The First Law of Thermodynamics

by Patrick Ford
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Hey everyone. So in earlier chapters in physics, we talked a lot about work, which is a transfer of mechanical energy. Remember if you push a box, you're doing work on it and it's going to gain some energy in more recent videos, we've talked about heat, which is a transfer of another type of energy called thermal. And in this video we're gonna put these ideas together and we're gonna talk about the first law of thermodynamics. What I'm going to show you is that it's just an equation and it relates heat and work with something called the internal energy of a system. This first law equation is gonna be super important for the rest of thermodynamics. So it's really important that you get this down, right? So let's check it out and we'll do a quick example together. So, the equation for the first law is that delta E is equal to q minus W. That's it. It's just three variables. What I want to mention though, however, is that you may also see this equation written as Q plus W. That's perfectly fine. That will also work Q minus W has always made a little bit more sense to me. So that's the one we're gonna use here at clutch. But if you do see this Q plus W. I'm going to cover that in a later video. Alright, so this delta E here refers to something called the change in internal energy of the system. And in most of your problems, the system will just be some kind of a gas. So these words are kind of interchangeable. This queue here refers to a heat and it's the heat added to the system. So I'm gonna I'm gonna write this as Q2 now, it also can be removed from the system, and I'll talk about that a little bit later as well. This w here refers to the work that gets done by the system. So I write this as W by. So I'm always going to write it out like this because it's really important that, you know, the definitions for these terms and what they refer to. Alright, so let's just go ahead and see this equation in action in our first example here. So in this example we have to calculate the change in internal energy of gas, we're told that the work done by the gas is 200 jewels and we're going to add 500 joules of heat to the gas. So in this problem we have internal energy work and heat, we're gonna start off with the first law equation. So that's delta. Internal of the system equals the heat added to the system minus the work done by the system. Right? It's just these three variables. If you want to calculate delta E, then you just need the other two variables. You need Q two and w by. It's as simple as that. So what is the heat added to the gas? Well, if you take a look at the problem here, we have two numbers 200 and 500, which one is it gonna be? Well, hopefully you guys realize we're adding 500 jewels of heats to the gas. That's a dead giveaway. That that's going to be this Q. Two. Alright, so this is gonna be the 500 over here. That's our that's our heat added now. We just have to figure out the work done by the system. What we're told here in this problem is that the work done on the environment, which is basically just everything else By the gas is equal to 200. So, that's what we're gonna plug in now for W by. So, this is gonna be 500 -200. And the overall change in energy is 300 jewels. Alright. It's pretty straightforward, basically it's happening. This problem is we're adding some heat to the system, but then it's doing some work. So, the overall change in internal energy is 300 jewels. All right, So, basically, that's all there is to it. Now, a lot of problems will try to trick you. They'll try to throw throw you off by making one of these numbers negative or you may have to go calculators or something like that. So, I want to walk through these two variables in a little bit more detail and give you some good rules to follow. Alright, so, basically, directly from this. First law equation, we can see that internal energy, right? This delta E. Can be changed in two ways. We have Q. And W. So you can change the internal energy through heat transfer. That's our cue, right? Or by some work. Which remember is just w. That's just the two variables involved here. So imagine that I had sort of a container of gas, right? And this container has some kind of a piston but the piston is locked so it can't move, but I turn on a little burner or a flame that's underneath here. What happens is this flame is going to transfer some thermal energy to the gas molecules inside. They're gonna start to move faster and faster and that means they're going to get more and more energy. So the rule is pretty simple for heat. Whenever you add heat, the queue in your equation here is going to be positive. And so therefore the internal energy will increase, that's pretty straight forward. And the opposite is true. If you remove heat, if you remove heat then this queue here, the heat added to the system has to be negative. It has to you have to subtract it and therefore the internal energy decreases. Alright, so that's pretty straightforward. Now, imagine that we have the same scenario here except we take away the flame and now the piston is able to move up and down. So what happens here? Well, if you imagine that the gas here exerts this kind of force on the piston, it can actually move the piston up and down. So it has its applying some force upwards. And if it applies some force over a distance delta X. Then that means it's doing work. Remember back from from forces, the work equals force times displacement. Now in these problems, you'll almost never see these variables F and delta X. So, I'm going to rewrite this in variables that are a little bit more relevant to gasses and that's just gonna be p times delta V. But it's gonna be p times delta V. For the gas. All right, now, I want to mention here that this equation here is only gonna be valid if the pressure is constant and most problems will tell you if the pressure is or isn't. All right. So, basically what's happening here is if the gas is expanding, like we have here in our diagram, then the volume is going to increase, right? It's gonna expand into a larger volume. The work done by the gas is going to be positive, and what that means is that the internal energy actually is decreasing. We can actually just see that from the equation here. Right. So, what happens is if the w in this equation is negative or sorry, if the w is positive and we're subtracting it in this equation. So the overall internal energy will decrease. Alright, And then what happens is the opposite is true. If you compress the gas, if it's compressed, the volume is lower than the work done is negative and the internal energy increases. Alright, so those are the rules for these variables. It's really important that you know them because it's gonna help out in your problems. So let's just finish off with our last example here. We're gonna put a gas instead of a sealed container with a movable piston. Kind of like we have in our diagram up here and we're going to remove 240 jewels of heat from it. The pressure is 100 pascal's and the volume expands from 1 to 3. And we want to calculate the delta E. Internal. So again, we're just gonna start off with the first law of thermodynamics equation, delta internal of the system is Q two minus W By we're gonna do the same exact thing here, calculate delta E. So I just need to know the heat and work. So do we have the heat the heat added to the system? Well, the heat is only mentioned in one place and it's right here, Right, we're removing 240 jewels of heat. So that means that according to our rules that if heat is removed, then this queue in our equation has to be negative. So we're going to write this as negative 240 jewels. Now, what about the work done? Well, we don't actually have the explicit value for work. All we have is that the pressure is 100 the gas expands from 1 to and the pressure is constant. So we don't have the value for work, but we can calculate it by using p times delta V for the gas. It's the change in the volume of the gas. Alright, so this just becomes negative. 240 minus the pressure is 100. And what about delta V? What? We're expanding from a volume of 123. So, what that means here is that delta V gas is final minus initial three minus one. And that's gonna equal to So this is gonna be too When you work this out, you're gonna get is be careful with the science here, negative 440 jewels. And so that's the overall change in the internal energy of the system. Alright, so what happens is you're removing some heats, Right? So the internal energy decreases and then the work done by the gas is positive. And so you're decreasing internal energy again. And that's why you end up with an overall negative number. Hopefully, this makes sense guys. Hopefully it's a good introduction to the first law. Let me know if you have any questions