Okay, what I did last time was
describe to you a certain heat engine called the Carnot engine.
So, let me refresh your memory on what that was.
The whole engine consisted of a gas at some pressure and some
volume. And the gas was subject to a
cycle, and the cycle went like this.
You start at some point A, you expand at a
certain temperature T_1,
isothermally, until you come to point
B. So, the way you should imagine
that is that this gas is placed on a reservoir whose temperature
is frozen, I mean stuck at T_1.
Reservoirs never change the temperature.
And what you do is you lift one grain after another of sand,
so you let the gas expand, but you don't let it cool down.
The minute it tries to cool some heat will go from below to
maintain the temperature and start at that volume and end at
that volume. During this process,
some amount of heat Q_1 enters the
gas. Next thing you do,
you let the gas expand some more, but now you let it expand
adiabatically. Adiabatic means you put the
whole thing in a heat insulating material, so no heat can go in
or out. Now, you can see if it expands,
it's going to have to pay for it with its own internal energy,
so the temperature will drop. So, you wait till it drops to a
lower temperature T_2,
then you go backwards like this;
that's when you start compressing the gas by putting
some grains of sand back, but this time the gas is kept
in a lower temperature on top of a temperature reservoir
T_2, and you drop more sand on it to
compress it, that is this part. During this portion,
heat will actually be rejected by the gas.
Why is heat rejected? Because normally,
when you compress a gas, you're doing work on it,
the energy should go up, the energy goes up,
the temperature should go up; but you're not letting the
temperature go up, you're forcing it to have the
temperature T_2.
So, it will reject heat into this heat bath.
And finally, you come back to the starting
point by further compression, but adiabatic,
and adiabatic means it's insulated from the outside
world; no heat flow.
So, there are four parts. There are two things you have
to notice about the Carnot engine.
In fact, let me draw a schematic of the Carnot engine.
This is the standard way people draw that.
The engine takes some heat Q_1 from the
furnace at T_1, it has some
Q_2 rejected at the exhaust, which is at
T_2 and it delivers a certain amount of
work which, by the Law of Conservation of
Energy, has to be Q_1 -
Q_2. Do you understand that?
The work done by the gas equal to that.
And who paid for it? It is the reservoir,
upper reservoir furnished the heat, some heat was rejected
downstairs and the difference between the two is the work. Now, we call this a cyclic
process because at the end of the day, the gases come back to
where it is. That's very important,
if you want to make an engine, it's not enough to convert;
the whole purpose of heat engines was to burn something
and get some work out of it. But you don't want it to be a
one shot thing. You want to be able to do it
over and over and over again and you can do that with this engine
because after the cycle, it's back to where it started
and you can do this any number of times.
Now, we define an efficiency for this engine,
which is what you get divided by what you pay for.
What you get from the engine is of course, the work W and
what you pay for is the coal that you burn which is
Q_1. Now, writing W as
Q_1 - Q_2 you can write
this as 1 minus Q_2 over
Q_1. This is true for any engine.
Any engine, even if it's not a Carnot engine;
with the Law of Conservation of Energy you can write something
like this. Anyway, let me say with the
Carnot engine I got this as the efficiency.
Now, why is the efficiency not 100%?
That's because Q_2 is not
zero. You can say why are we building
a stupid engine that rejects some heat?
Why not just use all of it in some other fashion,
and that's what we're going to talk about,
but assuming that the engine operates between two
temperatures, the furnace temperature and the
ambient atmospheric temperature, you can calculate the
efficiency for this engine by actually calculating
Q_2 and Q_1.
That's a simple problem of integrating the work done and
going from here to here. Do you remember that?
1 - nRT_1 log V_B over
V_A and downstairs you have NR
T_2 log V_C over
V_D. Again, you've got to remember
why that is true. When you go from A to
B there is no change in internal energy because
temperature is the same. Therefore, the work done and
the heat import are equal numerically, so
Q_2 is the work done and Q_1 is
the work done there. So, that's what I got.
Then, I did a little bit of calculation to show you by using
the fact that B and C are on an adiabatic
curve, A and D are on an
adiabatic curve, that these logs actually
cancel, these guys of course,
cancel to give the final result 1 minus T_2
over T_1. That's the bottom line.
I don't care what else you remember but this is something I
need for today's discussion. Efficiency of the Carnot engine
is 1 minus T_2 over T_1. Okay now, you can say why I'd
be interested in this very primitive engine containing a
cylinder and gas and so on. Okay, this efficiency of this
stupid engine, I'm sure people can build a
better engine. So, here is Carnot's claim.
Mr. Carnot say,s "No engine can
beat my engine." Okay that's his claim.
By "beat my engine" he means, no engine can be more efficient
than my engine. And he's going to demonstrate
that. I'm going to demonstrate that
for you. You don't get something for
nothing. It's based on the following
postulate, which I told you last time.
I'm just now going to write it; I will repeat it.
We are going to postulate, as the Second Law of
Thermodynamics, that it's impossible to find a
process, the sole result of which is to
transfer some heat from a cold body to a hot body.
We all know you can transfer heat from a cold body to a hot
body in a refrigerator. You take heat out of the
freezer and dump it into the room;
no one says that's wrong. But that's not the sole effect,
because you get an electric bill at the end of the month,
because there's a compressor doing a lot of work.
The claim is--not--no other agency in the end should be
affected in any fashion. If at the end of the day all
you had was heat flow up hill. That's not allowed;
that's a postulate and you have to grant him that postulate
which we accept to be phenomenologically valid.
But taking that postulate, Carnot will now show you that
no engine can beat his engine. The key to the whole Carnot
engine is that it is a reversible engine.
By reversible engine I mean on every step in the Carnot cycle,
let's say going from here to here, I can also go backwards.
I can put a grain of sand and compress a piston,
or I can take a grain it'll go back to where it was.
You're never far from equilibrium and if you can go
one way you can go the other way.
That also means the Carnot engine, starting at the point
A, can go backwards to D then C then to
B and then to A. If the Carnot engine went
backwards it would look like this: it will take heat
Q_1--I'm sorry it will take heat
Q_2, somebody will give it work
W and it'll reject Q_1 at this
temperature. This is the Carnot refrigerator
and that's the Carnot engine. The Carnot engine and the
refrigerator are the same machine;
you just can make it run one-way or the opposite way.
