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PROFESSOR BAWENDI: Last time you
talked about the first law
of thermodynamics.
And you talked about isothermal
expansion, the
Joule expansion.
You saw a very important result.
which is that for an
ideal gas, the energy content
is only dependent on the
temperature, nothing else.
Not the volume, not the
pressure, it just cares about
the temperature.
So, if you have an isothermal
process for an ideal gas, the
energy doesn't change. q plus
w is equal to zero for any
isothermal process.
And you also saw that du then
could be written as Cv dT for
an ideal gas always.
This is not generally true.
If you have a real gas and you
write du is Cv dT, and your
path is not a constant
volume path, then you
are making a mistake.
But for an ideal gas, you
can always write this.
And this turns out to be very
useful to remember.
OK, now most processes that we
deal with are not constant
volume processes.
So energy, which has this
wonderful property here, du is
Cv dt for constant volume
process, which happens to be
equal to d q, constant volume,
because there's no change.
there's no work if you've got
a constant volume process.
So du here is a very interesting
quantity, because
it's related to the heat that's
going in or out of the
system under constant
volume process.
But as I said, we're not
operating usually in a
constant volume environment.
When I flail my arms around
I generate work and heat.
This is not a constant
volume process.
If I'm the system, what's
constant when I do this?
Anybody have an idea?
What's the one function
of state?
I'm the system, the rest
are the surrounding.
What's the one function of state
that's constant when I'm
doing all my chemical
reactions to
move my arms around?
Temperature?
STUDENT: Pressure?
PROFESSOR BAWENDI:
Pressure, right.
Pressure is constant.
What is the pressure at?
One atmosphere, one bar.
So the most interesting
processes are the processes
where pressure is constant.
When I had have a vial on bench
top, and I do a chemical
reaction in the vial, and it's
open to the atmosphere, the
pressure is constant
at one atmosphere.
When you've got your cells
growing in your petri dish,
the pressure is constant at
one atmosphere, even if
they're evolving gas, pressure
is constant.
So we'd really like to be able
to find some sort of equation
of state, or some sort of
rather function of state
that's going to relate the heat
going in or out of the
system with that function of
state, because this isn't
going to do it. du only relates
to the heat under
constant volume.
And the heat is a really
important thing to know.
How much heat do you need to put
into a system, or how much
heat is going to come out of
a system when something is
happening in the system?
All right, this is a really
important quantity to know.
Your boiling water or whatever,
you want to know how
much heat do you need to boil
that amount of water under
constant pressure?
And this is where enthalpy
comes in.
You've all heard of enthalpy.
H we're going to write it as
the function of temperature
and pressure.
And the reason enthalpy was
invented was exactly for that
reason, because we need some
way to figure out how to
relate the heat coming in or
out of a system under a
constant pressure process.
Because it's so important.
And I should add and also under
reversible work, where
the external pressure is equal
to the internal pressure.
OK, so we're going to define
enthalpy as u + pV, these are
all functions of state here, So
H is a function of state,
and we're going to see now how
this is, indeed, related to
the heat flow in and
out of the system.
If you have a constant
pressure,
reversible work process.
Let's take a system.
Under constant pressure T1, V1,
going to a second -- this
is the system, so let me
write the system here.
And it's more dramatic if the
system is a gas, p, T2, V2,
And let's look at what happens
to these functions of state,
to H to u under this
transformation.
OK, so let's look at delta u.
Delta u is q plus w.
That's the first law.
And this is a constant pressure
path, so now I can
write, this q is actually q
under constant pressure.
Little p means the path is
a constant pressure path.
And I'm doing reversible work.
So that w is minus p, dV where
p is the pressure inside the
system, minus p delta V.
Rearrange that, delta u is
plus p delta V is
equal to q p.
All right, so this is the heat
flowing in or out of the
system, and these are all
functions of state.
This depends on the path.
It tells you right here, the
path is constant pressure.
These don't depend on
the path, right.
V doesn't care how you
get there. u doesn't
care how you get there.
In this case, p is a constant
because the path is constant.
So we can bring the p
inside, delta u plus
delta p V it's q p.
Take this delta outside again,
delta of u plus p V
is equal to q p.
And there you have it.
There is the H right there.
