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22. The First Law of Thermodynamics

PV Diagrams & Work


Work and PV Diagrams

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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 have to take these changes and you'll have to map them out or draw them and what's called a PV diagram. So in this video, I'm going to introduce you to what these PV diagrams are there 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. Basically what these processes do. What these diagrams do is the 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 expanding from a volume of 2-5 at a constant pressure. That's a thermodynamic process. The whole idea here is that now we're gonna take this text and this picture and we're actually gonna draw it out on this graph now, even though it's a graph of a lot of textbooks to refer to this as a PV diagram. So that's just what we're gonna call it. So let's get started here. We're gonna 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 gonna we have we're going from two m cubed. So that's sort of like our initial point right here and then we're gonna go to five m cube. So that's gonna 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 year will never change. So that's really what this process looks like. Notice how he indicated this arrow here because we're going from initial to final, That arrow is gonna 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 delta V. 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 gonna be 100 the change in volume is just gonna be from to 5. So this delta V here is just gonna be three, right? So this is gonna be three and you can work this out and you're gonna get 300 jewels. Pretty straightforward. That's thornton, P delta V. Right? 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 gonna be this rectangle right here. So all you 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 three. The height of this rectangle is 100. And so therefore, if you do three times 100 hopefully guys realize that you should get jewels. Notice how these two numbers are the same. And that's no coincidence here. So, what? 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 for this process. Because if you think about it, what happens is the base really is just equal to delta v. And the height really is just equal to p. So when you're doing the area, you're really just doing delta v times p. And that of course is just this equation over here. That's why you get 300. All right, let's move on to our second one here because we have other possibilities. Gasses could expand but they also could be compressed. So in our second example, we have a gas that's compressing from 5 to 2. But now, unlike the first problem, the pressure is rising steadily from 100 to 2 20. So we're gonna have to draw this process out first. So I'm going from five, I'm just going to call that over here five and this is gonna be too And I'm at five m cubed. But the pressure goes from 100 to 220. So I'm basically going here and then I'm also going up here like this. So my initial is going to be at five m cubed and 100, that's gonna be my initial And two m cubed into 20 is going to be my file. So that's my final The the process will look just like a straight line that goes from here to here. And remember I have to indicate the arrow. This is gonna be super important here. That leads me to an important point here, which which is that unlike other diagrams that we've previously looked at in physics, which only went from left to right because usually we had something like time on the X axis here. These thermodynamic processes can actually go in any direction on P. V diagrams, they can go right, they can go up down diagonal, whatever. So it's gonna be really important for you to indicate where it's going to pass the direction that's going with those arrows. If you don't do this, forget to do this, you may get the wrong answer. Alright, so that's what this process looks like. Let's go ahead and calculate now the work that's done. So this w by here, the only equation that we can use so far is this P delta v. So can we use this equation to solve this problem? Remember, this equation only works is if you have constant pressure, which we don't have in this problem number is rising steadily from 100 to 2 20. If you ever forget this, you could just look at this equation. Notice how this P. Doesn't have a delta symbol, it's just one P. So which one of these things would you even use if you were to try to use this equation? So it's not clear. So we actually can't use this equation, you cannot use this equation. And the only other way to calculate the work is to calculate the area that is under this curve. So that's gonna be this area over here. Alright, so we're gonna do is we're gonna split these areas up into sort of simpler shapes. I've got a triangle like this, I'm gonna call this a one and then I've got a rectangle that's a two. So my area, it's just gonna be a one plus a to the area of a triangle and a rectangle. Right. So let's just go and calculate those. So a one is gonna be one half of base times height. The base of this triangle here, I'm going from 5 to 2. So that's just gonna be three and the height of this is gonna be from 100 to 20. So that's gonna be 120. So this is gonna be one half of three. Oops. So I got one half of three times 120 when you work this out you're gonna get is 100 and 80 jewels. So is this the right value. Well if you think about it, this actually should not make sense to you remember with this gas is being compressed and we said that the rule is when the gas is compressed, the work that's done has to be negative. So what's going on here is that because we're going from 5 to 2 on our PV diagram, the base of this triangle has to pick up a negative sign. So you have to insert this negative sign. And then this work here just becomes negative 180 jewels. And that is going to be the right value for this area. So I have a really important rule for you guys to follow here. If the path of the process goes from left to right. Like it did in our first problem here, then that means that the work you calculate is going to be positive. In our second problem, the path of this process went from right to left, even if it kind of went up a little bit, that's perfectly fine, it's still overall going from right to left and therefore the area is going to be negative. That's why indicating the direction on those PV diagrams is super important. If you don't do it, you're gonna get the wrong answer. Alright, so now let's do the exact same thing here for area too. So this is just gonna be base times height, so we have the base of this and remember this is just negative three now and then the height of this is just equal to 100. So this is gonna be negative three, this is gonna be negative three times 100. Hopefully you guys should get three negative 300 jewels. So now all we have to do is just add these two areas together but they're both negative. So this is gonna be negative 1 80 plus negative 300. And you're just gonna get negative 480 jewels. So that is the right answer for the work done in this thermodynamic process. Alright guys, so that's it for this one, let me know if you have any questions

