Professor Ramamurti
Shankar: What I did yesterday, or Monday,
was to describe to you the First Law of thermodynamics.
So, let me remind you what the context was.
We're going to study a system, that's our thermodynamic
system, and for all our purposes you can imagine that to be some
amount of ideal gas trapped inside this piston and cylinder
combination. We put some weights or maybe
some sand on top to keep the piston where we want it.
The gas has a certain pressure and a certain volume and we
represent that by a point in this diagram.
That defines the state. That's called the state of the
gas. On a microscopic level,
the state is very complicated, right?
There are 10^(23) atoms or molecules zipping around.
We're not interested in such detailed properties.
We want to look at bulk macroscopic properties of which
pressure and volume is a complete pair.
That's all you need. I think I reminded you that you
don't need a third axis for temperature because we have this
equation PV = nRT. So, if you know P and V,
you don't need temperature as another variable.
That's called the equation of state.
You can use this all the time. Whenever the gas has got a
pressure and volume, feel free to use PV =
nRT. Alright, so here's the gas.
And then I want to change its state, and I explained to you
that unlike particles whose state can change as fast as you
like, you cannot change the state of
a gas abruptly because if you take half the weight off the
top, the piston will fly off;
there will be a period when the gas is not in equilibrium.
Different parts of it will have different densities,
different pressure. You cannot represent that
complicated state as a point in this plane.
So, what we want to do is to make changes which are slow
enough for us to follow. So, one way to do that is to
take one grain of sand. Take one grain of sand,
then you lower the pressure a little bit, the piston will move
up a little bit. Move up means what?
Bigger volume, lower pressure,
the gas will come there and then you can take off another
grain and by making the grain smaller and smaller,
you can make these dots closer and closer.
It's going to take you a long time but in principle you can
make the gas move on some path. That is a thermodynamic process.
This process is always near equilibrium.
You never deviate too much from equilibrium, and it's called a
quasi-static; that means not quite static,
but as slow as you like, and it is called reversible.
Reversible means that you can also go right back to where you
came. If you took a grain out,
piston expands. You put the grain back,
piston contracts. So, you can go down this way or
you can go up this way. I've just chosen to,
let's say, go down like this. When that happens,
we want to say the following thing about the state of the gas
before and after. At every point in the PV
diagram, the gas has something called the internal energy. The characteristic of internal
energy is it's called a state variable.
And this is a very important concept.
A sate variable depends on where you are in the PV
diagram, what your state is. It does not depend on the
history; it does not depend on all the
things the gas has gone through in its life.
If you tell me you are here, then we may associate with that
point an energy that doesn't look into the history of that
gas. That energy is nothing other
than the kinetic energy. You'll remember that it's 3/2
kT per atom, N atoms,
that's the internal energy. You can also write this as 3/2
nRT. If you count the molecules then
use Boltzmann constant, or you count the number of
moles then use the gas constant. It's the same thing.
But nRT is PV, so if you like,
you can also write it this way. That shows you very clearly the
internal energy is uniquely determined by PV,
and P and V determine the product PV.
So, at each point in this diagram, the gas is a certain
energy. But energy is simply that ½
mv^(2) for each molecule. You add it all up,
that's the total energy. Okay, so we can write it many
different ways. Now, the First Law of
Thermodynamics asks the following question.
If I want to change the energy of this gas, how can I do it?
There are two ways to change the energy of the gas,
and the total change in U,
meaning U at the end minus U at the beginning,
where these are two nearby points,
is either the heat input minus the work done by the gas--Let me
write it as capital PdV. Sometimes you write it as
∆Q - ∆W. That's the First Law.
So, you have to take some time to understand what it says. Notice from this formula here,
the energy of an ideal gas, and only an ideal gas because
that's the only thing we are going to talk about,
is a function only of temperature.
If you know the temperature of the gas, you know the kinetic
energy of the molecule, so if the number is constant,
the energy is determined. The energy depends only on the
temperature for an ideal gas. So basically,
when I say how can I change the energy of the gas,
I'm asking how can I raise the temperature?
One obvious way is this. Put it on a hotplate which is
slightly hotter than the gas. Then the heat will flow from
the hotplate to the gas. That's called heating up the
gas. There you don't see anything
happening, right? It's occurring at a molecular
level. The molecules in the hotplate
are, on average, moving faster than the
molecules in the gas and during the collision;
kinetic energy will be transferred from the hotplate to
the gas. That invisible flow people used
to think of as the caloric fluid, but now we know it's just
molecular kinetic energy transferred from the molecules
in the plate to molecules in the gas.
That's this guy. This ∆W,
the plus sign of the W is the work done by the gas,
and minus of that is the work done on the gas.
And that means if you want to increase the temperature of the
gas, you're also told you can do the following.
Push down on the piston. If you push down on the piston,
you are reducing the volume. ∆V is negative,
minus P∆V is positive, and that'll also increase the
energy of the gas. Or, if you take a grain of sand
out, and the piston moves upwards, the gas works against
the environment and it pays for it through its internal energy
because then P∆V will be positive and this thing will
be a negative contribution of the internal energy.
So, there are two ways to change the energy of the gas.
One is a way you can actually see by moving up and down the
piston. You can see somebody moving it.
It's a macroscopic motion of something you can see.
That's called the work done on the gas.
Other one is energy transferred at a molecular or microscopic
level, which is what we call heat.
And the First Law says the total change in the internal
energy is the sum of two numbers.
First one is the heat input and the second one is the work done. So, what I did for you last
time was a practice calculation in which I took a gas here,
went along an isotherm. That means temperature is a
constant. That's the meaning of the
isotherm. And I want to go from some
initial point to some final point.
So, this volume V_1 and
pressure P_1. This is P_2
and V_2, and I want to find the work
done by the gas during this expansion.
Well, P times ∆V, you can see,
is really that shaded area here because it's a rectangle of
height P and width ∆V.
