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Let's start with our classic
system that I keep using over

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and over again.

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And that's because it tends
to be very useful for

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instruction.

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It also tends to be the system
that is most covered in

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classrooms. So hopefully it'll
be productive for you and your

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school work.

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So I have this container.

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It has a movable piston
on top, or kind

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of a movable ceiling.

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Well, of course, inside of my
system I have a bunch of

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molecules or atoms bouncing
around, creating some type of

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pressure on the system.

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So let's say it has
some pressure, P1.

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This volume right here,
let's call that V1.

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And let's say it also has
some temperature that

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it starts off with.

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Everything is in equilibrium.

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Remember these are
macro states.

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The only way I can even tell
you what the volume, or the

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pressure, or the temperature
is, is if the system is in

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equilibrium, if everything
in it is uniform.

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The temperature is consistent
throughout.

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Fair enough.

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And in order to keep it placed
down, I have to put

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some rocks on top.

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And I've done this in multiple
processes so far.

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And, of course, I'm doing these
little pebbles, because

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I'm going to remove
them slowly.

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Because I want to approximate
a quasi-static process.

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Or I want to approximate a
system that's always close

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enough to equilibrium that I'm
cool with defining our macro

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states, our pressure, our
temperature, or our volume.

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Let me write V for volume.

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Now, in this video, I'm going
to study what's called an

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isothermic process.

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And really what it just means
is, I'm going to keep the

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temperature the same.

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Iso- just means the same.

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You probably remember when we
studied the periodic table.

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Isotopes, those are the same
element just with different

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mass numbers.

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So this is the same temperature
we're going to run

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our process.

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So my question is, how
can we do that?

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Because as I remove pebbles,
what's going to happen?

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If I just did this, without
any-- if it was completely

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isolated from the world.

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And actually I'll add in
a word right here.

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If it was an adiabatic
process-- If it was

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adiabatic-- fancy word.

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All that means is, completely
isolated from the world.

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So no heat is going into
or out of this system.

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If this was the case, what would
happen as I released or

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I took away some of these
little particles?

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Let me copy and paste it.

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Well, let me just redraw
it actually.

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So I have my one wall.

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I have another wall.

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I have another wall.

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As I release a couple of
pebbles, one at a time, my

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volume is going to increase.

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So I'm going to have a slightly
higher volume.

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My volume's going to go up.

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I have fewer pebbles here now.

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And since I have the same number
of molecules, they're

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going to bump into this less.

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So pressure is going
to go down.

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Volume is going to go up.

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And if I was adiabatic, if I had
no extra heat being added

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to the system, what do I know
is going to happen to the

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temperature?

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Well, think about it this way.

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Some work was done, right?

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Our old ceiling was maybe
some place around here.

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We pushed it up with some
force for some distance.

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So we did work.

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And so we changed some kinetic
energy, or we transferred some

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kinetic energy, out
of the system.

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That's essentially what
the work did.

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That kinetic energy was
turned into work.

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And temperature is just a
macro measure of average

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kinetic energy.

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In fact, we just-- well,
I won't go into it.

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But in the last video, the kind
of proof you want-- if

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you didn't watch it because you
didn't want to go through

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the math, which is completely
fair enough because it

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normally wouldn't be done in an
intro chemistry class-- I

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showed that the internal energy
is equal to the total

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kinetic energy, which was equal
to 3/2 times the number

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of moles, times R, times
temperature.

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So temperature is just, by some
scaling factor, a measure

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of kinetic energy.

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Now, when I do some work, it's
essentially a transfer of

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kinetic energy.

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And I can't replace that energy
with some heat because

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it's adiabatic.

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There's no heat going into
or out of the system.

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So in that situation, the
kinetic energy of

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the system went down.

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The average kinetic energy
of the system went down.

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So the temperature would've
also gone down.

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And actually, just as a bonus
point, what happened to the

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internal energy?

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Well, the internal energy is
the total kinetic energy of

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the system.

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And I could even right down
the original formula.

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Change in internal energy is
equal to change-- let me not

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do that, because I said I
shouldn't-- is equal to heat

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added to the system, minus
work done by the system.

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This is work done
by the system.

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That's why we're
subtracting it.

