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>> Okay, so Quiz 3 was by far the hardest quiz I think.
I don't have to tell you that.
>> It was.
>> It was much more difficult than 1 or 2
and I think 2 was harder than 1, 3 was harder than 2
but I think we've reached the point where we've titrated now
to the-- maybe past where we wanted to go.
So this is what the histogram looked like for Quiz 3.
Now there's a mistake on the key that I posted this morning
but on problem 5, I work out the answer and it's actually B not D
which D is indicated on the key but Gene Mark found
that error before I did,
so the quizzes should have been graded correctly.
The correct answer to 5 is B not D. So I'll try and make a change
to the key before the end of the day.
So here's where we are.
So far we've had three quizzes
and this is what the histogram looks like in other words,
all these folks here are getting As.
These folks are getting Bs and so on.
Okay, so going into Mid-Term 1
which is Friday you guys are doing really well
which is good that's the way we want to see it.
So here's what's going to happen Mid-term 1 is Friday.
It will cover all of Chapters 13 and 14.
It turns out that this is just perfect, accidentally,
this lecture that I'm going to give today will take us right
to the end of Chapter 14 then what's going
to happen is you're going to review Chapters 13 and 14
and lectures 1 through 10 on Wednesday.
We'll have a review for you.
Steven's going to give the lecture that I've written
and then we'll take the mid-term on Friday.
All right, there's a copy of last year's mid-term exam
and what you'll see, what Steven will do
on Wednesday is he'll review what's going to be
on the mid-term in some detail.
He will tell you how many questions there's going to be.
He'll tell you something about what those questions are going
to be, so Wednesday's lecture is a rather important one to attend
but let me just tell you that it's going to look a lot
like last year's mid-term and I already posted
that a couple of weeks ago.
All right, last year's mid-term is already
on the Announcements page so you can see what that is.
Turns out that last year's mid-term would have been a
perfect mid-term to give this year because there's nothing
about Chapter 15 on the mid-term.
There's nothing about entropy calculations on that mid-term.
It's really about Chapters 13 and 14.
So what we're going to do today is finish Chapter 14 and then
on Wednesday we're going to review and then
on Friday we'll take the mid-term exam.
Now the mid-term exam is going to be open-book, open-notes.
I'm not going to give you, you know, that shotgun page
of equations that I showed you on the screen earlier
where there was random equations all over?
No. All right open-book,
open-notes not open iPads or computers.
But you can use any calculator that you want.
Any questions about that?
Any questions about Mid-Term 1?
Okay, so Gene Mark, started to talk to you
about adiabatic processes
and an adiabatic process the heat flow is turned off, right?
The system is in a thermos bottle.
There's no flows of heat in or out
for any infinitesimal step of the process.
There can be work.
So we know normally the internal energy is given
by heat plus work.
In an infinitesimal change
in the internal energy is then an infinitesimal change in heat
or an infinitesimal change in work
but for an adiabatic process it's all about the work.
All right, the system is in a thermos bottle.
Another expression for D can be obtained from the definition
for the constant volume heat capacity
which we talked about some weeks ago.
The constant volume heat capacity is just the partial
derivative of the internal energy with respect
to temperature evaluated at constant volume.
That's why it's called the constant volume heat capacity
and so once I've got this heat capacity I can get,
I can express it as a derivative according
to to this equation here and so
for an ideal gas DW is simply given by,
we know the work is minus PDV and if I just substitute for P
from the ideal gas equation I get this equation right here
because P is just in NRT over V and since internal energy
and work are equal to one another for an adiabatic process
that means that these two things have to be equal to one another.
That equals that for an adiabatic process
and so I can just set these two things equal to one another
like this and then I can divide through by T
and I'll eliminate it then from this side and it will pop
up here on the left-hand side and then
if I integrate these two expressions now,
if I integrate this one from T1 to T2, this one from V1
to V2 here's what that integral looks like.
I'm just integrating from T1 to T2, V1 to V2.
The integral is the same, in other words
if here I'm integrating 1 over V, here I'm integrating 1 over T
so I get log T2 over T1, log V2 over V1
and then I can just divide through by this heat capacity.