The u plus p V. Delta
H is equal to q V.
And this is the reason why
enthalpy was invented, and why
it's so important.
Because we want to know this.
So this for a finite change.
If you want to have an
infinitestimally small change,
you end up writing
dh is dq sub p.
It's not always equal
to the heat.
It's only equal to the heat if
your process is constant
pressure reversible work.
OK, so this is the kind of, this
is the kind of concept
that needs to be branded into
your brain, so that if I come
into your bedroom in the middle
of the night and I
whisper to you delta H, you
know, you should wake up and
say q p, right?
Heat under constant pressure
reversible work.
This should become
second nature.
This is where the intuition
comes from.
This is why people
right tables and
tables of delta H's.
Why you have delta H's from all
these reactions, because
this is basically the heat, and
the heat is something we
can measure, we can control.
We can figure out how
much heat is going
in and out of something.
This is what we're
interested in.
OK, so last time you
looked at -- any
questions on this first?
Yes.
STUDENT: [INAUDIBLE] from the T
delta V to the delta p here?
What was the reasoning
behind that?
PROFESSOR BAWENDI: p
is constant here.
It's constant pressure.
OK, so now, last time you looked
at the Joule expansion
to teach you how to relate
derivatives like du/dV.
du/dV under constant
temperature. du/dT under
constant volume.
You use the Joule expansion
to find these quantities.
Now these quantities were
useful because you could
relate them.
The slope of changes, with
respect to volume or
temperature of the energy with
respect to quantities that you
understood, that you
could measure.
We're going to do the
same thing here.
So if we take as our natural
variables for enthalpy to be
temperature and pressure, and we
have some sort of change in
enthalpy, dH, and it's going
to be related to changes in
temperature and pressure through
the derivatives dH
through the slope of the
enthalpy in the T direction,
keeping pressure constant, dT
plus the slope of enthalpy in
the pressure direction, keeping
the temperature
constant, dp, and these are
knobs that we can turn.
We can change of temperature.
We can change the pressure.
These are physical knobs that
are available to us as
experimentalists.
And so when we turn these knobs
on our system, we want
to know how the enthalpy is
changing for that system.
Because eventually they will
tell us maybe things about how
heat is changing further on.
OK, but in order to relate
turning these physical knob to
this quantity here, which we
don't have a very good feel
for, we've got to have a
feel for the slopes.
If I keep the pressure
constant.
I change the temperature,
what does that mean?
What is dh/dT?
If I keep the temperature
constant, and just change the
pressure, dH is going
to change, but how
is it going to change?
What does this mean in terms of
something I can physically
understand?
That's the program now for
the next few minutes.
What are these quantities?
What is dH/dT as a function,
keeping pressure constant,
what is dH/dp, keeping
temperature constant?
All right, let's start with the
first one, dH/dT, keeping
the pressure constant.
And we're going to look at a
reversible process to help us
out, but the result is going to
be independent of whether
or not we have a reversible
process or
irreversible process.
Constant pressure, that means
dp is equal to zero.
So for reversible process,
constant
pressure, what do we know?
This is already branded
in your brain, right?
Reversible process, constant
pressure dH=dq.
So we can write that down,
dH=dq, constant pressure.
That's by definition
of enthalpy.
That's why we created
enthalpy.
What else do we know?
Well we can go look up here,
looking at the differential,
there are no approximations
here.
This is just an equality.
I have a constant pressure
process.
This term here is
equal to zero.
That means that dH is also
equal to dH/dT, constant
pressure dT.
All right, so now I've got
more dH/dT under constant
pressure. dH is equal to this,
and it's also equal to this.
All right, these two are
equal to each other.
Now, I know how to relate the
heat flow to temperature
change, through the heat
capacity. dq constant pressure
is that heat capacity, and I
have to tell you the path for
the heat capacity.
So it's C sub p, the heat
capacity under a constant
pressure path, dT, all right?
So these two are equal
to each other.
So, these two are equal to
each other as well, which
tells me that this derivative,
dH/dT constant pressure is Cp.
So now I have my first of my
two slopes, in terms of
something that's related to my
system, the heat capacity of
the system.
Something I can measure and I
can tabulate, and when I turn
my dT knob here I know
what's going to
happen to the enthalpy.
