to maintain the temperature and

1:05:13.110 --> 1:05:14.260
yet let it expand?

1:05:14.260 --> 1:05:18.820
Expansion is going to demand
work on part of the gas.

1:05:18.820 --> 1:05:21.840
That's going to require a loss
of energy unless you pump in

1:05:21.838 --> 1:05:22.878
energy from below.

1:05:22.880 --> 1:05:26.090
So, what I've done is that I
take grain after grain,

1:05:26.085 --> 1:05:29.475
so that the pressure drops and
the volume increases,

1:05:29.480 --> 1:05:32.760
but the slight expansion would
have cooled it slightly but the

1:05:32.755 --> 1:05:35.975
reservoir from below brings it
back to the temperature of the

1:05:35.978 --> 1:05:37.888
reservoir.
So, you prop it up in

1:05:37.891 --> 1:05:40.031
temperature.
So, we draw the picture by

1:05:40.027 --> 1:05:43.237
saying the gas went from here to
here, and we usually draw a

1:05:43.243 --> 1:05:46.463
picture like this and say heat
flowed into the system during

1:05:46.458 --> 1:05:50.088
that process.
Alright, now I'll come to this

1:05:50.092 --> 1:05:54.162
question that was raised about
specific heat.

1:05:54.159 --> 1:05:58.929
Now, specific heat,
you always say is ΔQ

1:05:58.928 --> 1:06:04.508
over ΔT or ΔT
divided by the mass of the

1:06:04.509 --> 1:06:07.819
substance.
Now, it turns out that for a

1:06:07.820 --> 1:06:11.740
gas, you've already seen that
what you want to count is not

1:06:11.735 --> 1:06:14.025
the actual mass,
but the moles.

1:06:14.030 --> 1:06:17.840
Because we have seen at the
level of the ideal gas law,

1:06:17.840 --> 1:06:21.160
the energy is controlled by not
simply the mass,

1:06:21.157 --> 1:06:22.637
but by the moles.

1:06:22.639 --> 1:06:25.919
Because every molecule gets a
certain amount of energy,

1:06:25.922 --> 1:06:28.562
namely 3/2 kT,
and you just want to count the

1:06:28.563 --> 1:06:30.433
number of molecules,
or the number of moles.

1:06:30.429 --> 1:06:34.249
Now, there are many,
many ways in which you can pump

1:06:34.246 --> 1:06:38.506
in heat into a gas and heat it
up and see how much heat it

1:06:38.511 --> 1:06:41.071
takes.
But let's agree that we will

1:06:41.072 --> 1:06:44.182
take one mole from now on and
not one kilogram.

1:06:44.180 --> 1:06:48.030
Not one kilogram.

1:06:48.030 --> 1:06:52.210
We'll find out if you do it
that way, the answer doesn't

1:06:52.211 --> 1:06:54.341
seem to depend on the gas.

1:06:54.340 --> 1:06:55.570
That's the first thing.

1:06:55.570 --> 1:06:59.370
Take a mole of some gas and
call the specific heat as the

1:06:59.367 --> 1:07:03.157
energy needed to raise the
temperature of one mole by one

1:07:03.164 --> 1:07:05.674
degree.
So, this should not be m.

1:07:05.670 --> 1:07:07.070
This should be the number of
moles.

1:07:07.070 --> 1:07:10.410


1:07:10.409 --> 1:07:13.079
If you take one mole,
you can say okay,

1:07:13.081 --> 1:07:17.651
one mole of gas I was told has
energy U = 3/2 RT.

1:07:17.650 --> 1:07:22.430


1:07:22.429 --> 1:07:25.099
Because it was there-halves
nRT but n is one

1:07:25.101 --> 1:07:27.571
mole.
Now, you want to put in some

1:07:27.569 --> 1:07:31.999
heat, and you really want the
ΔQ over ΔT,

1:07:32.000 --> 1:07:35.880
so I will remind you that
ΔQ is ΔU +

1:07:35.876 --> 1:07:36.936
PΔv.

1:07:36.940 --> 1:07:40.070


1:07:40.070 --> 1:07:43.600
The heat input into a gas is
the change of energy plus P

1:07:43.601 --> 1:07:45.501
Δv.
And that's just from the first

1:07:45.504 --> 1:07:48.354
law.
So, if I'm going to divide

1:07:48.347 --> 1:07:53.397
ΔQ by ΔT,
there's a problem here.