That's what we're going to use. So, let me give you a
demonstration that's for--as far as I can tell is good enough for
our present purpose, that you cannot beat the Carnot
engine. So, let us say a Carnot engine
takes some heat, say 100 calories,
delivers 20 calories in work, rejects 80 calories is
operating between some two temperatures
T_1 and T_2.
This is the Carnot. I am taking a particular
example where the efficiency is what, one-fifth.
But don't worry about the fact that, yeah, so it's one-fifth,
so one minus T_2 over
T_1 should be one-fifth.
Now, if you say you have a better engine,
you have a better efficiency. What you really mean is that
your engine can take a 100 calories and deliver more than
20 calories, maybe even deliver 40 calories
and reject only 60 calories, but it's also operating between
the same two temperatures. So, this is your engine.
U for--no, Y for your engine,
okay; your engine is better than
Carnot's engine, okay that's your claim.
You're now going to get shot down.
So, how am I going to shoot this down?
I am going to run the Carnot engine backwards first thing,
then I am going to get a Carnot engine which is twice as big as
this Carnot engine, it's not more efficient,
it has just got twice as much gas.
If you take a Carnot engine and do the following things,
run it backwards and make it twice as big.
What will that Carnot engine do? It will look like this;
it will take up 160 calories from downstairs,
it will want 40 calories input and it'll dump 200 calories
upstairs. So, this is a two-times Carnot;
let me put a star saying it's Carnot run backwards.
See, it doesn't take any special genius to make an engine
bigger. It won't be more efficient;
it'll have the same efficiency. But if you can build something
with one piston, I can make the same design with
a bigger piston or smaller piston.
I'm just saying I want it to be twice as big for the following
simple reason. Your engine's giving out 40
units of work; my refrigerator needs 40 to run.
So, what we do is we directly take the output from your engine
and feed it to my refrigerator. Okay, your heat engine makes
work, my refrigerator needs work and I've scaled mine so that its
appetite matches your output; that's all I've done.
Now, let's draw a box around these guys, don't look under the
hood and see what you've got. At the end of a full cycle,
when everything is done, all the gases,
all the pistons, everything has come back to
where it starts. But, no need to plug this
gadget into the wall. You don't have to plug this
into anything because this refrigerator is getting the
power from this heat engine, so it doesn't need external
power. I look at the lower reservoir;
I see 100 calories leaving. You see that 60 coming down and
160 going up? I look at the upper reservoir;
I see 200 calories out, in and 100 out,
so basically, the combined gadgets,
yours and mine are equal to this gadget.
It simply transfers heat from a cold body to hot body with no
other changes anywhere in the universe, and that is not
allowed. Therefore, you cannot have an
engine more efficient that the Carnot engine.
The logic is pretty simple. These numbers picked are
representative, but you can take any number as
long as your engine does better than mine,
instead of 40 calories, would have made 30 calories
needed. If it produced 30,
I'll get an engine which is one-half times as bigger than
the standard engine; run it backwards,
your 30 will feed my engine and you will find once again heat is
flowing from a cold body to hot body with no other changes
anywhere in the universe, that is forbidden by the
postulate. So, this is how Carnot's
engine, even though it's a very primitive engine,
is the standard for all engines.
No engine can be better than the Carnot engine.
Now, the key to the Carnot engine is that it's a reversible
engine. If I had more time I could show
you that all reversible engines operating between two
temperatures will have the same efficiency,
namely that of the Carnot engine.
Okay, now what has the Carnot engine got to do with what I
started saying earlier? You remember I started saying
earlier there are certain things in our world that seem allowed
but don't seem to happen. Like you know,
drop an egg it doesn't come back in your hand.
You let the weight go down and turn the paddle.
The paddle doesn't turn backwards and lift the weight.
I said many things are forbidden and there is a law
that's going to forbid all of them, so I am coming to that
law. But it all started with a
person asking a very practical question, "How good can I make
engines?" And what Carnot is telling you
is the best efficiency any engine can have is this.
Therefore, even today hundreds of years after Carnot,
no company in America or Japan, China, anywhere can build an
internal combustion engine for example, whose efficiency's
better than this one. It's a theoretical maximum.
In reality, efficiency will be less than this because most
engines have a loss; there is heat leaking,
there is friction that's good for nothing, so in the end
efficiency will be always lower than this.
Okay now, let me leave the practical domain and go to more
theoretical issues that follow from this.
What I want you to notice if the following:
for a Carnot engine Q_1 over
Q_2 turned out to be T_1 over
T_2. Right, I did that for you here.
Therefore, I am going to write it as follows:
Q_1 over T_1 minus
Q_2 over T_2 equal to
zero. That's just a simple rewrite.
But I am going to write this in another rotation.
My cycle had four parts; remember, I did this,
then I did this, then I did this,
then I did that. In this segment I had
Q_1 over T_1.
In this segment there was no Q_1 or
Q_2, there was no Q,
because this is the adiabatic process.
In this segment, I had minus
Q_2 over T_2 and the
last segment I had zero and whole thing is zero.
In other words, what I'm telling you is the
following. At every stage,
if you looked at the heat absorbed by the system,
in stage i, divided by the temperature of
stage i and you added all of them you get zero.
My process had four stages. Stage 1 is here,
stage 2 is adiabatic, state 3 is this,
and stage 4 is zero again. In this summation
Q_i is defined to be ∆Q_i is
the heat input in stage i in the cycle,
or in the process. So, we define
Q_2 to be positive, even though it was
rejected by the engine, but in this summation,
the agreement I make is ∆Q is positive if the
heat comes into the system, ∆Q is negative if heat
leaves the system. So, why is that important?
Now, this is the heart of the whole entropy concept.
Remember I told you there is nothing called the heat in a
system. You cannot look at a gas and
say that's the amount of heat in the system.
Why? Because if you take a point
there and say there's some amount of heat in the system,
if you go through some kind of cycle and come back to the same
point, then since you come back to the same point internal
energy doesn't change, this is the work done,
therefore, the work done is equal to the heat and it is not
zero. So, in this example what
would've happened is you'd added heat Q to the system.