So this is the first,
this is the one.
So it's very similar to what we
saw with the volume and the
energy, where du/dV under
constant temperature was equal
to Cv in this case here, right.
and you're going to
find that there's a lot of these
analogies between energy
and enthalpy.
You just change volume to
pressure and basically you're
looking at enthalpy under a
constant -- anything that's
done at a constant volume path
with energy, there's the same
thing happening under constant
pressure path for enthalpy.
So you can guess the answer
usually that way.
OK, so now we have the other
one, dH/dp constant
temperature.
How do we relate this to
something physical?
Well, it's going to be an
experiment, very much like the
Joule experiment.
The Joule experiment
was a constant
energy experiment, right.
Here we're going to have to
find a constant enthalpy
experiment, and that is going
to be the Joule-Thomson
experiment.
That's going to extract out
a physical meaning to this
derivative here.
OK, the Joule-Thomson
experiment.
This is going to get us dH/dp
constant temperature.
What is this experiment?
You take a throttle valve, which
consists of some sort of
porous plug between two
cylinders that is insulated.
There is insulation here.
Insulation on the bottom.
It's like a, think about a tube,
a large tube, insulated
tube, with a bunch of -- a
frit inside here, which
prevents flow of gas, which
slows down the flow of gas
from one side to the other.
It's a blockage in
this tube here.
Then you put two pistons, one on
that side here, and one on
this side here, and the external
pressure here, we're
going to call that p1.
The external pressure here,
we're going to call that p2,
and we're going to do it
slowly enough that the
pressure on this side of the
cylinder is in equilibrium
with the external pressure, and
the pressure on this side
of the cylinder is in
equilibrium with this pressure.
But not so slowly that these
two are in equilibrium with
each other.
So this is restricting
the flow, so there's
some sweet spot here.
When I'm pushing slowly enough
here that the pressure here is
equal to that one, but not so
slowly that the air flow from
here to here is fast, compared
to how fast I'm pushing.
You got the picture here?
Any questions on that?
All right, then as I push
through, I'm going to start
with all of my gas on this
side, and at the end, I'm
going to have all the gas
on the other side.
Let me first ask you this, is
this a reversible or in
irreversible process?
Right, let me add one more piece
of data here which I
said in words, but which is
actually important to write
down before doing the problem.
Is this a reverse
-- any guesses?
How many people vote for that
this is a reversible process?
I've got one vote back
there, two votes,
three votes, four votes.
Anybody else?
How many people think this
is irreversible?
It's about a tie, and everybody
else doesn't now.
All right, I'm going to give
you ten seconds, fifteen
seconds to make up your mind.
You're not allowed to be
on the fence here.
You've got to decide,
all right?
This is, you can talk to your
neighbors, you know, do a
little bit of thinking.
And I'm going to give you ten
seconds to figure this out,
what your vote is for that.
All right, let's try again.
How many people vote that
this is reversible?
That looks like a
majority to me.
Irreversible?
Let's look at the
show of hands?
All right, so this is
the majority here.
Good thing physics doesn't
work on the rule of the
majority, otherwise we'd
be in big trouble.
Wow, let's walk through that.
I'm sorry to say that this
is the wrong answer.
OK, why is that the
wrong answer?
Well, just think, you know,
think about it.
You're pushing through here.
p1 is greater than p2.
What does it mean for a process
to be in equilibrium
or reversible?
It means at any point you can
reverse the direction of time,
and it will look fine, right.
So now I'm pushing on this
plug here, with p1
greater than p2.
I'm pushing, I'm pushing,
I'm pushing. p1 is
greater than p2.
Then I want to reverse
direction of time.
I want the arrow of time to go
so that the gas goes from p2
to p1. p2 is less than p1.
Is that going to work out?
If p2, the pressure in p2, is
less than the pressure in p1,
is the gas going to want to go
from p2 to p1 and the whole
thing reverse back?
You've got to put more pressure
on one side than the
other if you want to push
that gas through
the throttle, right?
So this is where the time scale
issue comes into play.
I said let's do this slowly
enough that this p1 is in
equilibrium with this p1, but
not so slowly that this
pressure is equivalent
to that pressure.
