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PROFESSOR NELSON: Well, last
time we started in on a
discussion of entropy, a new
topic, and I started out by
writing out some somewhat
verbose written descriptions
that had been formulated that
indicate certain limitations
about things like the efficiency
of a heat engine,
and what can be done reversibly
and so forth.
And these verbal descriptions
lead to some pictures that I
put up and I'll put up again
about how you might try to
accomplish something like run an
engine or move heat from a
colder to a warmer body.
And what some of the
limitations are
on things like that.
I'll show those again, but what
I want to do mostly today
is try to put a mathematical
statement of the second law in
place that corresponds to
the verbal statements
that we saw last time.
So, just to review what we saw
before, we looked at a heat
engine where there is some hot
reservoir at some temperature,
T1, and it's connected to an
engine running in a cycle.
We could, write the work that we
generate that comes out as
negative w.
And then there's a cold
reservoir at some lower
temperature T2.
So the way I've got things
written here, q1 is positive.
Heat's flowing from the hot
reservoir to the engine that
runs in a cycle.
Work is coming out.
A positive amount of work is
being generated and is coming
out, so minus w is positive, and
this minus q2 is positive,
that is heat is also flowing
into a cold reservoir.
And what we saw last time is
that this was necessary to
make things work, this part
of it in particular.
You couldn't just run something
successfully in a
cycle and get work out of it,
using the heat from the hot
reservoir, without also
converting some of the heat
that came in to heat that would
flow into a cold reservoir.
All the heat couldn't
get successfully
converted into work.
That would be desirable,
but it's not possible.
And, similarly, if we try to run
things backward and build
a refrigerator, so now we have
our cold reservoir, and we're
going to remove heat from
it, and pump it
up into a hot reservoir.
And to do this, we saw that
you have to put work in,
right, you can't just remove
heat from a cold reservoir,
move it up to a hot reservoir,
without doing some work to
accomplish that.
So here, q2 is greater
than zero.
The work is greater than zero,
and minus q1 is greater than
zero, that is heat is
flowing this way.
So these are just the pictures
that we saw last time.
And then we had some statements
of the second law
of thermodynamics that I won't
re-write up here, but, for
example, Clausius gave us the
statement that it's impossible
for a system to operate in a
cycle the takes heat from a
cold reservoir, transfers it
to a hot reservoir, that is
acts like a refrigerator,
without at some, at the same
time, converting some
work into heat.
Work has to come in to
make that happen.
And we had the similar statement
by Kelvin about the
heat engine that required that
some heat gets dumped into a
cold reservoir in the process
of converting the heat from
the hot reservoir into work.
Fine.
Now what I want to do is put up
a specific example of the
cycle that can be undertaken
inside here in an engine, and
we can just calculate from what
you've already seen of
thermodynamics.
What happens to the
thermodynamics parameters, and
see the results in terms of the
parameters including entropy.
So let's try that.
So, the engine that I'm going
to illustrate is called a
Carnot engine.
And the cycle it's going to
undertake is called a Carnot
cycle, and it works the
following way: we're going to
do pressure volume work.
So this is something that,
by now, you're
pretty familiar with.
So we're going to start at one,
go to two and this is
going to be in isotherm at
temperature T1, and all the
paths here are going
to be reversible.
So that's our first step.
Then we're going to have
an adiabatic expansion.
So this is an isothermal
expansion.
Here comes an adiabatic
expansion, to point three.
So at this point the temperature
will change.
Then we're going to have another
isothermal step, a
compression to some
point four.
So this is an isotherm at some
different temperature T2, a
cooler temperature, because
this was an expansion.
We know in an adiabatic
expansion the
system's going to cool.
And now we're going to have
another adiabatic step, an
adiabatic compression.
And that's it.
That's going to take us back
to our starting point.
So that's our cycle.
Of course there are lots of ways
we can execute the cycle,
but this is a simple one, and
these are steps that we're all
familiar with at this point.
So this is the picture, and
this is a cycle that
undertakes it.
So let's just look step by
step at what happens.
So going from one to two, it's
an isothermal expansion a T1,
so delta u is q1, we'll
call it, plus w1,
for the first step.
Going from two to three, that's
an adiabatic expansion,
so q is equal to zero
in that step.
And delta u is equal to what
we'll call w1 prime.
So we'll use w1 and w1 prime to
describe the work involved.