Now I'm just doing a little algebra to clean this up
and when I do that and I take the exponential
of both sides I get this expression right here
which is an equation that we're going to use
to describe adiabatic processes.
It's your equation 14.37 so this NR over CV that's going to pop
up as an exponent when we take the exponential of both sides.
Now notice that I flip this over.
Here it's T2 over T1, V2 over V1.
I flip this over so now it's V1 over V2
so I can get rid of the minus sign.
That's all I'm doing there.
I'm not skipping any steps.
So this is an adiabatic reversible process involving an
ideal gas.
So we've got this expression
for an adiabatic process that we just derived.
Earlier we derived this expression
for an isothermal process.
We can use whichever one we need.
Gene Mark walked you through this slide on Friday.
We've got two isotherms here.
Here's an isotherm that applies at T1.
Here's an isotherm that applies at T2.
What we're plotting here is that pressure is a function of volume
as we do an expansion.
These are isothermal expansions here on these two isotherms.
We use the word isotherm because they're constant temperature.
If we want to look at an adiabatic process
that starts here for example, it's going to cross over.
It's not a constant temperature process.
What's constant is heat and when we do an expansion that means
that the temperature has to fall, so change in P
with an expansion is larger for an adiabatic process
than for an isothermal one because temperature decreases
for the adiabatic process.
The temperature is going the
down during this adiabatic process.
Obviously that's not happening with the isothermal process,
so in other words, if I look at T1 and I go
from this initial volume to this final volume here's the change
in pressure that happens at T1.
Everyone see that?
Here's T1.
Here's V1.
We're going from to here and the pressure difference would only
be this shown by this yellow bar here or if we wanted to look
at the pressure difference at T2 from the isothermal process,
here's T2 so here's its pressure difference given
by this yellow bar here even smaller and if we look
at what the adiabatic process does it's going from V1 to V2
over this much larger range of pressures, much larger range
of pressures here because temperature is changing too
in this process.
Now it's not obvious from these two equations
but there's a more profound implication
of this adiabatic business that we've sort of, we've said it
but we weren't very explicit about it.
DU equals DW because DQ is zero,
right and that has important implications.
Obviously, if I integrate DU
from some initial internal energy
to some final internal energy I'm just going
to get the final minus the initial
because U is a state function.
The difference between any integral is just going
to be the final minus the initial and we're just going
to call that delta U. In an adiabatic process that has
to be equal to W. If this process, if this change
in the internal energy is happening in a thermos bottle
where there can't be any flows
of heat only work can occur then this change
in the internal energy has got to be given by just the amount
of work that was done for as long as it was adiabatic work.
It was work that was done on the system when it was
in the thermos bottle, in other words the adiabatic work is
different from all the other types of works
because it is a state function just like you.
It would have to be.
It's equal to you.
All right, so that means for an adiabatic process--
you know what we saw earlier is if we take a normal process
and we do PV work, if we break the work
down into smaller chunks we can do less of it to get
from an initial state to a final state.
Remember that?
We took the brick and we ground it up to make it
into tiny particles and we can add tiny particles
and then we can do a minimal amount of work
but if the work is adiabatic there's only one way to do it.
It's a state function.
You can't, there's only one way to do the adiabatic work.
There's only one adiabatic work answer.
It's a state function just like you.
It does not depend on the path because all adiabatic paths
between Ui and UF have to be identical
so adiabatic work is special.
It's a state function just like the internal energy.
Now James Joule, we're going to come back
to this adiabatic work business
but we're going to take a detour.
This is James Joule.
He was born in 1918 in a little town outside
of Manchester, England.
Manchester, England is right here.
This is England, okay and he was born in Salford
which is right over here.
Manchester is a big city but in those days not so big.
Salford is actually a pretty good size town itself but not
when he, not in 1918, it's a tiny little town.
Every town in England had its own brewery, in fact that's true
of most of Europe at this point in time.
Every little town in Germany has its own brewery usually just one
and so every town has its own identity in terms of a beer
that you can drink there and that's still true to this day
to a large extent and let me just tell you
that making beer is not like making soda.
