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### Course: AP®︎/College Physics 2 > Unit 2

Lesson 3: Laws of thermodynamics- First law of thermodynamics
- First law of thermodynamics problem solving
- What is the first law of thermodynamics?
- PV diagrams - part 1: Work and isobaric processes
- PV diagrams - part 2: Isothermal, isometric, adiabatic processes
- What are PV diagrams?
- Kinetic molecular theory of gases
- Second law of thermodynamics

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# PV diagrams - part 1: Work and isobaric processes

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## Want to join the conversation?

- At10:14, the equation is written as (-PdeltaV) but the commentator said this reflected as work done ON the gas. Wouldn't this be work done BY the gas as this is negative work and it is going to the right (increase in volume)?(8 votes)
- This issue is very simple if you simply remember that work = -PDV. Why the negative sign in front of PDV?Well it is there to make sure that we are in line with the sign convention. First remember that work done by the gas (expansion) is negative work...this makes sense because DV is positive but the work is negative because of the negative sign in front of PDV.....Now if the gas is compressed (work done on the gas) the work is positive....because DV is negative and thus, -P(-DV)= +PDV= positive work.....(2 votes)

- Assuming the gas is the system here.

At8:42work done by the gas (positive delta V) it was marked in green that it's -W (i.e. negative work), but then at10:20it was stated that "work done by the gas is P delta V" and it's +W? I've watched this many times and I'm still confused. Isn't work done by the system negative and work done on the system positive? Halp plz(2 votes) - (9:48) Does the formula W=PdV only work for isobaric processes? How does using integration to solve for areas under curve (other processes) differ from this formula?(1 vote)
- At8:28, you say that if you're going to the left, then work will be positive. Shouldn't the work be negative because volume is decreases therefore, gas is being compressed? And gas compression is when the surrounding does work on the system?(1 vote)
- At7:31he explains that the area under the curve is "Work done BY the gas" [Wby]. In the previous video, he showed us "Work done ON the gas = Negative Work done BY the gas"[Won = - Wby]. Now, if the volume is increasing on the isobaric graph, that would result in a positive deltaV and using the equation Wby = PdV, Work done BY the gas [Wby] would be postive. When you plug in this +Wby into the First Law of Thermodynamics equation (dU=Q+Won), the Work would become negative since Won = - Wby giving us dU=Q + (-Wby). So if the volume is decreasing (going to the left on the graph) Vinitial = lets say 4 and Vfinal = lets say 2, would result in a negative volume change and therefore, a negative Wby using P(-V). When plugging this negative Wby into the First Law of Thermodynamics equation, remember Won = - Wby, so a negative of a negative Wby would result in positive work: dU=Q + (-(-Wby)).(3 votes)

- At07:31, why is pressure always positive?(1 vote)
- Pressure exists from 0-?kPa. There is no negative pressure. However, "negative" pressure is gauge pressure, local pressure (pressure reading on the gauge) in reference to atmospheric pressure.

Here is a website that defines a few terms

https://www.machinedesign.com/pneumatics/what-s-difference-between-gauge-absolute-differential-and-sealed-pressure(1 vote)

- I love how David was on the edge of inventing calculus at9:10, but he didn't even mention integrals. Its like wanting very badly to sneeze but it never happens.(1 vote)
- Why is it that when the bar goes to the right the work is seen as positive and when the bar is going to the left the work is seen as negative? should be around8:34(1 vote)
- Is it possible to calculate the work of an evolution if the pressure, temperature, and volume are all changing? I often encounter problems where a piston is moving (meaning the volume is not constant), but then so is everything else. And I'm not sure which type of evolution it counts as.(1 vote)
- Why isn't the work done by gas equal to the heat income in an isobaric process? Don't we have the same energy used by the gas to expand the voume that we got from the heat coming in?(1 vote)
- At6:39, wouldn't work done by the gas mean negative area since the energy is lost? So wouldn't the graph actually be showing work done on the gas?(1 vote)

