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OK, so we ended up last time, we
talked about Joule-Thomson
expansion, which is an
irreversible expansion through
a nozzle, through
a porous plug,
constant enthalpy expansion.
And I want to make sure
everybody figured out that it
really was an irreversible
expansion.
Anybody have any questions on
that, because when we voted,
the majority of the class
thought it was reversible Yes.
STUDENT: Just a quick question
on some of the constraints,
like isothermal, isobaric,
isovolumetric expansion.
For isothermal expansion, that
means that delta u does not
change, but delta q is
equal to delta w?
PROFESSOR BAWENDI: So the
question was, for an
isothermal expansion, delta u
does not change, therefore, dq
is equal to dw.
The answer is that's true
only for an ideal gas.
For a real gas, that
won't be true.
Ideal gas only depends on the
temperature, the energy only
depends on the temperature.
Therefore, if it's isothermal,
the energy of the ideal gas
doesn't change.
For a real gas it depends on
more than the temperature
STUDENT: Are there any other
constraints similar to that
[INAUDIBLE].
PROFESSOR BAWENDI: So, for an
ideal gas, the isothermal is
the easy one because the
energy doesn't change.
If you have an isobaric you're
going to have to calculate
where the energy changes, and
that's a calculation that's
likely on the homework
and very like on an
exam as well, too.
In fact, we're going to do
some of that today, OK,
calculate delta u.
Any other questions?
Good question.
OK, so we ended up last time
showing that for an ideal gas,
that there was was a
relationship between the heat
capacity under constant
pressure, and this is the
molar heat capacity, that's why
we have the bar on top,
and the molar heat capacity
for constant volume path.
They're related through
the gas constant.
OK, this is only true for an
ideal gas, and we went through
that mathematically where
the, with a chain rule.
And I wanted to do it a
different way which is a
little bit convoluted, but it
introduces the idea of a
thermodynamic cycle, and it's
something that we're going to
use a lot in the class.
Because anytime you're working
with the function of state,
and you're trying to connect two
points together in a p-V
or a T-V diagram, then it
doesn't matter which way
you're connecting those two
points, as long as you're
dealing with state functions
it doesn't matter.
It only matters if you're
looking at the work or the
heat right.
And so sometimes, when you want
to calculate what the
energy is at the endpoint, the
path of the experiment that
you're doing may not be an
easy path to calculate.
You may have to find a different
path which is an
easier one to calculate, and it
doesn't matter which path
you take, because in the end
all you care is what the
energy is at the end.
so we're going to use this
concept of the path to go from
the initial point to
the end point.
As long as you're dealing with
state functions it doesn't
matter to get at the
same results here.
So we're going to start with a
mole of gas, at some pressure,
some volume, temperature and
some mole so V, doing it per
mole, and we're going to
do two paths here.
We're going to take a constant
volume path.
So V is constant and we're going
to change the pressure
to some, a little change in
pressure, V is constant here,
and then temperatures is
also going to change.
So this is a path, which is
either a temperature change,
with the pressure changes at
the same time, going up or
down, constant volume.
And since the temperature is
changing, clearly Cv is going
to have something
to do with this.
So constant volume path, where
the temperature is changing.
So somehow this is going
to allow us to get Cv.
Cv is going to be related
to this path.
And if I draw a diagram on a T-V
diagram of what I'm doing
here, so there's temperature
axis.
There's the volume axis here.
I'm starting with
some V1 here.
And this is going to
be here V plus dV.
So I'm starting right here.
This is my initial point.
And the path that I'm describing
then, let's assume
that we're raising the
temperature up is this path
right here.
OK, this is my constant
volume path going
from T to T plus dT.
OK, we'll call this 1 here.
All right, now I want to make a
path where out of that path,
C sub p is going to fall out.
So that's going to have to be
a constant pressure path.
Temperature is going to
change but pressure
is going to be constant.
So what I can do is I can take
my initial point here and
create a new, different path,
where, which is going to be a
constant pressure path.
So at the end of this path
the pressure is constant.