1:07:53.400 --> 1:07:56.130
Did you allow the volume to
change or did you not allow the

1:07:56.126 --> 1:07:57.016
volume to change?

1:07:57.019 --> 1:08:00.039
That's going to decide what the
specific heat is.

1:08:00.039 --> 1:08:02.699
In other words,
when a solid is heated,

1:08:02.699 --> 1:08:06.829
it expands such a tiny amount,
we don't worry about the work

1:08:06.829 --> 1:08:10.679
done by the expanding solid
against the atmosphere.

1:08:10.679 --> 1:08:12.499
But for a gas,
when you heat it,

1:08:12.498 --> 1:08:15.958
the volume changes so much that
the work it does against the

1:08:15.959 --> 1:08:18.129
external world is
non-negligible.

1:08:18.130 --> 1:08:22.970
Therefore, the specific heat is
dependent on what you allow the

1:08:22.969 --> 1:08:24.529
volume term to do.

1:08:24.529 --> 1:08:27.789
So, there's one definition of
specific heat called

1:08:27.787 --> 1:08:31.237
C_V,
and C_V is the

1:08:31.244 --> 1:08:33.044
one at constant volume.

1:08:33.040 --> 1:08:34.640
You don't let the volume change.

1:08:34.640 --> 1:08:36.770
In other words,
you take the piston and you

1:08:36.771 --> 1:08:39.071
clamp it.
Now, you pump in heat from

1:08:39.073 --> 1:08:41.423
below by putting it on a
hotplate.

1:08:41.420 --> 1:08:44.300
All the heat goes directly to
internal energy.

1:08:44.300 --> 1:08:47.360
None of that is lost in terms
of expansion.

1:08:47.360 --> 1:08:49.080
So, ΔV is zero.

1:08:49.079 --> 1:08:52.489
In that case,
ΔQ over ΔT at

1:08:52.494 --> 1:08:56.564
constant volume,
we denote that in this fashion,

1:08:56.560 --> 1:08:58.970
at constant volume,
this term is gone,

1:08:58.969 --> 1:09:02.159
and it just becomes ΔU
over ΔT.

1:09:02.160 --> 1:09:03.220
That's very easily done.

1:09:03.220 --> 1:09:06.360
ΔU over ΔT is
3/2 R.

1:09:06.360 --> 1:09:11.360


1:09:11.359 --> 1:09:16.019
So, the specific heat of a gas
at constant volume is 3 over

1:09:16.016 --> 1:09:17.976
2R.
When I studied solids,

1:09:17.983 --> 1:09:20.673
I never bother about constant
volume because a change in the

1:09:20.674 --> 1:09:23.324
volume of a solid is so
negligible when it's heated up,

1:09:23.319 --> 1:09:26.639
it's not worth specifying that
it was a constant volume

1:09:26.635 --> 1:09:27.975
process.
But for a gas,

1:09:27.976 --> 1:09:31.086
it's going to matter whether it
was constant volume or not.

1:09:31.090 --> 1:09:35.310
Then, there's a second specific
heat people like to define.

1:09:35.310 --> 1:09:37.920
That's done as follows.

1:09:37.920 --> 1:09:39.080
You take this piston.

1:09:39.079 --> 1:09:41.429
You have some gas at some
pressure.

1:09:41.430 --> 1:09:45.650
You pump in some heat but you
don't clamp the piston.

1:09:45.649 --> 1:09:49.099
You let the piston expand any
way it wants at the same

1:09:49.103 --> 1:09:50.523
pressure.
For example,

1:09:50.518 --> 1:09:53.018
if it's being pushed down by
the atmosphere,

1:09:53.023 --> 1:09:56.343
you let the piston move up if
it wants to, maintaining the

1:09:56.343 --> 1:09:59.443
same pressure.
Well, if it moves up a little

1:09:59.438 --> 1:10:03.548
bit, then the correct equation
is the heat that you put in is

1:10:03.552 --> 1:10:07.942
the change in internal energy
plus P times ΔV,

1:10:07.939 --> 1:10:10.899
where now P is some
constant pressure,

1:10:10.896 --> 1:10:14.236
say the atmospheric pressure,
ΔV is the change in

1:10:14.237 --> 1:10:16.557
volume.
So, ΔQ needed now will

1:10:16.559 --> 1:10:20.459
be more, because you're pumping
in heat from below and you're

1:10:20.457 --> 1:10:24.417
losing energy above because
you're letting the gas expand.