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Now, it's adiabatic,
so there's no heat

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added to the system.

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So the change in internal energy
is equal to the minus

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the work done by the system.

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Well, in this situation,
the system did do work.

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It pushed this piston up by some
distance with some force.

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So, your delta U is negative.

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It's less than 0.

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So U went down, and
that makes sense.

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If temperature changed,
then the internal

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energy is going to change.

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And for our simple system,
where internal energy is

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represented by the kinetic
energy of these molecules,

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that's always going
to be the case.

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If temperature doesn't change,
internal energy won't change.

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If temperature goes up, internal
energy goes up.

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If temperature goes down,
internal energy goes down.

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And, of course, they're not
the same thing, though.

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The difference between
internal energy and

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temperature is the scaling
factor, 3/2 times the number

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of molecules, times our
ideal gas constant.

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So fair enough.

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I went through this whole
exercise just to show you that

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if I was completely isolated,
and if I removed a couple of

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these pebbles, that
my temperature

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is going to go down.

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Now, I told you already
that I want to do

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an isothermic process.

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So I want to do this process
while keeping the

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temperature the same.

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So how can I do that?

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Well, what I'm going to do is
I'm going to place my system

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on top of what we'll
call, a reservoir.

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So a reservoir, you can kind of
view as an infinitely large

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amount of something that is
the temperature that we

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started off with.

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So this reservoir is T1.

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So even though if, I took
two relative things

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of comparable size.

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That says, temperature A.

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This is temperature B.

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And I've put them next
to each other.

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They're going to average out to
A plus B over 2, whatever

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their temperatures are.

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But if B is massive-- if A is
just a speck of particle--

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let's say it's iron dust-- while
B is the Eiffel Tower,

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then essentially B's
temperature will

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not change a lot.

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A will just become
B's temperature.

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Now, a reservoir is

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theoretically infinitely large.

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It's an infinitely
large object.

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So if something is next to a
reservoir and is given enough

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time, it'll always assume the
heat of the reservoir, or the

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temperature of the reservoir.

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So what's going to happen?

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So this is adiabatic, but now
I'm actually putting it next

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to a reservoir.

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So this isn't going to happen.

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The adiabatic situation
isn't going to happen.

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Now, I'm going to have a
situation where I'm going to

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stay the same temperature.

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So what's that going to look
like on the PV diagram?

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So let me draw the PV diagram.

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This is my pressure.

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This is my volume.

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So this is my starting
point right here.

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And what I'm saying is, if I'm
doing an isothermic process,

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so I just keep removing
these pebbles.

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So I start at this
state right here.

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Let me copy and paste it
since I've done so

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much art already here.

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So I'm going from there to
here where I'm removing a

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couple of the particles.

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So let's say I've removed a
couple of them over here.

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And because of that, I've
increased the volume.

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So let's say the volume,
it's not there anymore.

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Let's say it's a little
bit higher.

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Let's say the volume is--
just for the sake of our

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discussion-- let's say the
volume has expanded a little

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bit, because I've remove some
particles, the little pebbles

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on the top keeping it down.

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So it's like the adiabatic
process, but instead of the

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temperature going down, my
temperature stays at T1.

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My temperature's at T1 the
entire time, because I'm next

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to this theoretical thing
called a reservoir.

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So because of that, I will
travel along what we'll call

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an isotherm.

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So this is my first state.

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When I'm done, I might end
up some place over here.

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And so this is state 2.

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So this is state 2,
this is state 1.

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What I'm claiming is that my
path along this is going to be

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00:09:18,530 --> 00:09:21,550
on some type of a rectangular
hyperbola, or at

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least part of it.

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00:09:22,550 --> 00:09:27,620
If I were to add rocks to it and
compress it, I claim that

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00:09:27,620 --> 00:09:29,920
my PV diagram would
go like this.

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00:09:29,920 --> 00:09:33,030
If I were to keep removing rocks
from this diagram, I

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00:09:33,030 --> 00:09:35,425
claim that my PV diagram would
keep going like that.

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00:09:35,425 --> 00:09:36,810
And so what's the intuition?

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That if I keep the temperature
constant, that I'm essentially

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moving along this hyperbola.

