Let's start with this classic system that I keep referring
to in our thermodynamics videos.
I have a cylinder.
It's got a little piston on the top of it, or it's got a
ceiling that's movable.
The gas, and we're thinking of monoatomic ideal gases in
here, they're exerting pressure onto this ceiling.
And the reason why the ceiling isn't moving all the way up,
is because I've placed a bunch of rocks on the top to offset
the force per area of the actual gas.
And I start this gas when it's in equilibrium.
I can define its macrostates.
It has some volume.
It has some pressure that's being offset by these rocks.
And it has some well-defined temperature.
Now, what I'm going to do is, I'm going to place this system
here-- I'm going to place it on top of a reservoir.
And I talked about what a reservoir was either in the
last video or a couple of videos ago.
You can view it as an infinitely large object, if
you will, of a certain temperature.
So if I put it next to-- if I put our system next to this
reservoir-- and let's say I start removing pebbles from
our system.
We learned a couple of videos ago that if we did it
adiabatically-- what does adiabatically mean?
If we removed these pebbles in isolation, without any
reservoir around, the volume would increase, the pressure
would go down, and actually the temperature would
decrease, as well.
We showed that a couple of videos ago.
So by putting this big reservoir there that's a lot
larger than our actual canister, this will keep the
temperature in our canister at T1.
You can kind of view a reservoir as-- say I had a cup
of water in a stadium.
And the air conditioner in the stadium is at 60 degrees.
Well, no matter what I do to that water, I could put it in
the microwave and warm it up, but if I put it back in that
stadium, that stadium is going to keep
that water at 60 degrees.
And you might say, oh, won't the reservoir's temperature
decrease if it's throwing off heat?
Well, it would, but it's so much larger that its impact
isn't noticeable.
For example, if I put a cup of boiling water into a super
large covered-dome stadium, the water will get colder to
the ambient temperature of the stadium.
The stadium will get warmer, but it will be so marginally
warmer than you won't even notice it.
So you can kind of view that as a reservoir.
And theoretically, this is infinitely large.
So the effect of this is, as we remove these little rocks,
we're going to keep the temperature constant.
And remember, if we're keeping the temperature constant,
we're also keeping the internal energy constant,
because we're not changing the kinetic
energy of the particles.
So let me see what happens.
So I keep doing that.
And so I get to a point-- let me see-- where my volume has
increased-- so let me delete some of my rocks here.
Delete some of the rocks.
So some of the rocks are gone.
And now my overall volume is going to be larger.
Let me move this up a little bit.
And then let me color this in black.
Oh, whoops.
Let me color this in, just to give an idea.
So our volume has gotten a bit larger.
And let me get my pen correctly.
So our volume has gotten larger by roughly this amount.
We have the same number of particles.
They're going to bump into the ceiling a little less
frequently, so my pressure would have gone down.
But because I kept this reservoir here, because this
reservoir was here the whole time during this process, the
temperature stayed at T1.
And that was only because of this reservoir.
And I want to make that clear.
And also, just as review, this is a quasi-static process,
because I'm doing it very slowly.
The system is in equilibrium the whole time.
So let's draw what we have so far on our famous PV diagram.
So this is the P-axis.
That's the V-axis.
You label them.
This is P.
This is V.
Let me call this-- I'm going to do it in a good color.
This is state A of the system.
This is state B of the system.
So state A starts at some pressure and volume-- I'll do
it like that.
That's state A.
And it moves to state B.
And notice, I kept the temperature constant.
And what did we learn in, I think it was one
or two videos ago?
Well, we're at a constant temperature, so we're going to
move along an isotherm, which is just
a rectangular hyperbola.
Because when your temperature is constant, your pressure
times your volume is going to equal a constant number.
And I went over that before.
So we're going to move over-- our path is going to look
something like this, and I'll move here, to state B.
I'll move over here to state B.
And the whole time, this was at a constant temperature T1.