The volume is going to change
though, to some new volume dV
and the temperature is going to
change also and I'm going
to make it the same endpoint
temperature as the endpoint
temperature here.
And out of this path, since the
pressure is constant but
the temperature is changing, I'm
hoping that C sub p will
fall out of that path
of that calculation.
All right?
This should give us something
about C sub p.
So this will give us something
about C sub v. This will give
us something about C sub p.
Now these two endpoints
here are different.
They have a different pressure,
a different volume.
They do have the same
temperature though.
So the connection between
this endpoint here,
and that one here.
Same temperature.
It's going to be an isothermal
path that's
going to connect them.
So I'm going to connect them in
an isothermal way, and as,
since I'm dealing with an ideal
gas, and as we saw from
the first question here,
isothermal always means delta
u equals zero.
So I'm going to write that
immediately here, delta u
equal zero, because I know the
answer to that path here.
All right, so let's go again
what our paths are.
Path number 1 I'm
going straight
up in the V-T diagram.
Path number 2 on my diagram it's
a reversible, this path
number 2, it's a reversible
constant pressure path.
So I'm going to be keeping the
pressure constant, but my
volume is going to go up.
My temperature is
going to go up.
Basically I'm putting
heat in this system.
That's going to be number
2 right here.
It's going to be the same
temperature as before but the
volume is V plus dV now.
That's 2, and then, so I'm
heating up the system in this
path here, and then to connect
the 2 endpoints here, I'm
going to connect them through
a constant temperature path.
Basically I'm going to cool down
the system so that the --
no, I'm not going to cool down,
I'm going to compresses
the system back to the smaller
volume V1 and the constant
temperature conditions.
So that's path number 3 here.
And I'm only going to
worry about energy
in this case here.
Because I know energy doesn't
care about the path.
Energy only cares about where
I am on the diagram.
And I know energy is related to
Cv through Cv dT etcetera.
So if I worry about energy I
have a pretty good chance of
extracting out these heat
capacities, right, and I don't
have to worry about exactly
which path and I can really
mix things up.
So let's start the process.
So the first path then, the
first path, constant volume
constant V, so I'm going
to, again, let's
just worry about energy.
So du for this first path here,
du for path 1, and write
dq plus dw what we
usually write.
Now it's constant volume when
the volume is constant.
There's no work being done
so I can immediately
ignore this dw here.
Constant volume dq = Cv dT.
Now it's an ideal gas.
So in some sense, I did too
much work by writing this,
because I knew already
du = Cv dT.
I could have written that down
immediately, but I just went
through the process of writing
the first law, what's dw, what
dq, Cv dT, etcetera.
I'm just writing what we
already know basically.
Let's look at path number 2.
Path number 2.
OK, let's look at the dw, du
again. du for path number 2.
dq like the first law down, dq
plus dw, then -- what's dq?
Well it's a constant pressure
path. dq is related to the
temperature, the change in
temperature. dT through the
heat capacity, happens to be
constant pressure, so this is
exactly what we want.
Cp, I forgot to put my little
bar on top here because it's
per mole Cp dT that's
my dq here.
The path is constant pressure.
And then the dw.
Well there's a change now
in volume and pressure.
So I need, well the pressure
is constant, but there's a
change in volume.
Minus p dV is the change,
is the work, right?
So I can write that down minus
p dV for that path there.
OK, now but it's an ideal gas,
so I know something about the
relationship between dV
and dT, because I've
got dT here the dV.
I don't want to have so
many variables around.
I've got three variables, T, p
and V. I know I only need 2,
so I can relate dV to dp through
the ideal gas law.
P dV is equal to R dT.
pV = RT for 1 mole, so I just
take dV here. dT here because
the pressure is constant, so
dV is equal to R over p dT.
Then pressure is constant.
So I can instead of the
dV here, I can insert
this R over p dT.
The p's cancel out, and now
I have Cp dT minus R dT.
This is equal to
Cp minus R dT.
All right, let's complete
the cycle now, the
path number 3 here.
Path number 3 is a constant
temperature path, and I
already wrote the answer.
