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PROFESSOR: So last time we
talked about the zeroth law,
which is the common-sense law,
which says that if you take a
hot object next to a cold
object, heat will flow from
the hot to the cold in a way
that is well defined, and it
allows you to define
temperature.
It allows you to define the
concept of a thermometer.
You have three objects, one of
them could be a thermometer.
You have two of them separated
at a distance.
You take the third one, and you
go from one to the other,
and you see whether heat flows,
when you touch one
object, the middle object,
between those two objects.
Let me talk to you about
temperature scales.
We talked about the
Celsius scale then
the Fahrenheit scale.
The late 1800's were a booming
time for temperature scales.
People didn't really realize
how important it was to
properly define the reference
points: Fahrenheit's
warm-blooded or 96 degrees,
and Romer's 7.5 degrees.
Romer because he didn't want
to go below zero degrees
measuring temperature
outside in Denmark
Those are kind of silly.
But they're the legacy that
we have today, and
that's what we use.
In science, we use somewhat
better temperature scales.
And the temperature scale that
turns out to be well-defined
and ends up giving us the
concept of an absolute zero is
the ideal gas thermometer.
So, let's talk about that
briefly today first.
The ideal gas thermometer.
It's based on Boyle's law.
Boyle's law was an empirical law
that Mr. Boyle discovered
by doing lots of experiments,
and Boyle's law says that the
limit of the quantity pressure
times the molar volume, so
this quantity here, pressure
times the molar volume, as you
let pressure go to zero.
So, you do this measurement, you
measure with the gas, you
measure the pressure and
the molar volume.
Then you change the pressure
again, and you measure the
pressure in the volume, and
you multiply these two
together, and you keep doing
this experiment, getting the
pressure smaller and smaller,
you find that this limit turns
out to be a constant,
independent of the gas.
It doesn't care where
the gas is.
You always get to the
same constant.
And that constant turns out
to be a function of the
temperature.
The only function it is -- it
doesn't care where the gas is.
It only cares where the
temperature is.
All right, so now we have the
makings of a good thermometer
and a good temperature scale.
We have a substance.
The substance could
be any gas.
That's pretty straightforward.
So now we have a substance,
which is
a gas, with a property.
So now the volume of mercury,
or the color of something
which changes with temperature,
or the
resistivity.
In this case here, our property
is the value of the
pressure times the volume,
times the molar volume.
That's the property.
The property is the limit as p
goes to zero of pressure times
molar volume.
It's a number.
Measure it.
It's a number.
It's going to come out.
That's the property that's going
to give us the change in
temperature.
Then we need some reference
points.
And Celsius first used the
boiling point of water, and
called that 100 degrees Celsius,
and the freezing
point of water and called that
zero degrees Celsius.
And then we need an
interpolation scale.
How to go from one reference
point to the
other with this property.
This property, which we're
going to call f(t).
There are many ways you can
connect those two dots.
If I draw a graph, and on one
axis I have this temperature.
The idea of temperature with two
reference points, zero for
the freezing point of water,
100 degrees for the boiling
point of water.
And on the y-axis I've got
the property f(t).
It has some value
corresponding to t equals zero.
So let's get some value
right here.
There's another value connected
to this property
here, when t is equal to 100,
a reference point here.
Now there many ways
I can connect
these two points together.
The simplest way is to
draw a straight line.
It's called the linear
interpolation.
My line is not so straight,
right here.
You could do a different
kind of line.
You could do a quadratic,
let's say.
Something like this.
That would be perfectly
fine interpolation.
All right, we choose to have
a linear interpolation.
That's a choice, and that choice
turns out to be very
interesting and really
important, because if you
connect these two points
together, you get a straight
line that has to intercept
the x-axis at some point.
Now what does it mean to
intercept the x-axis here?
It means that the value
of f(t) for this
temperature is zero.
That means that at this point
right here, f(t)=0.
That means the pressure times
the volume equals
zero, for that gas.
And if you're below this
temperature here, this
quantity, p times v it
would be negative.
Is that possible?
Can we have p v negative?
Yes?
No, it can't be.
Negative pressure doesn't
make any sense, right?
Negative volume doesn't
make any sense.
That means that this part
here, can't happen.
That means that this temperature
right here is the
absolute lowest temperature you
can go to that physically
makes any sense.