To make coke, Coca Cola there's some chemists who go to a lab.
They come up with this mixture of flavorings
and they make a flavor packet and if you go to Joe's
on the Green and you buy a Coke there's a thing
of concentrated syrup that gets diluted with carbonated water
and so to make Coke you just need to flavoring.
You need carbonated water and you need sugar.
It's that easy to mix them together.
Every time you get Coke out of a machine it's just mixing these
things together for you.
The syrup actually has the sugar in it.
In beer you start with barley and hops.
There's no flavor packet that you dilute with water
to get beer that has a particular flavor.
Every beer that you buy everywhere in the world has
to go through all of these different processes and so
if you think about it it's totally amazing
that when you buy Heineken it always tastes exactly
like Heineken.
You know what I'm talking about?
It's got that distinctive Heineken taste.
They're making billions of bottles of Heineken and every,
they have to do all of this to every single bottle.
So what that means is that there is exquisite control,
I mean at a level that you might not even believe--
>> Huge log books.
>> I'm sorry.
>> They usually have humongous log box
where their statistics come from.
>> Yes. So there's a lot of process, chemical engineering
that goes into making this absolutely reproducible
for every bottle.
It is really a chemical engineering feat
that this can be done and this was not something
that we've learned in the 20th century.
This was learned in the 18th century,
maybe even the 17th century how to reproducibly make beer
that tastes the same every time.
One of the keys is here, fermentation.
The temperature here has to be controlled and this is true
to a lesser extent in these earlier processes too
but in this fermentation process the temperature has
to be controlled with better than one degree C precision
and that's not enough to get beer
to taste the same every time.
A lot of other things-- if you ever want to--
I don't know if you're fascinated by such things
but if you've never been to a big commercial brewery all
of them will give you a tour.
You go to a Budweiser brewery and you can get a tour
of the Budweiser brewery.
You walk through that place.
It is spotlessly clean.
You can't believe it.
There is, it's like a hospital in there and you look
down at this 5 acres worth of stainless steel kettles
and bottles and there's like two guys running a factory
of 5 acres making billions of bottles of Budweiser.
It's all computer-automated and process control is taking care
of this whole, everything that happens here is happening
in an automated way and Budweiser is beechwood-aged.
You look down and there's a guy shoveling beechwood
into a stainless steel vat that is about the size of this room.
There's actually beechwood
in the stupid fermentation, beechwood.
So what does any of this have to do with thermodynamics?
Jim Joule, his dad was a brewer in Salford.
His dad made the beer in Salford
and like I said there was only one guy who did that
and he understood chemistry from being a brewer's kid.
He learned how to run the brewery,
in fact he did run the brewery
after his dad had some medical problems.
He went to school in Manchester for two years
with his brother Benjamin,
studied with a guy name James Dalton.
Anybody know that name?
What did he do?
Atomic theory.
He studied with James Dalton for two years
and then there was some sort of health problems back home and he
and his brother went back to run the brewery.
He never had any more college education than that
but he was interested.
He had an amateur's interest in science
and that really was driven by understanding how
to make a brewery run more efficiently.
He wanted to understand what are the limits and efficiencies
that we can achieve in this brewery
so we can maximize our profits.
He was thinking on a very practical level.
So one of the things he wanted
to understand is what is the relationship
between work and heat?
Now that is a pretty profound thing to want to understand
when you're brewing beer.
But he was smart enough to understand
that that was an important thing for him to appreciate in terms
of the brewing process.
And so the most famous experiment
that he did involved taking a weight using
that to drive an agitator inside a vessel of water
and imagine how hard this experiment would be?
You put a weight here.
You drop this weight.
This thing spins like crazy in a bucket of water
and that's a thermometer.
You're going to measure the temperature change.
Are you kidding me?
The temperature's going to change, not by much?
He could measure the temperature to 1/200th degree Fahrenheit.
He was the only guy in the world who could make
that measurement and he measured it.
He saw a reproducible temperature change
and he could correlate the temperature change
with the distance that this mass value and with the size
of this mass and the volume
of the water he figured all of that out.