Constant temperature isothermal
delta u is zero.
There's no thinking involved.
It's one of these things like
I mentioned last time, if I
tell you delta H in the middle
of the night what you say, q
Cp for reversible process.
Same thing here. isothermal
process, what do you say?
Delta u is zero, right?
It should be completely
second nature.
So delta u is zero here.
Delta u sub 3 is, for
an ideal gas.
I gotta put that constraint
in there for an ideal gas.
Delta u is equal to zero.
So now I'm back at
the same place.
I'm back here right.
So path number 1 went from i,
let's call this path up here.
went to f, and this is how much
energy change there was.
The sum of path number 2 and
path number 3 get me to the
same place, so the energy change
by going through this
time path, this intermediate
point here back all the way to
final state should be the
same the red path.
So what I'm saying is that du
through path 1 should be equal
to du through path 2 plus
du going through path 3.
All right, because energy
doesn't care about the path.
It only cares about
the end points.
OK, so now let's plug what
is du through path 1?
It's Cp dT and the du through
path 2 is Cp minus R dT and
then du through path
3 is zero.
It's the isothermal process.
The dT is cancelled out and
voila I have my answer.
Cv is equal, oh Cv plus R is
equal to Cp whichever way you
want to do it, it's a
relationship that we had up
here that we wanted to prove.
Now, I wanted to go to through
this just to go through one
cycle quickly because we're
going to be doing these all
the time, and the importance
of the fact that the path
doesn't matter, and you can
always connect things together
in a way, whatever you want.
Whatever is the easiest.
If you want to calculate
a change in delta u.
Now if you look at my derivation
here, there's one
spot where I could have just
stopped and proven my point
without going through
the whole thing.
So if you look at path number 2
here, the constant pressure
path du is Cp minus R
dT, du is Cp minus
r dT for that path.
What is du for an ideal gas?
It's -- what else do we know
that du is always equal to for
an ideal gas?
Cv dT right.
So I could have written right
here immediately equals Cv dT,
and that was the end
of my derivation.
I could have done that.
It would've been easier, but I
wanted to go through the whole
path right?
OK, any questions?
Cool.
All right, so we're going
to be doing more
thermodynamics processes.
We're building up to entropy
and to engines,
Carnot cycles, etcetera.
And so we need to build our
repertoire of knowledge in
manipulating these cycles like
the one I drew on the board
that's hidden now You've
already seen the simple
process, which is the isothermal
process, and you've
seen how to functionalize, or
what the function is that
describes the isothermal
expansion.
Let's say we start from some
V1 and p1 here, so high
pressure, small volume and we
end up with a high volume low
pressure, under constant
temperature condition.
So there's some path,
isothermal, so T is constant.
And we know the functional form
for this path, it's the
ideal gas law, pV
= RT, pV = nRT.
T is constant, so the functional
form here is pV is
equal to a constant.
All right, or p is equal to a
constant divided by volume.
If you want to write a function
that describes this
line here, it's pressure as a
function of volume related to
each other with this constant.
So we know how to do this.
So this allows us to calculate
all sorts of things to get
these functional forms.
That's the isotherm here.
We also know how to calculate
the work.
If it's a reversible process,
we know the work is the area
under the curve. w reversible.
in this case here let's see
this is it's a pressure
is going down.
It's an expansion, so we're
doing work against the
environment, or to
the environment.
So work is leaving the system.
So minus w reversible is going
to be a positive number.
V1 to V2 and RT -- you've done
this already last time or a
couple of times ago and you end
up with minus nRT log p2
over p1, OK?
So minus w reversible, we're
doing work to the environment.
This should be a positive
number. p2 is less than p1, so
p2 over p1 is less than one, log
of something less than 1
is negative times negative.
So in fact this is indeed a
positive number so we didn't
make a mistake in our signs.
You always want to check your
signs at the and because it's
so easy to get the sign wrong,
and I was hoping that it was
right here because I
wasn't sure, but it
turned out to be right.
Always check your
sign at the end.