So, P∆V is just the area under the graph that you
follow. So, the work done,
the sum of all the infinitesimal works,
is the pressure as a function of volume times dV from
the starting volume for the ending volume.
What is P as a function of V at a constant
temperature is nRT over V? Now, I'm using the factor at
every point PV is nRT, so P is
nRT over V, and this is a logarithm and we
all know how to do the integral, as log of the upper limit minus
log of the lower limit. So, that's the first successful
calculation of work done. You can take the gas on any
path you like, you know.
As long as you knew the pressure at each volume,
you integrated the function from the left limit to the right
limit; that's the work done.
For isothermal process, it's particularly easy because
the pressure is inversely proportional to volume,
and you do that integral. This is the work done.
For example, if the gas expands to double
the volume at temperature 300 Kelvin, the work done is number
of moles, gas constant, 300 log of 2.
Multiply all those things; that's the work done by the gas.
If that is the work done by the gas, we can also say the
following. The internal energy change is
zero. Right?
That's the trick question I told you to watch out for.
If you're going at constant temperature, you don't change
the internal energy. So, if the change in internal
energy is zero, it follows at every step,
∆Q = ∆W. This quantity we call
∆W. Therefore, the work done by the
gas is also equal to the total heat input of the gas.
We indicate that by drawing a little arrow like this;
it's just the way we like to show that heat flows into the
gas. Do you understand physically
what's happening? You take this gas,
you keep it on a reservoir at temperature T_1
[T_1 is just T].
It's stuck at that temperature. If you take a grain of sand,
gas is expanding. Normally, when the gas expands,
it does work and if energy will go down, it'll cool down.
But you're not letting it cool down because you're pumping in
heat from below, maintaining at that
temperature. So, this process you should
visualize as a piston with some weights sitting on a plate at
temperature T_1.
You take off a grain of sand, it moves up a little bit but
you don't let it cool down because you send in heat from
below, maintaining it at that
temperature. For example,
you got this computer air, right?
You blow air on your keyboard to blow off the dust.
You know, when you blow the air out, the can cools down.
It cools down because the compressed air is working
against the atmosphere as it comes out and,
therefore, it loses internal energy and it cools down.
That's because you're not giving it enough time to absorb
heat from the room. If you let the air out slow
enough so that in the process of expansion it tries to cool down
but the room pumps some more heat,
then, of course, the can will not cool down.
But, you know, that's not the purpose of
buying canned air. You don't want to buy canned
air so you can release it at room temperature.
You don't care if it cools down; you just want the gas to come
out really fast. So, if you don't allow heat to
come in, an expanding gas will cool down.
Okay, that's also why the universe cools down.
Process of expansion. Alright.
So, here's a simple problem where you can do many things.
You can find the change in energy from start to finish.
Zero. Work done you find by
integrating PdV and the heat input is equal to work done
because there's no change in energy.
This is all the stuff I did last time.
I also did a couple of other topics.
I'll just remind you what they are.
I said we can define a specific heat for a gas.
Now normally, specific heat is the number of
calories to raise one gram or one kilogram by one degree.
But for a gas, what matters is not the number
of grams but number of moles. You have already seen it here.
The energy of the gas is not proportional to the grams of gas
you have but to the number of atoms you have or the number of
moles you have. So, we define specific heat to
be the heat needed to produce a certain temperature change for
one mole. We take a mole of gas and you
heat the guy and you see how many calories do I need to get
an increase of one degree. But that turns out to be
insufficient. This derivative does not have a
unique value. Can someone tell me why it's
not sufficient information? You remember from last time?
Yes? Student: [inaudible]
Professor Ramamurti Shankar: In other words,
when I take a solid and I say find the specific heat,
take one gram of solid, pump in some heat so the
temperature goes up by one degree, keep track of the
calories. You are done.
For a gas, it turns out if you took this gas in the cylinder
and you pump in some calories, it's going to try to heat up
and when it tries to heat up, it may also expand.
When it's expanding, then some of the energy that
you pump in the form of heat, some of it's turning into work
to push the piston out. The part that goes through the
internal energy, the part that really heats up
the gas, is now less because you put in
some heat from below, you lose some heat to the
expansion, so what goes in to change the
internal energy, which is what controlled the
temperature, is less than the heat you put
in. So, if the gas expands during
the heating process, there's a certain energy lost
due to work you have to take into account.
On the other hand, another person can do the same
experiment of the gas but put a pin here so the piston cannot
move. Then, of course,
it cannot do any work and all the heat you put in goes into
internal energy. That gas will heat up more
readily. For a solid,
we don't worry about it because the expansion of a solid is such
a tiny fraction of its volume when the temperature changes
that the fact that a chunk of copper,
when you heat it, is also expanding against the
universe, against the atmospheric pressure,
is a negligible effect. The ∆V is so small for
a solid we ignore it. For a gas it's very important.
Therefore, for a gas there are two kinds of specific heat.
In fact, you can define many definitions.
The two common ones are C_V,
which is ∆Q over ∆T at constant volume,
and that's how you denote that in calculus [pointing to the
equation]. The other one is
C_P, which is ∆Q over
∆T at constant pressure.
So, let me derive one more time these two specific heats because
it's worth going over this again and again.
The key to all of this is the First Law.
The First Law, written after transposition,
looks like this. Let's put ∆Q on one
side and the other stuff on the right-hand side.
In the first case, at constant volume,
there is no P∆V term because ∆V is zero.
Therefore, ∆Q/∆T =
∆U/∆T, and for one mole,
you remember U is 3/2 nRT but n is 1,
for one mole. So, dU/dT is a simple
derivative of this function with respect to T, which is
3/2R. What do you notice about this
formula? It's the specific heat of one
mole of a gas doesn't depend on what the gas is.
As long as it's a monoatomic gas, it doesn't matter what the
mono atom is. It could be hydrogen,
it could be helium; it doesn't matter.