The quantity of work that must be expended at sea level
in the latitude of Greenwich in order
to raise the temperature one pound of water weight in vacuo
by one degree Fahrenheit from 60
to 61 degrees Fahrenheit is equivalent
to the mechanical force associated
with raising 772.55 pounds through one foot.
He measured it to 5 sig figs.
He measured it to one part in 10 thousand.
It is inconceivable how difficult it was to do that
and when he went around and gave talks in England and elsewhere
in Europe nobody believed him because he would show data,
he would say here's my data.
I'm measuring 1700ths of a degree change and they would go,
"There's no way you could do that reproducibly."
Nobody else can make item measurements with this level
of precision but where did he learn how to do that?
In the brewery, he'd been making precision temperature
measurements for years.
You've got to do that to run the stupid brewery.
He took it to a new level mind you,
so that number is right there on his gravestone.
He showed the equivalence between work and heat.
That's a pretty important thing to appreciate.
And I know what you're thinking, yes, you too.
This number is still available to you if you should want it.
[Laughter] Nobody, not everybody is going to honk at you
if you have this on your license plate.
Imagine what kind of nerd it has to be
to recognize, no, not too likely.
Okay, so that's not the experiment we care about.
He tried to do a harder experiment.
He's less famous.
He's really famous for the experiment
where the weight falls.
You get one degree change, 700.
He tried to do another experiment.
All right, you try to do this experiment right here.
He pressurized one bulb.
These tan things here are glass bulbs.
He pressured one with some nitrogen,
the other one evacuated and then he put a thermometer
in this water bath and he opened up this valve and vroom,
the gas goes from this bulb to this bulb and he's looking
for a temperature change here and nothing happened.
Okay, so the question is first of all what was he thinking?
He was convinced that if he did this he would see a
temperature change.
Why? This guy knew a lot of physical chemistry and he knew
that there should be a temperature change
if he did this carefully enough.
If the volume of water here was small enough--
it wasn't in his experiment.
He should see a temperature change.
So let's see if we can understand what he was thinking.
What could he have been thinking?
How many people have seen this before?
This is Leenard-Jones 6-12 potential.
Good. Where on earth did you see it?
Steven, don't put your hand up.
Who put their hand up?
Where did you the guys see this?
G-Chem? Okay.
What is it?
All right, it's the energy between two molecules
that are not going to form a covalent bond.
It's a non-covalent bonding interaction that we're talking
about here, okay and so there's an equilibrium bond distance.
Let's say we've got two neon atoms.
If they're far apart this is the distance
between them here on this axis here.
This is the energy.
For an ideal gas there's no energy.
An ideal gas this is the energy that you get
as function of distance.
In other words, in fact in an ideal gas we assume
that the gas is a point particle.
We don't even assume it has any volume.
So there is no interaction energy
as we bring the gas molecules close together nothing happens
but in a real gas there's long-range attraction.
That's what this is and then there's a short-range repulsion.
That's what this is and this repulsion is approximately R
to the 12th and this attraction's approximately R
to the 6th, so if I write this red line here is the sum
of the repulsive potential plus the attractive potential,
in other words if I add this dash line
to this dash line right here I get this red line
and that's what's given by this equation
and that's the Leenard-Jones 6-12 potential.
Okay, what does the Leenard-Jones 6-12 potential
tell us?
It tells us that there's an interaction energy, epsilon
and it tells us that there's an equilibrium bonds distance REQ.
All right, REQ is not in here but the minimum of this curve is
at REQ, equilibrium bond distance.
This is a Van der Waals bond that we're talking about here.
These energies, a hundred wave numbers,
how big is a hundred wave numbers?
How much thermal energy is there
in wave numbers at room temperature?
Two hundred, all right so is neon going to be a gas
or a liquid at room temperature?
A gas, all right because it's only got,
neon atoms are only held together
with a hundred wave numbers of binding energy.
If there's 200 wave numbers
of thermal energy they're going to be out the door.
They're going to be out here.
They're going to be a gas.
This is a liquid.
Are you with me?
Now, yes, for an ideal gas the intermolecular potential
everywhere is equal to zero.
That's this dash line right here.