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00:09:42,840 --> 00:09:45,000
Well, let's just take out
the ideal gas formula.

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Let me box off all this
stuff over here.

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If I just take the ideal gas
formula, PV is equal to nRT.

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If T is constant-- we know that
R is a constant, it's the

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00:09:57,530 --> 00:09:58,950
ideal gas constant.

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We know that we're not changing
the number of moles

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of particles.

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Then that means that PV is
equal to some constant.

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This whole thing is equal
to some constant.

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00:10:07,140 --> 00:10:09,260
And then, if we wanted to write
P as a function of V, we

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would just write P is
equal to K over V.

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Now, this might not look 100%
familiar to you, but if I

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wrote it in algebraic terms,
if I told you to graph y is

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equal to 1 over x, what
does that look like?

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That's a rectangular
hyperbola.

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00:10:24,580 --> 00:10:25,995
That looks like this.

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00:10:25,995 --> 00:10:29,350
And this is the y-axis, that's
the x-axis, at least in this

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00:10:29,350 --> 00:10:31,710
quadrant it looks like this.

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00:10:31,710 --> 00:10:33,980
It also looks like that in the
third quadrant, but we won't

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00:10:33,980 --> 00:10:35,550
worry about that too much.

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So whenever you hold temperature
constant, you're

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00:10:37,420 --> 00:10:41,260
on some rectangular hyperbola
like this, like an isotherm.

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00:10:41,260 --> 00:10:43,150
Now, if the temperature was a
different temperature, if it

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00:10:43,150 --> 00:10:47,180
was a lower temperature, you'd
be on a different isotherm.

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So you would be on an isotherm
that looked like

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this, maybe over here.

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It would also be a rectangular
hyperbola

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but at a lower state.

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00:10:55,130 --> 00:10:56,060
Why is that?

254
00:10:56,060 --> 00:11:04,020
Because, if you're at a lower
temperature, for any volume,

255
00:11:04,020 --> 00:11:05,910
you should have a lower pressure
and that works out.

256
00:11:05,910 --> 00:11:09,070
That's why this is some
temperature T2, that

257
00:11:09,070 --> 00:11:10,630
is lower than T1.

258
00:11:10,630 --> 00:11:12,480
So I want to do a couple of
things in this video.

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00:11:12,480 --> 00:11:15,240
I inadvertently explained to you
what an adiabatic process

260
00:11:15,240 --> 00:11:17,940
is, and why the temperature
would naturally go down on its

261
00:11:17,940 --> 00:11:21,870
own if you didn't have
this reservoir here.

262
00:11:21,870 --> 00:11:23,920
But the whole reason why I even
thought about doing this

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00:11:23,920 --> 00:11:26,240
video is because I wanted you
get comfortable with this idea

264
00:11:26,240 --> 00:11:30,720
of, one, that a reservoir will
keep you in kind of an

265
00:11:30,720 --> 00:11:31,880
isothermic state.

266
00:11:31,880 --> 00:11:33,610
It will keep the temperature
the same.

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00:11:33,610 --> 00:11:35,920
And that if you keep the
temperature the same, that you

268
00:11:35,920 --> 00:11:39,940
will travel along this isotherm,
these rectangular

269
00:11:39,940 --> 00:11:40,600
hyperbolas.

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00:11:40,600 --> 00:11:41,860
And that each temperature has

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00:11:41,860 --> 00:11:44,340
associated with it an isotherm.

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00:11:44,340 --> 00:11:46,740
So if you take that, let's
just do one more step.

273
00:11:46,740 --> 00:11:53,420
And let's think about the
actual work we did by

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00:11:53,420 --> 00:11:55,470
traveling from this state
to this state.

275
00:11:55,470 --> 00:11:57,720
Or if you just want to think
of it in visual terms, from

276
00:11:57,720 --> 00:12:01,220
removing our pebbles slowly and
slowly with this reservoir

277
00:12:01,220 --> 00:12:04,420
down here the whole time, from
this state to this state.

278
00:12:04,420 --> 00:12:07,050
Where our volume has increased,
our pressure has--

279
00:12:07,050 --> 00:12:08,080
So our volume has increased.

280
00:12:08,080 --> 00:12:10,300
Our pressure has gone down. but
our temperature has stayed

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00:12:10,300 --> 00:12:12,840
the same the entire time.