The work input during
the expansion steps.
Step three to four isothermal
compression, delta u is going
to be q2 plus w2.
And finally, we're going
to go back to one.
We're going to close the cycle
with another adiabatic step.
q is zero.
Delta u is w2 prime.
Now, the total amount of the
work that we can get out is
just given by the area
inside this curve.
We'll call it capital W. It's
just minus the sum of all
these things, because of course
these are just defined
as the work done by the
environment to the system.
So it's minus w1 plus w1 prime
plus w2 plus w2 prime.
The heat input is just q1, and
we'll define that as capital
Q. And I just want to write
those because what I really
want to get at is what's the
efficiency of the whole thing?
And in practical terms, we can
define the efficiency as the
ratio of the heat in
to the work out.
We would like that to be
as high as possible.
So it's just capital W over
capital Q, which is to say
it's minus all that
stuff over q1.
Now the first law is going to
hold in all of these steps,
and we're going around
in a cycle.
So what does that tell us about
a state function like u?
What's delta u going around
the whole thing?
I want a chorus in
the answer here.
What's delta u?
STUDENT: Zero.
PROFESSOR NELSON: Excellent.
MIT students, yes!
OK, so the first law, we know
that the, going around the
cycle the integral of
delta u is zero.
And that has to equal
q plus w, summed
up for all the steps.
Which is to say q1 plus q2 is
equal to minus w1 plus w1
prime plus w2 plus w2 prime.
So that means we can rewrite
this efficiency.
We can replace this sum by just
the some of q1 plus q2.
So sufficiency is q1
plus q2 over q1.
Or we can write it as
one plus q2 over q1.
Now that already tells us what
we know, which is that the
efficiency is going
to be something
less than zero, right?
Because we've got one and then
we're adding q2 over q1 to it,
but we've got negative q2
is a positive number.
Heat is flowing this way.
So q2, the heat in this way
is negative. q2 over q1 is
negative. q2 of course
is positive.
That is heat flowing from the
hot reservoir to the engine.
So that efficiency is something
less than one, and
we'd like to figure
out what that is.
So let's just state that.
Well, now let's go back to each
of our individual steps
and look, based on what we know
about how to evaluate the
thermodynamic changes that take
place here, let's look at
each one of the steps and
see what happens.
So from one to two
it's isothermal.
And now we're going to specify,
we're going to do a
Carnot cycle for an ideal gas.
It's an ideal gas.
It's an isothermal change.
What's delta u?
You know, it's like lots of
courses and all sorts of
things that you're seeing,
they're always the same.
What delta u?
STUDENT: Zero.
PROFESSOR NELSON?
Right.
Delta u is zero, and it's
also equal to this.
So that says the q1 is
the opposite of w1.
It's an isothermal expansion,
so dw is just negative p dV.
So it's, this is just
the integral from
one to two of p dV.
And it's an ideal gas,
isothermal, right.
RT over V. Were going to make it
for a mole of gas, so it's
R times T1, and then we'll have
dV over V. So that's just
the log of V2 over V1.
Let's have one mole,
n equals one.
Okay two going to three that's
this, it's adiabatic, right.
So delta u is just equal to the
work but we also know what
happens because the
temperature is
changing from T1 to T2.
So delta you is just Cv
times T2 minus T1.
du is Cv dT.
It's an ideal gas, and that's
equal to w1 prime.
Now, this is a reversible
adiabatic path, so there's a
relationship that I'm sure
you'll remember.
Namely, T2 over T1 is
equal to V2 over V3.
Don't be confused by
the subscripts.
We're talking about quantities
both starting at two
and going to three.
So it's V2 and V3.
They're different volumes.
The temperature though at two is
the same as it was at one,
so it's still a temp
at T1, and this
temperatures is at T2.
Anyway T2 over T1 is equal
to V2 over V3 to the
power gamma minus one.
So we're going to kind of store
that for the moment.
Now going from three to four
right, so we have another
isothermal process for an ideal
gas, so I won't try to
make you sing again so soon.
Delta u is zero.
And so just like here, now q2
is minus w2, that's integral
going from three to four p dV.
So it's R T2, right, now we're
at a lower temperature times
the log of V4 over V3.
So that's our work in
this path and heat.
And finally going from
four back to one.
So we've already seen
that q is zero.
So we know that delta u is just
Cv times T1 minus T2.
