Hey guys. So throughout our discussion on thermodynamics, we'll often be talking about gasses and specifically, we're talking about these things called ideal gasses. So in this video, I want to introduce you what an ideal gas is and the equation that we use for them, which is called the ideal gas law. So let's get started here. What is an ideal gas? Well, the definition I like to use is it's kind of like a simplified perfect gas. The analogy is that when we talk about forces and drew free body diagrams, it didn't matter whether it was a plane or a train or a car or whatever. We do everything as boxes because it was the simplest model for an object and similarly, the simplest model for a gas is called an ideal gas. Now, it's basically just satisfies a couple of conditions here and I want to go over them really quickly because you may need to know for a conceptual question on a homework or test. So let's get started. The first one is the gas has a low density. So what happens is that the particles in a container are going to be kind of spread out doing their own thing. They're not going to be super, super tightly crammed in together. So generally what this means is that the pressure is pretty low, but the temperature is pretty high. You won't see many extreme values or anything like that. The second condition is that there are no forces between the gas particles. So in real life and real gasses, you have particles that exist that exert some forces on each other, that you usually like electromagnetic forces. And for an ideal gas, we just consider that they don't exist. The third condition is that the particles have zero physical size. And what this means is that we're going to just treat ideal gas particles like points. So you basically just treat them as tiny little balls that are moving, they're very, very tiny, can treat them as point particles, whereas real gas molecules actually do take up space. But for an ideal gas we just assume that it's zero. The last one is that the particles are moving in straight lines and they collide elastically. This is the most important part here. So basically these tiny little balls, like these tiny little points are moving around. They move in perfectly straight lines until they collide with either the walls of the container or even each other, they can bounce off of each other. However, when they do collide, they collide elastically. And what that means is the energy in the container is conserved. There's no energy loss because these things are bumping around into each other. All right. So the night, the what you really need to know about everyday conditions and ideal gasses is that, you know, even under everyday conditions like in our atmosphere and things like that, most real gasses already behave very much like ideal gasses. So, you know, we can still use this model for an ideal gas, even when we talk about real gasses, like oxygen and nitrogen in our atmosphere. Alright, so let's get to the equation. Then the equation that governs ideal gasses is called the ideal gas law and it relates for related variables. We have pressure, volume, temperature and moles which is the amount of substance for an ideal gas. And it actually works for any ideal gas, doesn't matter what it is. Now, there's two different versions of this and you may see both of them. So I want to go over them. So the first one is PV equals N R. T. If you've ever taken an introductory chemistry course or something like that, you may be pretty familiar with this one, The other one is PV equals N K b T. Where the end is a big end. And the difference between them, they both work for ideal gasses is that you use this equation whenever you have the number of moles of substance and you use this one whenever you're when you're you're asked to find or you're given the number of particles in a gas and these things actually have a pretty simple relationship between them and the number of moles big. And this number of particles And they're related by this equation over here where moles is particles divided by avocados number around. That's 6.2 times 10 to the 23. Alright, so let's move on here. There's two constants that you need to know in both of these equations. The first one is big R and it's called the universal gas constant. It's sometimes called the ideal gas constant. And it's just this number here. 8.314 joules per mole kelvin. The other one is KB. This KB is called bolts means constant And it has a value 1.38 times 10 to the -23. Now, the last thing here is that we're working with absolute temperatures not changes in temperature. So the temperature actually must be in kelvin. That's really, really important. You have to plug in all your temperatures in kelvin. And the very last thing I want to mention here is that this is something you might see in your problems. So something called S. T. P. It's an acronym that stands for standard temperature and pressure. Basically, it's just a common set of conditions. That's very much like what we experience here at Earth's surface, which is that the temperature is zero degrees Celsius or 2 73 kelvin and the pressure is one atmosphere or 1.1 times 10 to the fifth. This is basically just whenever you see STP these are your numbers for temperature and pressure. Alright, so let's get to our example here, Which is we're going to calculate the volume of exactly one Mole of an ideal gas that is at STP So we're going to use those conditions here. So we have S. T. P, which means the temperature is zero Celsius or 2 73 kelvin and we also have that the pressure is 1.1 times 10 to the fifth pascal's. Now if we have one mole of substance that's N. Equals one, we want to figure out what is the volume. So we have these four variables. Remember they're all related using the ideal gas law. So which version are we going to use? Well remember we're given the number of moles of a substance. So that means we're gonna use N. R. T. So we're gonna use PV equals N. R. T. So what happens here is that there are five variables. There's five letters in this equation, but one of them is a constant. So there's really only four variables. You need to know if you ever have three, you can always figure out the other one. So we know what the pressure is, we know what the number of moles is and we have the temperature. Remember this is just a constant. So we can figure out what this volume is by just rearranging the equation. So what happens is this P. Goes to the other side like this and your V. Is going to be N. R. T. Divided by P. So I'm just gonna start plugging in one mole. We have 8.314, that's the gas constant to 73. That's the temperature. And then 1.1 times 10 to the fifth pascal's. So when you work this out or you're gonna get is 0.0224 and this is gonna be meters cubed. Now there's also a couple of other volume conversions that you may need to know for these types of problems. One of them is that one centimeter cubed is one mil leader and then sort of, if you work this out and you play around with the zeros, you'll figure out that one m cubed is 1000 liters. So one thing, when you might see this number represented 0.0224 is when they multiply it by this conversion factor, you're gonna multiply by 1000 liters divided by one meter cubed meters cubed cancels. And you're gonna get 22.4 liters and that's the answer. Either one of these will work. So either one of these uh, these numbers will be the answer and that's actually a really sort of special number here. So the idea is that one mole of any ideal gas at STP sort of at these special conditions has a molar volume of exactly 22.4 leaders notice how we didn't talk about what type of gas it is, all we had was these conditions. So what this means here is that any gas at STP, whether you might have one mole occupies this amount of space and this is sometimes called the molder volume of a gas at STP. Alright, so guys, so that's it for this one, let me know if you have any questions.
