Alright, you've seen the first law of thermodynamics. This is what it says. Let's see how you use it. Let's look at a particular example. This one says, let's say you got this problem, and it said 60 joules of work is done on a gas, and the gas loses 150 joules of heat to its surroundings.
What is the change in internal energy? Well, we're gonna use the first law. That's what the first law lets us determine.
The change in internal energy is gonna equal the amount of heat that's added to the gas, so let's see, heat added to the gas, well it says that the gas loses 150 joules of heat to its surroundings, so that means heat left the gas. So heat left... The gas, this must have been put into a cooler environment so that heat could leave.
And so it lost 150 joules. A lot of people just stick 150 here. It's gotta be negative 150, because this Q represents the heat added to the gas. If you lost 150 joules, it's negative 150. And then plus, alright, how much work was done. It says 60 joules of work is done on a gas.
So that's work done on the gas. That means it's a positive contribution. to the internal energy, that's energy you're adding to the gas, so 60 joules has to be positive.
And so this is plus, the work done is positive 60 joules. Now we can figure it out, the change in internal energy would be negative 90 joules. But why do we care?
Why do we care about the change in internal energy of the gas? Well here's something important. Whether it's a monatomic or diatomic or triatomic molecule, The internal energy of the gas is always proportional to the temperature. This means if the temperature goes up, the internal energy goes up. And it also means if the internal energy goes up, the temperature goes up.
So one thing we can say, just going over here, looking that the change in internal energy was negative. This means the internal energy went down by 90 joules. Overall, when all was said and done, this gas lost 90 joules of internal energy, that means the temperature went down.
That means this gas is going to be cooler when you end this process compared to when it started. Even though you added 60 joules of work energy, it lost 150 joules of heat energy. That's a net loss. The temperature's gonna go down.
So this is an important key fact. Whatever the internal energy does, that's what the temperature does, and it makes sense. Since we know that an increase in internal energy means an increase in translational kinetic energy, rotational kinetic energy, vibrational energy, that temperature is also a measure of that internal energy. Note that we cannot say exactly how low the temperature went.
This is a loss of 90 joules, but this doesn't mean a loss of 90 degrees. These are proportional, they're not equal. If I go down 90 joules, that doesn't mean I go down 90 degrees. I would have to know more about the makeup of this gas in order to do that.
But the internal energy and the temperature are proportional. Let's try another one. Let's say a gas started with 200 joules of internal energy.
And while you add 180 joules of heat to the gas, the gas does 70 joules of work. What is the final internal energy of the gas? Alright, so the change in internal energy equals Q, let's see, gas starts with 200 joules of internal energy, that's not heat. While you add 180 joules of heat, here we go.
180 joules, should it be positive or negative? It's gonna be positive, you're adding heat to that system. So positive 180 joules of heat are added, plus the amount of work done on the gas.
It says the gas does 70 joules of work. So most people would just do, alright, 70 joules. There we go. But this is wrong, this is wrong because this is how much work the gas does.
This W up here with the plus sign represents how much work was done on the gas. If the gas does 70 joules of work, negative 70 joules of work were done on the gas. You have to be really careful about that. So we can find the change in internal energy.
In this case, it's gonna equal positive 110 joules. But that's not our answer. The question's asking us for the final internal energy of the gas. This is not the final internal energy of the gas.
This is the amount. by which the internal energy changed. So we know the internal energy went up, because this is positive, and this is the change in internal energy.
Internal energy went up by 110 joules. That means temperature's also gonna go up. So what's the final internal energy of the gas?
Well, if the internal energy goes up by 110 joules, and it started, the gas started with 200 joules, we know the final internal energy, U final, is just gonna be 200 plus 110. is 310, or if you want to be more careful about it, you can write this out, delta U, we can call U final minus U initial. That's what delta U stands for. U final is what we want to find, minus U initial is 200. So positive 200 joules was what the gas started with, equals, that's the change and that's what we found, 110 joules. Now you solve this for u final, you would add 200 to both sides, and again you would get 310 joules as the final internal energy of the gas. Let's look at one more.
Let's say you got this one on a test and it said that 40 joules of work are done on a gas and the internal energy goes down by 150 joules. What was the value of the heat added to the gas? Note, we're not solving for the internal energy this time, or the change in internal energy. We're trying to solve for the heat. What's heat?
Heat is Q. So this time we're gonna plug in for the other two and solve for Q. What do we know? 40 joules of work are done on a gas. So this work has gotta be a positive 40 because the work is done on the gas and not by the gas.
And we know the internal energy goes down by 150 joules. It means the change in internal energy has to be negative 150. So if I plug in here, my delta U, since my internal energy went down by 150, delta U is gonna be negative 150. Q, we don't know. So I'm just gonna put a variable in there. Q, I don't know.
I'm just gonna put a Q in there. I'm gonna name my ignorance and I'm gonna solve for it. Plus the work done, we know the work done was 40 joules and it's positive 40. Positive 40 because work was done on the gas. Now we can solve for Q, the amount of heat, the value of the heat added to the gas is gonna be, if I move my 40 over here, I subtract it from both sides, I'm gonna get negative 190 joules.
This means a lot of heat left, 190 joules of heat left the system in order for it to make the internal energy go down by 150 joules, and that makes sense. 40 joules of work were added, but we said the internal energy went down. That means the heat has to take away not only the 40 that you added, but also another 150 to make the energy go down overall, so the heat taken away has to be negative 190 joules.
Alright, so those were a few examples of using the first law. Basically, gotta be careful with your positive and negative signs. You gotta remember what these things are, that Q is the heat, W is the work, delta U is the change in the internal energy, which, don't forget, That also gives you an idea of what happens to the temperature.