If I do it really, really
slowly, so that everything is
reversible, well I won't be able
to do it, because p1 and
p2 are different.
But suppose that I fix my
pistons here, with p1 greater
than p2, and I don't touch
it, eventually p1
will be equal to p2.
I'll come to some sort
of equilibrium.
So this is a system which
is out of equilibrium.
If I stop, if I move slowly,
if I move more slowly, then
these two will want
equilibrium.
So this is an irreversible
process.
The Joule-Thomson experiment
is irreversible.
OK, important -- if you are part
of this group here, think
about it and make sure
you understand that.
All right, so this is
the experiment.
Now what are we doing
with that?
The initial state, let's look
at what we are doing.
So initially, we're going to
have our piston, there's the
plug sitting here.
Our piston on the right side
here fully out, and the piston
on my right side, your left
side, fully inside.
There's no gas on
that side here.
So there's p2 sitting here.
There is p1 sitting here, and
all of the gas is sitting on
that side of the plug.
Then after we're done with the
experiment, we'll have
transferred all of the gas from
one side to the other. p1
here. p2 sitting here.
And there's going to be some
volume V2 and some volume V1,
but are not necessarily
the same.
Especially since the pressures
are different. we don't know
yet about temperature so I don't
know what to say about
these volumes because I don't
know what the temperatures'
are going to do.
OK, so let's go through this and
see what we would do which
is to calculate the
heat and the work.
This is well insulated.
So, what is the -- let's
do the first one here.
What's q for this process here?
Anybody?
STUDENT Zero.
PROFESSOR BAWENDI:
Zero, right.
This is an adiabatic process.
It's well insulated.
Heat is not going in or out,
adiabatic. q is equal to zero.
So all we need to find
out is the work now.
Let's divide it up into the two
sides, the work going on
on the left hand side, my left
hand side or your left hand
side, and the work going on
on the right hand side.
So let's first look at
the left hand side.
OK, so w, first of all, work is
being done to the system on
the left hand side here.
I'm pressing on the gas.
So I expect that to be
a positive number.
The pressure is constant, p.
The V goes from V1 to zero.
So we write down p1, V1.
On the right hand side, the
work, let's call this left
hand side, let's call this
right hand side.
Here there's an expansion going
on, so the system is
doing work to the
external world.
This piston is being brought
out, so we expect the work to
be negative, negative.
And we start out with
zero volume.
We end up with a volume of V2,
and the external pressure is
constant to p2.
Minus p2, V2.
Minus p delta V. So the total
work is the work from the left
hand side plus the work on the
right hand side, which is p1
V1 minus p2 V2.
Which I can rewrite
as minus delta pV.
Delta pV is p2 V2 minus p1 V1.
It's the pressure volume
multiplied together at the
final state, minus pressure
volume from the initial state,
with the minus sign
here because it's
the negative of that.
All right, what is delta u?
delta u is q plus w. q is
zero. delta u is just w.
So this is also just delta u.
delta u is minus delta pV, for
the process.
OK, delta H is delta
of u plus pV.
By definition, that's how we
define enthalpy up here.
H is u plus pV.
Delta H is delta of u plus pV,
which is equal to delta u,
plus delta pV.
Now delta u is minus delta pV.
So I have minus delta
pV plus delta pV.
This is equal to zero.
So this irreversible process,
this Joule-Thomson process, is
a constant enthalpy process.
Delta h for this process
is equal to zero.
Adiabatic q equal to zero.
It's also delta H
which is zero.
The two didn't necessarily
follow, because remember,
delta H is dq so p is only true
for a reversible constant
pressure process.
This is an irreversible
process.
So a priori, this was not
necessarily true.
It turns out after you do all
the math, it turns out to be
delta H equals zero.
All right, so this is
the experiment.
How do we go from that
experiment to the terms that
we're trying to get,
these slopes.
Remember, we're trying to get
delta H, we're trying to get
dH/dT constant pressure and
dH/dp constant temperature.
OK, these are the two things
were trying to get here.
OK, so let's write down,
what we know here.
I'm missing something.
Oh, we already know one thing.
We already know this guy here.
We already did that.
OK, dH/dT constant
pressure is Cp.
That was easy one.
So we already know that.
So now we can write or
differential dH as Cp dT plus
dH/dp, constant temperature,
dp.