All right so we know
how to do this.
Now we are going to do the same
thing with a different
constraint.
Our constraint is going to be
an adiabatic expansion or
compression.
Adiabatic meaning there's no
heat involved, and we're going
to see how that differs from the
isothermal expansion and
compression.
OK, so this is the
experiment then.
We're going to take a
piston here which
is going to be insulated.
It's going to have some volume,
temperature to begin
with, and then we're going
to do something to it.
We're going to change the
pressure above, right now
there's a p external, which is
equal to p on the inside.
OK, we're going to do this
reversibly, which means we're
going to slowly change the
external pressure very, very
slightly at a time, so that at
every point we're basically in
equilibrium, until the pressure
reaches a new smaller
pressure p2.
So that's going to results in
an expansion where the new
volume new temperature new
pressure and an external
pressure, which is p2 which
is a smaller pressure.
It's really important that this,
that you remember that
this is the reversible case.
We're going to also look at the
irreversible case briefly.
So on the p-V diagram, then,
there's a V1 here a V2 here, a
p1 here a p2 here.
This V2 is going to be different
than this one here.
A priori we don't
know what it is.
We don't know if it's
bigger or smaller.
We kind of feel like it's going
to be different because
it's a different constraint.
This is isothermal.
This is adiabatic,
there's no heat.
So there's going to be a line
that's going to connect the
initial point to the final
point, and that line
mathematically is not going to
be the same as this one here.
And our job is to find out
what is the mathematical
description of this path, this
line in p-V's case that
connects these two point.
And again, I want to stress
this is a reversible path.
If it was non-reversible, I
would be allowed to put an
initial point and a final
point, but I wouldn't be
allowed to put a path between
them like this, connecting
them together.
Because if it's irreversible,
it's very likely that I don't
know what the pressure inside
the system is doing while this
is happening.
It could be very chaotic.
The pressure could not be, might
not even be constant
throughout the system.
It could have eddies and
all sorts of things.
So for an irreversible process,
I wouldn't really be
allowed to put a path there.
I can do that because it's
reversible, and I can get a
functional form out.
What we're going to find in
this process is that the
functional form from this is
that p is related to the
volume through a constant, but
instead of being V to the
first power as it was for the
isothermal, it's now a V to
some constant gamma.
What we're going to find is that
gamma is greater than 1.
And if gamma is greater than
1, and you're changing the
pressure to the same final
point, what we're going to
find then is the volume that
you go to is going to be a
smaller volume than for
the isothermal.
We're going to go through these
points, so don't worry
about getting at all
straight right now.
All right, so let's do this.
Let's try to show this.
Let's try to show that in fact
this is the right functional
form, that pV to the gamma is
equal to constant is the right
form that describes the path in
an adiabatic expansion or
compression.
So let's do the experiment.
Let's write everything
that we know down.
This is where, the way that we
should start every problem set
or ever problem.
It's writing everything we
know and what it means.
OK, so adiabatic.
All the words mean things
mathematically.
Adiabatic means dq
equals zero.
Let's write that down.
dq equals zero.
Reversible.
Reversible means that if I
looked at dw it's minus p
reversible.
It's minus p external dV, that's
what it is, but in the
reversible process p external
is equal to p.
That's the only time you can set
p external equal to p is
if it's reversible.
So this is the external
pressure.
This is the internal pressure
of the system.
You can only do this because
it's reversible.
What else do I know?
It's an ideal gas.
So I can write du is Cv dT
and pV is equal to RT.
OK.
So what else can I write?
Well from the first law, du is
equal to dq plus dw, and I
wrote down everything I knew
at the beginning here.
I wrote dq equals zero
I wrote what dw was.
So du is just minus p dV.
Now from the ideal gas,
I have another
expression for du, Cv dT.
So now I have two expressions
for du.
That's a good thing, because I
can them equal to each other.
So if you get these two guys
together you get Cv dT is
minus p dV.
So we're connecting temperature
and changes in
temperature and changes
in volume together.
OK, now we don't really
want this p here.