At high pressures of course you're here.
In other words if you take the gas and you pressurize it
as hard as you can you eventually ***
up against this repulsive wall here and this pressure
that you measure for the system is higher
than you would expect based on VN and T. In other words,
if you measure V, the number of moles of gas and the temperature
and you calculate what the pressure should be in a real gas
if you press on it hard enough the pressure is higher
than you could ever achieve for an ideal gas
at those same volumes number of moles and temperature.
Likewise if you're at sort of normal pressures here,
all right if you measure this pressure it's actually lower
than you would expect for this volume, this number of moles
of gas and this temperature.
A real gas has a lower pressure here and a higher pressure here
because it's not ideal.
This is its non-ideality expressing itself.
Now let me recall for you
that there's something called the compressibility factor
which the just the actual pressure times the molar volume
divided by RT but that should be really big P not little P,
so in other words for an ideal gas this would just be one,
wouldn't it?
But for a real gas it's not one.
It could be higher than one at high pressure and lower than one
at moderate pressures.
So if I look at what this compressibility factor is,
here's that compressibility factor again,
come on don't fail me now.
All right, at high pressures,
this is pressure on this axis here.
All right, this is the compressibility factor here.
Here's one.
This black line is one.
At high pressures the compressibility factor is
greater than one.
Why? Because these gas atoms are banging into one another.
You can't compress the gas anymore
because it's got finite size
and in an ideal gas we assume the gas particles have no size
at all.
They're just point vertices in space.
At low pressure, the compressibility factor is less
than one because gas molecules are exerting an attraction
on one another at long distances and that lowers the pressure.
In the absence of those attractions there would be more
pressure on the vessel.
Do you see what I'm talking about?
But if the molecules are attracting one another they are
reducing the pressure that those gas molecules are applying
to the outside of the vessel.
Okay, so these two regions, this is a region
where repulsions dominate the gas behavior.
This is a region
where attractions dominate the gas behavior.
Got all that?
Now we're going to do a thought experiment.
What I just told you is what Jim Joule knew intuitively
and probably you knew it too,
so now he does this thought experiment in his head.
Let's say that we've got some gas molecules
at a normal pressure.
Now I think you can appreciate that in a gas atoms are moving
around and there's every possible intermolecular distance
because there's collisions occurring
but imagine the average nearest neighbor distance.
Let's say that you can calculate that.
How would you do that?
Take a bunch of snapshots of where all the molecules are
and then calculate what the nearest neighbor distance is
for each atom.
You'll need a super-computer.
Take the average of that.
The average nearest neighbor distance, are you with me?
Let's say it's here at a particular pressure
and now let's make the pressure lower.
The only way to do that would be to suck some of the gas
out of the container or to increase the volume,
one of the two, right?
Okay, take the molecules
and expand them to a normal pressure.
That puts us here.
This will now be the new average nearest neighbor distance.
I think you can see if I make the pressure lower
on average gas molecules are going
to be further apart, right?
Okay, so if I start here at high pressure and I end here
at low pressure this energy is the work required
to separate these molecules.
I've got to do that work for every molecule in the container.
Are you with me?
And the question is where does that energy come
from because I've got to put this energy into the system
because I've got to go from here to here?
That energy has got to come from somewhere.
Where does it come from?
Well, the first order it comes from the gas itself.
The gas can give up energy.
If the gas is at finite temperature
and that temperature is characterized by kinetic energy
of the gas molecules if the kinetic energy
of the gas molecules goes down the velocity
of the gas molecules decreases the temperature will go down.
That was Jim Joule's insight.
This is what he understood.
If I increase the pressure I should pull molecules apart
along this potential here and this should get colder.
Now the only question is am I good enough
at measuring temperature and designing the experiment
so that I can measure that delta T?
And it turned out he wasn't.
He got delta T equals zero which is the right answer
for an ideal gas but for
on ideal gas these interactions don't exist,
so we don't expect there to be any heat flows.
You expand an ideal gas, an ideal gas is here.
There's no difference between this point and this point
for an ideal gas in terms of its energy so there's no heat flows.
Are you with me on that?