So, if there is some notion of how much heat is there in the
beginning, you got that plus the Q that you added and yet
you are back to where you are. Therefore, you cannot define
something and say that's the amount of Q in the system
because you're able to add Q to the system and bring
it back to where it is. What happened,
of course, is you added Q and you did some work.
But the point is you cannot say the Q here is so and so.
But you can say the energy here is so and so.
Because if you come back to the same point P times
V is equal to 3 is equal to RT and the internal
energy of a gas is proportional to the T,
energy returns to the old value. But now I'm going to tell you
so listen very carefully. Energy is a state variable
because it comes to the starting value, if you come to the
starting point. It doesn't matter where you
wander off in the PV diagram.
Heat is not a state variable, there's nothing called a heat
at that point, because I am able to go on a
loop, change the value of Q or
add some Q to it, and I come to the same point.
So, there's no unique Q associated with that point.
But there is a new unique quantity S,
called entropy, which has a fixed value at a
given point and if you go for a little loop in the PV
diagram and you come to the same point,
S will return to the starting value.
So, who is this S? That S is defined by
saying the change in the entropy is the heat you add divided by
the temperature. In a tiny little process,
if you are at some temperature T, you add a little
amount of Q, keep track of all the changes,
and that change will be zero, as I showed you in this Carnot
cycle. You can show more generally
that if you take any path, not just the Carnot engine
bounded by adiabatic and isothermals,
but any path you take, you can show that the
∆Q added up is not zero,
but ∆Q/T, namely give a weighting factor
of 1/T to the heat you add,
then the positives and negatives cancel and give you
exactly zero. In other words,
Q_1 was not Q_2,
Q_1 is bigger than Q_2,
but Q_1/T _1 precisely
balances Q_2/T _2.
The heat absorbed at higher temperature if divided by
T_1, to compute the change in
entropy, then in the upper part of the
Carnot cycle here, and the lower part of the
Carnot here; the two cancel as far as the
entropy change is concerned. So, entropy is a new variable;
it's a mathematically discovered variable.
What we find is that if I postulate there's a variable
called entropy, the change in which,
in any process, is the heat transfer divided by
temperature, then that has the property that when you go around
on a loop you come back net zero change.
That means every point can be associated with a number you can
call entropy. Entropy is like a height.
Suppose you are walking around on some landscape,
each point is a height. You can walk here and there and
come back at every stage; you keep track of the change in
height if you come back to where you are, the change in height
will add up to zero. That's what I told you in the
case of a potential energy function.
It is just like a potential energy function.
Internal energy and the entropy are now state variables.
At every point the gas has a certain internal energy and an
entropy. They deserve to be called state
variables because when you go on a loop and come back they return
to the starting values. Now, you have no idea what this
quantity stands for; you agree it's bizarre.
Work done, we understand. Heat absorbed and heat rejected
we understand. What is ∆Q over
T? Why is dividing by T
make such a big difference? Why does it produce a new
variable? We can see it is true;
at least in the Carnot cycle you can verify in detail that
the change in the total of all the ∆Qs over T
is in fact zero. So, we'll develop up a feeling
for what it means, but historically this is what
happened. People realized,
hey there's another state variable.
We introduce this new variable, we have no idea what it means
but it is a state variable so we'd better take it very
seriously. So, I'm going to tell you what
it means. But first I want you to get
some practice calculating the entropy change for a couple of
processes. So, let us take for example,
one gram of water, let's say m grams of
some substance. It's got some specific heat
C. And I'll change the temperature
say from some T initial for some T final. I do that by keeping the
system--starting the system at T_i and I
should never be far from equilibrium.
That's one of the conditions on this.
In fact, you can say it should be near, always near
equilibrium. So, the system starts at
temperature T_i,
I put it on a heat bath, which is infinitesimally warmer
than this and I let it heat up at that point.
Then, I put it on another reservoir, slightly hotter than
this one. At all stages I want the system
to be almost near equilibrium. In the sense of calculus you
can make the difference as small as you like provided you do
enough number of times and slowly I raise this guy from
here to here. At every stage the system has a
well-defined temperature, has a well-defined heat input
and I wanted to add it all up. So, what's the change in
entropy? S final minus S
initial is equal to dQ over T,
summed over all the parts, I should write this as an
integral, mC ∆T over T. Do you see that, dQ is mC ∆T?
But I got to divide by T and that integral is done
between some initial and final temperature and we can do this
integral rather trivially in an mC log T final over
T initial. That is the increase in entropy
of some m grams of some substance of specific heat
C that's heated from T initial to T
final. We're just getting practice
calculating this. We still don't know what this
guy means. So, don't worry about that
right now. If it did not divide by
T, what are you calculating?
mC ∆T integrated is just mC times change in
temperature. That's the old calorimetric
problem you did. How many calories does it take
to raise the substance from initial to final temperature?
That you understand. When you divide by T,
something you don't understand, but anyway, let's make sure we
know how to calculate the increase in entropy when water
or something is heated from a lower to higher temperature.
If you cooled it from a higher to lower temperature,
you can use the same formula T_f over
T_i, but T_f will
now be smaller than T_i;
if you cooled it, if the log of a number less
than one is negative and the entropy changes it'll be
negative. So, let me do one more entropy
calculation that's going to be pretty important for us.
That entropy calculation is this.
Take a gas and watch it expand isothermally from some starting
point, at some fixed temperature T that goes from
V_1 to V_2.
What is the change in entropy when it goes from here to here? Again, the system must always
be in equilibrium or near equilibrium so I can plot it as
a point in the PV diagram,
so I slowly take grain after grain, do the whole thing I told
you, and find S_2 - S_1,
this is 2 and this is 1. That is equal to the sum of all
the heat transfers divided by temperature at every little step
of the way. But remember this is also the
same as P∆V over T.
Why? Because on an isothermal
∆U is zero, I repeated it many times but
you've got to know this, ∆Q is P∆V.
But P over T is nRT over V,
∆V. You also want to divide it by a
T. Do you understand P over
T is nR over V?