This is going to be a function,
I only need two
variables, and here I've got
three variables down.
So I know p is going to be a
function of T and V, and I
know how to relate p to
T and V because I know
the ideal gas law.
So instead of p, here I'm going
to put nRT over V. or RT
over V bar.
So now, this expression becomes
Cv dT over T is equal
to minus R dV over V. And I
also did a little bit of
rearrangement.
Yes?
STUDENT: [INAUDIBLE]
PROFESSOR BAWENDI: Well, let's
see. du is, for an ideal gas,
it's always equal to Cv dT.
Constant volume process tells
you that if you have a
constant volume, then
dq is Cv dT.
This is what the constant volume
path tells you, that
you can equate Cv dT with dq.
The ideal gas tells you that
this is always true here.
It's an ideal gas in adiabatic
expansion dq is equal to zero.
Adiabatic expansion means you
can't use this. dq is zero.
The that is adiabatic.
It's not constant volume.
You're allowed Cv comes out
here for this adiabatic
expansion, which is not a
constant volume only because
this is always true
for an ideal gas.
We didn't use a constraint
that V is constant.
Just use this as an
ideal gas here.
OK, so we've got dT
here dV here.
We want to get rid
of derivatives.
We want to integrate.
So let's take the integral of
both sides, going from the
initial point to the
final point.
We've got Cv integral from T1
to T2, dT over T is equal to
minus R from V1 to V2 dV over
V. OK, we do take that
integral, and that leads us to
Cv log of T2 over T1 is equal
to minus R log of V2 over V1,
and now we know how to
rearrange this, because the Cv
times a log of T2 over T1 is
the same thing as the
log of T2 over T1 to
the Cv power, right.
So by rearranging these
expression of the logs here,
this is the same thing is
writing log T2 over T1 to the
Cv power is equal to log of
V1 over V2 to the R power.
And I've reversed the fraction
here because of the
negative sign here.
OK, get rid of the logs, and
you get that T2 over T1 is
equal to V1 over V2 to the R
over Cv by taking this Cv and
bringing it on the
other side here.
OK, so now we've got the initial
and final points, a
relationship between the
temperature and volume for the
initial and final points.
Eventually that's not
what we want.
Eventually we want the same
relationship in the pressure
and volume.
We want a relationship in p-V
space, not in T-V space.
So we're going to have to
do something about that.
But first, it turns out that
now we have this R over Cv.
and remember, we have this
relationship between R Cv and
Cp here, which is going
to be interesting.
So let's get rid of R in
this expression here.
So R over Cv, R is
Cp minus Cv.
We proved that first thing this
morning, divided by Cv so
this is Cp over Cv minus 1 and
because I know the answer of
this whole thing here, I'm going
to call this thing here,
Cp over Cv.
I'm going to call it gamma.
Because I know the
answer, right.
I know the answer.
I'm going to call this gamma.
By definition I'm going to
define gamma by Cp over Cv by
definition.
OK, so that means that this is
really instead of R over Cv.
it's really gamma minus one.
So now I've got this gamma
placed in there, gamma minus
1, so I've ben feeling
a little bit better.
I've gotten V to the gamma
power almost to the gamma
power, it's minus 1 here.
But it's almost right and I'm
looking for something like
this here, V to the gamma power
pV to the gamma power is
a constant to describe
that line.
So I just got to get rid
of the temperature now.
Somehow I've gotta get rid of
the temperature and make it
pressure instead.
OK, so let's -- let me get rid
of this here, let's use the
ideal gas law to get rid
of the temperature.
So we have T is pV over R. So
now if I look at my V1 over V2
to the gamma minus 1,
that's T2 over T1.
I'll replace the ideal gas
law for those two T's.
That p2 V2 over R, and then I
have p1 V1 over R, the R's
cancel out.
Well let's see, there's a V1
over V2 to the gamma minus 1.
There's a V1 over V2
here or V2 over V1.
That's going to get rid
of this minus 1 here.
Then I'm going to get, be
left with a p2 over p1.
