PV Diagrams & Work: Study with Video Lessons, Practice Problems & Examples

PV diagrams plot pressure (P) against volume (V) to visualize thermodynamic processes. For constant pressure, work done (W) can be calculated using the equation $w=p\mathrm{\Delta v}$, where $\mathrm{\Delta v}$ is the change in volume. The area under the curve on a PV diagram represents work done, with positive work for expansion and negative for compression, emphasizing the importance of direction in these processes.

Work and PV Diagrams

Hey, everyone. So by now, we've studied the ideal gas law and the first law of thermodynamics, which, remember, relates variables like heat and work when you have a system or a gas that's changing from one state to another. Now in some problems, these changes will only be described in the text of the problem. But in others, you'll have to take these changes and you'll have to map them out or draw them in what's called a PV diagram. So in this video, I'm going to introduce you to what these PV diagrams are. They're going to be really useful for the rest of thermodynamics, so it's really important that you learn them very well. And I'll also show you how to calculate the work that's done by a system using these diagrams. So let's go ahead and check this out here. Basically, what a PV diagram does is it plots pressure, that's the p on the y-axis, versus volume, that's the v on the x-axis. We have p and v. And basically what these processes do or these diagrams do is they graph these things called thermodynamic processes. And this is just when any system or gas changes between a state. So let's just dive right into our first problem here. So we have that a gas is expanding from a volume of 2 to 5 at a constant pressure. That's a thermodynamic process. The whole idea here is that now we're going to take this text and this picture and we're actually going to draw it out on this graph. Now even though it's a graph, a lot of textbooks will refer to this as a PV diagram so that's just what we're going to call it. So let's get started here. We're going to draw this out on our PV diagram. So we have a constant pressure of 100 that's right here. And then what we have here is we're going to we have, we're going from 2 meters cubed, so that's sort of like our initial point right here, and then we're going to go to 5 meters cubed. So that's going to be right here. So the process will actually just look like a straight line that connects initial to final because it's a constant pressure. So what happens is this 100 here will never change. So that's really what this process looks like. Notice how we indicated this arrow here because we're going from initial to final. That arrow is going to be super important. That's the first part. Now let's jump into the second part here. We're going to calculate the work that's done by the gas. So that's just the equation w by. Now remember, if we have constant pressure, we can use this equation p times Δv. Now do we have constant pressure here? We do because we're told that in the problem. So we can totally use this equation here. So the constant pressure is going to be 100, and the change in volume is just going to be from 2 to 5. So this Δv here is just going to be 3. Right? So this is going to be 3 and you work this out and you're going to get 300 joules. Pretty straightforward. That's just the work done pΔv. Alright. So now let's jump into part c here, which is we want to calculate the area under the path of this process. What does that mean? Well, the path just goes from initial to final. The area underneath that path is just going to be this rectangle right here. So all we have to do to calculate the area is just calculate the area of a rectangle. Right? That's just base times height. So the base of this rectangle here goes from 2 to 5, so it's just 3. The height of this rectangle is 100. And so therefore, if you do 3 times 100, hopefully, you guys realize that you should get 300 joules. Notice how these two numbers are the same, and that's no coincidence here. So what it so the thing I want you to know here is that the work that is done in any thermodynamic process, no matter what it is, is always equal to the area under the curve. So this should make some sense to you over this process because if you think about it, what happens is the base really i

