WEBVTT

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- Something you see a lot
when doing thermodynamics

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especially problems
involving the first law

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are what are called PV Diagrams.

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Now, the P stands for Pressure

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and the V stands for Volume.

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And this gives you a diagram of what

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the pressure and volume
are in any given instant.

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So what does this mean?

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Well, imagine you had a container

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full of a gas and there's
a movable piston on top.

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Piston can move up or down, changing

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the amount of volume, right?

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This is the volume we're talking about,

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is the volume within here.

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So that movable piston can
change that amount of volume.

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And that would change the
amount of pressure inside,

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depending on what heat is
added, how much work is done.

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So say we started with a
certain amount of volume, right?

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Let's say we start with that much volume.

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And the pressure inside
is probably not zero.

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If there's any gas
inside, it can't be zero.

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And so we come over to here,

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let's say we start at
this point right here.

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Now, what do we do?

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I know if I push the piston
down, my volume decreases.

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And if I pull the piston
up, my volume increases.

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So if I push the piston down,
I know volume goes down.

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That means on this graph,
I'm going that way.

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Piston going down means decreasing volume.

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What about piston going up?

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Well, if the piston goes up,
then my volume's increasing

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and I know on my graph I'd
better be going to the right.

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Now maybe I'm going up and right.

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Maybe I'm going down and right.

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All I know is, my volume
better be increasing,

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so this is increasing volume,
that's increasing volume,

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that's increasing volume.

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This is not increasing volume,

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so I know if my piston goes up,

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my volume increases, I gotta be going

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to the rightward in
some way on this graph.

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And if my piston goes down, I better

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be going to the left
on this graph somehow.

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Now, what happens to the pressure?

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You gotta know a little
more detail about it.

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But just knowing the
direction of the piston,

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that lets you know which
way you go on this graph.

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So say I push the piston down.

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Say I push it down really fast.

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What do you think's gonna
happen to the pressure?

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The pressure's probably gonna go up.

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How would I represent that?

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Well, volume's gotta go down,
pressure would have to go up,

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so I might take a path that
looks something like this.

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Volume's gotta go down to the left.

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Pressure's gotta go up, so maybe

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it does something like that.

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There's really infinitely many ways

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the gas could get from
one state to another.

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It could take any possible range

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and unless you know the exact details,

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it's hard to say exactly
what's gonna happen.

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So there's infinitely many
possibilities on this diagram.

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You can loop around,
it's not like a function.

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You can do something like this.

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This gas can take some crazy
path through this PV Diagram.

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There's infinitely many ways it can take.

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But there are four thermodynamic processes

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that are most commonly
represented on a PV Diagram.

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Again, these are not the
only four possibilities.

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These are just the four that are kind of

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the simplest to deal with mathematically.

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And they're often a good representation

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and accurate approximation
to a lot of processes

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so the math's good, they work pretty well,

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we talk about them a lot.

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The first one is called
in isobaric process.

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Iso means constant, so
whenever you see iso

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before something, it means constant.

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Whatever follows next,
and this one's isobaric.

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Baric, well bars, that's
a unit of pressure,

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so baric is talking about pressure.

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Isobaric means constant pressure.

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So how do you represent
this on a PV DIagram?

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Well, if you wanna
maintain constant pressure,

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you can't go up or down, because if I were

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to go up, my pressure would be increasing.

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If I were to go down, my
pressure would be decreasing.

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The only option available is
to go along a horizontal line.

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So this would be in iso, well, sometimes

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they're called isobars,
and isobar for short.

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This is an isobar, this
is an isobaric expansion

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if I go to the right, cause
I know volume's increasing.

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And if I go to the left it would be

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an isobaric compression because
volume would be decreasing.

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But it doesn't have to be
in this particular spot.

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It could be anywhere on this PV Diagram,

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any horizontal line is gonna be an isobar,

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an isobaric process.

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Now, I bring up the isobaric process first

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because it allows me to
show something important

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that's true of every process
that's just easier to see

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for the isobaric process.

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In physics, the area under the curve

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often represents something significant.