So this guy here gets rid of
this, and I have my answer as
V1 over V2 to the gamma power
is equal to p2 over p1.
Or I can re-write this as p1, V1
to the gamma is equal to p2
V2 to the gamma.
That means that p1 and -- where
I am, the point number 1
and point number 2 were
completely arbitrary.
They happen to be on the path.
So any point I pick on that path
will be equal to p1, V1
to the gamma.
So if I take p times V to the
gamma, anywhere on the path,
it's going to be equal
to the same relation
from my first point.
This is going to be true for
any point on the path.
As long as I'm on the path, pV
to the gamma will be whatever
it was for the first point,
which is going to be some sort
of constant.
All right, so I've
proven my point.
I've proven what I was trying to
do, which was to show that
on this adiabat, everywhere on
the adiabat, if you take a
functional form that relates p
and V together, it's going to
have this relationship with
gamma as related to the heat
capacities.
Gamma is Cp over Cv.
OK, now there's more we can say
because we know that's Cp
is related to Cv through
this R here, and R
is a positive number.
So Cp is always bigger
than Cv.
In fact, if it's an ideal gas
Cv and Cp are well defined
numbers, it turns out.
If you have a monotomic ideal
gas, OK, so an atomic version,
not a molecular ideal gas, then
it turns out that Cv is
equal to 3/2 R, and therefore,
Cp which is just R plus this
is five halves R, which means
that gamma for a mono atomic
ideal gas is 5/3.
So if you have a gas of argon
atoms, you know what gamma is.
This is something that you're
going to prove in statistical
mechanics, and so we're not
going to worry about where
this comes from.
We're just going to take
it for granted.
All right, so gamma is for ideal
gas, is bigger than one.
In fact we have a number
for it if it's an
atomic ideal gas.
So what it means then is if I
look at if this relationship
here, gamma is bigger than 1,
so gamma something bigger
than 1 minus 1.
This is, so this is
positive here.
It's positive.
So if I have an expansion where
V2 is greater than V1,
so V2 is greater than V1, so
this is a number which is less
than one, then I expect then T2
is going to be less and T1,
T2 is less than T1.
OK, I have this number here to
a power, which is positive, a
positive power.
And its number is
less than one.
This is going to give
me something which
is less than one.
So T2 is less than T1, and I
have a cooling of a gas.
So whenever you have an
adiabatic expansion, the
temperature cools,
it gets colder.
It gets colder, and if I look at
my graph here, then if the
volume of the expansion, if
the temperature of the
adiabatic expansion was colder,
that means that --
this is the wrong graph here.
If I want to compare it with
the isothermal expansion,
which is sitting here.
All right, so gamma, the gas is
cooling so V2 is going to
be less than it what
would be if the
temperature kept constant.
So finally, on the same graph,
I put down what an isothermal
expansion would be, if my final
pressure is the same
pressure, my volume for an
isothermal expansion will be
bigger than for an adiabatic
expansion.
So when I expand this gas
adiabatically and it cools
down, why do you think
it might cool down?
It cools down because I'm
doing work through the
environment, but the energy has
to go somewhere, right?
There's an interplay between
the energy inside the gas
which is the heat energy which
is allowing me to do all that
work to be outside, and so I'm
using up some of the energy
that's inside the gas to do
the work on the outside.
And here, in an isothermal
expansion, The bath is putting
back the energy that the gas
is expanding or using to
expand, and so the energy is
flowing back into the gas
through the environment in
the isothermal expansion.
In the opposite case, if you
have a compression, then it's
the opposite of expansion.
Compression you're expected
to heat up, right?
So in the compression, you're
going to have V2
is less than V1.
So that T2 is going to
be greater than T1.
That's going to be your
heating up of the gas.
And that really is
the bicycle pump.
That example I gave last time
with the bicycle pump was not
quite the right example.
But this time this really
is the bicycle pump.
That you're pushing
really hard on it.
So you start with the
bicycle pump at one
bar, let's say, right?
And you compress the air
in the bicycle pump.