That's the absolute zero.
So the concept of an absolute
zero, a temperature below
which you just can't go, that's
directly out of the
scheme here, this linear
interpolation scheme with
these two reference points.
If I had taken as my
interpolation scheme, my white
curve here, I could go to
infinity and have the
equivalent of absolute
zero being at
infinity, minus infinity.
So, this temperature, this
absolute zero here, which is
absolute zero on the
Kelvin scale.
The lowest possible temperature
in the Celsius
scale is minus 273.15
degrees Celsius.
So that begs the notion of
re-referencing our reference
point, of changing our
reference points.
To change a reference point
from this point here being
zero, instead of this point
here being zero.
And so redefining then the
temperature scale to the
Kelvin scale, where t in degrees
Kelvin is equal to t
in degree Celsius,
plus 273.15.
And then you would get
the Kelvin scale.
All right, it turned out that
this thermometer here wasn't
quite perfect either.
Just like Fahrenheit measuring
96 degrees being a
warm-blooded, healthy
man, right,
that's not very accurate.
Our temperature probably
fluctuates during the day a
little bit anyways, it's
not very accurate.
And similarly, the boiling
point, defining that at a 100
degrees Celsius, well that
depends on the pressure.
It depends whether you're in
Denver or you're in Boston.
Water boils at different
temperatures, depending on
what the atmospheric pressure
is; same thing for the
freezing point.
So that means, then, you've
got to define the pressure
pretty well.
You've got to know where
the pressure is.
It would be much better if you
had a reference point that
didn't care where the
pressure was.
Just like our substance doesn't
care where the gas is.
It's kind of universal.
And so now, instead of using
these reference points for the
Kelvin scale, we use the
absolute zero, which isn't
going to care what
the pressure is.
It's the lowest number
you can go to.
And our other reference point
is the triple point of water
-- reference points become zero
Kelvin, absolute zero,
and the triple point.
The triple point of water is
going to be defined as 273.16
degrees Kelvin.
And the triple point of water
is that temperature and
pressure -- there's a unique
temperature and pressure where
water exists in equilibrium
between the liquid phase, the
vapor phase, and the
solid phase.
So the triple point is liquid,
solid, gas, all in
equilibrium.
Now you may think, well
I've seen that before.
You take a glass of ice
water and set it down.
There's the water phase, there's
the ice cube is the
solid phase, and there's
some water, gas, vapor,
and that's one bar.
Where am I going wrong here?
The partial pressure of the
water, of gaseous water, above
that equilibrium of ice
and water is not one
bar, it's much less.
So the partial pressure or the
pressure by which you have
this triple point, happens
to be 6.1 times 10
to the minus 3 bar.
There's hardly any vapor
pressure above
your ice water glass.
So this unique temperature and
unique pressure defines a
triple point everywhere,
and that's a
great reference point.
Any questions?
Great.
So now we have this ideal gas
thermometer, and out of this
ideal gas thermometer, also
comes out the ideal gas law.
Because we can take our
interpolation here, our linear
interpolation, the slope
of this line.
Let's draw it in degrees
Kelvin, instead
of in degrees Celsius.
So we have now temperature
in degrees Kelvin.
We have the quantity
f(t) here.
We have an interpolation scheme
between zero and 273.16
with two values for this
quantity, and we have a linear
interpolation that defines our
temperature scale, our Kelvin
temperature scale.
And so the slope of this thing
is f(t) at the triple point,
which is this point here, this
is the temperature of the
triple point of water,
divided by 273.16.
That's the slope of that line.
The quantity here, which is
f (t of the triple point),
divided by the value
of the x-axis here.
So that's the slope, and the
intercept is zero, so the
function f(t), you just
multiply by t here.
This is the slope. f(t)
is just the limit.
As p goes to zero of
p times v bar.
And so now we have this
quantity, p times v bar, and
the limit of p goes to zero is
equal to a constant times the
temperature.
That's a universal statement.
It's true of every gas.
I didn't say this is only true
of hydrogen or nitrogen, This
is any gas because I'm taking
this limit p equals to zero.
Now this constant is
just a constant.
I'm going to call it r.
I'm going to call it r.
It's going to be the gas
constant, and now I have r
times t is equal to the limit,
p goes to zero of p r.