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And that's gonna be true here as well.

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Let's try to figure out
what the area under this

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curve represents.

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So first of all, to find the area

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of this rectangle, we know it's gonna be

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the height times the
width, what's the height?

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The height's just the pressure, right?

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The value of this pressure over here

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is gonna be the height and the width

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is the change in volume so if I started

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with V initial and I ended with V final,

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let's say it was the expansion
instead of the compression.

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This V final minus V initial, this delta V

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is going to represent the
width of this rectangle.

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So we know area is going to
be the value of the pressure

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times the change in the volume.

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Well, what does that mean?

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We know that pressure, we know
the definition of pressure,

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pressure is just the force per area.

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So on this gas, even on a
force exerted on it per area,

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and the change in volume,
what do we know is the volume?

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How could I represent the volume in here?

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I know this piston has some area,

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so there's some area that this piston has.

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And then there's a certain height.

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This inner cylinder of volume in here

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has a certain height
and then a certain area

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so we know the volume is
just height times area.

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So it would be height times
the area of the piston.

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Which of these is
changing in this process?

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Well, the area is not changing.

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If the area of this piston changed,

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it either let some of the gas out

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or it would bust through
the sides of the cylinder,

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both of which we're
assuming is not happening.

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So I can pull area out of this delta sign

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since the area is constant.

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And what I get is F times A over A times

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the change in the height.

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Well the A is canceled, A cancels A

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and I get F times the
change in the height.

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But look at, this is just
force times a distance.

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Times the distance by
which this height changes.

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So delta H will be the amount by which

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this piston goes up or down.

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And we know force times the distance

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by which you apply that
force is just the work.

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So now we know the area
under this isobaric process

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represents the work done either on the gas

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or by the gas depending
on which way you're going.

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So this area is the work, this
area, the value of this area

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equals the amount of work
done on the gas or by the gas.

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How do you figure out which?

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Well, technically this area represents

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the work done by the gas, because if we're

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talking about a positive area,

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mathematically that means
moving to the right,

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like on a graph in math class.

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The area, positive area,
you're moving to the right.

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So if we want to be
particular and precise,

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we'll say that this is a
process moving to the right.

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And we know if the volume is going up

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like this graph is going to the right,

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which means volume is increasing,

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we know that gas is doing work.

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So technically, this area
is the work done by the gas.

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You can see that as well
since this is P delta V.

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If your delta V comes out positive,

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pressure is always
positive, if your Delta V

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comes out positive, the
volume is increasing.

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That means work is being done by the gas.

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So you have to be careful.

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If you calculate this P delta V

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and you go to your first law equation,

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which remember, says delta U is Q plus W,

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well you can't just plug
in the value of P delta V.

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This is the work done by the gas,

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so you have to plug in negative that value

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for the work done, and
also correspondingly,

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if you were to go to the left,

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if you did have a process
that went to the left.

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That is to say the volume was decreasing.

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If you find this area and you're careful,

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then you'll get a negative delta V

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if you're going leftward
because you'll end

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with a smaller value for the
volume than you started with.

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So if you really treat
the left one as the final,

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cause that's where you end
up if you're going left,

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and the rightward one as the initial,

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your leftward final point will be smaller

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than your initial point, you
will get a negative value here.

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So again, you plug in negative
of that negative value.

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You'll get your positive work,

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cause positive work is
being done on the gas.

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That sounds very complicated.

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Here's what I do, quite honestly.

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I just look at the shape, I find the area,

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I do the magnitude of the height, right,

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the size of it, no negatives.

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The size of the width, no negatives.

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I multiply the two and then I just look.

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Am I going to the left?

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If I'm going to the left,
I know my work is positive.

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If I'm going to the right,
I know my work is negative

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that I plug into here, so I
just add the negative sign in.

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Makes it me easier for me to understand.

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So I said that this works for
any process, how is that so?

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If I take some random process,

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I'm not gonna get a nice
rectangle, how is this true?

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Well, if I did take a random process

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from one point to another, say I took

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this crazy path here.

