First Law of Thermodynamics - Online Tutor, Practice Problems & Exam Prep

The first law of thermodynamics relates internal energy change (ΔE) to heat (q) and work (w) through the equation ${\mathrm{\Delta E}}^{=}(q-w)$. Here, q represents heat added to the system, while w denotes work done by the system. Understanding the signs of q and w is crucial: adding heat increases internal energy, while work done by the system decreases it. This foundational concept is essential for analyzing thermodynamic processes.

The First Law of Thermodynamics

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 going to put these ideas together and we're going to 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 going to 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 ΔE=q-w. That's it. It's just 3 variables. What I want to mention though, however, is that you may also see this equation written as q+w. That's perfectly fine. That will also work. q-w has always made a little bit more sense to me, so that's the one we're going to use here at Clutch. But if you do see this q+w, I'm going to cover that in a later video. Alright. So this ΔE 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 q here refers to a heat and it's the heat added to the system. So I'm going to write this as q₂. 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. 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 joules, 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 going to start off with the first law equation. So that's ΔE_internal of the system = heat added to the system - work done by the system. Right? It's just these three variables. If you want to calculate ΔE, then you just need the other two variables. You need q and w. 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, 500 and 200. Which one is it going to be? Well, hopefully, you guys realize we're adding 500 joules of heat to the gas. That's a dead giveaway that that's going to be this q. Continue with more details...

A gas in a cylinder held at a constant pressure 1.80×10⁵ Pa expands from a volume of 1.2 m³ to 1.6 m³. The internal energy of the gas decreases from 4.40×10⁵ J to 3×10⁵ J. How much heat was transferred to the gas?

• A. –6.8 × 10⁴ J
• B. 2.12 × 10⁵ J
• C. 4.72 × 10⁵ J
• D. 2.28 × 10⁵ J

Alternate Equation of the First Law of Thermodynamics

Hey, guys. So in an earlier video, I mentioned that here at Clutch, we'll always write the first law equation as ΔE=Q-W, but I mentioned that you may see this written as Q+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-W, which we're always gonna use, that w, this "work" here is defined as the work done by the system or by the gas. However, when you see this as Q+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 problem 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 pascals, so I've got that \( p = 100 \) in this container. So we're gonna put our hand 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 meters cubed. Alright. So we're gonna compress the gas. So let's jump right into part a. We're gonna 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×ΔV, the change in volume for the gas. That's exactly what we're gonna do here. So this is gonna be p×ΔVgas. Now the pressure, we have 100, that's 100 pascals. What about the change in the volume? We're going from 5 to 3, so the change in volume is gonna be negative 2 and that's ΔVgas. So this is gonna be 100×(-2) and this is gonna be negative 200 joules. 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 gonna 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×ΔV equation? That's the only one that we have. Well, if you do that, if you use p×ΔV, you're just gonna end up with the exact same number. Right? You're gonna do 100 times negative 2, and you're just gonna get negative 200 joules. So is that the answer? Well, it cannot be the answer. Right? How do both things decrease by 200 joules? The work done by the gas on your hand and the work done on the gas by your hand. What ends up 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 b 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 joules. It's also kind of like money. Right? If I hand you $5, we're both not losing or gaining $5. I'm losing 5 and you're gaining 5. So they have to be opposite sign here. So the work that's done on the gas has to be positive 200 joules. So hopefully that makes sense here. Again, these problems are gonna be pretty straightforward in terms of the equation. So they'll try to throw you off with some of these, you know, the by and on and whether it's positive or negative. That's really, like, the complication, that's the tricky part. Alright. 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 ΔEinternal using our standard form, which is the Q-W by so let's check this out. Right? If I wanna calculate ΔEinternal, 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. So this is gonna be 300 joules. And now what's the work done by? Well, we actually calculated that in part a here. This is negative 200 joules. 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 gonna end up with is 500 joules. So that's the change in internal energy. Now what if we use the other equation, the alternate? What if we use Q+W? Well, if you do this, what's gonna happen is you're gonna have ΔEinternal is gonna be Q+W. 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 200 in here, and you just end up with the exact same thing over here. Right? When you subtracted these negative, you ended up with a positive, and that's exactly what happens here. So you ended up with 500 joules, 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 joules. Alright? So the last thing I want to 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 it for this one, guys. Let me know if you have any questions.