It's true for any gas, and if I
remove this limit here, r t
is equal to p v bar, I'm going
to call that an ideal gas.
See, this is the property
of an ideal gas.
What does it mean, ideal gas?
It means that the molecules or
the atoms and the gas don't
know about each other.
They effectively
have no volume.
They have no interactions
with each other.
They occupy the same
volume in space.
They don't care that there
are other atoms
and molecules around.
So that's basically what
you do when you
take p goes to zero.
You make the volume infinitely
large, the density of the gas
infinitely small.
The atoms or molecules in the
gas don't know that there are
other atoms and molecules in the
gas, and then you end up
with this universal property.
All right, so gases that have
this universal property, even
when the pressure is not zero,
those are the ideal gases.
And for the sake of this class,
we're going to consider
most gases to be ideal gases.
Questions?
So now, this equation here
relates three state functions
together: the pressure, the
volume, and the temperature.
Now, if you remember, we said
that if you had a substance,
if you knew the number of moles
and two properties, you
knew everything about the gas.
Which means that you can
re-write this in the form,
volume, for instance, is equal
to the function of n, p, t..
In this case, V = (nRT)/P. Have
two quantities and the
number of moles gives you
another property.
You don't need to
know the volume.
All you need to know is the
pressure and temperature and
the number of moles
to get the volume.
This is called an equation
of state.
It relate state properties
to each other.
In this case it relates the
volume to the pressure and the
temperature.
Now, if you're an engineer, and
you use the ideal gas law
to design a chemical plant or
a boiler or an electrical
plant, you know, a steam plant,
you're going to be in
big trouble.
Your plant is going to blow up,
because the ideal gas law
works only in very small range
of pressures and temperatures
for most gases.
So, we have other equations
of states for real gases.
This is an equation of state
for an ideal gases.
For real gases, there's a whole
bunch of equation the
states that you can find in
textbooks, and I'm just going
to go through a few of them.
The first one uses something
called a
compressibility factor, z.
Compressibility factor, z.
And instead of writing PV = RT,
which would be the ideal
gas law, we put a fudge
factor in there.
And the fudge factor
is called z.
Now we can put real instead of
ideal for our volume. z is the
compressibility factor, and z is
the ratio of the volume of
the real gas divided
by what it would be
were it an ideal gas.
So, if z is less than 1, then
the real gas is more compact
then the ideal gas.
It's a smaller volume.
If z is greater than 1, then
the real gas means that the
atoms and molecules in the real
gas are repelling each
other and wants to have
a bigger volume.
And you can find these
compressibility factors in tables.
If you want to know the
compressibility factors for
water, for steam, at a certain
pressure and temperature, you
go to a table and you find it.
So that's one example of a
real equation of state.
Not a very useful one for our
purposes in this class here.
Another one is the
virial expansion.
It's a little bit more useful.
What you do is you take that
fudge factor, and you expand
it out into a Taylor series.
So, we have the p v real
over r t is equal to z.
Now, we're going to take z and
say all right, under most
conditions, it's pretty close to
1, when it's an ideal gas.
And then we have to add
corrections to that, and the
corrections are going to
be more important, the
larger the volume is.
Remember, it's the limit of p
times v goes to zero, so if
you have a large volume with a
large pressure, then you're
out of the ideal gas regime.
So let's take Taylor series in
one over the volume, it's
going to be one over the volume
squared, etcetera.
And these factors on top, which
are going to depend on
the temperature, are the virial
coefficients, and those
depend on the substance.
So you have this p B(t) here.
This is called a second
virial coefficient.
And then, so you can get,
you can actually find a
graph of this B(t).
It's going to look something
like this.
It's the function of
temperature, as B(t).
There's going to be some
temperature where B(t) is
equal to zero.
In that case, your gas is
going to look awfully
like an ideal gas.
Above some temperature is going
to be positive, below
some temperature is going
to be negative.
Generally, we ignore the
high order terms here.
So again, if you do a
calculation where you're close
enough to the ideal gas, and
you need to design your, if
you have an engineer designing
something that's got a bunch
of gases around, this is
a useful thing to use.
Now, the most interesting one
for our class, the equation of
state that's the most
interesting, is the Van der
Waals equation of state,
developed by Mr. Van
der Waals in 1873.
And the beauty of that equation
of state is that it
only relies on two parameters.