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Even though it's not a perfect rectangle,

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I can break it up into small rectangles

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so I can take this, break
this portion up into,

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if I make the rectangle small enough,

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I can approximate any
area as the summation

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of a whole bunch of little rectangles.

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And look at, each one of these rectangles,

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well, P delta V, that's the
area underneath for that one,

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add them all up, I get
the total area undeneath.

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So even though it might be
difficult to find this area,

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it's always true that if
I could find this area

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under any process, this area
does represent the work done.

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And again, it's by the gas.

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So in other words, using the formula

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work done by the gas
that we had previously

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equals P times delta V, that works

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for one small little rectangle
and you can add all those up,

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but it work for the entire process.

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If you tried to use the,
say, initial pressure

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times the total change in volume,

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and that's not gonna
give you an exact answer,

00:09:55.842 --> 00:09:58.338
that's assuming you
have one big rectangle.

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So this formula won't work
for the whole process.

00:10:01.182 --> 00:10:04.480
But we do know if you
have an isobaric process,

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if it really is an isobaric process,

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then we can rewrite the first law.

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The first law says that delta U equals Q

00:10:13.199 --> 00:10:15.637
plus work done on the gas?

00:10:15.637 --> 00:10:18.479
Well, we know a formula for
the work done by the gas.

00:10:18.479 --> 00:10:20.617
Work done by the gas is P delta V.

00:10:20.617 --> 00:10:21.975
So the work done on the gas is just

00:10:21.975 --> 00:10:25.078
negative P times delta V.

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Here's a formula for the first law

00:10:27.060 --> 00:10:31.540
if you happen to have an isobaric process.

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So an isobaric process is pretty nice.

00:10:34.002 --> 00:10:37.020
It gives you an exact
way to find the work done

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since the area underneath
is a perfect rectangle.

00:10:40.201 --> 00:10:42.082
But how would you physically set up

00:10:42.082 --> 00:10:44.219
an isobaric process in the lab?

00:10:44.219 --> 00:10:45.960
Well, imagine this, let's say you heat up

00:10:45.960 --> 00:10:49.037
this cylinder, you allow heat to flow in.

00:10:49.037 --> 00:10:51.299
That would tend to increase the pressure.

00:10:51.299 --> 00:10:54.133
So the only way we could
maintain constant pressure,

00:10:54.133 --> 00:10:58.056
cause an isobaric process
maintains constant pressure,

00:10:58.056 --> 00:11:00.774
if I want the pressure to stay
the same as heat flows in,

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I better let this piston move upwards.

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While I add heat I can
maintain constant pressure.

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In fact, you might think
that's complicated.

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How are you going to do that exactly?

00:11:10.213 --> 00:11:11.860
It's not so bad, just allow the piston

00:11:11.860 --> 00:11:14.856
to come into equilibrium with whatever

00:11:14.856 --> 00:11:18.130
atmospheric pressure plus
the weight of this piston is.

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So there's a certain pressure
down from the outside

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and then there's the weight of the piston

00:11:21.915 --> 00:11:23.854
divided by the area
gives another pressure.

00:11:23.854 --> 00:11:25.850
This heat will try to make
the pressure increase,

00:11:25.850 --> 00:11:27.579
but if you just allow this system

00:11:27.579 --> 00:11:30.355
to come into equilibrium
with the outside pressure,

00:11:30.355 --> 00:11:32.618
the inside pressure is always gonna equal

00:11:32.618 --> 00:11:35.475
the outside pressure
because if it's not equal,

00:11:35.475 --> 00:11:37.889
this piston will move
up or down accordingly.

00:11:37.889 --> 00:11:39.515
So if this piston can move freely,

00:11:39.515 --> 00:11:41.639
it'll maintain a constant pressure

00:11:41.639 --> 00:11:44.019
and that would be a way
to physically ensure

00:11:44.019 --> 00:11:46.016
that the pressure remains constant

00:11:46.016 --> 00:11:49.290
and you have an isobaric process.

00:11:49.290 --> 00:11:52.076
I'll explain the next three
thermodynamic processes

00:11:52.076 --> 00:11:53.549
in the next video.