Now we're going from
T2 up to T1.
And this is equal to w2 prime.
And now, just like we had
before, again we've got a
reversible adiabatic path, so
T1 over T2 has to equal V4
over V1, to the gamma
minus one, okay.
All right, now, if this is the
case, of course, this is just
the inverse of this.
So this just must be the
inverse of this.
We can combine these two to see
that V4 over V1 must equal
V3 over V2.
So now we have a relationship
between the ratios of these
volumes that are reached during
these adiabatic paths.
Now, let's just look
at our efficiency.
So we saw that efficiency
is one plus q2 over q1.
Let's just use our expressions
that we've
found for q2 and q1.
We're going to have
q2 over q1.
R is going to cancel.
We're going to have the ratio of
temperatures and the ratio
of these logs.
So it's T2 over T1 times the
log of V4 over V3, over the
log of V2 over V1, but we just
arrived at this relationship
between these volumes.
And this of course tells us that
V4 over V3 is V1 over V2.
Right.
I'm just inverting these.
So here it is.
Here's the V4 over
V3, oops, sorry.
V2 over V1, this is equal
to the inverse of this.
So the ratio of the logs
is just minus one.
So this is just, sorry, I forgot
about the one here.
One plus this, but this is the
same as one minus T2 over T1.
That's our expression
for the efficiency.
All right, isn't
that terrific?
Now, we have an expression.
We can figure out
the efficiency.
All we need to know is
the temperatures.
What's the temperature
of the hot reservoir?
What's the temperature of
the cold reservoir?
We're done.
That's the efficiency.
As we expected, it's less than
one, and of course if we build
an engine, we want it to be as
high as possible, as close to
one as possible.
What that means is we want to
run the hot reservoir as hot
as possible and the cold
reservoir as cold as possible.
In principle, this value, this
efficiency, can approach 1 as
the low temperature approaches
absolute zero.
If this were to be an absolute
zero Kelvin, then we could we
can have something,
wait a minute.
Sorry, it's T2.
As T2 goes to zero, the cold
reservoir, then this goes to
zero and our efficiency
approaches one.
So that would be the
best we can do.
And that basically make
sense, right?
The whole thing is being powered
by sending heat from
the hot to the cold reservoir.
The colder that cold reservoir
is, the hotter that hot
reservoir is, the better
off we are.
The more work we can
get out of it.
The closer the efficiency
will get to one.
So in some sense, the first law
would suggest you can sort
of break even.
That is, it gives us the
relationship between energy
and work and heat.
It might suggest that
you could convert
all the work to heat.
The second law says that really,
you can't do that.
The only way you could possibly
contemplate it is to
be working at absolute
zero Kelvin.
Guess what the third law
is going to tell us?
You can't get to zero Kelvin.
We'll see that shortly.
But this is those closest you
could come at least by trying
to do that to having your
efficiency approach one.
Now, we've seen that q2
over q1 is equal to
negative T2 over T1.
So let me just rewrite
that as q1 over T1 is
minus q2 over T2.
Or in other words, q1 over T1
plus q2 over T2 is zero.
But q1 and q2, each one of those
is just the integrated
amount of heat that was
transferred going along a
reversible constant
temperature path.
Which means that q1 over T1,
that's this delta S thing that
we saw before.
And what we can say about this
is it's saying that if we go
around in a cycle and look at
dq reversible over T it's
zero, because that's what
these quantities are.
Now, of course, we can run the
engine backward and build a
refrigerator, and if you've got
your lecture notes from
last period, your 8-9, well,
they're labeled 8-9 lecture
notes, I made an attempt to
define something which was a
little bit misguided.
And so instead of defining
efficiency the way you've got
it written there, I'm going to
define what's called something
different for a refrigerator
which is called the
coefficient of performance.
So it's the following: so we can
define the coefficient of
performance written as eta
as q2 over minus w.
In other words, you know, it's
how much heat can we pull out
of the cold reservoir, for
whatever amount of work that
we're going to put in
in order to do that?
Well of course, we'd like that
to be as big as possible.
But of course, this is just
q2 over minus q1 plus q2.
Just to be clear, so it's heat
extracted over the work in.
So let me just rewrite that as,
I just want to divide by
q1 everywhere.
So I'm going to write this as q2
over q1 over minus one plus
q2 over q1.