Finding Value of V on Axis

Hey, guys. So I hopefully got a chance to look at this problem here. This one's kind of interesting. So what we have in this problem is we're given this thermodynamic process, and we're told what the work done is. It's \(2 \times 10^5\). What we want to do is figure out the value of \(v_1\) indicated on the axis. So really what we're trying to find here is we have some kind of a thermodynamic process. It changes from this pressure to this pressure, but what we want to do is figure out what the volume is. Basically, what do the tick marks represent on our \(v\) axis? That's kind of what we're interested in. So let's go ahead and check this out here. If we're given this thermodynamic process and we're asked to find something about the work that's done, we're going to start off with our work equation. Now, what we can't do in this problem is we can't use \(p \times \Delta v\) because the pressure does not remain constant. Remember, we can only use \(p \Delta v\) when we have flat processes, but this one goes up like this. So the pressure is changing, we can't use it. So instead, what we're going to have to do is relate the work done to the area that's under the curve or under the process. So what's happening here is that we have this process that goes from here to here, and really the work that's done is going to be the area that's under this shape right here. So this \(w\) is equal to \(2 \times 10^5\), and what we're going to have to do is relate this work to one of the area equations that we have for shapes like a rectangle or a triangle or trapezoid or whatever. So what happens here is if we look through this shape, this kind of looks like a trapezoid. Right? So what we have here in the trapezoid is I'm going to call this base 1. This is going to be base 2, and then this is going to be my height of my trapezoid. The trapezoid doesn't always have to look like this; you could have this as a trapezoid as well. So here's what's going on. We're going to use the area for a trapezoid, which is going to be \(\frac{1}{2}\) (base 1 plus base 2) times the height. Now, what we're told here is that this is equal to \(2 \times 10^5\). So what is base 1, base 2? What is height? What do all that stuff mean in terms of the variables that we have here? So what's going on here is that this base one represents \(1 \times 10^5\). This base 2 represents \(3 \times 10^5\). So what I'm going to do here is I'm going to replace this with \(\frac{1}{2}(1 \times 10^5 + 3 \times 10^5)\), and then times the height. Well, what is this height here? Well, if you look at this graph, the height is going to be the difference between this piece and this piece or this part and this part here on the \(v\) axis. So the height here of my trapezoid actually represents the difference between \(v_1\) and \(3 v_1\). So the height here is actually going to be \(2 v_1\). Right? We're going 1 and then 2 of whatever unit that is. So that's going to be my height here. It's going to be \(2 v_1\). So this is going to equal \(2 \times 10^5\). So now what I can do here is all I have to do is solve for this \(v_1\). Okay? So here's what's going on. We're going to have I'm just going to combine these two things in this parenthesis here. So we've got is \(4 \times 10^5 \times 2 v_1 = 2 \times 10^5\), and so now what we're going to do is one half of \(4 \times 10^5\) is just going to be \(2 \times 10^5\). So \(2 \times 10^5 \times 2 v_1 = 2 \times 10^5\). So what happens here is we can divide out this \(2 \times 10^5\) from both sides, and what happens is this goes away, and this just becomes 1. So we've got \(2v_1\) is equal to just 1. Right? And so therefore, \(v_1\) is equal to \(0.5 \text{ meters}^3\). So that's basically what the first little tick mark represents. So what's going on here is this is \(0.5\), this is \(1\), and this is \(1.5 \text{ meters}^3\) on the \(x\)-axis. And so what happens is if you go through and double-check real quick, if you basically just go ahead and plug these numbers back into this trapezoid equation, what you'll get here is \(2 \times 10^5\). Alright? So that's it for this one. Let me know if you have any questions.

Here’s what students ask on this topic:

What is a PV diagram and how is it used in thermodynamics?

A PV diagram is a graphical representation that plots pressure (P) on the y-axis against volume (V) on the x-axis. It is used in thermodynamics to visualize the changes in a system, particularly gases, as they undergo various processes. These diagrams help in understanding how pressure and volume change during processes like expansion and compression. By analyzing the area under the curve in a PV diagram, one can calculate the work done by or on the system. For instance, the work done during a constant pressure process can be calculated using the equation $W=P\mathrm{\Delta V}$, where $\mathrm{\Delta V}$ is the change in volume.

How do you calculate work done in a PV diagram for a constant pressure process?

To calculate the work done in a PV diagram for a constant pressure process, you can use the equation $W=P\mathrm{\Delta V}$, where $P$ is the constant pressure and $\mathrm{\Delta V}$ is the change in volume. For example, if a gas expands from 2 m³ to 5 m³ at a constant pressure of 100 Pa, the work done is calculated as $W=100(5-2)=300 J$. The area under the curve in the PV diagram, which is a rectangle in this case, also represents the work done.

What does the area under the curve in a PV diagram represent?

The area under the curve in a PV diagram represents the work done by or on the system during a thermodynamic process. For a constant pressure process, this area is a rectangle, and the work done can be calculated using the equation $W=P\mathrm{\Delta V}$. For processes where pressure changes, the area might be more complex, such as a combination of rectangles and triangles. The sign of the work depends on the direction of the process: positive for expansion (left to right) and negative for compression (right to left).

How do you determine the direction of a process on a PV diagram?

The direction of a process on a PV diagram is indicated by arrows that show the transition from the initial to the final state. This is crucial because the direction affects the sign of the work done. If the process moves from left to right, it indicates expansion, and the work done is positive. Conversely, if the process moves from right to left, it indicates compression, and the work done is negative. Always ensure to mark the direction with arrows to avoid errors in calculations.

Can you use the equation W = PΔV for processes with changing pressure?

No, the equation $W=P\mathrm{\Delta V}$ is only valid for processes with constant pressure. For processes where the pressure changes, you need to calculate the work done by finding the area under the curve in the PV diagram. This might involve breaking the area into simpler shapes like rectangles and triangles and summing their areas. For example, if the pressure increases linearly, the area under the curve will be a combination of a rectangle and a triangle.