1

concept

## The First Law of Thermodynamics

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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 gonna 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 is cute too. 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 gonna write it out like this, because it's really important that, you know, the definitions for these terms and what they refer to. All right, 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 gonna 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 heat to the gas. That's a dead giveaway. That that's gonna 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 minus 200. And the overall change in energy is 300 jewels. Alright, it's pretty straightforward, basically what'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 gonna 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 Q. 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. Alright, So basically what's happening here is if the gas is expanding like we have here in our diagram, then the volume is gonna 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, 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 we're gonna 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 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 gonna write this as negative 240 jules. Now, what about the work done? But 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. 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, it's 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

2

Problem

A gas in a cylinder held at a constant pressure 1.80×10^{5} Pa expands from a volume of 1.2 m^{3} to 1.6 m^{3}. The internal energy of the gas decreases from 4.40×10^{5} J to 3×10^{5} J. How much heat was transferred to the gas?

A

–6.8 × 10

^{4}JB

2.12 × 10

^{5}JC

4.72 × 10

^{5}JD

2.28 × 10

^{5}J3

concept

## Alternate Equation of the First Law of Thermodynamics

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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.

4

Problem

The internal energy of a system decreases by 500 J, and 230 J of work is done on the system. What is the heat transfer into or out of this system?

A

730 J

B

270 J

C

– 270 J

D

– 730 J

5

example

## Calculating Work Done on Monoatomic Gas

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everybody. So, let's check out this problem here. So, we have the internal energy of a mono, atomic, ideal gas is given by this equation here, three halves. Nrt. Now, some of you may have already seen this equation, that's totally fine if you haven't. That's also fine. Just know that this is an equation for the internal energy where this end here is the number of moles in the gas, and t is the temperature. So, here's what's going on in this problem, We're gonna add 1300 joules of heat to some amount of gas, and its temperature is going to increase from 2 72 3 20. We want to calculate how much work was done by the gas. So, ultimately, we're trying to figure out what is w what's the w by the gas here. So, how do we start things off? Well, we're gonna have some heats, some works some internal energy. So, we're going to use the first law of thermodynamics. So, we're gonna write out this equation here. This change in internal energy of the gas equals the heat added to the gas minus the work done by the gas. Alright, so, if you want to calculate what this work done by the gas is, I'm gonna go ahead and rearrange some stuff. So, basically what I'm gonna do is I'm gonna move this to this side on the left side, and I'm gonna move the internal to the right side. What you end up with here, is that the work done by the gas is equal to the heat added to the gas minus the change in the internal energy of the gas. Okay, so that's really what's going on here, just moving some stuff around. So what is first the heat that's added to the gas here? So I'm gonna go over here and write out my variables. Well, if you look at the problem here we're gonna add 1300 joules of heat to the gas, so therefore Q two is equal to 1300. And we have what that variable is Now, what about this change in internal energy of the gas? That's really sort of what's going on in this problem? We're gonna have to go figure this out. Well remember that the change in the internal energy, the change in anything really is always equal to final minus initial. So it's the final minus initial here. So what are this the final and initial? What what happens is now we have a new equation to solve for this, it's just three halves N. R. T. So what's going on here is that the final minus E. Initial is really just equal to three halves N. R. T. Final minus three halves and R. T. Initial. Right so just final minus initial. The end is going to stay the same because it's just the amount of gas that I have. So that doesn't change. But what happens here in this problem is that we have some kind of increase in temperature. So this is my t. Initial and this is my t final. So that's what's going on. We have to use this equation to figure out what the change in internal energy is. And then once we figure this out, we can just plug that back into this equation and solve for the work done by the gas. Okay, so this is just gonna be three halves and we're gonna just go ahead and start plugging some stuff in. So this is gonna be 1.5, that's the number of moles that I have, This are constant, remember is just 8.314, you may remember that and then T final. So this is just going to be 320 Then we have -3/2, 1.5, 8.314 times to 70. Now you could have rearranged some stuff, you could have condensed this and shorten this equation, but I just plugged everything in, but which basically you're gonna get here, is that the change in the internal energy is equal to 935 jewels. So, remember this number here is what we plug into this in our original equation. So now, basically we're just gonna have to go and plug this last step in here, which is the work done by the gas is equal to Q to remember. This is just the 1300, that's the heat added minus the change in the internal energy here, which was just 935. So what you end up with here is that the work done by the gas is equal to 365 jewels. And that is the final answer here. Alright, so that's it for this one. Guys, let me know if you have any questions and I'll see you the next one.

Additional resources for First Law of Thermodynamics

PRACTICE PROBLEMS AND ACTIVITIES (9)

- An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from 10.0°C to 25.0°C at constant pr...
- Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2....
- A gas in a cylinder is held at a constant pressure of 1.80 * 10^5 Pa and is cooled and compressed from 1.70 m^...
- Five moles of an ideal monatomic gas with an initial temperature of 127°C expand and, in the process, absorb 1...
- Propane gas (C3H8) behaves like an ideal gas with g = 1.127. Determine the molar heat capacity at constant vol...
- Heat Q flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fr...
- Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2....
- A gas in a cylinder expands from a volume of 0.110 m^3 to 0.320 m^3 . Heat flows into the gas just rapidly eno...
- Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature o...