You do it so quickly that the
heat flow between the inside
of the bicycle pump and the
outside is too slow compared
to the speed at which
you compress.
But you don't compress it so
quickly that you're not in
there reversible process so you
compress that if you were
to feel the temperature of the
air and you can feel it
through the nozzle gives you an
idea of the temperature of
the air inside your
bicycle pump.
You'll feel that it's warmer.
You've just done an adiabatic
compression of the ideal gas,
you can pretend there
is an ideal gas.
And that gives rise to
the heating that you
can actually measure.
OK, any questions on
this part here.
All right, so we've just gone
through the reversible
adiabatic, and I'm not going
to do this in detail, but I
want to go through the
irreversible adiabatic fairly
quickly, and see where the
difference is here.
OK, so now we're going to do the
same kind of experiment,
but irreversibly.
An irreversible adiabatic.
OK, so the experiment is going
to be to take my cylinder now,
and the external pressure and
the internal pressure are not
going to be in equilibrium
with each
other during the process.
I'm going to start with my
cylinder here, I'm going to
put little pegs here so
it doesn't fly up.
There's going to be some
external pressure.
I'm going to set that
equal to p2.
There is going to be an internal
pressure where p1 is
less than p2 and there's
V1 and T1 here.
The whole thing is going
to be insulated.
Then I'm going to release these
little pegs here and my
piston is going to shoot
up now because p2
is less than p1.
So p2 is less than p1, the
external pressures is less
than the internal pressure.
So it's going to shoot up until
the internal pressure
and the external pressure
are in equilibrium.
So they're both p2. external is
p2 and I have p2, V2, T2,
on the other side.
OK, so in my diagram now, I have
p1, V1 for my gas. p2, V2
for my gas, so I know that I'm
starting here and I know that
I'm ending here, but I can't
connect this path here.
I don't know how to do that.
It's an irreversible
expansion.
I don't know what the
pressure is doing in
there, doing that expansion.
It could be quite chaotic.
It could be non-spacially
constant.
OK, so but you still know
a number of things.
We're now going to be able to
write you know dw is minus p
dV, but we're still going
to be able to
write a bunch of things.
We're still going to be able to
write that it's an adiabat
so that dq equals zero.
We're still going to be able
to write dw instead of pV
where p is the internal
pressure. dw now is actually
much easier.
It's minus p external.
Well p external is actually p2,
so that makes it actually
easier. dV we're still
going to write that
it's an ideal gas.
So du is still going to be
equal to Cv dT, and we're
still going to be able to use
the first law, all these
things don't matter
where the path is.
Still know that du
is dq plus dw.
dq is zero. dw is minus p2 dV.
So this is what we know, just
like what we wrote there.
We write what we know.
I'm not going to turn
the crank here.
I'm going to give
you the answer.
OK, you use the ideal gas law,
etc., then you get a
relationship that connects the
pressure and the temperature,
like here we got a relationship
that connected
the temperatures and the
volumes together.
You can get a relationship that
connects the temperature
so this is T2 times Cv plus R
is equal to T1 on Cv plus p2
over p1 times R.
This is going to be the connect,
what connects the
pressures and the temperature.
And p2 is less than p1, so
this number right here
is less than 1.
So and Cv plus R, Cv plus
something less than R, so T1
better be bigger than T2.
OK, so T2 is less than T1.
Same qualitative result
as before.
You have an expansion.
It's and adiabatic expansion.
You get a cooling of the gas.
OK?
Now here's a question for you
guys, which we're going to
vote on, so you better start
thinking about it.
Is the temperature T2 in this
process smaller or larger than
if I were to do the process
reversibly with the same
endpoint pressure.
So now, instead of having p
external equals to p2 here, I
put p1 and I let this whole
thing go reversibly.
And I compare T2 irreversible
to T2 reversible.
They're both less than T1.
They're supposed, there's
a cooling
happening in both cases.
One is going to be colder
than the other, maybe.
Maybe they're going
to be the same.
I don't know.
I'm going to let you try to
figure that out qualitatively
without doing any math.
See if you can figure it out.
OK, so which is colder?