So, P = nRT over V, I'm just saying,
PV is nRT, so P over T is
NR over V. Now, dV over V
when you sum or integrate will give you nR log
V_2 over V_1. That is the change in entropy
of this gas when it went from volume V and went to
volume V_2, at a given temperature.
You don't have to use calculus to do this.
Does everybody understand why this could have been done a lot
easier? You don't need to do the
integral here. Because I showed you in
studying the heat engine that the heat transfer Q is
nRT log V_2 over V_1.
Since the whole process takes place at fixed temperature,
instead of dividing by T at every step,
you can just divide by T overall.
Just bring the T here; that's why this is a change in
entropy. In other words,
during the whole process you feel at one temperature then
integral of ∆Q over T is just 1 over T
times integral of ∆Q, which is the total heat
transfer. Anyway, this is the heat,
this is the change in entropy, S_2 -
S_1 is nR log V_2 over
V_1. Okay, so what have I done so
far? Let's collect our thoughts here.
I went to the Carnot engine today and reminded you what the
efficiency of the Carnot engine was.
Then, I showed you no engine can be better than the Carnot
engine. That's a separate story.
That has to do with how efficient things can be.
Then, a byproduct of the Carnot engine was this great
realization that if you go on a closed loop and at every stage
you add, you compute the heat absorbed,
but divide by the temperature at that point,
the sum of all those is zero. What that means is that,
there is a quantity S whose change,
if it is defined to be ∆Q over T,
has a property the total change in S is zero and you go
around a loop. That means at every point you
can associate an S. S is a property of the
point. Another example--so then,
I said let's get used to computing change in entropy.
I took one example of heating a substance of some mass and
specific heat C by a temperature dT and the
change in entropy was mC log TF over TI.
You got the log because of integral of dT over
T was logarithm. Then, I took an ideal gas,
I let it expand isothermally, I found the entropy change,
I got this answer. By the way, here's an
interesting exercise, I don't have time to do it,
but you can ask the following question.
If you tell me that every point there is a unique entropy,
then the entropy difference between 2 and 1 should be
independent of how I go from 1 to 2.
Do you understand? If you're walking on a
mountain, you take two points, they have a height difference,
and I can find the height difference by going on this path
keeping track of the change in height or any other path.
In the end, the height difference between two points is
the height difference. So, I could find the entropy
another way. Let me show you another way
that's easy for you guys to work out.
Come down like this and go to the right like this.
And find the entropy change. You'll get the same answer as I
got on this thing. Okay, it's too tempting for me
to just leave it here. Let me tell you how it works.
Call this intermediate point, give a subscript zero for all
its parameters. Then in this step,
when you come from here to here, the entropy change will
be--let's take one mole of a gas when I go from here to here.
Then for one mole of a gas the heat transfer dQ is
C_VdT and I divide by T and I do the
integral from T_1 to this
point T_0. That's the entropy change here.
Then in the horizontal part, since I'm going at constant
pressure, I go C_PdT over
T from T_0 up to back
to T_1, because this
T_1 and T_2 are the
same temperature. I have to add all these to get
the entropy change. Now it's a two-step process,
you come down because you're at constant volume and you're doing
something to the gas, dQ is
C_VdT. Horizontally,
you're at constant pressure, dQ is
C_PdT. That's the definition of
specific heat at constant volume, constant pressure.
But notice the following: C_P is equal
to C_V + R.
So, put C_V + R here and look what you
get, then you get C_V times
log T_0 over T_1 plus
log T_1 over T_0,
which we can write as the product here,
plus R log T_1 over
T_0. In other words,
this log is log of T_0 over
T_1, the next log is log of
T_1 over T_0,
log A plus log B is log of AB.
But when you do this look what happens here.
This all cancels, log of 1 is zero;
the total entropy change is R log T_1 over
T_0. But for a gas at constant
pressure, using PV = RT, this ratio of temperature is
also the ratio of the volumes. So this one--I'm sorry I should
write it more carefully. When I say T_1
I really meant the temperature at this point,
second point. So, that temperature,
or this temperature is also that volume divided by the
initial volume, which is V_1.
See, this is a confusing problem because
T_1 happens to be T_0.
The correct way to do this for me would be to write
T_0 over T_1 times
T_2 over T_0,
then realize the fact that T_2 and
T_1 are equal because I am on an isotherm.
That's what I should have done. So, I should cancel this factor
because T_2 is T_1,
but here T_2 over T_0 is
V_2 over V_0,
which is V_2 over V_1.
That's, of course, the entropy change I got in one
shot here. So, you can find entropy change
anyway you like. You usually pick the easiest
path. Alright, so we have now learned
how to find entropy change. We have no idea what it means.
It seems as if when you heat something entropy goes up;
when you cool something entropy goes down.
That's certainly correct, because ∆Q is positive
for heating and you divide by T,
and you make it a logarithm, but we know at every
infinitesimal portion we're adding positive numbers.
And when I cooled things it goes down.
Why not just call it temperature?
Why do you need this new concept called entropy?
So, that's what the rest of the lecture is about.
It's a very, very powerful and beautiful
concept so I wanted to explain it, make sure I get it right.
Here is now the result for all this hard work.
Remember I told you there are many, many, many phenomena that
seem forbidden in our world, and we're not quite sure what
law to invoke to prevent all of them from happening.
Do we want a law for each one that says if you drop a bottle
and it shatters it won't come back?
If you drop an egg it won't come back.
If you let hot and cold mix, they're not going to unmix.
Or if you let a gas trapped in half a room, escape to the whole
room, it'll never untrap and go back to half the room.
A lot of things happen one way but not the other and I said I
am looking for a mega law that will prevent all these things
from happening. Now, I'm ready to state that
law. This is the third law--I mean,
the Second Law of Thermodynamics.
Carnot said it one way, but I'm going to say it in a
way that's very, very general.
The Second Law of Thermodynamics says ∆S
for the universe is either zero or positive.
There you have it, that's the great law.
The law says, take any process,
if at the end of the process the entropy of the universe is
bigger than it was before, that will happen.
If the entropy of the universe is smaller than it was before,
it will not happen. Now, we have to make sure that
this law has anything whatsoever to do with all the other things
we have studied and I will show you;
this will forbid everything that should be forbidden and
allow everything that should be allowed.