That's why we like to think about the molar-specific heat or
specific heat per mole. You get an answer that's
universal. If you took specific heat for a
gram, then one gram of hydrogen and one gram of helium won't
have the same specific heat because they don't have the same
number of atoms. But if you count in moles,
the beauty of the formula is this number is true for all
ideal gases, whereas if you took the
specific heat of copper or silver or iron,
it varies from material to material.
Specific heat of ideal gases is very simple.
You don't need to look at a table.
Here's the answer for all ideal gases.
It's very important it's a monoatomic gas,
which is a point-like atom which only moves but doesn't
have any internal structure so it can spin and rotate and
vibrate. If it can do that,
there are other contributions to energy, but we're not going
there. Now, let's calculate
C_P. C_P is
∆Q over ∆T at constant pressure.
So now, you cannot ignore this term.
So, ∆U, from what I did before,
will be C_v times ∆T,
plus there is a new term. P∆V is ∆ of
PV because P is constant.
If P is a constant, the change in PV is the
same as P times change in V.
But PV = RT, so if C_v
∆T plus R∆T. If you divide every side by
∆T and take ∆Q over ∆T,
then C_P will be C_v plus
R and that's a very famous formula.
And you should understand why specific heat at constant
pressure is more than specific heat at constant volume.
Think about that. If the gas is stuck at a given
volume, it's easier to heat it up because all the heat goes
into internal energy. If it is at constant pressure,
you're letting the piston move up and some of the heat is being
paid to move the piston, only the rest is going to
change the internal energy. That's why C_P
is always bigger than C_v,
and for an ideal gas, C_P is
C_v plus R and this number,
γ, is defined as C_P over
C_v. This is always the definition
of γ, where there is an ideal gas or
not. For an ideal gas,
we can calculate C_p over
C_v, which is 3/2 R plus
another R, divided by 3 /2 R,
which is 5 over 3. So, γ is not always 5 over 3.
You should be aware of that. But for ideal gas,
γ is 5 over 3. So, this is a review of what I
did last time, but I'm very eager that you
guys should be on top of this and it's worth doing it one more
time. But now, I'm going to consider
some more processes where I want to find work done. So, maybe I'll start fresh here. Let's take the following
process. I start here.
I go on an isothermal here [pointing at graph],
then I go backwards at constant pressure until I'm right below
this point, then I go straight up. You can ask,
"What is the work done in this process?" The work done in the curvy part
of it is all of this. That's in the part AB.
In the part BC, work done is the area of this
rectangle but it's counted as negative, because a gas is being
compressed. Or, if you like,
it's at some constant pressure but the change in volume is
negative for every step. So, even though the work done
is said to be the area, you've got to keep track of
sign. If you're compressing the gas,
P∆V is negative. Therefore, in the part
BC, we must take away the shaded region and CA,
no work is done because volume is not changing.
Area under the vertical line is zero.
So, you can see that in this full cycle, when I go from
A to B to C back to A,
the work done is the area enclosed by the loop in the
PV plane. That's a very useful result.
We say the work done in a cycle, there's a symbol we like
to use. This curly thing on the
integral means taken around a closed loop. But you guys have to be little
careful when you write this, because somebody else can do
the following thing to gas, do exactly the opposite.
Come down like this, go down like this,
and go down like that. Go up like that.
The opposite way. In this case,
the work done is considered positive because the part of it,
AB, is taller than the BC at
point Y, so the net work is positive.
Here, it's the same amount of work but the net work is
negative. So, you've got to be careful
that you describe the sense in which you do the loop.
So, the same loop, as a geometrical figure,
as the same area, but the work done by the gas is
positive in this case and negative in this case,
and you can use common sense to find out if it's positive or
negative. But here is another very
interesting result. Let's go to the full cycle and
ask, "What's the change in U?" Yes?
Student: [inaudible] Professor Ramamurti
Shankar: Right. His answer was PV is the
same. And more generally,
even if it's not an ideal gas, it's a gas made up of whatever
you like. It's any system.
If you return the system back to the same state,
the internal energy returns to that value.
That's what I meant by internal energy doesn't depend on the
history of the gas. For example,
you can take the cylinder, you leave it alone,
I measure the energy, I go outside the room and come
back. In the meantime,
you took it on a loop and brought it back.
I will not know, and I don't care because as far
as I'm concerned, if P and V are
back to where they are; the gas is back to where it is;
energy is back to where it is. So, the change in the internal
energy is zero. That means the work done and
the heat input are the same. So, let's be very clear.
If you guys do problems, I don't want you to lose points
for sign mistake. W is always the work
done and Q is always the heat into the gas. So, this is a gas here which
has done some work and some heat that's been put into the gas,
and the two have to be equal because the energy doesn't
change in the end. Now, that leads to a very,
very subtle and interesting point.
There is a very clear notion of what is the energy in the gas.
There's a unique answer. There's no notion of what is
the heat in the gas. You look at a gas and you
cannot say this is the heat in the gas.
That makes no sense. Try to understand why.
Because if you say there's some amount of heat in the gas,
okay, whatever you want, some number,
I do the cycle and I come back to exactly where I am.
I'll put in some amount of heat Q, which is non-zero.
So, if there's anything called the heat in the gas,
it has changed by an amount of Q.
That's why there's no notion of heat in the gas or work in the
gas. There's only energy in the gas.
A state variable returns to its original value and when you go
on closed loop, the change should add up to
zero. That's not true for work,
that's not true for heat, but it's true for the
difference. That defines an internal state
variable. Okay, now I'm going to consider
a new process which is of special importance.
So, the processes I've considered so far are
isothermal, at constant pressure is called isobaric,
and I don't know the name for constant volume,
but no work is done at constant volume.
Now, I'm going to consider the last process very,
very important. It's called adiabatic. Adiabatic process is in which
the gas changes its volume but it's completely thermally
isolated, so ∆Q is zero.
You wrap this guy in a blanket and you do things to it.
No heat can flow into the gas or out of the gas.