Now we want to find out
what this guy is here.
Now for this experiment, this
is a constant enthalpy
experiment for the Joule-Thomson
experiment, this
is equal to zero.
So I can rearrange this to get
this dH/dp in terms of things
that I can either measure, like
the heat capacity, or
that I have control of,
like dT and dp.
So in this case, dH/dp constant
temperature is minus
Cp dT/dp, and this is under
constant temperature, no, not
constant temperature.
Whatever the experi -- for
that the experiment is.
For that experiment, the
constraint, so we need a
constraint here, right, we
need a constraint here.
Right?
We need a constraint here.
The constraint isn't constant
temperature because the
temperature is going
to be changing.
It's not constant pressure,
because we have a
delta p going on.
It's not constant
volume either.
The constraint is the constraint
of the experiment,
and the constraint of the
experiment is that the
enthalpy is constant.
So the constraints
we have here, is
the constant enthalpy.
It's the constant enthalpy
process that we're looking at.
This we can do experiments on.
It's tabulated in books, and
this we can measure in the
experiment.
Delta p here is the change in
pressure from the left side to
the right side, and we can put
a thermometer, measure the
temperature before the
experiment, and measure the
temperature after
the experiment.
So this is something
we can measure.
So now we have this derivative,
in terms of
physical quantities, things
that we can measure.
Things that we can relate
to the properties of the
substance that we're doing
the experiment on.
So this is basically delta
T and delta p and the
Joule-Thomson experiment.
And so Joule and Thomson did
these experiments, and they
measured lots of gases, and they
found that, in fact, this
was something that they
could measure.
Sometimes it was positive,
sometimes it was negative, and
it was an interesting number.
And so they defined them, after
many experiments, the
limit of this, delta T delta p
and the limit of delta p goes
to zero as the Joule-Thomson
coefficient.
So, basically dT/dp, constant
enthalpy is equal to mu, by
definition, Joule-Thomson, where
mu Joule-Thomson is the
Joule-Thomson coefficient, just
like you saw last time
eta sub j was the Joule
coefficient for dT/dV under
constant energy.
So there's, again, total analogy
here between what
we're doing with enthalpy to
what you did last time with
energy, replace p with
T and H with u.
Flip those two and you
get the same thing.
OK, so now we have dH/dT is
equal to Cp, and we can also
write, then, dH/dp under
constant temperature is equal
to minus Cp mu Joule-Thomson.
We have our two derivatives in
terms of physical quantities,
which is going to allow us,
then, whenever we have a
change to go back and, when we
have a change where we adjust
the temperature and the
pressure, we'll be able to
know what the enthalpy
change is.
OK, now let's take two cases.
Let's first start talking
about ideal gases.
The last time you saw that for
an ideal gas, the energy only
cared about the temperature.
It didn't care what the volume
was doing. du/dV under
constant temperature was equal
to zero for an ideal gas.
And by analogy, we expect the
same thing to be true here,
because enthalpy and energy
have all this
analogy going on here.
So let's look at an ideal gas.
So for an ideal gas, we saw that
u was only a function of
temperature.
We also have the equation
of state for an
ideal gas, pV = nRT.
We can write our definition of
enthalpy, h is u plus pV.
This only depends on the
temperature. pV= nRT.
So we have u only depends
on temperature plus nRT.
The only valuable now on this
side is temperature.
Pressure and volume
have dropped out.
So enthalpy, for an ideal gas,
only cares about temperature.
Pressure has dropped out of the
picture completely here.
So there is no p dependence
here.
H for an ideal gas is only a
function of temperature.
This is not true for a real
gas, fortunately, but it's
true for an ideal gas.
So for an ideal gas then,
dH/dp under constant
temperature, that has
to be equal to zero.
Because temperature is constant
H only cares about
temperature. and that's
equal to zero.
And if that's equal to zero,
that means that the
Joule-Thomson coefficient
for an ideal gas is
also equal to zero.
We're going to actually prove
this later in the course.
Right now, you're taking
it for granted.
Right now we told you Joule did
all these experiments and
he found out that for an ideal
gas, that the limit in and
ideal gas case was that the
eta J was equal to zero.