Now, we've done a bunch of videos now.
We said, OK, how much work was done on this system?
Well, the work done on the system is the
area under this curve.
So some positive work was-- not done on the system, sorry.
How much work was done by the system?
We're moving in this direction.
I should put the direction there.
We're moving from left to right.
The amount of work done by the system is
pressure times volume.
We've seen that multiple times.
So you take this area of the curve, and you have the work
done by the system from A to B.
Right?
So let's call that work from A to B.
Now, that's fair and everything, but what I want to
think more about, is how much heat was
transferred by my reservoir?
Remember, we said, if this reservoir wasn't there, the
temperature of my canister would have gone down as I
expanded its volume, and as the pressure went down.
So how much heat came into it?
Well, let's go back to our basic internal energy formula.
Change in internal energy is equal to heat applied to the
system minus the work done by the system
Now, what is the change in internal
energy in this scenario?
Well, it was at a constant temperature
the whole time, right?
And since we're dealing with a very simple ideal gas, all of
our internal energy is due to kinetic energy, which
temperature is a measure of.
So, temperature didn't change.
Our average kinetic energy didn't change, which means our
kinetic energy didn't change.
So our internal energy did not change while we moved from
left to right along this isotherm.
So we could say our internal energy is zero.
And that is equal to the heat added to the system minus the
work done by the system.
Right?
So if you just-- we put the work done by the system on the
other side, and then switch the sides, you get heat added
to the system is equal to the work done by the system.
And that makes sense.
The system was doing some work this entire time, so it was
giving energy to-- well, you know, it was giving
essentially maybe some potential
energy to these rocks.
So it was giving energy away.
It was giving energy outside of the system.
So how did it maintain its internal energy?
Well, someone had to give it some energy.
And it was given that energy by this reservoir.
So let's say, and the convention for doing this is
to say, that it was given-- let me write this down.
It was given some energy Q1.
We just say, we just put this downward arrow to say that
some energy went into the system here.
Fair enough.
Now let's take this state B and remove the reservoir, and
completely isolate ourselves.
So there's no way that heat can be transferred to and from
our system.
And let's keep removing some rocks.
So if we keep removing some rocks, where do we get to?
Let me go down here.
So let's say we remove a bunch of more rocks.
So let me erase even more rocks than we had in B.
Maybe I only have one rock left.
And obviously, the overall volume would have increased.
So let me make our piston go up like that, and I can make
our piston is maybe a lot higher now.
And let me just fill in the rest of our, just so that we
don't have some empty space there.
So if I fill that in right there-- OK Let
me fill that in.
And then I just use the blue-- I should be talking about
thermodynamics, not drawing.
But you get the idea.
And then I have some more-- you know, I
shouldn't add particles.
But my volume has increased a good bit.
My pressure will have gone down, they're going to bump
into the walls less.
And because I removed the reservoir, what's going to
happen to the temperature?
My temperature is going to go down.
This was an adiabatic process.
So an adiabatic just means we did it in isolation.
There was no exchange of heat from one system to another.
So let me just-- this arrow continues down here.
I'll say adiabatic.
Now, since I'm moving from one temperature to
another, this is at T2.
So I will have moved to another isotherm.
This is the isotherm for T1.
If I keep my temperature constant, I
move along this hyperbola.
And I would have kept moving along this hyperbola.
But now that we didn't keep our temperature constant, we
now move like this.
We move to another isotherm.
So let's say I have another isotherm at T2.
It looks something like this.
So let me draw like that.
So let's say I have another-- it should actually curve up a
little bit.
So let's say, everything at temperature T2, depending on
its pressure and volume, is someplace along this curve
that asymptotes up like that, and then goes to
the right like that.
Now, I would have moved down to this isotherm, and my
pressure would have kept going down, and my volume would have
kept going down.
So this move, from B to state C, will look like this.
Let me do to it in another color.