That's called an adiabatic process.
Isothermal is quite different. Isothermal, you keep it on a
hotplate at a given temperature and as the volume changes,
say as the volume increases, heat comes in from below.
Or, when you compress it and you don't want it to heat up,
heat is rejected below. But this is not the case here.
In an adiabatic process, if you let the gas expand you
thermally isolate it from everything.
So, you should think about what'll happen to a gas if I
start here. That's what isothermal is.
What do you think will happen? So, let's draw a few isotherms.
This is temperature, say T = 300. This is T equal to what?
More than 300 or less than 300? Student: Less.
Professor Ramamurti Shankar: Less.
And why do I say less? You take any point here;
it's got the same volume but lower pressure,
so P times V is less.
So, this may be T = 200. If I start here,
now I let the gas expand adiabatically.
That means I pull out the grains of sand,
the gas expands against the atmosphere, but no heat is
allowed to come in. Think about what that means.
Gas is pushing the piston. Gas is doing some work.
It'll pay for it through its own internal energy.
Internal energy will go down and that means temperature will
go down. So, what you will be doing is
you'll be cascading down from one isotherm to another,
plummeting down in temperature until you come and stop
somewhere at the lower temperature.
So, an adiabatic process will cross from one isotherm to the
next to the next to the next, changing temperature. Another way to say this is that
the drop in pressure for a given drop in increase in volume,
it would be more precipitous for adiabatic because it's
expanding but there's no energy coming into the form of heat so
pressure drops more precipitously.
So, it won't be PV equal to constant, and the question
is, "What is it?" What is the equation for an
adiabatic process? What is P as the
function of V? That's what we want to ask now. So, whenever you have any such
question in thermodynamics, up to this point the only law
we know is the First Law of Thermodynamics.
So, you have to go back to that. So far, I've driven home the
point that every question asked in mechanics can be traced back
to Newton's Law. There are no new laws.
But when you come to thermodynamics and you study
heat and temperature, there is new stuff and the
First Law of Thermodynamics is called a law,
because there is no way you can derive it from Newtonian laws.
It's a new concept. So, let's ask what is P
as a function of V, given that ∆Q = 0?
So, what's our strategy going to be?
How would you even begin this? Do you have some idea of what
you might do? This is a tough question so
even if you give a wrong answer, that's fine.
How do I calculate the relation of P to V given
that ∆Q = 0? Yes?
Student: [inaudible] Professor Ramamurti
Shankar: Okay. Let me write down everything he
said. When I said ∆Q = 0,
your first reaction has to go back to the First Law and write
∆Q as ∆U + P∆V,
and set that equal to zero. But ∆U is
C_v times ∆T, plus P∆V
is zero. By the way, ∆U is
C_v ∆T for one mole.
I'm going to consider one mole of gas.
Now, this is another thing you people should understand.
The way in which pressure falls for temperature is going to be
the same, whether I consider one mole or two moles or ten moles.
The rate at which it changes will not be dependent on how
many moles I took, so if you want you can take
n moles and put an n here.
You'll find everywhere an n will come and cancel
out. Now, P is nRT
over V. So, maybe I'll do it this way
for you guys. If you want,
put an n back there, but you will find n
cancels part of the equation because nC_v
∆T, plus nRT ∆V over
V equals zero; so the answer doesn't depend on
how many moles you took. So, I will take one mole for my
calculation. Think about what this says.
Divide everything by T. You get C_v
over R∆T over T plus ∆V over V
is zero. This is telling you that as you
change your volume and you're coming down this graph of
adiabatic expansion, when you go from there to
there, there's a change in temperature such that the change
in temperature by the temperature,
plus the change in volume divided by the volume,
should add up to zero. That is a condition of no heat
flow. So, it relates a change in
T to a change in V.
What I would have ideally liked is a change in P due to a
change in V, but what comes more naturally
is a change in T over a change in V.
But let's worry about that later, because we've got PV =
RT. We can always swap the
temperature for pressure in the end.
So, let's just take this definition.
So, every change, remember now,
you understand you go from one point to the neighboring point;
there's a plunge in temperature. That's why ∆T will be
negative and ∆ will be positive and these numbers will,
of course, cancel. Now, let's add all the changes.
Let's call it doing the integral.
Do the integral of that from start to finish,
do the integral of that from start to finish,
and that integral of zero is just zero. So, this tells me
C_v over R times log of
T_2 over T_1,
plus log of V_2 over
V_1 is zero. Now, you've got to go back and
think about your logarithms and realize--and I can write it as
log of T_2 over T_1 raised to
the power C_v over R,
times V_2 over V_1 is equal to
zero. We just combine all the
logarithms and realize the log of X to the power
n is n log X.
So, that is the same as this. Now, the log of something is
zero means that something is 1, because log of 1 is zero.
So, we rewrite the final expression I got up there by
saying T_2 over T_1 raised to
C_v over R times
V_2 over V_1 is equal to
1. And we usually write it as
follows. We say T_2
raised to the C_v over
R, V_2 to the
power of 1 is T_1 raised to
the C_v. I'm sorry, T_1
raised to C_v over R times
V_1. So, this is the relation
between the temperature and the volume.
Yes? Student: [inaudible]
Professor Ramamurti Shankar: You could do that.
That's correct. You could write down
three-halves if you like. At the moment,
I'm trying to keep it general. If it was a gas for which
C_v was not three-halves R,
all of this would still be true.
But he is quite right. For an ideal monoatomic gas,
C_v over R would be just 3 over 2.
So, what it's telling you is that when the gas expands,
the temperature falls. I already told you that.
It's telling exactly how it falls.
It tells you that T to some strange power times
V is constant during the process.
Now, if you want to know what happens to the pressure,
you can do the following. You remember PV = RT. There are many ways to write
this. I'm going to write it in
another way and you guys see if you can follow this.
T_2V _2 to the power
R over C_v is equal to
T_1V _1 to the power
R over C_v.