## Video transcript

- Something you see a lot
when doing thermodynamics especially problems
involving the first law are what are called PV Diagrams. Now, the P stands for Pressure and the V stands for Volume. And this gives you a diagram of what the pressure and volume
are in any given instant. So what does this mean? Well, imagine you had a container full of a gas and there's
a movable piston on top. Piston can move up or down, changing the amount of volume, right? This is the volume we're talking about, is the volume within here. So that movable piston can
change that amount of volume. And that would change the
amount of pressure inside, depending on what heat is
added, how much work is done. So say we started with a
certain amount of volume, right? Let's say we start with that much volume. And the pressure inside
is probably not zero. If there's any gas
inside, it can't be zero. And so we come over to here, let's say we start at
this point right here. Now, what do we do? I know if I push the piston
down, my volume decreases. And if I pull the piston
up, my volume increases. So if I push the piston down,
I know volume goes down. That means on this graph,
I'm going that way. Piston going down means decreasing volume. What about piston going up? Well, if the piston goes up,
then my volume's increasing and I know on my graph I'd
better be going to the right. Now maybe I'm going up and right. Maybe I'm going down and right. All I know is, my volume
better be increasing, so this is increasing volume,
that's increasing volume, that's increasing volume. This is not increasing volume, so I know if my piston goes up, my volume increases, I gotta be going to the rightward in
some way on this graph. And if my piston goes down, I better be going to the left
on this graph somehow. Now, what happens to the pressure? You gotta know a little
more detail about it. But just knowing the
direction of the piston, that lets you know which
way you go on this graph. So say I push the piston down. Say I push it down really fast. What do you think's gonna
happen to the pressure? The pressure's probably gonna go up. How would I represent that? Well, volume's gotta go down,
pressure would have to go up, so I might take a path that
looks something like this. Volume's gotta go down to the left. Pressure's gotta go up, so maybe it does something like that. There's really infinitely many ways the gas could get from
one state to another. It could take any possible range and unless you know the exact details, it's hard to say exactly
what's gonna happen. So there's infinitely many
possibilities on this diagram. You can loop around,
it's not like a function. You can do something like this. This gas can take some crazy
path through this PV Diagram. There's infinitely many ways it can take. But there are four thermodynamic processes that are most commonly
represented on a PV Diagram. Again, these are not the
only four possibilities. These are just the four that are kind of the simplest to deal with mathematically. And they're often a good representation and accurate approximation
to a lot of processes so the math's good, they work pretty well, we talk about them a lot. The first one is called
in isobaric process. Iso means constant, so
whenever you see iso before something, it means constant. Whatever follows next,
and this one's isobaric. Baric, well bars, that's
a unit of pressure, so baric is talking about pressure. Isobaric means constant pressure. So how do you represent
this on a PV DIagram? Well, if you wanna
maintain constant pressure, you can't go up or down, because if I were to go up, my pressure would be increasing. If I were to go down, my
pressure would be decreasing. The only option available is
to go along a horizontal line. So this would be in iso, well, sometimes they're called isobars,
and isobar for short. This is an isobar, this
is an isobaric expansion if I go to the right, cause
I know volume's increasing. And if I go to the left it would be an isobaric compression because
volume would be decreasing. But it doesn't have to be
in this particular spot. It could be anywhere on this PV Diagram, any horizontal line is gonna be an isobar, an isobaric process. Now, I bring up the isobaric process first because it allows me to
show something important that's true of every process
that's just easier to see for the isobaric process. In physics, the area under the curve often represents something significant. And that's gonna be true here as well. Let's try to figure out
what the area under this curve represents. So first of all, to find the area of this rectangle, we know it's gonna be the height times the
width, what's the height? The height's just the pressure, right? The value of this pressure over here is gonna be the height and the width is the change in volume so if I started with V initial and I ended with V final, let's say it was the expansion
instead of the compression. This V final minus V initial, this delta V is going to represent the
width of this rectangle. So we know area is going to
be the value of the pressure times the change in the volume. Well, what does that mean? We know that pressure, we know
the definition of pressure, pressure is just the force per area. So on this gas, even on a
force exerted on it per area, and the change in volume,
what do we know is the volume? How could I represent the volume in here? I know this piston has some area, so there's some area that this piston has. And then there's a certain height. This inner cylinder of volume in here has a certain height
and then a certain area so we know the volume is
just height times area. So it would be height times
the area of the piston. Which of these is
changing in this process? Well, the area is not changing. If the area of this piston changed, it either let some of the gas out or it would bust through
the sides of the cylinder, both of which we're
assuming is not happening. So I can pull area out of this delta sign since the area is constant. And what I get is F times A over A times the change in the height. Well the A is canceled, A cancels A and I get F times the