Therefore, from experiments,
u is only a function of
temperature for an ideal gas,
and therefore from these
experiments, we come
out with delta H
dH/dp is equal to zero.
The Joule-Thomson coefficient
is equal to zero.
Later we are going to prove it
exactly. but right now you're
going to have to take
it for granted.
So, if the Joule-Thomson
coefficient is equal to zero,
just like we wrote, du = Cv dT
for an ideal gas, we're going
to have dH = Cp dT for
an ideal gas as well.
dH is Cp dT.
This term goes away.
dH = Cp dT.
That's the only thing
that's left behind.
So if you know the heat
capacity, you know the change
in temperature, you know
what enthalpy is doing
for an ideal gas.
This needs to be stressed that
this is the ideal gas case.
Now regular gases, real gases,
fortunately as I
said, don't obey this.
This is important because we
use this all the time for
when, when we do technology.
One example of Joule-Thomson
coefficient being not equal to
zero for a real gas that you've
like experienced is if
you take a bicycle pump, take
a bicycle pump, and you're
pumping up your tire, you're
working pretty hard, so you're
getting hot.
But if you touch the valve going
into your tire, which
basically measures the
temperature of the air going
into your tire, that is
getting hot, right.
So if you've got to pump that
tire really a lot, then you're
going to you're going
to really feel
a lot of heat there.
The compression of the,
basically it's an adiabatic
compression.
You're taking the air inside
of the pump and you're
compressing it.
You're doing it so fast that
there's not enough time for
heat to come out of the gas
that's inside the pump towards
the walls of the pump.
So your time scale it just
fast enough that this is
basically an adiabatic
compassion.
Your compressing it really fast,
all right, so you're
changing the pressure.
You're changing the pressure,
and the temperature is going
up. dT/dp is positive. dT/dp
is positive. dT/dp is
positive, well that's mu
JT. dT/dp is mu JT.
So for a real gas like air,
this is a positive number.
It's not zero.
Air is not an ideal gas.
It's one simple example of --
The fact that mu JT is not zero
for real gases is how we
are able to liquify things like
hydrogen and helium, by
compressing them and pushing
them through a nozzle, and the
expansion through the nozzle
cools the gas.
Right, in this case here it
wouldn't happen if it
was an ideal gas.
Or in many kinds of gas
refrigerators where you push a
gas through a nozzle close to
room temperature, what you
find is that the gas coming out
on the other side under
lower pressure is cooler than
the gas that went through on
the other side.
Real refrigerators actually work
with liquids that go into
gases so use the latent heat of
the liquid, so it doesn't
really work like the
Joule-Thomson expansion.
So this is real.
This is real, unlike the Joule
coefficient which is very
small so that most gases have
tiny Joule coefficients.
So if you do a Joule experiment,
you hardly measure
a temperature change.
With real gases, here you
do actually measure it.
You can feel it with your finger
on your bicycle tire.
OK, so we're going to see this
using a Van der Waal's gas.
Let's look at a Van der Waal's
gas and see what happens in
the Van der Waal's gas.
Any questions, first?
What we've been talking about,
the Joule-Thomson experiment,
constant enthalpy process?
OK, so let's take our
Van der Waal's gas.
Remember the equation of state
for Van der Waal's gas is not
pV is equal to nRT, but p plus
the attraction term.
And then V minus the excluded
volume term is equal to RT.
Two parameters, this is the
attraction between two atoms
or molecules in the gas phase.
This is the repulsion,
not the repulsion.
This is the fact that we occupy
a finite volume in
space, because they're little
hard spheres in this molecule.
OK, in a few weeks, you're going
to find out that we can
calculate dH/dp from this
equation of state, and you're
going to find out that dH/dp
from that equation of state is
proportional to b
minus a over RT.
This is going to be probably
a homework at
some point to do this.
For now, let's take
it for granted.
Let's take it for granted that
we know how to calculate this
derivative from an equation
of state like this.
But now we're going
to use that.
OK.
So since dH/dp under constant
temperature is proportional to
minus mu JT, then we have that
dT/dp under constant enthalpy
than it is related to the
negative of this, a over RT
minus b
All right, this is how the
temperature changes when you
change, when you have
pressure changing.
So when you do an expansion or
compression, my adiabatic
compression of my bicycle
pump is dT/dp.