Let me do it in the orange color of this arrow.
So it will look like this.
And now we're at state C.
Now, this was adiabatic.
Which means, there is no exchange of heat.
So I don't have to figure out how much heat got transferred
into the system.
Now, there's something interesting here.
We still did do some work.
We can take the area under this curve.
And we're going to leave it to a future video to think about
where that work energy-- well, the main thing is, is what was
reduced by that work energy.
And, well, if you think [UNINTELLIGIBLE]
to leave it to future video.
Our internal energy was reduced, right?
Because our temperature went down.
So our internal energy went down.
We'll talk more about that in the future video.
So now that we're at state C, and we're at temperature T2.
Let's put back another sink here.
But this sink, what it's going to have is a reservoir.
So let me put two things right here.
So I'm going to add-- let me erase some of
these blocks in black.
So now I'm going to add blocks back.
I'm going to add little pebbles back into it.
But I'm going to do it as an isothermic process.
I'm going to do it with a reservoir here.
But this reservoir here, it's not going to be the same
reservoir that I put up there.
I swapped that one out.
I got rid of any reservoir when I went from B to C.
And now I'm going to swap in a new reservoir.
Actually, let me make it blue.
Because it's going to be--
Because here's what's happening.
I'm now adding pebbles in.
I'm compressing the gas.
If this was an adiabatic process, the gas would
want to heat up.
So what I'm doing is, I need to put a reservoir to keep it
at T2, to keep it along this isotherm.
So this is T2.
Remember, this reservoir is kind of a cold reservoir.
It keeps the temperature down.
As opposed to here.
This was a hot reservoir.
It kept the temperature up.
So you can imagine.
The heat generated in the system, or internal energy
being generated in the system-- well, no, I
shouldn't say that.
The temperature of the system will want to go up, but it's
being released, because it's able to transfer that heat
into our new reservoir.
And that amount of heat is Q2.
So I move along this.
This is right here.
I'm moving along another isotherm, I'm moving along
this isotherm.
Until I get to state D.
We're almost there.
This is state D.
So state D will be someplace here, along this isotherm
right here.
Maybe this is state D.
And once again, you can make the argument that we moved
along an isotherm Our temperature did not change
from C to D.
We know that our internal energy went down from B to C,
because we did some work.
But from C to D, our temperature stayed the same.
It was at temperature-- let me write it down-- T2, right?
Because we had this reservoir here.
It stayed the same.
If your temperature stays the same, then your internal
energy stays the same.
At least for the system we're dealing with, because it's a
very simple gas.
It's actually the system you'll deal with most of the
time, in an intro thermodynamics course.
So.
Given our internal energy didn't change, we can apply
the same argument that the heat added to the system is
equal to the work done by the system.
Right?
Same math as we did up here.
Now, in this case, the work wasn't done by the system.
The work was done to the system.
We compressed this piston.
The force times distance went the other way.
So given that work was done to the system, the heat added to
the system was negative, right?
We're just applying the same thing.
If our internal energy is 0, the heat added to the system
is equal to the work done by the system.
The work done by the system is negative.
Work was done to it.
So the heat added to the system would be negative.
Or another way to think about it is that the
system gave away heat.
We put that with Q2.
And where did it give that heat?
It gave it to this reservoir that we put here, this kind of
cold reservoir.
You could almost view it as a-- well, it's
accepting the heat.
OK.
We're almost there.
Now, let's say we remove this reservoir from under our
system again, so it's completely isolated from
everything else, at least in terms of heat.
And what we do is, we start adding-- so state D, we still
had a few less pebbles.
But we start adding more pebbles again.
We start adding more pebbles to get it to state A.
So let me change my pebble color.
So we start adding more pebbles again to
get it to state A.
So that's this process right here.
Let me do a different color.
Let's say this is green.
So as we add pebbles, that's this movement right here.
We're moving from one isotherm up to another isotherm at a
higher temperature.