282
00:12:12,840 --> 00:12:17,120
So, several videos ago, we
learned that the work done is

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00:12:17,120 --> 00:12:19,810
the area under this graph.

284
00:12:19,810 --> 00:12:23,410
It's the area under
that graph.

285
00:12:23,410 --> 00:12:26,480
Or, if we want to do in calculus
terms-- and I'm about

286
00:12:26,480 --> 00:12:29,290
to break into calculus, so if
you don't want to see calculus

287
00:12:29,290 --> 00:12:32,300
cover your eyes or ears.

288
00:12:32,300 --> 00:12:33,510
It would be the integral.

289
00:12:33,510 --> 00:12:35,730
And the rest of this video will
be a little bit mathy,

290
00:12:35,730 --> 00:12:39,430
and I guess I should make
that statement on the

291
00:12:39,430 --> 00:12:41,450
title of the video.

292
00:12:41,450 --> 00:12:44,130
But if I want to calculate
what this area is,

293
00:12:44,130 --> 00:12:45,230
I can now do this.

294
00:12:45,230 --> 00:12:49,130
The isotherm assumption makes
our math a little bit easier.

295
00:12:49,130 --> 00:12:55,410
Because we know that PV is equal
to nRT, ideal gas law.

296
00:12:55,410 --> 00:12:59,150
Or we could say P, if we divide
both sides by V, is

297
00:12:59,150 --> 00:13:02,740
equal to nRT, divided by V.

298
00:13:02,740 --> 00:13:03,580
So there we have it.

299
00:13:03,580 --> 00:13:05,800
We have P as a function of V.

300
00:13:05,800 --> 00:13:10,470
This function right here, this
graph right here, is this.

301
00:13:10,470 --> 00:13:14,450
We could write P as a function
of v is equal to nRT over V.

302
00:13:14,450 --> 00:13:16,990
So if we want to figure out the
area under the curve, we

303
00:13:16,990 --> 00:13:21,570
just integrate this function
from our starting, our V1 to

304
00:13:21,570 --> 00:13:24,340
our ending point, to our V2.

305
00:13:24,340 --> 00:13:26,190
So what is that going to be?

306
00:13:26,190 --> 00:13:29,695
Well, we're going to integrate
from V1 to V2.

307
00:13:29,695 --> 00:13:31,860
Actually, that shouldn't
be an equal.

308
00:13:31,860 --> 00:13:36,510
The work is going to be the
integral from V1 to V2, times

309
00:13:36,510 --> 00:13:42,050
our function, P as a function
of V, times dV.

310
00:13:42,050 --> 00:13:45,560
We're summing up all of the
little rectangles here.

311
00:13:45,560 --> 00:13:47,550
We did that a couple
of videos ago.

312
00:13:47,550 --> 00:13:49,420
So what's P as a
function of V?

313
00:13:49,420 --> 00:13:56,870
So work done is equal to,
from V1 to V2, nRT

314
00:13:56,870 --> 00:14:01,700
over V, times dV.

315
00:14:01,700 --> 00:14:03,620
Now, this is our simplifying
assumption.

316
00:14:03,620 --> 00:14:07,070
We said we're sitting on
top of a reservoir.

317
00:14:07,070 --> 00:14:09,150
That this reservoir keeps
our temperature the

318
00:14:09,150 --> 00:14:10,160
same the whole time.

319
00:14:10,160 --> 00:14:12,360
And we're going to learn in a
second, it's doing that by

320
00:14:12,360 --> 00:14:14,130
transferring heat
into the system.

321
00:14:14,130 --> 00:14:16,270
And we're going to calculate how
much heat is transferred

322
00:14:16,270 --> 00:14:17,770
into the system.

323
00:14:17,770 --> 00:14:20,970
So, if we look at this right
here, temperature-- since

324
00:14:20,970 --> 00:14:23,310
we're assuming we're on an
isotherm, is a constant. n and

325
00:14:23,310 --> 00:14:24,940
R are definitely constants.

326
00:14:24,940 --> 00:14:29,330
So we can rewrite this integral
as the integral from

327
00:14:29,330 --> 00:14:34,380
V1 to V2 of 1 over V, dV.