So let's build it up.
Let's see where it comes from.
Let me just first write it
down, the Van der Waals
equation of state. p plus a over
v bar squared times v bar
minus b equals r t.
All right, if you take a equal
to zero, these are the two
parameters, a and b.
If you take those two equal
to zero, you have p v
is equal to r t.
That's the ideal gas law.
Let's build this up.
Let's see where this comes from,
where these parameters a
and b comes from.
So, the first thing we're going
to do is we're going to
take our gas in our box,
let's build a box
full of gases here.
We've got a bunch of gas
molecules or atoms.
OK, there's the volume
of a box here.
While these gas molecules
or atoms through first
approximation, are like
hard spheres.
They occupy a certain volume.
Each atom or molecule occupies
a particular volume.
And so, we can call b is the
volume per mole of the hard
spheres, volume per mole that is
the little sphere that the
molecules are.
So that the volume that is
available to any one of those
spheres is actually smaller than
v. Because you've got all
these other little spheres
around, so the actual volume
seen by any one of those spheres
is smaller than v. So
when we take our ideal gas law,
p v bar is equal to r t
we have to replace v bar by the
actual volume available to
this hard sphere.
So instead of v bar, we write
p v bar minus b, equal r t.
OK, that's the hard sphere
volume of the spheres.
Now, those molecules or atoms
that are in here, also feel
each other.
There are a whole bunch of
forces that you learn in
5.112, 5.111 like with Van
der Waals' attractions
and things like this.
So there are attractive forces,
or repulsive forces
that these molecules feel, and
that's going to change the
pressure that the
molecules feel.
For instance, if I have,
what is pressure?
Pressure is when you have one
of these hard spheres
colliding against the wall.
There's the hard sphere.
It wants to collide against the
wall to create a force on
the wall, and I have a couple
of the hard spheres that are
nearby, right, and in the
absence of any interactions, I
get a certain pressure.
This thing would but careen
into the wall, kaboom!
You'd have this little force,
but in the presence of these
interactions, you've got these
other molecules here that are
watching this, you know, their
partner sort of wants to do
damage to themselves, like
hitting that wall,
and they say, no!
Come back, come back, right?
There is an attractive force.
There are no other molecules
on that side of the wall.
So there's an attractive force
that makes the velocity within
not quite as fast.
The force is not quite as strong
as it was without this
attractive force.
So the real pressure is not
quite the same because of this
attractive force as it was,
as it would be without the
attractive forces.
The pressure is a little bit
less in this case here.
So instead of this p here.
Now if I re-write this equation
here as p is equal to
r t divided by v bar minus
b, just re-writing this
equation as it is.
So the pressure is going to
depend on how strong this
attractive force is.
So the pressure is going to be
less if there's a strong
attractive force.
And the 1 over v squared is a
statistical, is basically a
probability of having another
molecule, a second molecule in
the volume of space.
So, if the molar volume is
small, then one over v bar is
large, there's a large
probability of having two
spheres together in
the same volume.
If the molar volume is large,
that means that there's a lot
of room for the molecules, and
they're now going to be close
to each other, and
so this isn't
going to be as important.
So, a is the strength of the
interaction, v bar is how
likely they are to be
close to each other.
And that's going to
affect the actual
pressure seen by the gas.
And a is greater than zero when
you have the attraction.
And that gives use the Van der
Waals' equation of state, with
two parameters, the hard
sphere volume and the
attraction.
You don't have to go look
up in tables or books.
You don't have to have all the
values of the second virial
coefficient, or the fudge
factor, just two variables
that make physical sense, and
you get an equation of state
which is a reasonable equation
of state, and that's the power
of the Van der Waals' equation
of state, and that's the one
we're going to be using
later on this class to
describe real gases.
Question?
OK, so we've done
the zeroth law.
We've done temperature,
equations of state.
We're ready for the first law.
We're just going to go to
through these laws pretty
quickly here.
Remember, the first law
is the upbeat law.
It's the one that says,
hey, you know,
life is all rosy here.
We can take energy from fossil
fuels and burn it up and make
it heat, and change that
energy into work.
And it's the same energy, and
we probably can do that with
100% efficiency.
We can take heat from the air
surrounding us and run our car
on it with 100% efficiency.
Is this possible?
