22. The Boltzmann Constant and First Law of Thermodynamics

22. The Boltzmann Constant and First Law of Thermodynamics


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

Posts created 19620

Leave a Reply

Your email address will not be published. Required fields are marked *

Begin typing your search term above and press enter to search. Press ESC to cancel.

Back To Top