You believe that? That's just a mathematical
trick. If that is true, this is true.
You know how I went from here to here? How did I go from here to here?
Do you have an idea? Student: You take
R over C_v to the
power of [inaudible] Professor Ramamurti
Shankar: Exactly. Raise both sides to the power
R over C_v.
It's cooked up so that C_v over
R goes away and R over C_v falls
on V_2. There is another way to write
it, if you like. I'll tell you why I write it
this way. Because T_2 is
just P_2V _2.
But let me combine that one V_2 with this
and write the one plus R over C_v.
Then, I can show it as P_1V
_1 to the power 1 plus R over
C_v. You may object that PV
is not T but is RT, but R will
cancel so I didn't bother with that.
And this--don't draw the box around this yet.
I will write it in a way that's very, very standard and if you
remember in your high school days,
you may remember is written as P_2V
_2 to the γ,
is equal to P_1V _1 to the
γ, but γ is just 1 plus R over
C_v, which you can see is equal to
C_v plus R over
C_v and that is C_P over
C_v. So, this γ is really
C_P over C_v.
For an adiabatic process, P_2V
_2 to the γ is P_1V
_1 to the γ.
That's one way to write it, but I will also expect you to
know that you can also write it in terms of temperature,
as temperature times volume raised to a strange power is
equal to a constant, before and after. Any questions?
So, what we have done is found pressure as the function of
volume on an adiabatic curve. So, I'm now going to do one
last ingredient for the rest of the stuff, which is the
following. I'm going to calculate the work
done in an adiabatic process from here to here.
This is one, this is two. This is not an isothermal,
which looks like that, it's an adiabatic process.
What's the work done in adiabatic process?
So, let's calculate that. Work done is equal to integral
P(v)dV. But remember,
PV to the γ is a constant.
Do you guys understand the relation of writing PV to
the γ equal to constant? It's the same as saying
P_1V _1 to the
γ is P_2V _2 to the
γ. That means you can pick any
point on the path and P at that point times V at
that point to the power of γ will be a fixed
number. That number is C.
In other words, this C can be
P_1V _1 to the
γ if you like, or P_2V
_2 to the γ if you like or
P_3V _3 to the
γ if 3 is another point on this trajectory.
It's convenient to call that invariant product of PV
raised to γ as C. Because then we can do the
integral as dV over V to the γ with a
constant C here from V_1 to
V_2. And this is a pretty simple
integral. You guys know that the x
to the n integral is x to the n plus 1
over n plus 1, such that V to the
γ minus 1, over γ minus 1. Because this is really V
to the minus γ. Now, did I get it wrong here?
I think I got it wrong. Sorry.
It's 1 minus γ, 1 minus γ,
over 1 minus γ. Because V to the minus
γ dV, is equal to V to the
minus γ plus 1 over minus γ plus 1.
Right? x to the n
integral is X to the n plus 1 over n
plus 1, n happens to be minus γ here.
Okay, now watch this. This constant I said can be
written as any of these numbers, but let me write when it
multiplies this; let me write as
P_2V _2 to the
γ. Can you see what happens if I
do that? Try to do this in your head.
C multiplies all of this. When C multiplies this
guy, let me write the constant as P_2V
_2 to the γ;
that gives me a P_2
,V_2 to the γ combines with
V_2 to the one minus γ,
to give me just V_2.
Likewise, C multiplying this one, when the same C
multiplies this, let's write C as
P_1V _1 to the
γ, and we'll find there's that. And people like to rewrite this
by changing the sign everywhere as P_1V
_1 minus P_2V
_2 over γ minus 1,
because γ minus 1 is the positive quantity.
So, this is the work done in an adiabatic process. So, what's been going on?
Let's think about what I've been doing.
I told you that there is a new process called adiabatic,
a really important process, in which the system is not
allowed to exchange heat with the outside world.
∆Q = 0 is a condition relating ∆U and
P∆V. And the changes are correlated
and the effect of adding up all the changes, namely doing the
integral, is this one,
namely temperature to the C_v over
R times V_2 does not
change along the path. So, if you take any two points,
that combination of temperature and volume doesn't change.
Then, since PV equal to essentially T,
you can also write it as pressure times volume to the
power of γ doesn't change.
So, I'm saying you can call all of that equal to some constant
C. The C is not a universal
constant, like velocity of light.
For a given gas, on a given experiment,
it does not change as it goes through adiabatic expansion.
So, don't think of it as a universal constant.
On the trajectory, on the adiabatic curve,
it's a constant. It's like saying kinetic energy
plus potential energy is a constant.
It's not a constant you can look up in a book.
It depends on what your particle is doing.
But for that particle in those conditions, kinetic with
potential adds to a particular constant we call total energy.
Likewise, PV to the γ is a constant for this
particular gas. Okay, so all this is actually a
preparation for something else. And that's what we are going to
discuss now. So, between now and the end of
this course we are going to discuss the Second Law of
Thermodynamics. So, we have done all we want to
do with the First Law. So, you guys should be able to
be given various graphs on the PV plane,
find the work done in going from here to there,
find the change in internal energy,
find the heat added and so on. I hope you can do such
problems, but they're all going to be one of these formulas.
Say there's going to be adiabatic or constant pressure
or constant temperature, so you just plug the numbers
in, keep track of the signs by using common sense.
Alright. So now, we come to the last
part of thermodynamics, which is to me one of the most
beautiful parts of thermodynamics,
which starts out with asking the following question.
There are certain things in this world that are perfectly
allowed but don't seem to happen.
Let's take many examples. Take the Joule experiment.
You take a cylinder of water with a paddle that can spin and
you put some weight around the pulley and the weight goes down.
Paddle spins; the water heats up.
You take a movie of that, and the movie is so fine it can
even catch individual molecules. Now, you play the movie
backwards. So, you watch the movie being
played back and what do you find?
Suddenly, the weight starts moving up, paddle spins the
other way, and the water cools down.