I want to do that because we
know that q2 over q1 is
negative T2 over T1.
So this is negative T2 over
T1 over negative one
minus T2 over T1.
I can cancel those, and
then I'm going to
multiply through by T1.
So this is just going to be T2
over T1 minus T2, that's our
coefficient of performance.
So it's a measure of how well
we can do to run the
refrigerator.
So you know, what the second law
is doing, in words, it's
putting these restrictions on
how well or how effectively we
can convert heat into work in
the case of the engine, or
work into heat extracted in the
case of a refrigerator.
And it follows qualitatively
from just your ordinary
observations about the direction
in which things go
spontaneously, right.
You know the heat isn't going to
flow from a cold body to a
hot body without putting some
work in to make that happen.
And so, we'll be able of follow
that further and see
really how to determine the
direction of spontaneity for a
whole set of processes, really
in principle for any
processes, went analyzed
properly.
So the first step to doing
that is I want to just
generalize our results so
far for a Carnot cycle.
You might think well
okay, but this is a
pretty specialized case.
We've formulated one particular
kind of engine, and
seen how we can analyze what
it does, come up with
relations that seem of value
for efficiency and other
quantities.
How general is it?
Well, let's take a look.
So, what I want to do is make a
new engine which really just
consists of two engines,
side by side.
One is our Carnot engine as
we've seen it, and the other
is just any other reversible
engine.
So generalize our Carnot
engine results.
So let's take our hot reservoir
and draw it bigger.
We know it has to big anyway,
since we can extract heat from
it without changing
the temperature.
And on this side, we're going
to write out an engine, and
we're going to say this is a
Carnot engine. so on the side
of it we have q1 prime.
We're going to have w prime.
And we're going to have q2
prime, and we're going to draw
the arrows in the positive
direction in these cases.
Or in the defined directions
I should say.
Here is our colder reservoir.
OK, so here is just an engine
like what we've already seen,
and I'm going to specify that
this is a Carnot engine which
is to say all the results
that we just derived
hold for this case.
And side by side, we're going
to run another engine.
So it's got q2, and it's
got a cycle w.
And it's got q1.
So this is some other
engine that runs
using reversible processes.
So I can define the efficiency
of each one of them.
The efficiency for the one on
the left is minus w over q1,
The efficiency prime for the
one on the right is minus w
prime over q1 prime.
Now I want to just assume that
in some way they're different.
So let's assume that epsilon
prime is greater than epsilon.
So in other words, this engine
is running less efficiently
than my Carnot engine.
It's also reversible.
And now since it's reversible,
we can run
it forward or backward.
So we can run this one backward
and we can use the
work that comes out as the
input, well sorry, use this
work that comes out of the one
of the right to run this one
backward, which is to
say we'll move heat
from cold to hot.
So use work out of right-hand
side to
run left-hand backward.
Why not?
We need work to come
in, we might as
well get it from here.
So, the total work that we can
get out must be zero, out of
the whole sum of them.
After all, we're taking the
output work that we get from
the right, and using it
all to drive the left.
So that means that minus
w prime must equal w.
And w is greater than zero. that
is in this one we have
the environment doing
work on the system.
OK, now we've assumed that
epsilon prime is
greater than epsilon.
So we can just write out what
those are, minus w prime over
q1 prime is greater than
minus w over q1.
But we know that minus w prime
is the same thing as w.
So w over q1 prime is greater
than minus w over q1.
Which is the same thing just
to be a little bit maybe
pedantic because w over minus
q1, and the only reason I'm
writing that is to illustrate
that this says that q1 must be
less than q1 prime.
So q1 is less than zero.
Remember this one's running
backward, we're pumping heat
up. q1 prime is greater
than zero, it's
running as an engine.
It's taking heat from
the hot reservoir.
Well, that's interesting.
That says that minus q1 prime
plus q1 is greater than zero.
Well, it's a pretty interesting
result.
There's no work being done on,
out of the whole thing.
This is providing work that's
being used in here, but if you
take the whole outside of the
surroundings and this whole
thing is the system, no net
work, these things cancel each
other, and yet heat's
going up.
How did that happen?
Well it happened because we
clearly must have a faulty
assumption underlying what
we've just done.
This can't possibly be true.
This is impossible.
So what this says is the
efficiency of any reversible
engine has to be one
minus T2 over T1.