Okay, so he had to, he didn't have to but he met,
when he was giving one of these talks where everybody sort
of laughed him out of the room this guy William Thompson was
at one of these talks.
He was a hot shot chemical physicist over in Scotland
which is not too far from Manchester, a few days on horse.
This guy's name, he was knighted.
He became Lord Kelvin.
Yes, that Kelvin, so together he goes and he talks to this guy
after he gives one of these, James Joule gives one
of these talks where he's measuring 17 parts per hundred
temperature change and people are just shaking their heads
and Thompson's in the audience and he comes and sees him
and says, "Look, I think I can help you do this experiment
in a way that we can measure this temperature change.
I agree. I think your right.
It's happening.
What we're going to do is we're going
to make the gas its own thermal bath."
What he was doing before is counting on the transmission
of this thermal energy in these two spheres to this water tank
and water has got an enormous heat capacity,
okay and so you have to put a lot of heat into the water
to get the temperature to change very much.
But if you're clever you can use the gas as its own thermal bath,
so here's what they did.
Here's the defined pressure on this side,
a defined pressure on this side.
This is high pressure.
This is low pressure.
This is an orifice, a tiny needle pin hole
between these two chambers.
I've drawn it big here but it's just a pin hole.
It's got to be small.
You'll see why in a second
because what we do is we transfer this gas
at high pressure to this low pressure region and we make sure
that these pressures remain constant during the process.
So this is nontrivial experiment.
Even today you would need feedback control
of the pressure to do this properly.
I don't know how they did it in those days.
They must have had some sort of pressure transducer.
I don't know what that could have been.
This is an insulated container
so there's no heat flows outside of the container.
All the heat has got to be transferred from here
to here or vice versa.
That's key to making this experiment work
and so you blow this gas through this orifice.
It's hissing as this happens.
Can you imagine that?
And you coordinate the motions of these two pistons
so these two pressures remain constant
and you measure temperature in both of those two compartments.
All right, and James Joule
by golly knew though measure temperature.
So you see how the experiment is different?
Now there's a tiny thermal mass here compared
to what he was doing before.
He had a water tank, you know, good luck.
Here this has got a tiny heat capacity
by comparison, so does this.
Now if there's a temperature change we're much more likely
to be able to measure it.
So here's how the math works.
Work on the left-hand side, P1 times delta V we know that,
minus P1 mind you on the right hand side, same thing.
For simplicity let's just say we push the piston all the way in.
Then this V is just equal to that difference it's
that whole volume there and this V is just
that whole volume right there.
All right, so that's V1, that's V2.
Here the final volume is bigger than the initial volume.
Here the final volume is smaller than the initial volume
so the total work is the sum of the work on the left side
and the work on the right side.
That's P1, V1.
This is positive because that's smaller than that.
Here it's negative because that's bigger than that,
okay and so we get minus delta PV, so that difference.
Okay, since the apparatus is insulated this is all insulation
here Q equals zero and so U is equal to W
and that's minus delta PV
and rearranging this equation we notice
that this is just equal to that.
I move that over to the left-hand side
and they're both equal to Q. So this is an isenthalpic process.
Delta H or Q, either one is equal to zero.
And what they measured is the change in temperature
at constant pressure for a process that involved no change
in enthalpy, no heat flows.
All right, that's the Joule constant coefficient.
The change in temperature with pressure and what Joule measured
in his first experiment was zero.
He didn't see any change in temperature with pressure.
And that's the right answer for an ideal gas,
so later on we're not going to derive this equation now
but suffice it to say, later on we will derive it.
It's not an issue for a Mid-term 1 and so if you just believe me
for the time being that this equation is correct, look,
we can evaluate this derivative right here.
What's the partial derivative of volume with respect
to temperature for an ideal gas?
Well it's just NR over P, right?
Even I can do that derivative.
Plug that into DV DT at constant P and that's zero.
If I substitute for V, I just get this whole thing again.
All right, so for an ideal gas the Joule constant coefficient
should be zero and that's what he measured
in his first experiment.
It shouldn't have measured zero but that's what he measured.
All right, so they were very happy.
