Learn about Thermodynamics - part 2 Video

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Learn about Thermodynamics - part 2

Welcome back! In the last video, I told you that pressure times volume is a constant that if you increase the pressure or if you increase the volume, you’re going to decrease the pressure and hopefully, you got an intuitive sense why or likewise, if you squeeze the balloon or the box and there are no openings there then the pressure within the box would increase. So with that said, let’s see if we can do a couple of fairly typical problems that you’ll see. So, let’s say that I have a box or a balloon or something and let’s say it has a volume so let me call this the initial volume. So, let’s say the initial volume is 50m3. And let’s say my initial pressure is 500 Pa. And just so you remember, what’s a Pascal? That’s 500 N/m3. And then let’s say that I take that box or balloon or whatever and I compress it down to 20m3. So, let’s say I compress it. So, I squeeze it. So, that was like the first example I gave the last time. Well, it’s the same container and I squeeze it down to 20m3. What’s going to be the new pressure? Well, you should immediately have an intuition that what happens when you squeeze a balloon, it becomes harder to do it. So, what is going to be the new pressure? Well, it’s definitely going to be higher. When you decrease the volume, the pressure increases. They’re inversely related. So, the pressure is going to go up and let’s see if we can calculate it. Well, we know that P1 × V1 = K. And since we have no aggregate change in energy, I am just telling you that the box squeezed. I am not telling you whether it did any work or anything like that. The same constant is going to be equal to the new pressure times the new volume. It’s going to be equal to P2 × V2. So, you could just have the general relationship, P1 × v1 = P2 × V2 assuming that no work was done and there was no exchange of energy from the outside of the system. And the most of these cases when you see this in the exam; that is the case. So, the old pressure was 500 Pa × 50m3. And one thing to keep in mind, because this equivalence is not equal, we’re not saying it has to equal to some necessary absolute number. For example, we don’t know exactly what this K is although we could figure it out right now. As long as you’re using a one unit for pressure on this side and one unit for volume on this side, you just have to use the same units. So, we could have done this the same exact problem, the exact same way if instead of m3, they said liters. And as long as we want it, we had liters here. You just have to make sure using the same units on both sides. So in this case, we have 500 Pa of pressure. The volume is 50m3. That is going to be equal to the new pressure P2 times the new volume, 20m3. And so, let’s see what we can do. We can divide both sides by 10. So, we could just take 10 out of there and then we could divide both sides by two. So, that becomes a 250. And so we get 250 × 5 = P2. And so, P2 = 1250 Pa. And if we kept with the units, you would have seen that. So, when I decreased the volume by roughly 60% as how much I decrease the volume, I have the pressure actually increased by 2 ½. So, that gels what we talked about before. Now, let’s add another variable into this mix. Well, let’s talk about temperature and like pressure and like volume and like at work and a lot of concepts that we talked about in physics, temperature is something that you probably are at least reasonably familiar with. Well, how do you view temperature? Temperature; I think if a high temperate means something is hot and a low temperature means something is cold. And I think that also gives you intuition that a higher temperature object has more energy. The sun has more energy than an ice cube. I think that is fair enough. And I think you also have the sense that what would have more energy? A 100° couple of tea or 100° barrel of tea. I want to make them equivalent in terms of what they’re holding. So, I think you have the sense. Eventhough they are the same temperature, they are both pretty warm. Let’s say this was 100°C, so they’re both boiling. That the barrel, because there is more of it, it’s going to have more energy. It’s equally hot and there are just more molecules there and so that’s what temperature is. Temperature in general is a measure roughly is equal to some constant times the average kinetic energy per molecule. So, the average kinetic energy of the system divided by the total number of molecules we have. So, I guess another way we could talk about it is temperature is essentially energy per molecule. So, something that has a lot of molecules where n is the number of molecules. So, another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature and this is just a constant, so times 1/K but 1/K, we don’t even know what this is, so we could say that’s still a constant. So, the kinetic energy of this system is going to be equal to some constant times the number of particles times temperature. And we don’t know what this is and we’re going to figure this out later. So, this is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system; the aggregate if you take all of the molecules and combine their kinetic energy. And these are the same case. I mean I could put another constant here like all that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature. So, if you think about it, you could also say that this is proportional to this and this is proportional to this. So, you could say that pressure times volume is proportional to the number. And these are all different proportional constant. We’ll figure out this exact constant later. But we could say the pressure times volume is proportional to the number of molecules we have times temperature. And we said the kind of temperature is we can kind of view it as energy per molecule; or another way we could say, if this constant is constant which it is by definition. And the number of molecules is constant. We have PV/T. Pressure × Volume over Temperature is going to be equal to something times the number of molecules. So, we could that’s some other constant. So, this is another interesting thing to think about. We said pressure times volume is equal to pressure times volume, now we added temperature into the mix. So, we could say P1 × V1/ T1 = P2 × V2/T2. And does this make sense to you? Let’s say what happens if I have another box and I have my particles bouncing around like always and I have some volume and some amount of pressure, what happens when the temperature goes up? What am I saying? Well, I am saying that the average kinetic energy per molecule is essentially going to go up. So, they’re going to bounce against the walls more. So, if they bounce against the walls more, the pressure is going to go up assuming volume stays flat. Another way you could think about it. Let’s say the temperature goes up and let’s say the pressure stays flat. So, what did I have to do? Well, I just said if the temperature goes up, the average kinetic energy of each molecule will bounce more. So, in order to make them bounce against the sides of the walls as often, I’d have to increase the volume. So, if you hold pressure constant, the only way you can do that is by increasing the volume while you increase the temperature. So, let’s keep this in mind and we will use this to solve some pretty typical problems in the next video.