This is not in violation of any of the laws you have learned
including the First Law of Thermodynamics,
because when the weight went down and the water heated up,
some amount of work was done on the water,
and the energy of the water went up.
In the reverse process, the energy of the water went
down and instead the weight went up, so work was done by the
system. But that doesn't seem to happen.
If the weight comes down you can wait all day.
Well, let me give you another example.
Take a chunk of wood, you slide it there.
Sorry. It was not a piece of wood and
it didn't stop. See, this is why I don't do
demonstrations because even the simplest demonstration I have
done doesn't work. So, that's why some of us
become theoretical physicists. I did a demonstration where
some trolley is supposed to shoot a little marble.
All the marble had to do was come down and even that didn't
work, so I don't do demonstrations.
Well, let's try this. Here's an eraser,
and give it a push and it stops.
Take the movie of that. Play it backwards.
You will find suddenly, you know the table heats up a
little bit due to friction. The table can cool down and the
eraser can move backwards picking up speed [laughter].
Right. So, why did we laugh?
You have no reason to laugh right now because it doesn't
violate anything you have learned,
other than your daily experience, because in the
reverse process every atom, you know, every atomic thing on
the desk that started moving, is made to stop,
turn around, and move in the backwards
direction. And every collision between
atoms on the desk and atoms on the eraser will obey all the
laws of mechanics. For example,
if you have a planet going around the Sun,
if you stop the planet and you reverse the velocity,
it will go the opposite way and the opposite motion is in
complete accord with Newton's laws.
In fact, there may be somewhere a solar system where the planet
actually goes the other way. It doesn't violate anything.
So, in all the movies which are played backwards,
at the very microscopic level there's nothing that says it
cannot happen, and yet it does not happen.
Here's another thing that does not happen.
I take a box and I put some gas molecules on one side and
there's a partition holding them in that side.
Then I remove this partition and I wait a little bit;
then I know the gas fills up. That's like perfume leaking out
of the bottle. Now, take a movie of these
molecules. When they go from here to here,
and play the movie backwards--What'll happen when
you play the movie backwards? Let's say this guy was moving
like this the instant we played the movie back.
Well, in the backwards movie it started going the other way and
slowly the whole thing will untangle itself and end up here.
That also will not happen. We all know that if you go to a
room and you release some gas from one side,
it can fill the room but it's not ever going to come back,
and yet it doesn't violate any laws because if I made a movie
and played it backwards, in every microscopic
interaction between molecules, there'll be no violation of any
of the laws of nature. But that doesn't happen.
Here's another thing that doesn't happen.
I take a chunk of some hot copper and I take a chunk of
some cold copper. Hot, cold.
I just put them on top of each other, come back in an hour,
it's become, say, lukewarm. That's fine.
But now maybe, if I wait long enough,
the lukewarm will automatically spontaneously separate into hot
and cold. That doesn't seem to happen.
The heat simply seems to flow from hot to cold but never back
from cold to hot. But flowing from cold to hot
will not violate the Law of Conservation of Energy.
As long as the same number of calories go from cold to hot,
it doesn't violate anything. And that doesn't seem to happen.
So, this can go on and on and on with some large number of
things that are allowed but don't happen.
So, we cannot explain them with any known laws of physics so we
elevate that to a new law. The new law will say these
things cannot happen. But you don't have a law that
says these things cannot happen, right?
That's not a good enough law. You've got to say what things
cannot happen. I've got a list a mile long.
Hot and cold, when mixed, will never separate
back into hot and cold. A gas, that genie that got out
of the bottle, won't go back,
there's gas that came out won't go back.
There's a whole lot of processes which we call
irreversible. They never seem to happen
backwards. They only happen forwards,
and we cannot list all of them. So, you wonder if there's any
law--see, anytime you cannot explain something from prior
things, you elevate that to a law.
But you like an economical law. You would not like a law that
says the following million things cannot happen.
And it's really amazing that there is one law,
a single law, that not only is qualitative,
but is quantitative, that tells you exactly when
some things can happen and when some things cannot happen,
and according to the one single law, all the things that we say
don't happen won't happen. Will be forbidden by the law,
and that's a Second law of Thermodynamics,
but what form does it take? Well, we will see it introduces
a certain quantity called entropy, about which let's say
right now we don't know anything,
and the law will simply say the entropy of the universe will
always increase. You don't have to write it down
now. I'm going to define the concept.
It'll arise naturally from more thinking.
The one law that entropy will always increase,
in the universe as a whole, is enough to forbid any of
these forbidden processes. In other words,
if the paddle spun the other way and the weight went up and
the water cooled down, you can show the entropy of the
universe would have actually gone down.
That's why that's not allowed. If you drop a piece of egg,
it comes and splatters all over the floor.
If it rejoined and rose back to your hand, it won't violate any
other principle other than the law of entropy,
because the shattered egg has more entropy than the egg that
comes back to your hand. And if hot and cold mix to
become lukewarm, you can show the entropy goes
up this way, but if you went the other way,
entropy will go down and that's now allowed.
So, one great law will take care of all these things.
So, you've got to ask yourself, "What is the one great law and
how do we get to it?" Well, it turns out this great
law was discovered by an engineer called Carnot.
So, here is another thing. People don't wake up and say,
"I'm going to discover a great law."
People just go out and do their business and all you need is
have enough sense to realize when you've stumbled on
something really great. So, Carnot had a very practical
question. Carnot was working on engines.
At the time they were working on something like the steam
engine. So, a steam engine--You know
what happens in a steam engine. You take some coal,
you set fire to it, then you boil some water,
it turns into steam, then the steam pushes the
pistons and the wheels turn and the train goes forward and
that's your steam engine. What happens in a steam engine
we can schematically as follows. There is a hot reservoir
T_1 that's the furnace of your steam engine.
Then, you have an engine in which you've got some steam.
You've got some water that's going to be boiled and it's
going to boil only because you take in some heat.