I'm going to let you think
about it for about thirty
seconds or so.
You can talk to each
other if you don't,
can't figure it out.
All right, let's take a vote.
How many people say that T2
irreversible is colder than T2
reversible?
One, two, don't be shy, three
four, anybody else, five.
OK, I've got five votes here.
Professor Nelson?
PROFESSOR NELSON: Oh,
I'm abstaining.
PROFESSOR BAWENDI: You're
abstaining, OK, good choice.
OK, what about T1
cooler than T2?
OK, I've got a lot
of votes here.
So the majority of the
people, so this is
much bigger than five.
OK.
So remember last time the
majority was wrong, right?
The majority was wrong
last time.
That doesn't mean that the
majority is wrong here.
It could be right.
Let me give you another piece
of information here that you
already know.
Minus w irreversible, this is
the work which is done to the
environment by the system, minus
w irreversible is always
smaller than minus
w reversible.
You learned that a
little while ago.
OK, so this is a little hint.
Let's vote again.
Let's think about it for ten
seconds why this could be
interesting and relevant, and
I want to ask if anybody
changed their mind.
Anybody change their mind,
change their vote?
Yes You want to change
your vote?
To which one?
STUDENT: From T2 reversible is
greater than T2 irreversible,
saying that T2 reversible
is [UNINTELLIGIBLE].
PROFESSOR BAWENDI: So
the question then,
which one is colder?
STUDENT: T2 reversible
should be colder.
PROFESSOR BAWENDI: T2
irreversible should be colder?
STUDENT: Yes.
PROFESSOR BAWENDI: All right.
OK.
I'm going to keep you there
then, the majority.
Yes?
STUDENT: I'm switching
to [INAUDIBLE].
PROFESSOR BAWENDI: You're
switching this way here?
So now we have six here.
Anybody else want to switch?
All right, let's take the
example of, the extreme
example, let's go to the
extreme example where p
external is really small.
And I have this adiabatic
expansion where p external is
really small. it's kind
of like the Joule
expansion, an ideal gas.
What happened to the temperature
in a Joule
expansion in ideal gas?
Anybody remember?
Anybody remember?
Joule expansion, this is where
we proved that delta u was
only dependent on temperature.
Bob Field taught that lecture.
Nobody remembers?
Well, all right, delta u is
equal to zero for that.
What happened to the temperature
in that expansion?
It's an adiabatic expansion.
Well, we'll revisit that.
Let me ask you another
question here.
So w, the work is less for the
irreversible process than the
reversible process.
There's less work done
to the outside.
So there's less energy expanded
by the system.
The energy expanded by the
system is smaller for the
irreversible process.
It's not doing as much work.
It's not pressing against
as much pressure.
Right? p external equals
p2 is less than p
external equals p internal.
So it doesn't have to
do as much work.
It doesn't have to expend as
much energy, therefore, the
majority in this
case is right.
The temperature is not going
to cool as much for the
irreversible process.
The reversible process is going
to be colder because it
has to expend more energy.
It's going to have a
higher pressure.
The pressure is going to
decrease along the way, but
it's going to have to
do more work because
of this right here.
So the majority in this
case is right.
Now the Joule expansion, the
temperature doesn't change, T
equals zero. eta sub J, the
Joule coefficient for an ideal
gas, is equal to zero and
eta sub J was dT/dV
under constant energy.
All right, any questions?
Yes?
STUDENT: [INAUDIBLE].
PROFESSOR BAWENDI: The only
-- a priori, that's true.
But it's hard to imagine an
irreversible compression the
way we can just imagine an
irreversible expansion.
You have to do it awfully fast
to get your irreversible
compression.
I don't know how to
do it really.
I think it's, but it's more
complicated, but you're
basically right, basically
right.
That's why we always talk about
irreversible expansion,
because it's easy
to write down an
irreversible expansion.
A compression is a little
bit more complicated.
Any other questions?
OK, great.
Monday Professor Nelson who is
sitting here will be taking
over for a few weeks
and then you'll see
me again after that.