3 moles of an ideal gas fill a cubical box with a side length of 30cm. If the temperature of the gas is 20°C, what is the pressure inside the container?
2.7 × 105 Pa
1.85 × 104 Pa
8.12 × 103 Pa
Hydrogen gas behaves very much like an ideal gas. If you have a sample of Hydrogen gas with a volume of 1000 cm3 at 30°C with a pressure of 1 × 105 Pa, calculate how many hydrogen atoms (particles) there are in the sample.
2.42 × 1022
2.39 × 1022
2.39 × 1028
Solving Ideal Gas Problems With Changing States
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Hey guys, so now that we've been introduced to the ideal gas law and some of the problems you're gonna run across, you're gonna have to compare an initial and final state of a gas in the example, we're gonna work out down here, we have some amount of moles of gas in a container and were given with the initial pressure and volume are then what happens is we're going to compress the gas using a piston or something like that and we're gonna try to figure out is what's the final pressure. So basically what happens is we're gonna try to we're gonna take a gas in its initial state and we're gonna change it. We're gonna figure out which one of the variables changes as a result. Now, in order to do that, we're gonna use an approach kind of similar to how we use energy conservation. We know the equation for ideal gasses is PV equals N. R. T. Now what happens is this is sort of like sometimes called an equation of state, it's sort of like a snapshot of these four variables at a specific moment in time. If you change the gas however, from an initial to final, this PV equals NRT has to remain sort of conserved if you will. So we're gonna do is we're gonna stick subscript on each one of these letters here, so we have P initial, the initial equals and initial R. T. Initial. Now remember this R. Is the universal gas constant. Right? So it doesn't actually need a subscript and then we have p final the final equals and final R. T. Final. So what happens is these two things are equal to each other? We're gonna set these two equations equal. And in doing so what happens is that the R. Is going to cancel? Right? It's just a constant that cancels out from both sides and this equation has a lot of equal signs. So we're gonna do is we're gonna divide by N. S. And T. S. On each side. And what you end up with is this equation over here, P initial the initial over an initial T. Initial equals P final the final over and final T. Final. So this is the one equation that you need to solve any kind of problem or an ideal gasses change from one state to another. Now let me show you how to use this. Using a step by step approach, that's gonna get you the right answer every single time here. So the first step, if we're just gonna use this equation is to actually write the equation. Now we're just gonna write this equation every time. It's always gonna work. So the first step here is P initial the initial over an initial T initial equals P. Final the final over end. Final T final notice how again the R. Is missing because it actually gets canceled out when you set these things equal to each other. Alright, so the next thing we have to do is we're gonna have to cancel out the constant variables. So what ends up happening here is that in most of these problems, two out of the four variables are going to remain constant, which is actually just gonna let you cancel it even more things inside of this equation. So let's go through the our variables and figure out which ones we can cancel out. Now. The first part of the problem tells us that we have two moles of an ideal gas. So we know that N equals two, but we're also told that no gas can leak out or in. So what that means here is that the change in number of moles is just zero. That's the same amount of gas. So that means that N can just be canceled out. So, for instance, if we put two for two moles inside of the left and right and sides of the equation, it's just going to cancel out, right, Those things are gonna cancel each other out. Alright. So the other thing we can do is we can say that the container is then at constant temperature, so that's the other variable that has remained constant. So we know that T. I. Is equal to T. F. Whatever that number is, it doesn't really matter that we know it or not because it's just gonna be canceled out. So we're left with is really just these four variables over here. Now, we know that the initial pressure is going to change to some final pressure and we're told that the initial volume goes from 0.05-0.01. So that means here that we have the initial pressure with the initial volume. We have the final volume and we want to calculate what PF is. So all these variables are going to change, we can't cancel them out. And now the last thing we do is we just go ahead and solve. So what happens here is we just end up with p initial, the initial over the final equals p final. So we just move this down to the other side and now we just start plugging in. So the initial pressure is one times 10 to the fifth. The initial volume is 0.5 and the final volume is 0.1. When you work this out, what you're gonna get is five times 10 to the fifth pascal's and that is the answer. All right. So that's how to go through these steps here. Now, if you think if you think about what's happened here, what we've had is that the initial volume went from 0.05 to 0.01. So, what ended up happening was we had the volume that decreased by a factor of five. And