With Thompson's strategy
for doing this experiment together they got Joule-Thompson
coefficients that were not zero.
They could see exactly the physics
that they hoped they would see.
The gas got cooler when it jetted through this orifice
and they could measure it for different gases.
This is what the data looks like today
when you have all the modern conveniences
to make these measurements with high precision.
All right, these Joule-Thompson coefficients are positive.
Okay, we can look them up.
There are tables for Joule-Thompson coefficients
for different gases correlating temperatures.
The Joule-Thompson coefficient of air for example,
300 degrees Kelvin and 25 degrees atm is equal
to 0.173 Kelvin per atm.
It's really known to very high precision.
If a Joule-Thompson expansion is carried out from a pressure
of 50 atm to a pressure
of one atm estimate the final temperature
if the initial temperature is 300 degrees Kelvin.
Okay so we're starting out with 300 degrees Kelvin and 25 atm
and we're going to expand into a pressure of one atm.
That's a pretty big pressure change, factor of 50.
We should be able to measure something.
Now the easiest way to do this calculation which is the way
that we'll do it right now is just by assuming
that this Joule-Thompson coefficient is invariant
over this range of temperatures.
That's not true.
The Joule-Thompson coefficient does depend weakly
on temperature.
But let's just assume for the time being that it does not
in this case because we want to get a feeling
for how big these temperature changes are.
Fifty atm-- so the delta T that we're going
to measure is just the Joule-Thompson coefficient times
delta P because we're just going
to linearize this partial derivative,
so that's really easy because then it's just the
Joule-Thompson coefficient times delta P, by golly
and that's going to give us directly our delta T
and there's the Joule-Thompson coefficient.
Change in pressure is 50 atm.
I got minus 8 degrees Kelvin so I'm going to go 300 K to 292 K
which is a cooling of 8 degrees.
That's not a lot of cooling is it?
It's really not a lot of cooling and you know when you think
about the experiment that Joule and Thompson had
to do this device that's got to two pistons the reason it works
so well is because it doesn't have very much thermal mass.
The gas inside these two pistons doesn't have a lot
of thermal mass and so a small heat change changes the
temperature quite a bit but the downside is
that if you stick a thermometer
in those pistons there isn't great thermal contact
between the gas and the thermometer.
The thermometer has got a lot of heat capacity.
To change its temperature it's going--
so somehow they figured all this out.
Somehow they had thermometers,
they must have been tiny thermometers otherwise there
would be no way they would be able to see anything
because this would be a pretty heroic experiment.
Fifty atm is enormous pressure isn't it?
And so we're only measuring 8 degrees Kelvin here.
That would be easy for us to measure today
but these guys were really very, very good experimentalists.
So what am I showing you here?
This is actually a plot from your Chapter 14
and what it shows is the temperature as a function
of pressure and what it shows is the temperature goes
and then it goes back down
and the Joule-Thompson coefficient is the slope
of these isenthalps.
These are iso-enthalpy traces,
so to Joule-Thompson co-efficient is negative here,
positive here and negative here
so at any particular pressure there are two
inversion temperatures.
The sign of the Joule-Thompson coefficient changes here
and then it's positive.
This is the region that we were just talking about.
This is the so-called normal region for the behavior
of the gas and then as we go through this bottom part
of this sideways parabola it inverts again.
Is there a nice intuitive way to think about this?
No, not for me and if you have one I would
like to know what it is because this has never been an intuitive
concept to me.
Why does the Joule-Thompson coefficient change signs
like this?
How could it be positive at high pressure, sorry,
negative at high pressure, negative at low pressure
but positive in between?
The fact that it's positive here that part I can understand.
That's the explanation that I gave earlier.
We got this potential.
We're going to move from here to here.
We've got to put energy into the system to make that change.
It has to work this way.
Those physics apply here, other physics here and here.
So now what we've just learned is how the
to design a refrigerator it turns out and I'm not going
to walk you through this because we're out of time
and you don't need to know this for the mid-term exam but it's
in your book actually.
Your book walks you through how a Linde refrigerator works
and maybe we'll talk about it after the mid-term exam.
Okay? ------------------------------8c987f682f7c--