Alternate Equation of the First Law of Thermodynamics

by Patrick Ford
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Hey guys. So in an earlier video, I mentioned that here at clutch will always write the first law equation as delta E equals Q minus W. But I mentioned that you may see this written as Q plus W. What I want to do in this video is explain the differences between these two different forms of the first law, so that these videos will still help you, no matter which one you see it really all has to do with how books define this W. Here. So I'm just gonna explain the difference real quick and then we'll do a quick example together. So let's check it out. So remember that I mentioned here that when you use Q minus W. Which we're always going to use that that w this work here is defined as the work done by the system or by the gas. However, when you see this as plus w what's going on here? Is that books are defining the work as the work done on the system or on the gas? That's really all there is to it. Is it by or is it on that really explains the difference between these two different forms. So let's just jump into our problems so I can actually show you real quick what's going on here. So in this problem we're gonna fill a container with some gas, we're gonna add 300 joules of heat. We have a constant pressure of 100 pascal. So I've got that P equals 100 in this container. So we're gonna put our hands on this movable piston, right? I've got my hand like this and you're gonna push down on the piston, you're gonna compress the gas from 5 to 3 m cubed. Alright, so we're gonna compress the gas. So let's jump, jumping right into part A. We're going to calculate the work done by the gas on your hands, right? So the work that's done by the gas. Remember we have an equation for this, assuming that we have constant pressure. We can always just use this P times delta V. The change in volume for the gas, that's exactly what we're gonna do here. So this is going to be p times delta V gas. Now the pressure, we have 100 that's 100 pascal's What about the change in the volume where we're going from 5 to 3? So the change in volume is going to be negative two and that's delta v. Gas. So this is gonna be 100 times negative two and this is gonna be negative one, negative 200 jewels. Now this is consistent with our rules for the work. Remember if you're compressing a gas, the volume decreases and the work done is negative. So this makes total sense for us. Now let's move on to part B. Which is we're going to calculate the work done on the gas by your hands. So which equation do we use for that? How do we calculate w on. Do we just use the same p delta V equation, that's the only one that we have. Well, if you do that, if you use P delta V, you're just gonna end up with the exact same number, right? You're gonna do 100 times negative two And you're just going to get negative jewels. So is that the answer? Well, it cannot be the answer, right? How do both things decrease by 200 jewels? The work done by the gas on your hands and the work done on the gas by your hands? What is happening here is if you just look at these two equations in order for them to be equal to each other, the work done by the gas has to have the opposite sign as the work done on the gas. These two things have to be equal and opposite. The way I like to think about. This is kind of like when we talked about forces, if a pushes on B Then be pushes back on a with the same force but negative sign, it's the same idea here. So what you have to do here is whatever the equation is that you normally would see for work done by, you just have to stick a negative sign or you just have to change the sign. So this actually becomes positive 200 jewels. It's also kind of like money, right? If I hand you $5, we're both not losing or gaining $5, I'm losing five and you're gaining fives. They have to be opposite sign here. So, the work that's done on the gas has to be positive, 200 jewels. So hopefully that makes sense here again. These problems are gonna be pretty straightforward in terms of the equation. So, we'll try to throw you off with some of these, you know, by and on and whether it's positive or negative, that's really like the complication. That that's the tricky part. All right, so that's basically all there is to it. Let's go ahead and now move on to the last part here and finish this off. We're gonna calculate the change in the internal energy of this gas using both of the forms of the first law that we just talked about. Right. So, we're just gonna calculate delta E. Internal Using our standard form, which is the Q two W. By. So, let's check this out. Right? If I want to calculate delta internal, I have to figure out the heat added to minus the work done by. So, do we have the heat added to? Well, if you look through our problem here, we're gonna add 300 joules of heat. So that's gonna be the Q two. So, this is gonna be 300 jewels. And then what's the work done by? We actually calculated that in part a here. This is negative 200 jewels. Now be careful, we have a minus sign out here, but this also has a minus sign and you have to keep both of them. So you're gonna subtract a negative 200 And what you're going to end up with is 500 jewels. So that's the change in internal energy. Now, what if we use the other equation? The alternate. What if we use Q. Plus. Well if you do this, what's gonna happen is you're gonna have delta internal is going to be cute too. Plus the work done on the gas. And what you end up here is the heat doesn't change. It's still 300. But then what's the work done on the gas? We calculated in part B. That's just 200. So you're just gonna stick to 100 in here and you just end up with the exact same thing over here. Right? When you subtracted these, this negative um you ended up with a positive and that's exactly what happens here. So you end up with 500 jewels and that makes sense. Right? The two forms should agree with each other, we should get the same number no matter what. And so we just get 500 jewels. All right. So the last thing I wanna mention here is that throughout these videos, if you are using this form of the equation, you may see that some of the equations might be slightly different from yours and that has to do with the fact that you know your work is actually negative opposite sign. Alright, So that's this one guys, let me know if you have any questions.