328
00:14:34,380 --> 00:14:36,140
And then we could put
the nRT out here.

329
00:14:36,140 --> 00:14:37,950
I should have done that
first. nRT, it's

330
00:14:37,950 --> 00:14:39,910
just a constant term.

331
00:14:39,910 --> 00:14:42,920
Now, what's the antiderivative
of 1 over V?

332
00:14:42,920 --> 00:14:44,780
It's the natural log of V.

333
00:14:44,780 --> 00:14:53,970
So our work is equal to nRT
times the natural log-- this

334
00:14:53,970 --> 00:14:59,180
is the antiderivative-- of V,
evaluated at V2 minus it

335
00:14:59,180 --> 00:15:00,890
evaluated at V1.

336
00:15:00,890 --> 00:15:08,730
So that is equal to nRT times,
evaluated at V2, so the

337
00:15:08,730 --> 00:15:10,370
natural log of V2.

338
00:15:10,370 --> 00:15:12,890
Minus the natural log of V1.

339
00:15:12,890 --> 00:15:15,210
Now we know from logarithm
properties is the same thing

340
00:15:15,210 --> 00:15:26,010
as nRT times the natural
log of V2 over V1.

341
00:15:26,010 --> 00:15:27,310
So there you have
it, we actually

342
00:15:27,310 --> 00:15:28,670
calculated the real value.

343
00:15:28,670 --> 00:15:32,760
If we know our starting volume
and our finishing volume, we

344
00:15:32,760 --> 00:15:34,840
can actually figure out
the work done in

345
00:15:34,840 --> 00:15:36,960
this isothermic process.

346
00:15:36,960 --> 00:15:41,760
The work done in this isothermic
process is the area

347
00:15:41,760 --> 00:15:42,730
under this.

348
00:15:42,730 --> 00:15:44,560
And we figured out
what it was.

349
00:15:44,560 --> 00:15:46,630
By pushing up that piston,
it was nRT.

350
00:15:46,630 --> 00:15:49,800
These are the number of moles
we have, ideal gas constant.

351
00:15:49,800 --> 00:15:51,460
Our temperature that
we're sitting on.

352
00:15:51,460 --> 00:15:53,140
It would be T1 in this case.

353
00:15:53,140 --> 00:15:55,730
And the natural log of our
finishing volume divided by

354
00:15:55,730 --> 00:15:57,570
our starting volume.

355
00:15:57,570 --> 00:15:59,920
Now, let me ask a follow-up
question.

356
00:15:59,920 --> 00:16:06,350
How much heat was put into the
system by this isotherm here?

357
00:16:06,350 --> 00:16:10,210


358
00:16:10,210 --> 00:16:12,540
It put in heat to keep the
temperature up, otherwise the

359
00:16:12,540 --> 00:16:13,980
temperature would have
gone down, right?

360
00:16:13,980 --> 00:16:15,660
Heat was going into the system
the entire time.

361
00:16:15,660 --> 00:16:17,190
How much was it?

362
00:16:17,190 --> 00:16:20,600
Well, since it's an isotherm,
since the temperature did not

363
00:16:20,600 --> 00:16:22,370
change, what do we know about
the internal energy?

364
00:16:22,370 --> 00:16:23,910
Did the internal
energy change?

365
00:16:23,910 --> 00:16:26,060
The temperature not changing
told us that the kinetic

366
00:16:26,060 --> 00:16:26,875
energy didn't change.

367
00:16:26,875 --> 00:16:29,590
If the kinetic energy didn't
change, then the internal

368
00:16:29,590 --> 00:16:31,820
energy did not change.

369
00:16:31,820 --> 00:16:37,950
And we know that the change in
internal energy is equal to

370
00:16:37,950 --> 00:16:41,570
the heat put into the
system minus the

371
00:16:41,570 --> 00:16:44,100
work done by the system.

372
00:16:44,100 --> 00:16:48,250
Now, if this is 0-- so we know
that this didn't change,

373
00:16:48,250 --> 00:16:50,180
because the temperature
didn't change.

374
00:16:50,180 --> 00:16:55,700
So that means 0 is equal
to Q minus W, or that

375
00:16:55,700 --> 00:16:59,500
Q is equal to W.