1:10:24.420 --> 1:10:27.430
Because you were letting the
pressure be controlled from the

1:10:27.429 --> 1:10:28.959
outside at some fixed value.

1:10:28.960 --> 1:10:35.870
So now, ΔQ--this
ΔU will be 3 over

1:10:35.873 --> 1:10:39.403
2RΔT.
Now, what's the change in

1:10:39.403 --> 1:10:40.933
P times ΔV?

1:10:40.930 --> 1:10:42.890
Here is where you should know
your calculus.

1:10:42.890 --> 1:10:46.200
The P times change in
V is the same as the

1:10:46.200 --> 1:10:49.150
change in PV,
if P is a constant.

1:10:49.150 --> 1:10:51.570
Right?
Remember long back when I did

1:10:51.571 --> 1:10:53.971
rate of change of momentum is
d/dt of mv,

1:10:53.970 --> 1:10:55.460
it's m times
dv/dt,

1:10:55.463 --> 1:10:57.053
because m doesn't
change.

1:10:57.050 --> 1:10:58.830
You can take it inside the
change.

1:10:58.830 --> 1:11:01.930
But now we use PV = RT.

1:11:01.930 --> 1:11:03.650
I'm talking about one mole.

1:11:03.650 --> 1:11:07.500
PV = RT.
That's a change in the quantity

1:11:07.495 --> 1:11:11.245
RT, R is a
constant, that's R times

1:11:11.249 --> 1:11:15.139
ΔT so I put in here
R times ΔT.

1:11:15.140 --> 1:11:19.680
This is the ΔQ at
constant pressure.

1:11:19.680 --> 1:11:23.630
So, the specific heat of
constant pressure is ΔQ

1:11:23.630 --> 1:11:27.150
over ΔT,
keeping the pressure constant.

1:11:27.149 --> 1:11:30.039
You divide everything by
ΔT you get 3 over

1:11:30.041 --> 1:11:32.871
2R, plus another
R, which is 5 over

1:11:32.873 --> 1:11:33.703
2R.

1:11:33.700 --> 1:11:39.970


1:11:39.970 --> 1:11:43.460
So, the thing you have to
remember, what I did in the end,

1:11:43.457 --> 1:11:46.637
is that a gas doesn't have a
single specific heat.

1:11:46.640 --> 1:11:48.690
If we just say,
put in some heat and tell me

1:11:48.688 --> 1:11:51.018
how many calories I need to
raise the temperature,

1:11:51.022 --> 1:11:52.072
that's not enough.

1:11:52.069 --> 1:11:55.609
You have to tell me whether in
the interim, the gas was fixed

1:11:55.605 --> 1:11:57.725
in its volume,
or changed its volume,

1:11:57.727 --> 1:11:59.787
or obeyed some other condition.

1:11:59.789 --> 1:12:03.329
The two most popular conditions
people consider are either the

1:12:03.332 --> 1:12:06.412
volume cannot change or the
pressure cannot change.

1:12:06.409 --> 1:12:09.269
If the volume cannot change,
then the change in the heat you

1:12:09.266 --> 1:12:12.216
put in goes directly to internal
energy, from the First Law of

1:12:12.220 --> 1:12:14.930
Thermodynamics.
That gives you a specific heat

1:12:14.928 --> 1:12:16.178
of 3 over 2R.

1:12:16.180 --> 1:12:19.560
If the pressure cannot change,
you get 5 over 2R.

1:12:19.560 --> 1:12:21.860
You can see
C_P is bigger

1:12:21.860 --> 1:12:25.000
than C_V
because when you let the piston

1:12:25.002 --> 1:12:28.082
expand,
then not all the heat is going

1:12:28.077 --> 1:12:31.407
to heat the gas.
Some of it is dissipated on top

1:12:31.405 --> 1:12:33.585
by working against the
atmosphere.