So, let me start with the most obvious formulation.
Mr. Carnot's version of the Second
Law was that you cannot have a process in which some heat
Q goes from a hot body to a cold body.
I'm sorry this allowed, right, according to Carnot,
this is not allowed. Right, heat can flow downhill,
cannot flow up hill. Let's see the entropy change
the universe with the two cases. In this case,
with this guy, the change in entropy is the
following: the upper reservoir lost some amount Q,
so ∆Q is a negative number, add some temperature
T_1, the lower reservoir gained
Q at temperature T_2.
Now, what's the overall sign of this?
Think about it. T_1 is bigger
than T_2, so this negative number is
smaller than this positive number,
so the whole thing is positive. That means it's allowed;
it's okay. By the same token,
if you take this process when heat flows up hill,
then this reservoir loses some heat Q at temperature
T_2, the other one gains heat
Q at temperature T_1,
but this is clearly less than zero because this positive
number is smaller than this negative number.
So, there you have a very simple example where if heat
flows the wrong way, the entropy of the universe
goes up. And it's very simple,
it's only--it's because, it's true that the heat loss of
this guy is the heat gain of that guy,
that's the Law of the Conservation of Energy.
So, energy doesn't change, but entropy changes,
because for entropy it's not the heat transferred,
it's heat transferred divided by temperature.
Therefore, a heat loss at one temperature and a [equal]
heat gain at a lower temperature don't cancel when it
comes to entropy. In one case the entropy goes up
and is allowed; other case entropy goes down,
that's not allowed. Okay so let's try,
that's certainly one process that you know should not be
allowed, but it's forbidden by the Third Law,
by a computation of entropy. So, the reason when you put a
hot and cold body together, heat flows from the cold to the
hot--I'm sorry from the hot to the cold,
is because that's the way the entropy will go up. So, let's take one more example.
Suppose I have some mass of water at some temperature
T_1 and an equal mass of water at
temperature T_2.
One is hot and one is cold. I put them together,
what will happen? We believe it will then get
this big mass, all at some common temperature
T star which is T_1 +
T_2/2; this is just by symmetry.
It's equal mass, equal specific heat,
where will they meet? They will meet half way.
They will meet here. Energies, of course,
is conserved, and that's how we determine in
fact where they will meet. But look at the entropy.
What happens with entropy change?
With entropy change, you should imagine this water
being steadily heated by putting it in contact with a lot of
reservoir, so it's never far from
equilibrium and slowly bringing it to this point.
Then, the ∆S total will be M specific heat
is 1, then ∆T over T starting from
T_1 to T star for one thing and dT
over T from T_2 to T
star for the other one. Do you understand that?
They both meet at T star, upper limit is T
star, lower one is T_1 for one and
T_2 for the other,
so the change in entropy becomes M log T
star squared over T_1T
_2. Now, we have to ask,
okay you got an entropy change from start to finish,
but how do you know it's positive? Can anybody give some reason
why this has to be a positive without looking into properties
of logarithms and so on? Any of these in, what--yes?
Student: [inaudible] Professor Ramamurti
Shankar: Yeah, that's one way to prove that.
But I am saying, I'm going to prove it to you
that way. But can you think of a reason
why at every step of the process, when I brought this
down and when I pushed that up, the entropy change is always
positive in each step. So, let me explain why.
Imagine cooling this down a little bit, by putting,
taking some heat ∆Q out of this guy.
That's happening at some temperature here,
this one is gaining some ∆Q, but at a lower
temperature. Okay.
Therefore, the gain has got the same ∆Q,
but at a lower temperature, this is always at an upper
temperature. Throughout the process,
at every stage, this guy's hotter than this
guy. So, every calorie of heat it
gains at a higher temperature, every calorie it loses is
divided by higher temperature, every calorie this gained is
divided by lower temperature. At every increment,
at every infinitesimal step you can see the total entropy change
is positive. Now, to prove it,
we will do what he just said. If you want to prove this is
positive, you want to show that the fellow inside the logarithm
is bigger than 1, so I'm asking is
T_1 plus T_2^(2) over 4
bigger than T_1T _2?
Or I'm saying is T_1 plus
T_2^(2) bigger than 4T_1T
_2? If you rearrange this,
you can show the left-hand side becomes T_1 -
T_2^(2); that's of course,
a positive number because this will be T_1^(2) +
T_2^(2) + 2T_1T
_2. When I bring this guy to the
other side, it'll become minus 2T_1T
_2 and so it's clearly positive.
So, when hot and cold meet and create lukewarm,
entropy of the universe has gone up.
It follows, therefore, if lukewarm spontaneously
separated into hot and cold, the entropy would go down and
that's why that doesn't happen. That's why if you leave a jar
of water at one temperature, it doesn't spontaneously
separate into a part on top which is cold and a part on
bottom, which is hot.
Such a separation would not violate anything.
It would not violate the Law of Conservation of Energy,
but it will violate the law that entropy has to always go
up. So, one way is allowed another
way is not allowed. So, you see over and over
again, that anything that's allowed is in the direction of
increasing entropy; anything that is forbidden is
in the direction of decreasing entropy.
But who is this entropy? What does it mean?
Well, that we have not understood at all from any of
these calculations. Because dQ over T
doesn't speak to us the way dU or dQ or
PdV does. So, that last part of what I'm
going to do is going to explain to you what is the microscopic
basis of entropy. Why is the entropy going up?
Why is--why do we understand the tendency of things to go the
direction of increasing entropy? All I've shown you now is that
it's a mystical quantity called entropy, circumstantially you
find every time something is forbidden,
it's because had it taken place entropy would have gone down.
Do you remember the examples? Heat flowing from hot to cold
entropy goes up, allowed.
Flowing from cold to hot, entropy goes down,
not allowed. Hot and cold mix and become
lukewarm; allowed, entropy goes up.
Luke warm separates into hot and cold;
not allowed, entropy goes down. So, there's definitely a
correlation. And yet we don't know what it
means. So, we're going to talk about
what it means. For that, I'm going to consider
the following process. I take a gas and I put it in a
room in a box where there's a partition, half way.