376
00:16:59,500 --> 00:17:01,520
So this is the work done
by the system.

377
00:17:01,520 --> 00:17:03,920
You'll end up getting
something in joules.

378
00:17:03,920 --> 00:17:06,839
And this is also equal to the
heat put into the system.

379
00:17:06,839 --> 00:17:09,240
It's also equal to Q.

380
00:17:09,240 --> 00:17:12,450
So when you look at this, if we
were to just draw this part

381
00:17:12,450 --> 00:17:17,118
of the curve-- let me redraw it
just to make things neat.

382
00:17:17,118 --> 00:17:19,389
I want to give you a little bit
of the convention of what

383
00:17:19,390 --> 00:17:23,723
people in the thermodynamics
world tend to do.

384
00:17:23,723 --> 00:17:26,159
I'll make a neat drawing here.

385
00:17:26,160 --> 00:17:28,470
We started here, at state 1.

386
00:17:28,470 --> 00:17:31,210
And we moved along this
rectangular hyperbola, which

387
00:17:31,210 --> 00:17:34,550
is an isotherm, to state 2.

388
00:17:34,550 --> 00:17:37,380


389
00:17:37,380 --> 00:17:41,770
And now we calculated the area
under this, which is the work

390
00:17:41,770 --> 00:17:44,000
done, which was this
value right here.

391
00:17:44,000 --> 00:17:45,750
Let me write it there.

392
00:17:45,750 --> 00:17:50,750
It's nRT natural log
of V2 over V1.

393
00:17:50,750 --> 00:17:52,430
This is V2.

394
00:17:52,430 --> 00:17:52,990
This is V1.

395
00:17:52,990 --> 00:17:56,040
This whole axis, remember, was
the V-axis, volume axis.

396
00:17:56,040 --> 00:17:58,840
This axis here was the
pressure axis.

397
00:17:58,840 --> 00:18:02,400
And the convention is that
because we did work, but we

398
00:18:02,400 --> 00:18:05,340
were constant temperature, so
our internal energy didn't

399
00:18:05,340 --> 00:18:08,510
change, we had to add energy to
the system to make up for

400
00:18:08,510 --> 00:18:09,700
the work we did.

401
00:18:09,700 --> 00:18:12,170
So some heat must have been
added to the system.

402
00:18:12,170 --> 00:18:13,760
And then what they do is they
just put this little downward

403
00:18:13,760 --> 00:18:15,720
arrow and they write
a Q right there.

404
00:18:15,720 --> 00:18:21,450
So some heat was put into the
system during this isothermic

405
00:18:21,450 --> 00:18:22,450
process right there.

406
00:18:22,450 --> 00:18:26,840
And the value of that Q is
equivalent to the work we did.

407
00:18:26,840 --> 00:18:29,740
We put the exact amount of heat
into the system as the

408
00:18:29,740 --> 00:18:31,190
work that was performed.

409
00:18:31,190 --> 00:18:34,590
And because of that, our
internal energy didn't change.

410
00:18:34,590 --> 00:18:36,660
Or you could say our temperature
didn't change.

411
00:18:36,660 --> 00:18:37,380
Or you could go the other way.

412
00:18:37,380 --> 00:18:40,350
Because our temperature didn't
change, these two things have

413
00:18:40,350 --> 00:18:41,530
to be equal.

414
00:18:41,530 --> 00:18:42,860
Anyway, I want to
leave you there.

415
00:18:42,860 --> 00:18:45,140
Hopefully I gave you a little
bit more of an intuition of

416
00:18:45,140 --> 00:18:48,410
how PV diagrams work, a little
bit more intuition behind what

417
00:18:48,410 --> 00:18:50,290
isotherms and adiabatic mean.

418
00:18:50,290 --> 00:18:52,680
And the most important thing is,
once we get a little bit

419
00:18:52,680 --> 00:18:55,630
mathy, this result can be useful
for coming up with

420
00:18:55,630 --> 00:18:59,720
other interesting things about a
lot of these thermal systems

421
00:18:59,720 --> 00:19:01,000
that we're dealing with.

422
00:19:01,000 --> 00:00:00,000
See you in the next video.