That's what the first law says,
it's possible; work is
heat, and heat is work, and
they're the same thing.
You can break even, maybe.
So let's go back and
see what work is.
Let's go back to our
freshman physics.
Work, work is if you take a
force, and you push something
a certain distance,
you do work on it.
So if I take my chalk here and
I push on it, I'm doing work
to push that chalk.
Force times distance is work.
The applied force times
the distance.
There are many kinds of work.
There's electrical work, take
the motor, you plug it into
the wall, electricity makes
the fan go around, that's
electrical work.
There's magnetic work.
There is work due to gravity.
In this class here, we're going
to stick to one kind of
work which is expansion work.
So expansion work, for instance,
or compression work,
is if you have a piston
with a gas in it.
All right, you put a pressure
on this piston here, and you
compress the gas down.
This is compression work.
Now the volume gets smaller.
p external here.
Pressure, the piston goes
down by some volume l.
The piston has a cross-sectional
area, a, and
the force -- pressure is
force per volume area.
So the force that you're pushing
down on here is the
external pressure
times the area.
Pressure is force
per volume area.
That's the force you're
using to push down.
Now the work that's it is
calculated when you push down
with the pressure on this piston
here, that work is
force times distance, f times
I. f is p external times a,
times the distance l.
So that's p external times
the change in the volume.
The area times this distance is
a volume, and that is the
change in volume from going
to the initial state
to the final state.
Now we need to have
a convention.
We've got force.
Work is force times distance,
it's p external times delta v,
and I'm going to be stressing
a lot that this is the
external pressure.
This is the pressure that you're
applying against the
piston, not the pressure
of the gas.
It's the pressure the external
world is applying on this poor
system here.
OK, but we need a
convention here.
The convention, and then
we need to stick to it.
And this convention,
unfortunately, has changed
over the ages.
But we're going to pick one,
and we're going to stick to
it, which is that if the
environment does work on the
system, if we push down on this
thing and do work on it,
to compress it, then we call
that work negative work.
No, we call that work
positive work.
All right, so that means we need
to put a negative sign
right here, by convention.
So if delta v is negative, in
this case delta v is negative,
OK, delta v is negative,
pressure is a positive number,
negative times negative
is positive, work is
greater than zero.
We're doing work on the
system, to the system.
In this case here,
work is positive.
If you have expansion on the
other side, if the system is
expanding in the other
direction, if you're going
this way, right, you're going to
do work to the environment.
There might be a mass here.
This could be a car.
Pistons in the car, right,
so the piston goes up.
That's going to drive
the wheels.
The car is going
to go forward.
You're doing work on
the environment.
Delta v is going to
be negative. w
is going to be negative.
Sorry, I got it backwards
again.
Delta v is positive
in this direction
here, the work is negative.
So work on the system
is positive.
Work done by the system
is negative.
Convention, OK, this negative
sign is just a pure
convention.
You just got to use
it all the time.
If you use an old textbook,
written when I was taking
thermodynamics, they have the
opposite convention, and it's
very confusing.
But now we've all agreed on this
convention, and work is
going to be with the
negative sign here.
OK, any questions?
This is an example where the
external pressure here is kept
fixed as the volume changes,
but it doesn't
have to be kept fixed.
I could change my external
pressure through the whole
process, and that's the path.
We talked about the path last
time being very important.
Defining the path.
So if I have a path where my
pressure is changing, then I
can't go directly from
this large volume
to this small volume.
I have to go in little steps,
infinitely small steps.
So, instead of writing work is
the negative of p external
times delta v, I'm going to
write a differential. dw is
minus p external dv, where this
depends on the path, it
depends on path and is changing
as v and p change.
Now I'm going to add a
little thing here.
I'm going to put a little
bar right here.
And the little bar here means
that this dw that I'm putting
here is not an exact
differential.
What do I mean by that?
I mean that if I take the
integral of this to find out
how much work I've done
on the system, I
need to know the path.
That's what this means here.
It's not enough to know the
initial state and the final
state to find what w is.
You also need to know
how you got there.
This is very different from the
functions of state, like
pressure and temperature.
There's a volume, there's
a temperature, than
the pressure here.
There's other volume,
temperature and pressure here,
corresponding to this
system here.
And this volume, temperature and
pressure doesn't care how
you got there.
It is what it is.