The molecules are stuck on the left side, gas has reached
equilibrium and it's done what it can, which is to spread out
over this volume. Now, I suddenly remove the
partition, rip it out. So, let's follow this gas.
Initially, it is a point here. Then, there is a period when it
goes off the radar because I told you when you suddenly
remove the wall, the gas is not in a state of
equilibrium. It doesn't even have a
well-defined pressure. The minute you remove the wall,
pressure is something here, pressure is zero in the vacuum.
So, you got to wait a bit, so gas has gone off the radar.
We cannot talk about what's happening.
Then, if you wait long enough, I get a gas that looks like
this after some seconds, or whatever milliseconds.
That is again gas in equilibrium.
It's got a certain state; I can call it 2.
What is the entropy change now? That's what I want to ask.
What's the change in entropy here?
Because I know that this is allowed, that is forbidden,
so I want to ask did the entropy go up?
Now, how do we do the entropy calculation?
Here's the wrong way to do the calculation.
You go back and say ∆S at ∆Q over T,
but this whole box I forgot to mention--I'm sorry,
this whole box is thermally isolated.
It's not in contact with anything.
I just ripped out the partition in the middle;
that's it. So, you might say,
well, if it's thermally isolated, ∆Q = 0 so you
can divide by T, you can do what you want,
so this entropy change is zero. But that is a wrong argument. Can you think about why that's
not the way to do the entropy change?
Why that's the wrong analysis? First of all,
could that be the right answer? Yes.
Student: It can't because then that would mean
that going backwards is not a valid process.
Professor Ramamurti Shankar: No,
that is correct, but I'm saying this computation
of entropy. The one line calculation I did,
namely there is no heat inflow, the ∆Q = 0,
so ∆Q over T summed up is also zero.
That's not how you do the entropy change. Is there any condition I made
on computing? Yes.
Student: Well, uhm… there's a gas
[inaudible] Professor Ramamurti
Shankar: Wait in fact, here's the interesting thing.
What's going to be the temperature difference between
before and after? Which is going to be hotter or
cooler? Any views on this?
Yep. Student: [inaudible]
Professor Ramamurti Shankar: Ah,
well, you can say it is cooler because it was insulated and
expanded, but you cool down because you
expand against an external pressure, right?
If, when--If I were to move this piston here,
there is no pressure pushing this gas back.
Do you understand that? Doesn't do any work,
doesn't take any heat. So, what does that mean?
In the end when you settle down, what can you conclude?
No heat input, no work done,
so what does that mean in terms of temperature?
Student: Same. Professor Ramamurti
Shankar: Same. Because the internal energy
cannot change because ∆Q = 0,
work done is zero. So, in fact, this is a surprise.
This gas when it expands will be at the same temperature.
You guys are thinking of the computer air,
where if you let it expand it cools down because that is
expanding against the atmospheric pressure.
This gas is expanding into a complete vacuum.
It's got no piston to push against, nothing from outside
pushing, so these two points are the same temperature. So, the correct answer,
maybe you didn't catch on, but I will tell you the correct
answer is: There is an entropy increase in this problem,
because the final state definitely has a well-defined
entropy you can calculate, because it's an equilibrium
state. Initial one has a well-defined
entropy, in between stages here, are not on the PV
diagram because they were not equilibrium states.
So in this experiment, the way to calculate the
entropy change is not to do ∆Q over T as it
happened in this experiment. What you really want to say is
my gas was here in the beginning, my gas is there in
the end, both are states of equilibrium,
both have well-defined temperature and I can find the
entropy change in going from here to here for any process I
want. As long as that process keeps
the system near equilibrium, so I can follow dot by dot
where I'm moving and add the ∆Q over Ts.
The correct rule for entropy change is ∆Q/T computed
on a path in which the system never strays from equilibrium.
Now, your system strayed from equilibrium all the time.
But we're not interested in how it got from here to here.
We are saying what's the entropy here and what's the
entropy here. Since this at the same
temperature, I know one way to go from there to there is to
follow the isotherm because that'll connect the two points
in this particular problem. Then I know the entropy change
is S_2 - S_1 is nR log
V_2 over V_1. So, here is the subtle point I
want to explain to you. This is worth understanding.
A lot of people do all the problems and get everything
right, but don't appreciate this particular question.
The free expansion of a gas into a vacuum takes it from one
equilibrium state called 1 to another equilibrium state called
2. In the actual expansion,
the system went through a stage which cannot be even shown in
the PV diagram, they were not equilibrium
states, they didn't have well defined pressure,
didn't have well defined temperature.
How do you find a temperature when half the box is empty?
Doesn't have it, you cannot it T
temperature of the gas. So, what do you mean by
∆Q over T, there is no T.
Only when it settles down, there is a T.
Every settled down equilibrium state is a well-defined entropy.
What we're trying to do is to compute that difference.
The way to compute the difference is to forget about
what actually happened and instead do the following.
Take the gas like that and turn it this way, [90 degree
rotation] if you like.
Just for convenience, put a piston there at the
halfway point, put some weights on it so that
the pressure equals the pressure inside and slowly remove little
by little, and keeping it on a reservoir a
temperature T_1.
Which is the temperature at which all of this happens,
or at some temperature T.
As you remove the grain of sand, what happens is the gas
gets some energy, that's the ∆Q over
T, the reservoir loses some heat;
in fact, the reservoir loses the same amount of heat
∆Q at the same temperature T.
So in fact, what happens in this process is entropy of the
universe does not change at all. The entropy gain of the gas;
this is an entropy loss of the reservoir.
But I used this as a device for finding the entropy gain of the
gas in the other process. So, this is the part that is
absolutely central to this whole argument how entropy is
calculated. In a real spontaneous expansion
of a gas, where there was no reservoir, no nothing,
we just broke the partition and expanded, there is a change in
entropy. That's an irreversible process.
But to find the change in entropy in this problem,
since 1 and 2 have well-defined entropies,
we can find the change between 1 and 2 by going to a different
experiment in which it was brought from 1 to 2,
not in this crazy irreversible way, but by putting it in
contact with the heat bath at that temperature T,
removing grain after grain, never straying far from
equilibrium, that's when we find this change.