as a result, our pressure went from one times 10 to the fifth to 10 5 times 10 to the fifth. So it increased by a factor of five. So what we say here is that P and V are sort of indirectly or in uh yeah, sorry, in inversely proportional to each other. And that brings me to this point here, which is that when scientists were studying these gasses 100 years ago, hundreds of years ago, they actually sort of came up with these three relationships that are historically called the gas laws and they're called Charles or boils, Charles and gala sacks law. Now, the thing is, these are actually special cases of this ideal gas law, which we now know how to use. So I want to go over them quickly just in case you need to know them. Basically what they did is they held the number of moles and one of the other variables as fixed. And they found that these relationships between the gas, the gas variables. Now, what we actually just saw was Boyle's law. Boyle's law says that if the temperature is constant, basically, if there is no change in the temperature of the gas, then what happens is that p is inversely proportional to V. That's what we just saw here, and basically how you get to these equations is you just cancel out the constant variables and we end up with, is this relationship which we just saw, right, if one goes down by a factor of five, the other one goes up by a factor of five. So they're inversely proportional. Now, the other one's called Charles law. And basically, if you held the pressure constant, V is directly proportional to T. So if you have N and P. That gets sort of canceled out like this, then what happens is you just end up with V over T on both sides. Now these are directly proportional even though it's a ratio and think about what's happening is that if this increases by a factor of five, Then this also has increased by a factor of five in order to keep that number, whatever it is constant. The same works by the way for gala sacks law. If V is constant, basically cancel out these two things here, then you just get P over T and you have the exact same relationship so you might need to know that. Um, but basically those are just the gas laws. So now that we know how to solve the ideal gas law problems, let me know if you have any questions. That's it for this one.
A balloon contains 3900cm3 of a gas at a pressure of 101 kPa and a temperature of –9°C. If the balloon is warmed such that the temperature rises to 28°C, what volume will the gas occupy? Assume the pressure remains constant.
Doubling Pressure & Temperature
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Hey everybody. So let's get started with our problem here. So we have an ideal gas in a sealed container, we're told the the initial volume pressure and temperature are, and we're told that the pressure and the volume are both going to double. So let me just go ahead and draw out sort of a sketch of what's going on here. So imagine that I have this kind of like sealed container like this, but I have a sort of plunger that I can move up and down, so this thing can sort of go up and down like this. So what happens here is that this plunger? Imagine it's kind of like halfway through this container, which means that the gas is sort of contained inside of just this area over here. So this is the gas, we have an initial volume pressure and temperature. Now, what happens here is I'm gonna move the plunger up so that it goes like this, so that's basically at the top of the container. So then I have all the gas has now effectively sort of doubled in size. Right? The container is now sort of bigger than it was before. So that's what's going on in this problem here, we have a final volume, final pressure and final temperature. What we're trying to find this first part here is what is the temperature final Alright, so we've got an ideal gas, we're going to change some of the characteristics or variables like pressure and volume. We want to figure out what's the final temperature. So in order to do that, we're just going to stick to the steps here. The first step we're gonna do is write out our changing ideal gas equation. So I'm gonna do P initial v initial divided by n initial T initial equals P final the final over and final T final. So we're looking for is T final here. Now what I want to do is we want to cancel out any constant variables because they'll just cancel out from both sides of the equation. So we'll look through our variables, we'll pressure and volume clearly don't remain constant because we're both told that they both double. However, this is a sealed container. So if it's a sealed container, what happens is that nothing is allowed to go in or come out. So what that means here is that the number of moles, the amount of gas that's inside the container does actually remain constant. So that's what happens any time you have a sealed container just means nothing goes in or out and remains constant. Alright, so now that that steps down, we're gonna go ahead and solve for our target variable. So there's a couple of ways to do this, but you can basically just you can flip the equation or what you can do is you can move this over to the other side and what you'll get is p initial the initial times T final equals p final v final uh sorry, I also have a T initial over here. So then what you have to do is you have to move everything else over to the other side. So the P initial the initial comes down, T initial comes up like this. And what you end