It defines the state
of the system.
The amount of work you've
put in to get here
depends on the path.
It's not a function of state.
It's not an exact
differential.
So the delta v here is an
exact differential,
but this dw is not.
That's going to be
really important.
So if you want to find out how
much work you've done, you
take the integral from the
initial state to the final
state of dw minus from one to
two p external dv, and you've
got to know what the path is.
So let's look at this path
dependence briefly here.
We're going to do two different
paths, and see how
they're different in terms of
the work that comes out.
So we're going to take an ideal
gas, we can assume that
it's ideal.
Let's take argon, for
instance, a nice,
non-interacting gas.
We're going to do
a compression.
We're going to take argon, with
a certain gas, certain
pressure p1, volume V1, and
we're going to a final state
argon, gas, p2, V2.
Where V1 is greater than V2,
and p1 is less than p2.
So if I draw this on a p
v diagram, so there is
volume on this axis.
There's pressure on this axis.
There is V1 here.
There's V2 here.
There's p1 here, and p2 here.
So I'm starting at p1, V1.
I'm starting right here.
And I'm going to
end right here.
Initial find -- there are many
ways I can get from one state
to the other.
Draw any sort of line
to go here, right?
There are a couple obvious ones,
which we're going to --
we can calculate, which
we're going to do.
So, the first obvious one
is to take V1 to V2
first with p constant.
So take this path here.
I take V1 to V2 first, keeping
the pressure constant at p1,
then I take p1 to p2 keeping
the volume constant at V2.
Let's call this path 1.
Then you take p1 to p2
with V constant.
An isobaric process followed by
a constant volume process.
You could also do a
different path.
You could do, let me draw p v,
there's my initial state.
My final state here, I could
take, first, I could change
the pressure, and then
change the volume.
So the second process, if you
take p1 to p2, V constant, and
then you take V1 to V2
with p constant.
This is path number two.
Both are perfectly fine paths,
and I'm going to assume that
these paths are also
reversible.
Let's assume that both are
reversible, meaning that I'm
doing this pretty slowly, so
as I change, let's say I'm
changing my volumes here, V1
to V2, it's happening, I'm
compressing it slowly, slowly,
slowly so that at any point I
could reverse the process
without losing energy, right?
It's always an equilibrium.
All right, let's calculate the
work that's involved with
these two processes.
Remember it's the external
pressure that's important.
In this case, because it's
a reversible process, the
external pressure turns out to
be always the same as the
internal pressure.
It's reversible, that means
that p external, equals p.
I'm doing it very slowly so that
I'm always in equilibrium
between the external pressure
and the internal pressure so I
can go back and forth.
So, let's calculate w1.
The work for path one.
First thing is I change the
volume from V1 to V2 The
external pressure is kept
constant, p1, so it's minus
the integral from 1,
V1 to V2, p1, dv.
And then the next step here is
I'm going from -- the pressure
is changing.
I'm going from V2 to V2 dv
-- what do you think this
integral is?
Right, so this is easy
part, zero here.
This one is also pretty easy.
That's minus p1 times V2 minus
V1. p1 times V2 minus V1.
What that turns out to be,
this area right here.
It's V1 minus V2 times p1.
This is w1 here.
OK, I can re-write this as p1
time V1 minus V2 and get rid
of this negative sign here.
Now V1 is bigger than V2,
so this is positive.
So I am compressing, I'm doing
work to the system, positive
work everything follows
our convention.
Number two here, OK, the first
thing I do is I change the
pressure under constant volume,
V1, V1 minus p dv, and
then I change the volume
from V1 to V2 and
then this is p2, dv.
This first integral is zero V1
to V1, then I get minus p2
times V2 minus V1 or p2
times V1 minus V2.
Again, a positive number.
I'm doing work to the system to
go from the initial state
to the final state.
But it's not the same as w1.
In this case, I have p1 times
delta V. In this case here, I
have p2 times delta V. And p2
is bigger than p1. w2 is
bigger than w1.
The amount of work that you're
doing on the system depends on
the path that you take.
All right, how do
I, practically
speaking, how do I do this?
Anybody have an idea?
How do I keep p1 constant while
I'm lowering the volume?
STUDENT: Change the
temperature?
PROFESSOR: Change the
temperature, right.