So, you replace the dotted line, where you cannot compute
anything, by a solid line where you can.
But since the change in entropy between 1 and 2 doesn't depend
on how you got there, that's the answer.
But this process, where you can actually see it
is the one in which ∆S of the universe is actually at
zero. Why?
Because you are putting in contact with the reservoir at
the same temperature, so ∆Q over T
for the gas and ∆Q over T for the reservoir
exactly cancel because they're at the same temperature and
∆Q for one is minus ∆Q for the other.
So, don't confuse the two processes.
One is a reversible process in which a gas is slowly allowed to
expand from the initial state of the final stage by removing
grain after grain of some externally applied pressure in a
process where the entropy of the universe doesn't change,
gas and reservoir change by equal amounts.
But the punch line is I don't care what happens in reservoir,
I want to know how much the gas gained,
because that is the gain I need to find S_2 -
S_1 for this process.
So, this is how we always find entropy change.
In real life processes happen in a way in which entropy of the
universe goes up. Because take this box,
in the real experiment, remember there is no reservoir,
there's only the box of gas. It expanded,
its entropy went up, so entropy of the universe
actually went up. That's why it's allowed.
The entropy of the universe went up by an amount equal to
the increase in entropy of the gas when it goes from 1 to 2.
To find the increase in entropy of the gas when it goes from 1
to 2, I create a second process in which I can compute the
change by letting it go in a sequence of connected dots and
finding the change. So, the way you use to find the
change in entropy is not the way the entropy increase took place
in the real process. So, I don't think I want to
repeat it anymore, but it's an important point and
you should really think about that.
Alright, now the last point is, what does--Let's analyze this
final formula I got for entropy change for free expansion of a
gas. Let's take a simple case where
the gas expanded from some volume V to volume 2
times V. What is S_2 -
S_1? S_2 -
S_1 = nR log 2.
2 happens to be V_2/V
_1; the volume is doubled.
Let me re-write this as Nk log 2, because the number of
moles times R is the number of molecules times
k. Let me re-write that as log of
2 to the N times k. Now, I will tell you what all
these factors mean and that comes from the following.
Thermodynamics only told you the formula for the change in
the entropy of a gas, didn't tell you what the
entropy itself is. So, the formula for the entropy
of a gas, without talking about the change and also this formula
∆S is ∆Q over T,
doesn't even believe in atoms. You don't need atoms,
you don't need molecules, you don't need to know what
anything is made up of. In fact, it was discovered long
before atoms were proven to exist.
Boltzmann is the guy who gave the formula for what the entropy
of a gas is in terms of what the individual molecules are doing.
That's the punch line of what's called statistical mechanics,
which gives you the microscopic origin of thermodynamics.
Thermodynamics to this stage talk about fluids and solids and
gases and nothing there requires that everything be made up of
atoms, you don't have to know what
it's made of. But if you tell me it's made up
of atoms, you have a better idea of what entropy means and here
is Boltzmann's formula for entropy.
I am going to write it down. Boltzmann constant times log
Ω. I'll tell you what omega is in
a minute. This is such an important
formula it is written in Boltzmann's tombstone.
So, when physicists go to Vienna, we skip the orchestras
and everything else, we go to Boltzmann's tomb and
we read this formula once more with some passion.
That's a summary of a lifetime of work.
But it's a great formula and you will understand once you
know the formula what all this means.
So, let me tell you what omega means.
I am going to say it in words and then I am going to apply it
to this problem. If you took any gas,
let it be under certain conditions that macroscopically
has some pressure and some volume and so on.
So, here is the gas at one instant;
now we believe in atoms, here are these atoms scattered
all over the place. Suppose I wait a little bit,
what did the gas look like? It looks like that.
If you can see every atom, the atomic configuration has
changed. But macroscopically nothing has
changed. Instead of atom A being
here, maybe A has gone there and B has come
here. In particular,
if you divide the volume into some number of little cells that
you can count, on the average the gas' uniform
density, there are some number of atoms, per every little cube
you can form, a little later I move on and
somebody else moves in. Macroscopically,
it looks the same to me. Microscopically,
stuff is going on. Omega is the number of
different microscopic arrangements that agree with
what is seen macroscopically. When I say microscopic
arrangements, I really mean the following:
give every molecule a name, A, B,
C, D, whatever.
Then put A here and B here,
C here and D there;
that's one arrangement. Then permute them,
put D here and B here and C here and
A there, that's another arrangement.
To my naked eye they, all look the same,
but they all produce the same macroscopic effect,
but they're different microscopic arrangements.
And you're supposed to take the log of the number of these
arrangements multiply by this constant,
called Boltzmann's constant, to get the entropy of the
state. So, now let's apply it to this
gas here. Now, the number of microscopic
arrangements, if you really want to get down
to it, requires dividing the box into
tiny little cells, maybe 1 cubic millimeter and
saying how many atoms are in each region.
But we're going to do a crude calculation in which we'll only
say the following: that either a molecule is on
the left or is on the right. We are saying these are the
only two positions open to the gas;
left or right. Of course, in the left,
there is left corner top and left corner bottom,
but don't worry about it. Just say that only two things
it can do, it can be on the left or it can be on the right.
Now, let's ask ourselves if the gas looks like this. How many ways can it be in this
arrangement? As compared to how many ways it
can be in this arrangement? It is like the following:
molecules are moving randomly, a given molecule has got 50/50
chance of being on the left or being on the right.
You are demanding that all the N molecules on their own,
there's no partition now, on their own be on the
left-hand side. You can see it's like tossing a
coin N times and wanting heads every time.
Because you want every guy in a random process to end up picking
left. So, if you want them all to be
left, there's only one way to do that, and that's like saying I
tossed a million coins, I wanted all heads.
There's only one way to do that. So, that arrangement,
if you like, we will say looks like this,
left, left, left, left, left;
for everybody, that's the arrangement here.
Take another one where one molecule is over here,
and all the others are here on the left side.
That looks like left, left, left, right,
left, left, left. But that's one arrangement.
But have I done my complete homework here?
Are there more arrangements? I want you to think about it.
Student: [inaudible] arrangements Professor
Ramamurti Shankar: Pardon me.