Here's my engine, it's got some substance inside,
it could be a gasoline engine or a steam engine,
we don't care what happens. Some amount of heat
Q_1 is taken from a hot reservoir and some
amount of work is delivered. Now, here is an engine which,
as drawn here, doesn't violate any of the laws
of physics, provided Q_1 = W,
then energy is conserved. In practice you find that's not
how engines work. The engine also rejects some
amount of heat Q_2 to a lower
temperature T_2.
The lower temperature, if you look at a steam
engine--if you've gone--Have you guys seen a steam engine?
Okay, if you really go see a steam engine you will find
there's a lot of hot steam coming out of the back or in a
car, you've got exhaust gases coming
out. That's the hot gas emitted at
atmospheric temperature. Inside the engine is a very hot
place. So, the car emits some heat at
a lower temperature. The steam engine emits some
steam at a lower temperature. So, the work that you can get
is really Q_1 - Q_2 is what you
can get. At least in this engine,
you get Q_1 and -Q_2.
So, we define a quantity called efficiency η.
It's called the efficiency. It's what you get divided by
what you pay for. You pay for burning the coal
and you get the work out. The work is,
of course, the thing that makes the engine move forward.
Turns into kinetic energy of the locomotive.
That, then, you can see is Q_1 -
Q_2. What you get divided by what
you pay for, you can write it as 1 minus Q_2
over Q_1. So, in every engine,
some heat is taken in. Some of it is converted to work
and some of it is rejected. To the extent heat is rejected,
the efficiency is less than 1, because Q_2
minus--over Q_1 and is subtracted from 1.
So, you can ask yourself why not build an engine in which you
just don't reject any heat downstairs?
Why not take all the heat and convert it to work? So, Carnot gave a great
argument on the most efficient engine you can build.
In other words, there is an upper limit to the
efficiency and it's not 1. So, what is it and how are you
going to find out? Well, he's going to find out
what it is, but he also needs a postulate.
Carnot's postulate is the old version of the Second Law of
Thermodynamics. He didn't talk about entropy.
You can show in the end they're all equivalent,
but Carnot's law, which is his own version of the
Second Law, which says the following.
So, Law 2 according to Carnot. Let me just draw a figure;
then, we'll say T_1 is not
allowed. Carnot says you cannot build an
engine whose sole effect at the end of the day is to transfer
some heat from a cold body to a hot body,
whose sole effect--that's very, very important.
You cannot have a process whose sole effect is this.
I mean, I don't feel like writing what this is.
It's what the picture says. All right, it doesn't take a
genius to have the heat flow the other way.
You understand? If you want heat to flow from a
hot body to a cold body, that's trivial.
Just connect them with a metal rod and it'll flow.
The sole effect of that is transfer of heat from hot to
cold. That's the way of the natural
order of things, but Carnot is saying you can
never build a set of gears and wheels and teeth and whatever
you like so that at the end of the day,
all that has happened is heat has flowed from cold to hot.
That is going to be taken as a postulate, and people are
willing to take that as a reasonable postulate,
and we want to see what we can get out of that postulate.
It turns out it is that postulate that's going to
control the efficiency of heat engines, put a bound on the
efficiency of heat engines. So, don't deduce this.
This is a law. Now, you might say,
well, this is how the law began, and the law assumed more
and more sophisticated forms, so Carnot was not thinking
entropy. That came later,
but his postulate was, please grant me this,
which seems to be a fact in nature.
Then from that let me talk about heat engines and show you
that there is an upper limit to the efficiency of an engine.
In particular, the upper limit is not 1.
It's clear to everybody that the limit cannot be more than 1.
If it's more than 1, you are taking some amount of
heat and giving even more amount of work delivered.
That violates the First Law of Thermodynamics.
Conservation of energy. But the fact that you must
necessarily reject some heat, which you cannot use in this
experiment, is the content of Carnot's result.
So, here is the engine that Carnot built.
This is called a Carnot engine. By the way, look,
don't rely on my graphics. I mean, you have to go look at
just about any book you can find on the market.
It'll talk about Carnot engines. So, I'm here to merely tell you
the words and not the pictures. So, Carnot's engine is like
this. Take two isothermals,
T_1 and T_2.
You remember the adiabatic curves are much steeper.
Draw two adiabatic curves like that.
Then take a gas from A to B to C to
D and back to A. That was the Carnot engine. So, let's see what we are doing.
In the process AB, take the piston with the gas in
it; slowly lift the grains of sand
so it expands to a volume B at the same
temperature. Then having reached B,
you thermally isolate the gas. It's not on top of a reservoir
of temperature T_1.
It's completely isolated. Take out even more grains of
sand. Now, it's expanding without any
energy coming in. It cools down to
T_2. This is the point C.
Now, you put the grains of sand back, but it's still on a hot
reservoir at T_2.
I mean, the cold reservoir of T_2.
Put the grains of sand back, temperature is locked in at
T_2 until you come here.
Then, you isolate the gas and put more grains of sand back
until you come back to A. Obviously, such a path exists
because of the way isothermals and adiabatics are in the
PV plane. That's the Carnot cycle,
and the important part of the Carnot cycle is that it's
reversible. It's reversible because at
every stage, when you took a grain of sand out you can put it
back and you can run it backwards.
So, it can also run backwards. The Carnot engine looks like
this. Here is the engine,
it takes some heat Q_1 from a
reservoir of T_1 that
delivers some work W; it rejects some heat
Q_2, of temperature
T_2. This is the model of a steam
engine. Suppose you run the Carnot
engine backwards. By the way, the work done by
the Carnot engine is this shaded region here and it absorbs heat
in the stage AB. It doesn't have any heat
absorption in the state BC because it's
adiabatic. In CD,
it rejects some heat and DA again,
there's no heat transfer. So, heat comes in here,
Q_1, heat is rejected here,
Q_2. If you run the Carnot engine
backwards, let me show you what it looks like.