up with his here is T final is equal to P final. The final over P initial V initial times. This is going to be T initial. Alright, so that's what the equation works out to now. Notice how we haven't plugged in any numbers yet and if you look the numbers are kind of strange. We've got you know, volumes that are in different units like leaders instead of meters cubed. We've got pressures in terms of atmospheres and things like that. So there are a couple of instances a couple of problems where you may be given in units of leaders and units of pressure and atmosphere. So there's a couple of conversions but we're actually gonna go ahead and shortcut this for just a second here because we can actually sort of solve this problem in an easier way because we know that the pressure and volume of both double. So what that means here is I can sort of rewrite the initial pressure of the final pressure is as multiples of the initial. What do I mean? It means that p final if it's just double the initial pressure, I can write this as two P initial right. I can do the same exact thing for the volume. The volume also doubles. So the tv final is just two times the initial. All right. Now. The reason that this is really helpful is because then I can just cancel out the P initial V initials. Right. I can just cancel out both of those things in the numerator and denominator and then I just end up with basically are just the multiples on the outside. So really what happens here is a T final just becomes four times T initial and you can kind of like sort of reason through this when you look at PV equals Nrt remember this is an equation. Whatever you do to one side you have to do to the other. So what's going on here is you're gonna multiply both of these things by two P initial and V initial. So then what happens is in order for this equation to stay balanced. The temperature has to increase by four times. That's what's going on here. All right. So that basically all I have to do is just take the initial temperature which is 40 degrees Celsius and then multiply it. Right, so what happens here is my T initial here is Celsius. So I'm gonna add to 73 to it and I'm gonna get 313. So really mighty. Final just becomes four times This is going to be 3:13 and I'm gonna get a final temperature. That is 1252 Kelvin. Alright, so you could have actually written all the numbers outdone all the conversions and you'll actually have just gotten the same exact answer over here. This is just a shorter way to do this. All right. So, let's get started now with part B. So, that's basically what the temperature is. It jumps by four. So, let's look at part B. We're supposed to figure out how many moles of gas there are. So remember the moles of gas is just gonna be little N There's actually a couple of different ways. I can do this very if I'm looking for little end that was the only variable that I canceled that in my equation. I can actually just go ahead and use just PV equals Nrt to solve for this. So, you can use PV equals N R T. Now, here's a question. Do you use the initials or the finals? It actually doesn't matter. You can use P initial, the initial equals an initial R. T. Initial or you could use P final v final equals and final. Our final t final. Right. So basically these things are gonna be the same. Remember this end remains constant. So it never changes. So, you can use either set of variables. The choice is up to you, but you'll still get the exact same answer. So, I'm just gonna go ahead and use the initial because for whatever reason. Right. So, I'm just gonna use the initial. So, what happens here is your P initial over here? The initial divided by r Times T initial is going to equal N all right. And then if you chose final you just use all the final versions. So the initial pressure here is going to be 0.15 atmospheres. So I'm gonna, what I'm gonna do is I'm gonna come over here and I'm going to convert this. My initial temperature are pressure is 0.15. And what I'm gonna do here is multiplied by a conversion factor. And I've actually listed a couple of common ones that you might need to know. So one atmosphere is 1.01 times 10 to the fifth. So I'm gonna take 0.15 and I'm gonna multiply it by 1.01 times 10 to the fifth. And what you'll get is let's see, let's scoot us over here. So have a little bit more room Which you end up with here is 1.515 times 10 to the 5th. In pascal's. Alright, so it's a little small but what about the initial volume? But the initial volume is in terms of leaders. But remember we need this in terms of uh in terms of meters cubed because of our our constant over here. Remember so a common one that you'll need to know is that one leader is equal to 0.1 m cubed. So if we're given 2.8 liters and we're just gonna have to multiply by Um let's see the conversion is one leader divided by zero Over 0.001. Actually, I'm sorry. This is gonna be is 0.1. This is gonna be meters cubed divided by one leader. So then the leaders cancel what you'll end up with here is 0.0 um 0 to 8. Alright, So 0.0 to eight. Alright, So if you work this out here, what you're gonna get here is 1.515 times 10 to the fifth, Then you'll have 0.00-8 divided by the R. Which is 8.314 times the initial temperature which we already know is 313. So, if you go ahead and work this out, what you'll get is uh 0.016. And that's mold. So that's your final answer. All right. So these are your true answer choices over here and let me know if you guys have any questions