Student: [inaudible] arrangements.
Professor Ramamurti Shankar: This is only one of
the arrangements for the molecule on the right.
There's another arrangement in which I can have the second
molecule, molecule Moe or Joe or Poe, whoever's here.
You can pick them in N ways.
Already when one is on the left and N - 1 on the right,
there are more ways in which that can happen.
So, your eye sees one on the right, N - 1 on the left.
There are N ways in which that can happen.
It's like saying if I toss coins, I want 1 head and 99
tails. Well, it turns out there are a
100 ways in which it can happen, because the one head I got
could have occurred in one of the hundred different turns.
Now, suppose I want 2 heads and 98 tails, then as you know the
way to do that is 100 times 99 divided by 1 times 2.
These are some combinatorics you learn in high school.
So, you find that if you plot number heads over the total
number, and ask for how many ways in which that can happen,
all heads can occur in only way, every coin has to be head,
1 head less, 1 tail and 99 heads can occur
in N ways, then it'll become more and more
and more. And you will find the 50/50
chance of half the number of heads and half the number of
tails has the biggest occurrence rate, because that's the most number
of ways to get 50/50 than anything else.
And 100 heads is as bad as 100 tails.
Again, there's only one way in which you can get all heads in
one way and which you can get all tails.
But as you start scrambling them there are more and more
ways. That means for the molecule,
if you have a 100 molecules, the number of states omega is
largest when they're equally distributed.
In fact, as the number molecules goes from 100 to 1000
to million to 10^(23), this function is so sharply
peaked that the approximate formula for omega,
when it's at the 50/50 thing, is in fact 2 to the N.
Namely, all the arrangements more or less belong to this
configuration. Therefore, if you go back to
the Boltzmann formula and, wait let me keep the formula
here, and look at S = K log
Ω, then you find S_1 is when
everybody is on the left, that is k log 1,
and S_2 is 50/50 mix, half on the left and
half on the right, that entropy is k log
2^(N),. Therefore, you can see that's
what I wrote down here, K log 2^(N). So, the way to understand
entropy is the following. If I give you a gas and ask,
"Will they ever like to be in the left half of the box on
their own?" Of course they won't.
But if you start them out that way, you force them to be on the
left they'll be on the left. The question is if you remove
the partition, what would they likely do?
Unlike coins, which once they land on the
head, they have to stay on the head, these molecules can move
around. The question is if they were
all left to begin with, how long can they possibly
last? If they start moving randomly,
they will always end up doing this because there are many,
many more ways to do this, than to do this. So, things go from one
arrangement, macroscopic to another one, left to their own
device, because the final state can be
realized in many, many more ways than the initial
state. Similarly, if you combine a hot
gas and cold gas--This is hot and this is cold,
with a thermally insulating partition, then all the hots are
on the left and all the colds are on the right.
If you remove the thermal insulation, but not even remove
it, just replace it by a heat-conducting thing,
eventually the temperatures would equalize.
So, what you find every time is, there's many,
many ways in which you can pack your atoms--if you put all the
hot on the left and all the cold on the right,
that can be achieved in fewer ways than in which you let them
go wherever they like. [He should have said,
"If you put all hot ones on one side and cold on the other,
there are fewer ways of doing that, than if you let them go
wherever they liked"] So,
what you find is entropy is the direct measure of how disordered
your system is. One technical measure of
disorder is to ask how many microscopic arrangements can
lead to what I see, and take the log of that
number. And what you find is this has a
very low entropy; this is a very high entropy.
Because this configuration--In other words, if you took a gas
in a whole room and asked will it ever go to this
configuration, there is a slight chance it
will. That chance is one part and
2^(N), where N is 10^(23),
so it's not something you should wait for,
but it can happen. So, the Second Law of
Thermodynamics is a statistical law.
Microscopically, it's perfectly allowed for a
gas, for suddenly all the gas in the whole room to come to where
I am. It's allowed but you don't hold
your breath, because that's not likely to happen.
The odds for that is again one part in 1 over 2 to some huge
number. On the other hand,
if in this part of the room we release some gas,
it'll very quickly spread out, because there are more ways to
do it. So, we understand completely
now why certain things occur and why they don't.
Because if you took hot water and cold water,
you are separating them to fast molecules and slow molecules,
as long as they're in different insulated containers,
that's the best they can do. But if you put them in contact,
so that they no longer have to be separated,
the question is will they remain separated.
It's like saying I have bunch of coins which are all left and
all right, but they're allowed to flip as a function time,
but they'll never remain that way.
Okay, so what'll happen is the cold molecules and hot will
mingle, and soon the container will have cold everywhere and
hot everywhere and you will get some intermediate temperature.
Or if you took two different dyes, you know water here,
on top of it you pour some red paint,
they'll be initially separated after a while they will mix,
because there's no reason for the hot--for the red molecules
to stay on the top forever; they would like to occupy the
whole box and so would the colorless ones,
and eventually it will become pink.
Because there's more ways to remain pink than to remain
separated. Likewise, if pink spontaneously
separated into red and colorless, that is very,
very improbable, because entropy for that state
would be lower. So, you have to understand the
Second Law of Thermodynamics is saying certain things occur
because if you go in that direction you can realize that
arrangement in more ways. And since microscopic motion is
random, it's like tossing coins to ask for N heads many
thousand times; it's like asking all the
molecules to be on the left side of the room, to ask for 50/50
head and tails, is like asking for molecules to
populate the room equally. Okay, so that's called the
arrow of time. But you've got to be careful
about one thing; the entropy of a part of the
universe can go down. It's just the entropy of the
whole universe will not go down. So, all of life is an example
of lowering entropy because, you know, the creation of life
or the creation of tomatoes out of mud is a highly organizing
process so the entropy there is really going down.
But if you kept track of the rest of the world,
you will find there's some corresponding increase of
entropy somewhere else. Or take your freezer,
you know, your refrigerator sucks heat out of your freezer,
∆Q over T for that is negative,
but somewhere there's bigger heat emitted outside from the
exhaust of the refrigerator. If you took care of all of
that, entropy of the universe will go up or remain same;
it really doesn't go down.