All right.
Or when I do an expansion of
hydrogen or helium at low
temperature, through a
Joule-Thomson experiment, when
I want to liquify hydrogen
or helium.
I want to cool a gas with a
Joule-Thomson experiment, what
temperature do I
have to be at?
So this tells you that you
have to be careful what
temperature you're at, because
depending on how high, how big
this temperature here is, you
could either be, have a
negative dT/dp, if this first
term is small enough, meaning
if temperature is very high,
then you end up with a
negative term.
If the temperature is very
small, then one over the
temperature is large, and the
first term wins and you have a
positive number.
So there's some special
temperature which is going to
depend on the gas where the
first term is going to be
equal to the second term, where
Joule-Thomson is zero,
or it's going to behave
like an ideal gas.
So when that is the case,
we're going to call that
temperature the inversion
temperature or T inv.
We call that inversion because
on one side you end up cooling
if you compress.
And on the other side of that
temperature you end up heating
if you compress.
OK, so there's some temperature,
t inversion minus
b where the gas behaves
like an ideal gas.
The Joule-Thomson coefficient
is zero and that inversion
temperature, you can
solve for it.
Conversion temperature is equal
to a over R times b.
OK, so when you are at that
temperature, everything looks
like an ideal gas,
as far as the
enthalpy changes are concerned.
Now if you're at the temperature
which is higher
than the inversion temperature,
in that case
here, a over RT is small
compared to b, and this is
going to turn out
to be negative.
So if you had a high
temperature, this a small
compared to b.
If you're negative, which means
that dT/dp at constant H
is less than zero.
So that means that if you
compress something
it's going to cool.
The temperature rises when the
pressure drops, right?
Or in this case here, if I do my
Joule experiment delta p is
negative, p2 is less than p1,
that means that delta T is
positive, right?
So in this experiment here, this
side is going to heat up.
So for materials where T
inversion is low, lower than
room temperature, then you would
end up heating up in
this expansion.
You're basically expanding the
gas from one side to the other
and the expansion causes the
temperature to rise.
If T is less than T inversion,
you have the opposite case,
and dT/dp is greater
than zero.
So in this experiment here,
delta p is less than zero.
You need to have this whole
thing greater than zero.
So delta T is less than
zero as well.
So if you're below the inversion
temperature and you
do the Joule-Thomson experiment,
you're going to
end up with something that's
colder on this
side than that side.
Ideal gas would be the
same temperature.
But now, so this is where the
refrigeration comes in.
So if you take a gas, and you're
below the inversion
temperature and you make it go
through this irreversible
process, the gas comes
out colder from that
side than that side.
So the work that you're doing to
expand, to go through this
experiment, ends up
cooling the gas.
OK, for most gases, T inversion
is much greater than
300 degrees Kelvin.
Much greater than room
temperature.
For more most gases, if you
do this experiment at room
temperature, you end up cooling
the gas, and you can
cool it measurably, which is why
also, your bicycle pump,
you know, you push down,
you compress, this is
an expansion here.
You're expanding from this
side to that side.
Bicycle pump you are compressing
the gas, which is
the opposite, you end up heating
up the air in the
bicycle pump in your
compression.
There are two exceptions to this
rule that most gases have
a T inversion which is greater
than 300 degrees Kelvin, and
that's hydrogen.
The T inversion for hydrogen
turns out to be 193 degrees
Kelvin, and the T inversion for
helium turns out to be 53
degrees Kelvin.
And so when, there's a lot of
liquid helium that's being
used on campus.
We use a liquid helium.
And so in order to make a liquid
helium, you can't take
helium at room temperature and
do this, because if you did,
you would just heat it up,
because the room temperature
is above the inversion
temperature, so Joule-Thomson
would heat up the helium.
So you need first to take the
liquid helium and cool it
below 53 degrees Kelvin before
you can do the Joule-Thomson
to cool it even further
to make liquid helium.
So you have to do
it in stages.
You take your room temperature
liquid helium, and you cool it
with liquid nitrogen to 77
degrees Kelvin, the new,
you're not quite there yet,
unfortunately right?
Then you take hydrogen you cool
it would liquid nitrogen
to 77, then you can use
your hydrogen gas.