So what I'm doing here is I'm
cooling, and then when I'm
sitting at a fixed volume and
I'm increasing the pressure,
what am I doing?
I'm heating, right?
So I'm doing cooling
and heating cycles.
So in this case here, I
cool and then I heat.
In this case here, I heat
and then I cool.
All right, so I'm burning some
energy, I'm burning some fuel
to do this somehow, to get
that work to happen.
All right, now suppose that I
took these two paths, and
coupled them together.
So in this case, it's the amount
of work is the area
under that curve.
And in this case here, the
amount of work is bigger, w2
is bigger, and it's the
area under this curve.
Now, suppose I took this two
paths, and I took -- couple
them together with one the
reverse of the other.
So I have my initial state, my
final state, my initial state,
my final state here.
And I start by taking
my first path here.
I cool, I heat.
So there's w1.
So the w total that I'm going
to get, is w1, and then
instead of the path from V1
to, from 1 to 2 going like
this as we had before, I'm going
to take it backwards.
If I go backwards, to work --
everything is symmetric, the
work becomes the negative from
what I had calculated before,
so this becomes minus what I
calculated before for w2.
The total work, in this case
here, is p1 times V1 minus p2
times V1 minus V2, it's p1 minus
p2 times V1 minus V2.
This is a positive number,
p1 is smaller than p2.
This is a negative number.
The total work is
less than zero.
That's the work that the
system is doing to the
environment.
I'm doing work to
the environment.
The work is negative, which
means that work is being done
to the environment.
And that work is the area
inside the rectangle.
What you've built
is an engine.
You cool, you heat, you heat,
you cool, you get back to the
same place, but you've just done
work to the environment.
You've just built
a heat engine.
You take fuel, rather you take
something that's warm, and you
put it in contact with the
atmosphere, it cools down.
You take your fuel, you
heat it up again.
It expands.
You change your constraints on
your system, you heat it up
some more, then you take the
heat source away, and you put
it back in contact with
the atmosphere.
And you cool it a little bit,
change the constraints, cool
it a little bit more, and heat,
and you've got a closed
cycle engine.
We're going to work
with some more
complicated engines before.
But the important part here is
that the work is not zero.
You're starting at one point.
You're going around a cycle and
you're going back to the
same point.
The pressure, temperature, and
volume are exactly the same
here as when you started out.
But the w is not zero.
The w, for the closed path, and
when I put a circle there
on my integral that means a
closed path, when you start
and end at the same point,
right, this is not zero.
If you had an exact
differential, the exact
differential around a closed
path, you would get zero.
It wouldn't care where
the path is.
Here this cares where
the path is.
So, work is not a function
of state.
Any questions on work before we
move on to heat, briefly?
So heat is a quantity that
flows into a substance,
something that flows into a
substance that changes it's
temperature, very
broadly defined.
And, again, we have a sign
convention for heat.
So heat, we're going
to call that q.
And our sign convention is
that if we change our
temperature from T1 to T2, where
T2 it's greater than T1
then heat is going
to be positive.
Heat needs to go into the
system to change the
temperature and make it go up.
If the temperature of the system
goes down, heat flows
down heat flows out of the
system, and we call that
negative q.
Same convention is
for w, basically.
Now, you can have a change of
temperature without any heat
being involved.
I can take an insulated box,
and I can have a chemical
reaction in that
insulated box.
I can take a heat pack,
like the kind
you buy at a pharmacy.
Break it up.
It gets hot.
There's no heat flowing from the
environment to the system.
I have to define my terms.
My system is whatever's
inside the box.
It's insulated.
It's a closed system.
In fact, it's an isolated
system.
There's no energy or
matter that can go
through that boundary.
Yet, the temperature goes up.
So, I can have a temperature
change which is an adiabatic
temperature change.
Adiabatic means without heat.
Or I could have a non-adiabatic,
I could take
the same temperature change, by
taking a flame, or a heat
source and heating
up my substance.
So, clearly q is going to
depend on the path.
I'm going from T1 to T2, and
I have two ways to go here.
One is non-adiabatic.
One is adiabatic.
All right, now what we're going
to learn next time, and
Bob Field is going to teach the
lecture next time, is how
heat and work are related, and
how they're really the same
thing, and how they're related
through the first law, through
energy conservation.
OK, I'll see you on
Wednesday then.