It's not a result that's
specific to the one
cycle that we put up.
And you know, you could have a
reversible engine with lots
and lots of steps, but you could
always break them down
into some sequence of adiabatic
and isothermal steps.
So you know your cycle, you
know, you could have a whole
complicated sequence on a p v
diagram of steps going back.
As long as it's reversible, you
know what the efficiency
has to be, and in principle, you
could break it down into a
bunch of steps that you
could formulate as
isothermal and adiabatic.
They might have to be formulated
as very small
steps, in order to do
that, but you could.
What this says too, is this
result that we found, right,
now of course that's our
integral dS is zero.
It's general.
Entropy S is a state function,
generally.
Now remember, we went through
before how it's a state
function but to calculate
it, you'd need to find a
reversible path, along which
you can figure this out.
But the fact is it's a state
function, in a general way.
Now, if we go back to our Carnot
cycle which is a set of
reversible paths, it's useful
to compare this to what
happens in an irreversible
case.
In a real engine, of course,
you can approach the
reversible limit.
Every step won't be perfectly
reversible.
And of course it's not hard
to see what happens.
Let's just take our reversible
engine and
modify it a little bit.
Let's imagine that instead
of all the steps being
reversible, let's just put
in one irreversible step.
Let's do this.
Instead of this reversible
isothermal step, Let's make it
an irreversible isothermal
step.
We can have a different
isothermal step.
Of course in the reversible
case, you're always pushing
against an external pressure,
which is essentially equal to
the internal pressure.
Let's not do that.
Let's imagine maybe instead we
just immediately dropped the
pressure and let the
system expand
against the lower pressure.
Now we know we're going
to get less work out
of it in that case.
You've seen that before.
And the work in this case
is the area inside here.
The work in this step is just
the area under this curve.
So in some way we're going to
have a difference here between
the irreversible case and
the reversible one.
So if we do that, then what I
want to do is just see what
happens to dq over T, so in our
irreversible engine, one
to two, what we know for sure
is that minus w in the
irreversible step, that's the
work out, extracted in that
step, is going to be smaller
than minus w in
the reversible case.
So w irreversible is bigger
than w reversible.
And of course, in either case,
delta u is q plus w, so it's q
irreversible plus w
irreversible, but of course, u
being a state function it's
the same in either case.
So it's the same as q reversible
plus w reversible.
And we've just seen that this
is bigger than this, but the
sums are equal.
So this has to be
less than this.
q irreversible is less
than q reversible.
In other words, and irreversible
isothermal
expansion takes less heat from
the hot reservoir than a
reversible one does.
It makes sense, right, because
you know we got less work out
and delta u is the same right,
so it must be that less heat
got transferred.
So the expansion against lower
pressure draws less heat from
the hot reservoir right.
So now let's look at the
efficiency of our
irreversible engine.
So it's one plus q2.
Now q2 was in this step, and
we're going to leave that
reversible, right, but
q1 is irreversible.
And this has got to be less than
one plus q2 reversible
over q1 reversible, which is
to say it's less than our
efficiency in the
reversible case.
Why?
This is smaller than this, but
it's a negative number.
So this negative number has a
bigger magnitude than this
negative number.
We're subtracting them from one
and they're less than one,
so this is bigger than this.
And all we know about this is
that it's really for some
irreversible reversible step.
All we really needed to know
about it is that this is
smaller right.
So it's going to be the case for
any irreversible engine.
And that's the point.
So it's that the irreversible
efficiency is lower than the
reversible one.
But, of course, since we saw
that this is smaller than it
was in the reversible case,
we can also write that dq
irreversible over T is less
than dq reversible over T.
Which is to say if we go
around a cycle for dq
irreversible over T, that's
less than zero.
It's only in the reversible case
that dq over T around a
cycle is equal to zero.
So to write that in a general
way, it's actually formulated
by Clausius.
Going around in a cycle the
integral of dq over T is less
than or equal to zero.
Never greater.
OK, what we'll see shortly is
that this will allow us to see
that for an isolated system the
entropy never decreases.
It only can go up.
And in fact any spontaneous
process will make it go up.
Only in the case of reversible
processes.
Of course, then you can see
that this will be zero.
Anything else, which is to say
any spontaneous process, it'll
be less than zero.
We'll see that next time, and
then we'll generalize in a
broader sense to look at the
direction in which spontaneous
processes go.