Everything is backwards in a Carnot engine if you run it the
other way. Some work is put in here,
some heat Q_2 is taken from the lower
reservoir and the heat Q_1 is
delivered to the upper reservoir.
You know what that gadget is. That's a refrigerator.
So, in a refrigerator, that's what happens.
You might say hey, you violated the Second Law.
You have transferred heat from a cold to hot body,
but it was not the only thing that happened,
because some compressor somewhere did some work.
That's certainly allowed. What is not allowed is to have
it flow from cold to hot with nothing involved.
The second thing to notice is the gas, after doing its thing,
comes back to the starting point.
That means you can do it over and over and over again.
In one cycle, the gas has some heat coming
in. Q_1 heat
rejected Q_2, some work done,
W, and it comes back to where it
is and you can go on doing it many, many times.
That's why it's a useful engine. If the substance returns to the
starting point, having done some work in return
for some heat, you can do it over and over
again. And Carnot's question was,
"What's the efficiency of this engine?"
We're going to calculate that now.
So, the efficiency of the engine is η,
is Q_1 - Q_2 over
Q_1. Or 1 - Q_2
over Q_1. Well, Q_1 and
Q_2 are pretty easy to calculate.
Let me tell you why. In the process AB,
remember isothermal expansion? The heat input is the work
done, because change in internal energy is zero.
And the work done, you remember,
is nRT log V_B over
V_A. Similarly,
Q_2, as a positive number,
is nRT_2 log V_c over
V_D. Remember, Q_2
here is defined as a positive number, even though it's the
heat rejected, so if I want to get a positive
number, I write the log of the initial over V final,
so I get a--sorry, V_c over
V_D. So, the efficiency now becomes
1 - Q_2 over Q_1,
becomes 1 minus T_2 over
T_1 log V_c over
V_B divided by log--I'm sorry,
V_c over V_D divided by
a log V_B over V_A. Again, if you cannot follow my
script here, you may have to go back and get the subscript
right. But don't worry too much about
all the details. What I want you to know is the
heat input here and the heat rejected here are all a familiar
quantities from what I did earlier.
But now, here is the punch line; η, I claim is 1 -
T_2 over T_1.
In other words, I'm saying forget this ratio of
logarithms. It is just 1.
I will show that to you, of course.
It's not going to be written without proof,
but the final answer for efficiency of this engine is 1 -
T_2 over T_1.
So, we need to show why V_c/V
_D is the same as V_B/V
_A. Well, I will derive that here,
then for a couple of minutes I'll talk about it,
then we'll come back and resume the discussion. I'm going to show you that
V_c/V _D is the same as
V_B/V _A. I'm going to show that to you.
Then you agree, if I take the logs then,
the logs will cancel. How do I know this?
Well, go back and use the fact that that's an isothermal
process. That is an isothermal process.
That is an adiabatic process and that's an adiabatic process
[drawing graph on board]. This is temperature
T_1, this is temperature
T_2, this is A,
B, C, D.
Now, somewhere in your notes is a formula that in an adiabatic
process, volume A times temperature 1 to some strange
power, I don't care what the power
is--it's not going to matter. It's the same.
Sorry, volume B times temperature to some power is the
same as volume C, the same, temperature
T_2 to some power. Because in the process of
adiabatic, we have seen PV to the γ is
constant, but I told you can also write
it as volume times temperature to some power is a constant.
So, B and C are correlated this way.
Likewise, A and D are connected by an adiabatic
process so you can be sure that V_A,
T_A, that happens to be
T_1 , raised to some power,
is the same as V_DT
_2 raised to same power. We don't care what the power is.
It's probably C_v over
R or R over C_v.
I'm not interested in that. In fact, if you go back to this
top right-hand corner you can see it is volume times T
to the power of C_v over
R. But it's not important.
The fact that B and C lie on an adiabatic
curve means that the volume at B and the temperature at
B, which happens to be
T_1, and the volume of C and the
temperature of C, which happens to be
T_2, are connected by this result.
Once I write it down for this adiabatic and this adiabatic,
now divide that by that, and that by that,
and the point is that T_1,
to whatever power it is, cancels top and bottom,
top and bottom, and I get the result I wanted,
which is V_B over V_A,
is V_c over V_D.
Once you put that in, you get the final result
η is 1 - T_2 over
T_1. Now, there are some things you
should know all the time and something you should know some
of the time, and something you should've
seen at least once. The derivation of the result
you should've seen at least once in your life.
Whether you carry it in your head or not, I'm not interested
in. But what have I really done?
I've taken an ideal gas and taken it over a cycle and I
found out that in that cycle, the ideal gas does function
like a very primitive heat engine because it takes some
heat Q_1, rejects some heat
Q_2, and does some work given by the
shaded area. And the ratio of work done to
heat absorbed in the end happens to be 1 - T_2/T
_1. It does not depend on the gas.
It depends only on the upper temperature and the lower
temperature. Now, the question you can ask
is, "Why are you computing the efficiency of this really
primitive and stupid engine when we're interested in real engines
that GM is going to make or some steamship is going to make or
some engineer is going to make?" We will show next time that
this efficiency is a theoretical maximum.
I will show you next time that no engine can beat the Carnot
engine. No engine can be as good.
It can be as good but cannot be better than the Carnot engine,
and that's what makes the result important.
It's a lot like a relativity example, but to show you why
time slows down in a moving frame I took a very stupid clock
where the clock ticks because light beam goes up and down.
That's not your idea of a clock, but in that clock,
if you can see why it slows down,
you know that every clock has to slow down because all clocks
in a moving rocket must run at the same rate.
Otherwise, you can compare the two clocks and find out you are
moving. Likewise, if you take a very
primitive engine, but you can show that that
engine is the most efficient engine,
you are done because that's going to put an upper limit on
the efficiency. So, I'll come back and I'll
tell you next time why no engine can beat this engine and why
hidden in this result is the notion of entropy.