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Welcome back! In the last video, I told you that pressure times volume is a constant that if you increase the pressure or if you increase the volume, you’re going to decrease the pressure and hopefully, you got an intuitive sense why or likewise, if you squeeze the balloon or the box and there are no openings there then the pressure within the box would increase. So with that said, let’s see if we can do a couple of fairly typical problems that you’ll see. So, let’s say that I have a box or a balloon or something and let’s say it has a volume so let me call this the initial volume. So, let’s say the initial volume is 50m3. And let’s say my initial pressure is 500 Pa. And just so you remember, what’s a Pascal? That’s 500 N/m3. And then let’s say that I take that box or balloon or whatever and I compress it down to 20m3. So, let’s say I compress it. So, I squeeze it. So, that was like the first example I gave the last time. Well, it’s the same container and I squeeze it down to 20m3. What’s going to be the new pressure? Well, you should immediately have an intuition that what happens when you squeeze a balloon, it becomes harder to do it.

So, what is going to be the new pressure? Well, it’s definitely going to be higher. When you decrease the volume, the pressure increases. They’re inversely related. So, the pressure is going to go up and let’s see if we can calculate it. Well, we know that P1 × V1 = K. And since we have no aggregate change in energy, I am just telling you that the box squeezed. I am not telling you whether it did any work or anything like that. The same constant is going to be equal to the new pressure times the new volume. It’s going to be equal to P2 × V2. So, you could just have the general relationship, P1 × v1 = P2 × V2 assuming that no work was done and there was no exchange of energy from the outside of the system. And the most of these cases when you see this in the exam; that is the case.
So, the old pressure was 500 Pa × 50m3.

And one thing to keep in mind, because this equivalence is not equal, we’re not saying it has to equal to some necessary absolute number. For example, we don’t know exactly what this K is although we could figure it out right now. As long as you’re using a one unit for pressure on this side and one unit for volume on this side, you just have to use the same units. So, we could have done this the same exact problem, the exact same way if instead of m3, they said liters. And as long as we want it, we had liters here. You just have to make sure using the same units on both sides. So in this case, we have 500 Pa of pressure. The volume is 50m3. That is going to be equal to the new pressure P2 times the new volume, 20m3. And so, let’s see what we can do. We can divide both sides by 10. So, we could just take 10 out of there and then we could divide both sides by two. So, that becomes a 250. And so we get 250 × 5 = P2. And so, P2 = 1250 Pa. And if we kept with the units, you would have seen that. So, when I decreased the volume by roughly 60% as how much I decrease the volume, I have the pressure actually increased by 2 ½. So, that gels what we talked about before.

Now, let’s add another variable into this mix. Well, let’s talk about temperature and like pressure and like volume and like at work and a lot of concepts that we talked about in physics, temperature is something that you probably are at least reasonably familiar with. Well, how do you view temperature? Temperature; I think if a high temperate means something is hot and a low temperature means something is cold. And I think that also gives you intuition that a higher temperature object has more energy. The sun has more energy than an ice cube. I think that is fair enough.

And I think you also have the sense that what would have more energy? A 100° couple of tea or 100° barrel of tea. I want to make them equivalent in terms of what they’re holding. So, I think you have the sense. Eventhough they are the same temperature, they are both pretty warm. Let’s say this was 100°C, so they’re both boiling. That the barrel, because there is more of it, it’s going to have more energy. It’s equally hot and there are just more molecules there and so that’s what temperature is. Temperature in general is a measure roughly is equal to some constant times the average kinetic energy per molecule. So, the average kinetic energy of the system divided by the total number of molecules we have.

So, I guess another way we could talk about it is temperature is essentially energy per molecule. So, something that has a lot of molecules where n is the number of molecules. So, another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature and this is just a constant, so times 1/K but 1/K, we don’t even know what this is, so we could say that’s still a constant.
So, the kinetic energy of this system is going to be equal to some constant times the number of particles times temperature. And we don’t know what this is and we’re going to figure this out later.

So, this is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system; the aggregate if you take all of the molecules and combine their kinetic energy. And these are the same case. I mean I could put another constant here like all that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature.
So, if you think about it, you could also say that this is proportional to this and this is proportional to this. So, you could say that pressure times volume is proportional to the number. And these are all different proportional constant. We’ll figure out this exact constant later. But we could say the pressure times volume is proportional to the number of molecules we have times temperature. And we said the kind of temperature is we can kind of view it as energy per molecule; or another way we could say, if this constant is constant which it is by definition. And the number of molecules is constant. We have PV/T. Pressure × Volume over Temperature is going to be equal to something times the number of molecules. So, we could that’s some other constant.

So, this is another interesting thing to think about. We said pressure times volume is equal to pressure times volume, now we added temperature into the mix. So, we could say P1 × V1/ T1 = P2 × V2/T2. And does this make sense to you? Let’s say what happens if I have another box and I have my particles bouncing around like always and I have some volume and some amount of pressure, what happens when the temperature goes up? What am I saying? Well, I am saying that the average kinetic energy per molecule is essentially going to go up. So, they’re going to bounce against the walls more. So, if they bounce against the walls more, the pressure is going to go up assuming volume stays flat.

Another way you could think about it. Let’s say the temperature goes up and let’s say the pressure stays flat. So, what did I have to do? Well, I just said if the temperature goes up, the average kinetic energy of each molecule will bounce more. So, in order to make them bounce against the sides of the walls as often, I’d have to increase the volume. So, if you hold pressure constant, the only way you can do that is by increasing the volume while you increase the temperature. So, let’s keep this in mind and we will use this to solve some pretty typical problems in the next video.

Transcription by: Scribe4you Transcription Services