And remember, this whole time we went
this clockwise direction.
So a couple of interesting things are going on here.
Because we're assuming an ideal scenario, nothing was
lost of friction.
This piston just moves up and down.
No heat loss due to that.
What we can say is that we've achieved-- we are back at our
original internal energy.
In fact, this is one of the properties of a state
variable, is that if we're at the same point on the PV
diagram, the same exact point, we have
the same state variable.
So now we have the same pressure, volume, temperature,
and internal energy as what we started with.
So we've done here is completed a cycle.
And this particular cycle, it's an important one, it's
called the Carnot cycle.
It's named after a French engineer who was trying to
just optimize engines in the early 1800s.
So Carnot cycle.
And we're going to study this a lot in the next few videos
to really make sure we understand entropy correctly.
Because in a lot of chemistry classes, they'll
throw entropy at you.
Oh, it's measure of disorder.
But you really don't know what they're talking about, or how
can you quantify it, or measure it anyway.
And we really need to deal with the Carnot cycle in order
to understand where the first concepts of entropy really
came from, and then relate it to kind of more
modern notions of it.
Now, a system that completes a Carnot cycle is called a
Carnot engine.
So our little piston here that's moving up and down, we
can consider this a Carnot engine.
You might say, oh, Sal, this doesn't seem
like a great engine.
I have to move pebbles and all of that.
And you're right.
You wouldn't actually implement an engine this way.
But it's a useful engine, or it's a useful theoretical
construct, in order for understanding how heat is
transferred in an engine.
I mean, if you think about what's happening here, is this
first heat sink transferred some heat to the system, and
then the system transferred a smaller amount of heat back to
the other reservoir.
Right?
So this system was transferring heat from one
reservoir to another reservoir.
From a hotter reservoir to a colder reservoir.
And in the process, it was also doing some work.
And what was the work that it did?
Well, it's the area under this curve, or the area inside of
this cycle.
So this is the work done by our Carnot engine.
And the way you think about it is, when you're going in the
rightward direction with increasing volume, it's the
area under the curve is the work done by the system.
And then when you move in the leftward direction with
decreasing volume, you subtract out the work done to
the system, and then you're left with just
the area in the curve.
So we can write this Carnot engine like this.
It's taking, it's starting-- so you have a reservoir at T1.
And then you have your engine, right here.
And then it's connected-- so this takes Q1
in from this reservoir.
It does some work, all right?
The work is represented by the amount of-- the work right
here is the area inside of our cycle.
And then it transfers Q2, or essentially the remainder from
Q1, into our cold reservoir.
So T2.
So it transfers Q2 there.
So the work we did is really the difference
between Q1 and Q2, right?
You say, hey.
If I have more heat coming in than I'm letting out, where
did the rest of that heat go?
It went to work.
Literally.
So Q1 minus Q2 is equal to the amount of work we did.
And actually, this is a good time to emphasize again that
heat and work are not a state variable.
A state variable has to be the exact same value when we
complete a cycle.
Now, we see here that we completed a cycle, and we had
a net amount of work done, or a net amount of heat added to
the system.
So we could just keep going around the cycle, and keep
having heat added to the system.
So there is no inherent heat state variable right here.
You can't say what the value of heat is at
this point in time.
All you could say is what amount of heat was added or
taken away from the system, or you can only say the amount of
work that was done to, or done by, the system.
Anyway, I want to leave you there right now.
We're going to study this a lot more.
But the real important thing is, and if you never want to
get confused in a thermodynamics class, I
encourage you to even go off on your
own, and do this yourself.
Kind of-- you can almost take a pencil and paper, and redo
this video that I just did.
Because it's essential that you understand the Carnot
engine, understand this adiabatic process, understand
what isotherms are.
Because if you understand that, then a lot of what we're
about to do in the next few videos with regard to entropy
will be a little bit more
intuitive, and not too confusing.