Do a Joule-Thomson hydrogen gas
which you first cool with
liquid nitrogen.
Liquid nitrogen, 77, that's
below 193, so you can do
Joule-Thomson on hydrogen, cool
the hydrogen to below 53,
then use your cold hydrogen to
cool the helium, and then you
can do the Joule-Thomson on the
helium to cool it further
until it liquifies.
So that's what you do when
you make liquid helium.
All right, any questions on
this lecture or any of the
concepts that we talked about?
The last thing that we're going
to do today then is to
look at a relationship which
is going to turn out to be
very useful.
It's a relationship for ideal
gases which relates the heat
capacities at constant pressure
and constant volume.
Cp = Cv + R. This is very useful
because often you just
have lots of tables of Cp's but
sometimes you want to know
what the energy change is going
to be for the ideal gas,
and you know that du
is Cv dT not Cp dT.
So, you need to know what Cv
is, and if this is true
always, then there's a
very easy way to go
from one to the other.
We're going to do it two ways.
Today we'll do the first way,
and then next time we'll do
the second way.
The first way is just to turn
the crank on the math, and the
second way is to do
a little bit more
imaginative about the process.
OK, let's just turn the
crank on the maths.
What do we know about
the Cp and Cv?
Well Cp, we already know how to
relate it to dH/dT through
the slope of the enthalpy, and
we also related Cv to the
slope of the energy with respect
to temperature under
constant volume.
And we also know that
H is u plus pV.
Right, H is u plus pV.
So we're going to take the
derivatives of both sides of
this equation here, by, with
respect to temperature,
keeping pressure constant.
So we have dH/dT keeping
pressure constant, is du/dT
keeping pressure constant.
Got to keep track of what's
being constant, kept constant,
plus dPV/dT, keeping pressure
constant. dH/dT keeping
pressure constant, that's just
Cp, and then we have du/dT
keeping pressure constant.
The p is constant here, comes
out of the equation, so we
have p and then we have dV/dT
well it's an ideal gas, an
ideal gas for dV/dT for an ideal
gas is equal to R over p
because pV is equal
to RT, right.
So dV/dT is Rp.
So dV/dT is times R over
p, the p's are going
to cancel out here.
OK, so now it's very tempting
at this state to say, oh
there's the answer right
here. du/dT is Cv.
There I have it Cp is equal
to Cv plus R, right?
But even though it looks like
that's the right way to do it,
it's actually not right because
it turns out to be
right by sort of by accident.
But here you've got pressure
constant. du, this is du, not
H here. du/dT is only equal
to Cv when the volume is
constant, not when the
pressure is constant.
So if you're going to turn the
crank on the math correctly,
you're going to have to change
this p into a V somehow.
Because this isn't Cv
mathematically speaking.
We don't know what it is yet.
In order to change this from a
p to a V, you have to use the
chain rule.
So let's use the chain rule.
And then we'll be done.
OK, so u is actually a function
of temperature and
volume, which in this case here
could be a function of
pressure and temperature.
So if we want du/dT under
constant pressure, you have to
use the chain rule.
There's the pressure
sitting right here.
It's going to be du/dT under
constant volume, plus du/dV
dV/dT under constant pressure.
All right, chain rule. du/dT
constant pressure is the
direct derivative with respect
to temperature here, which is
sitting by itself under constant
volume, keeping this
constant but there
is temperature
sitting right here too.
That's where that term comes
from, du/dV dV/dT.
Now, for an ideal gas, du/dV
under constant temperature is
equal to zero.
It doesn't care what the
volume is doing.
It only cares what
temperature is.
If temperature is constant,
there's no change in energy.
For an ideal gas,
this is zero.
It's not zero for a real gas.
Right, so this whole term
disappears and for an ideal
gas, it turns out that du/dT
constant pressure is equal to
du/dT constant volume, but
this is equal to Cv
for an ideal gas.
It wouldn't be true for a real
gas, and this is a common
mistake that people make for
real gases to equate this.
This is only true for
an ideal gas.
Since it's true for an ideal
gas, then we can go ahead and
replace this with Cv, and then
we have Cp with Cv plus R,
which is what we were after.
OK, next time we'll do the other
way of getting to the
same answer.