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

Calculating Work Done on Monoatomic Gas

Hey, everybody. So let's check out this problem here. We have the internal energy of a monoatomic ideal gas is given by this equation here, 32nRT. 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 n here is the number of moles in a gas and T is the temperature.

Here's what's going on in this problem: We're going to add 1300 joules of heat to some amount of gas, and its temperature is going to increase from 270 to 320. We want to calculate how much work was done by the gas. Ultimately, we try to figure out what W is, the work done by the gas here. How do we start things off? Well, we're going to have some heat, some work, some internal energy, so we're going to use the first law of thermodynamics. We're going to write out this equation here, the change in internal energy of the gas equals the heat added to the gas minus the work done by the gas.

To calculate what the work done by the gas is, I'm going to go ahead and rearrange some stuff. What I'm going to do is move this to the left side and move the internal energy 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.

Now, what about the heat that's added to the gas here? Well, if you look at the problem here, we're going to add 1300 joules of heat to the gas. The change in the internal energy of the gas is really what's going on in this problem. We're going to have to figure this out. Remember that the change in anything really is always equal to final minus initial. So it's efinalmomeinitial.

Now, what happens is we have a new equation to solve for this: it's just 32nRT. The n 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 is that we have some kind of increase in temperature. We have to use this equation to figure out what the change in the 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.

We are going to start plugging some stuff in. This is going to be 1.5, that's the number of moles that I have. This R constant, remember, is just 8.314. Then T final. So this is just going to be 320 minus 3/2 1.5 * 8.314 * 270. Now you could have rearranged some stuff, condensed this equation, but I just plugged everything in. What you're basically going to get here is that the change in the internal energy is equal to 935 joules.

Remember, this number here is what we plug into our original equation. So now we just have to go and plug this last step in here, which is the work done by the gas is equal to q2, 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 joules, 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 in the next one.

Here’s what students ask on this topic:

What is the first law of thermodynamics?

The first law of thermodynamics states that the change in internal energy of a system (ΔE) is equal to the heat added to the system (q) minus the work done by the system (w). Mathematically, it is expressed as:

$\mathrm{\Delta E}=q-w$

This law is a statement of the conservation of energy, indicating that energy can neither be created nor destroyed, only transferred or converted from one form to another. In thermodynamic processes, understanding the signs of q and w is crucial: adding heat increases internal energy, while work done by the system decreases it.

How do you calculate the change in internal energy using the first law of thermodynamics?

To calculate the change in internal energy (ΔE) using the first law of thermodynamics, you use the equation:

$\mathrm{\Delta E}=q-w$

Here, q represents the heat added to the system, and w represents the work done by the system. For example, if 500 joules of heat are added to a gas and the gas does 200 joules of work, the change in internal energy would be:

$\mathrm{\Delta E}=500-200=300$ joules

This means the internal energy of the gas increases by 300 joules.

What is the difference between q-w and q+w in the first law of thermodynamics?

The difference between the forms q-w and q+w in the first law of thermodynamics lies in the definition of work (w). In the equation ΔE = q - w, w represents the work done by the system. In the equation ΔE = q + w, w represents the work done on the system. The choice of form depends on the convention used in the textbook or course. Both forms are correct but require consistent use of the definitions of work.

How does adding or removing heat affect the internal energy of a system?

Adding heat to a system increases its internal energy, while removing heat decreases it. In the first law of thermodynamics, this is represented by the term q in the equation ΔE = q - w. If q is positive (heat added), the internal energy (ΔE) increases. If q is negative (heat removed), the internal energy decreases. For example, adding 300 joules of heat to a system increases its internal energy by 300 joules, while removing 300 joules of heat decreases its internal energy by 300 joules.

What is the significance of the first law of thermodynamics in thermodynamic processes?

The first law of thermodynamics is significant because it establishes the principle of energy conservation in thermodynamic processes. It allows us to analyze how energy is transferred and transformed within a system. By relating internal energy change (ΔE) to heat (q) and work (w), it provides a framework for understanding how different processes, such as heating, cooling, expansion, and compression, affect the energy state of a system. This foundational concept is essential for studying and solving problems in thermodynamics.