change in the height. But look at, this is just
force times a distance. Times the distance by
which this height changes. So delta H will be the amount by which this piston goes up or down. And we know force times the distance by which you apply that
force is just the work. So now we know the area
under this isobaric process represents the work done either on the gas or by the gas depending
on which way you're going. So this area is the work, this
area, the value of this area equals the amount of work
done on the gas or by the gas. How do you figure out which? Well, technically this area represents the work done by the gas, because if we're talking about a positive area, mathematically that means
moving to the right, like on a graph in math class. The area, positive area,
you're moving to the right. So if we want to be
particular and precise, we'll say that this is a
process moving to the right. And we know if the volume is going up like this graph is going to the right, which means volume is increasing, we know that gas is doing work. So technically, this area
is the work done by the gas. You can see that as well
since this is P delta V. If your delta V comes out positive, pressure is always
positive, if your Delta V comes out positive, the
volume is increasing. That means work is being done by the gas. So you have to be careful. If you calculate this P delta V and you go to your first law equation, which remember, says delta U is Q plus W, well you can't just plug
in the value of P delta V. This is the work done by the gas, so you have to plug in negative that value for the work done, and
also correspondingly, if you were to go to the left, if you did have a process
that went to the left. That is to say the volume was decreasing. If you find this area and you're careful, then you'll get a negative delta V if you're going leftward
because you'll end with a smaller value for the
volume than you started with. So if you really treat
the left one as the final, cause that's where you end
up if you're going left, and the rightward one as the initial, your leftward final point will be smaller than your initial point, you
will get a negative value here. So again, you plug in negative
of that negative value. You'll get your positive work, cause positive work is
being done on the gas. That sounds very complicated. Here's what I do, quite honestly. I just look at the shape, I find the area, I do the magnitude of the height, right, the size of it, no negatives. The size of the width, no negatives. I multiply the two and then I just look. Am I going to the left? If I'm going to the left,
I know my work is positive. If I'm going to the right,
I know my work is negative that I plug into here, so I
just add the negative sign in. Makes it me easier for me to understand. So I said that this works for
any process, how is that so? If I take some random process, I'm not gonna get a nice
rectangle, how is this true? Well, if I did take a random process from one point to another, say I took this crazy path here. Even though it's not a perfect rectangle, I can break it up into small rectangles so I can take this, break
this portion up into, if I make the rectangle small enough, I can approximate any
area as the summation of a whole bunch of little rectangles. And look at, each one of these rectangles, well, P delta V, that's the
area underneath for that one, add them all up, I get
the total area undeneath. So even though it might be
difficult to find this area, it's always true that if
I could find this area under any process, this area
does represent the work done. And again, it's by the gas. So in other words, using the formula work done by the gas
that we had previously equals P times delta V, that works for one small little rectangle
and you can add all those up, but it work for the entire process. If you tried to use the,
say, initial pressure times the total change in volume, and that's not gonna
give you an exact answer, that's assuming you
have one big rectangle. So this formula won't work
for the whole process. But we do know if you
have an isobaric process, if it really is an isobaric process, then we can rewrite the first law. The first law says that delta U equals Q plus work done on the gas? Well, we know a formula for
the work done by the gas. Work done by the gas is P delta V. So the work done on the gas is just negative P times delta V. Here's a formula for the first law if you happen to have an isobaric process. So an isobaric process is pretty nice. It gives you an exact
way to find the work done since the area underneath
is a perfect rectangle. But how would you physically set up an isobaric process in the lab? Well, imagine this, let's say you heat up this cylinder, you allow heat to flow in. That would tend to increase the pressure. So the only way we could
maintain constant pressure, cause an isobaric process
maintains constant pressure, if I want the pressure to stay
the same as heat flows in, I better let this piston move upwards. While I add heat I can
maintain constant pressure. In fact, you might think
that's complicated. How are you going to do that exactly? It's not so bad, just allow the piston to come into equilibrium with whatever atmospheric pressure plus
the weight of this piston is. So there's a certain pressure
down from the outside and then there's the weight of the piston divided by the area
gives another pressure. This heat will try to make
the pressure increase, but if you just allow this system to come into equilibrium
with the outside pressure, the inside pressure is always gonna equal the outside pressure
because if it's not equal, this piston will move
up or down accordingly. So if this piston can move freely, it'll maintain a constant pressure and that would be a way
to physically ensure that the pressure remains constant and you have an isobaric process. I'll explain the next three
thermodynamic processes in the next video.