1:12:33.590 --> 1:12:39.940
Then, notice that I've not told
you what gas it is.

1:12:39.939 --> 1:12:44.329
That's why the specific heat
per mole is the right thing to

1:12:44.334 --> 1:12:48.734
think about because then the
answer does not depend on what

1:12:48.728 --> 1:12:50.848
particular gas you took.

1:12:50.850 --> 1:12:54.230
Whether it's hydrogen or
helium, they all have the same

1:12:54.233 --> 1:12:55.803
specific heat per mole.

1:12:55.800 --> 1:12:59.260
They won't have the same
specific heat per gram,

1:12:59.260 --> 1:13:01.530
right?
Because one gram of helium and

1:13:01.534 --> 1:13:04.794
one gram of hydrogen don't have
the same number of moles.

1:13:04.789 --> 1:13:07.249
So, you have to remember that
we're talking about moles.

1:13:07.250 --> 1:13:09.350
The final thing I have to
caution you--very,

1:13:09.349 --> 1:13:12.909
very important.
This is for a monoatomic gas.

1:13:12.910 --> 1:13:16.560


1:13:16.560 --> 1:13:21.630
This is for a gas whose atom is
the gas itself.

1:13:21.630 --> 1:13:24.040
It's a point.
Its only energy is kinetic

1:13:24.042 --> 1:13:26.062
energy.
There are diatomic gases,

1:13:26.063 --> 1:13:28.413
by two of them [atoms]
joined together,

1:13:28.409 --> 1:13:30.939
they can form a dumbbell or
something;

1:13:30.939 --> 1:13:33.679
then the energy of the dumbbell
has got two parts,

1:13:33.684 --> 1:13:35.144
as you learned long ago.

1:13:35.140 --> 1:13:39.140
It can rotate around some axis
and it can also move in space.

1:13:39.140 --> 1:13:42.430
Then the internal energy has
also got two parts.

1:13:42.430 --> 1:13:45.290
Energy due to motion of the
center of mass and energy due to

1:13:45.287 --> 1:13:48.087
rotation.
Some molecules also vibrate.

1:13:48.090 --> 1:13:50.140
So, there are lots of
complicated things,

1:13:50.138 --> 1:13:52.338
but if you got only one guy,
or one atom,

1:13:52.340 --> 1:13:55.560
whatever its mass is,
it cannot rotate around itself

1:13:55.560 --> 1:13:57.960
and it cannot vibrate around
itself,

1:13:57.960 --> 1:13:59.820
so those energies all disappear.

1:13:59.819 --> 1:14:03.249
So, we have taken the simplest
one of a monoatomic gas,

1:14:03.250 --> 1:14:07.060
a gas whose fundamental entity
is a single atom rather than a

1:14:07.061 --> 1:14:08.651
complicated molecule.

1:14:08.649 --> 1:14:10.449
And that's all you're
responsible for.

1:14:10.450 --> 1:14:11.700
I'll just say one thing.

1:14:11.699 --> 1:14:13.299
C_P over
C_V,

1:14:13.301 --> 1:14:15.461
I want to mention it before you
run off to do your homework.

1:14:15.460 --> 1:14:16.630
I don't know if it comes up.

1:14:16.630 --> 1:14:19.400
It's called γ,
that's five-third for a

1:14:19.402 --> 1:14:21.772
monoatomic gas.
You can just take the ratio of

1:14:21.768 --> 1:14:24.078
the numbers.
If in some problem you find

1:14:24.077 --> 1:14:26.907
γ is not five-thirds,
do not panic.

1:14:26.909 --> 1:14:30.679
It just means it's a gas which
is not monoatomic.

1:14:30.680 --> 1:14:32.860
If it's not monoatomic,
these numbers don't have

1:14:32.862 --> 1:14:33.932
exactly those values.

1:14:33.930 --> 1:14:35.880
We don't have to go beyond that.

1:14:35.880 --> 1:14:37.650
You just have to know there's a
ratio γ,

1:14:37.651 --> 1:14:39.311
which is five-thirds in the
simplest case,

1:14:39.310 --> 1:14:41.460
but in some problem,
somewhere in your life,

1:14:41.455 --> 1:14:43.995
you can get a γ which
is not five-thirds.