Finding Value of V on Axis

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Hey guys, so 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 giving this thermodynamic process here, and we're told what the work done is, it's two times 10 to the fifth. And what we wanna do is we want to figure out the value of V. One indicator 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 wanna do is we want to figure out what is the volume, basically, what do the tick marks represent on RV access? 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 gonna 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 and we have flat processes, but this one goes up like this, so the pressure is changing, we can't use it. So instead 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, It's going to be the area that's under this shape right here. So this w is equal to two times 10 to the fifth, and we're gonna have to do is we're going to have to relate this work to one of the area equations that we have for shapes like a rectangle or 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 we have here in the trapezoid is I'm gonna call this base one, This is gonna be base two, and then this is gonna be my height of my trapezoid, right? The trapezoid doesn't always have to look like, like this, like the shape here, because you could have this as being as as a trapezoid as well. So, here's what's going on, we're gonna use the area for a trapezoid, which is going to be one half, this is gonna be base one plus base two times the height. Now we're told here, is that this is equal to two times 10 to the fifth. So what is base 1? Base to? What does height? What does all that stuff? I mean in terms of the variables that we have here. So what's going on here? Is that this base one represents one times 10 to the fifth. This base to represents three times 10 to the fifth. So what I'm gonna do here is I'm gonna replace this with one, half, one times 10 to the 5th, Plus three times 10 to the 5th, and then times the heights. Well, what is this height here? Well, if you look at this at this graph, the height is going to be the difference between this piece and this piece or this part of this part here on the V axis. So the height here of my trapezoid actually represents the difference between V one and three times V one. So the height here is actually gonna be two times V one, right, we're going one and then two of whatever unit that is. So that's sort of gonna be my uh my height here, it's gonna be two times V one, so this is going to equal two times to the fifth. So now what I can do here is all I have to do is solve for this V one. Okay, so here's what's going on, we're gonna have, I'm just going to combine these two things in this parentheses here. So we've got is four times 10 to the fifth Times two. V 1 is equal to two times 10 to the 5th. And so now we're gonna do is one half of four times 10 to the fifth is just gonna be two times 10 to the fifth. So two times to the fifth times to be one equals two times 10 to the fifth. So what happens here is we can divide out This two times 10 to the 5th from both sides. And what happens is this goes away and this just becomes one. So we've got two V one is equal to just one, right? And so therefore the V one is equal to 0.5 m cubed. So that's basically what the first little tick mark represents. So what's going on here is this is 0.5, this is one And this is 1.5 m cubed on the x axis. And so what happens is if you go ahead through and double check real quick if you basically just go ahead and plug these numbers back into this trapezoid equation, which you'll get here is two times 10 to the fifth. Alright, so that's over this one. Let me know if you have any questions.