Lecture 20. Dalton's Law of Partial Pressures
Brindley, Amanda
UCI Chem 1A is the first quarter of General Chemistry and covers the following topics: Atomic structure; general properties of the elements; covalent, ionic, and metallic bonding; intermolecular forces; mass relationships. 
 Okay, so today we're gonna pretty much see. All right, I think that's better. Okay, so today if you guys can be quiet, we will. Okay, we're gonna kind of start back and talk a little bit more about Dalton's partial pressures. I think really we'll spend most of today on doing the various versions of this in the various examples because it is a little bit long. We may make it into a fusion. So I'm hoping to finish up the gas laws on next Monday and then be able to spend Wednesday and Friday on review for you guys. I think two review lectures would be best, but it really just depends on if we can make it through or not. I'm not 100% sure if we will or not. So without further ado, we'll get going. So review a little bit from the last five minutes last time. So remember that partial pressure, the way that this works is that you can take the pressures of each individual gas. So let's say you have a mixture of three or four gases. You can take the pressure of each one of them and then add them together and that'll give you your total pressure. So when we did this sort of with an equation form, this is it written out. I'm sure you'll be having some computer issues. So this is it written out in just pressure of each gas equals pressure of the total. So using the fact that we know that pressure is equal to nRT over V, ideal gas law, we can take and we can solve the pressure for each individual gas and then we can add them all up together. And we know that if we have a container with a mixture of gases in it, that the temperature and the volume are going to be the same. So we can go ahead and we can pull that out. We can factor the moles out and then we can get this. So what this really says is that you add up all of the moles and you can add that up and then you get your total pressure. So your moles of all your gas, you can fill it right into your ideal gas law as well and solve for your total pressure that way. So two different ways of doing it. Now there's a very important concept here that we need to talk about and that's called the mole faction. So this X, this is actually the Greek letter chi, so that's how you say that. And this says that if you have the, all this really is a fraction just like any other fraction, right? It's whatever you're talking about over top of the total. So if you want to know the mole fraction of A, you take the moles of A and you put it over top of the moles total. It's no different really than saying the percent of A, like the mole percent of A, except that you don't have the times 100% there. It's just the decimal point or the fraction. So that's what mole fraction is. Now what this says is it kind of takes this concept, combines it with this one, and says well we can take that chi, we can take that mole fraction, right, our part over our whole, and we can just times that by the total pressure. This isn't any different than the sort of how you work with percents your entire life, right? If you want to know what percent something is of a total, you take the fraction and you multiply it by the total, okay? So this is your total pressure. So if I tell you that something has a total pressure of five atmospheres, but the mole percent or the mole fraction of it is only 0.2, take 0.2 and times it by five, okay? So don't let this seem like a new equation to you. This is just using percents like you've always used them, except what's the first thing you do if I say you have 75% of something? You turn it into a decimal, right? Well that's all this is. It's just the decimal of the part over top of the whole. So that's all the mole fraction is. It's like a percent, except not with the percent. I'm not sure why they go through the, you know, special aspect of naming it and different and calling it chi and all of that, but you know, that's fine. So don't let this be new to you, just kind of a different way of naming it. Okay. So before we do partial pressures with actual numbers, let's just go through and do it with a quick picture. So if I have this and I have this mixture of molecules, so I have some purples and some greens of aqua or cyan and some pinks. If I tell you that this has a pressure of 10, it doesn't matter what unit, I say 10. And we want to know the partial pressure of each gas. What do we do? You have to find the mole fraction, right? So let's start with the purple ones. So first of all, let's figure out how many, what's our total number of molecules? Our total mole of molecules. So our total number of molecules we just count. One, two, three, four, five, six, seven, eight, nine, ten. Now if we want to know our mole fraction of the purple ones, what do we do? Well how many purple ones do we have? And I apologize for the colorblind people. Statistics say there's at least a few of you and I do apologize. One, two, three, four. I won't do that on an exam obviously. So there's four out of ten, right? So your mole fraction is four out of ten. So what would be our mole fraction of the pink ones? Would it be six out of ten or three out of ten? Good, three. It doesn't matter that it's a diatomic, right? We're treating all of these molecules and all of these atoms and all of these gases as just like they're just spherical balls that are all the exact same. They have no volume, they have no interaction, they have nothing. They're just ideal gases. So it doesn't matter if I drew these as diatomics and these as single. So it would be three out of ten, right? Now so what does that mean the pressure of the purple ones would be? Four out of ten. So just four, right? So and you can go through, you can do that with each one. So the blue would just be one out of ten. That would be your fraction, it would be point one. The greens would be two out of ten, point two. So all mole fraction is is a count, right? It's how many moles of something you have, or in this case how many atoms works the same either way? Over top of the total. So now if we were to go through and say well what if the total pressure of the box is thirty units then what would it be? So that changes things a little bit, right? But not too much. So I'm going to have to flip over to the document screen for this so we can write. So let's just do this for a couple of them. So we talked about the pink and the purple ones. And we talked about chi. So let's write it down. I would suggest it's doing this right on the slide. That's why I didn't put it on the worksheet. So we have our pink, we have our purple. What is our chi for each one? So you're looking at your picture. And for the pink we had said, well there's three out of ten. And for the purple we had said four out of ten. Alright so let's just do it for those two. I'll say that you can do the green and the blue at home. So now I said well what if the total pressure is thirty? Thirty whatever. Because we're talking about molecules and not moles, it's kind of odd to use normal units. So we're going to call them thirty of man units. What is going to be our pressure, our partial pressures? How do we figure this out? Well it's just a percent, right? You're just treating this like a percent. So if we want to know our partial pressure of each one, we take our mole fraction, we multiply it by the total to get our answer. If we took our purple we would do four out of ten. In times if I had thirty to get our answer. It doesn't matter that I drew one as sort of kind of your ideal gases, right? Healing of neon, whatever, because it's just mono atomic and your other is diatomic. Just take the mole fraction, multiply it by the total and you're done. So part, Dolan's law is really nice. It makes life kind of easy for us compared to if all of this wasn't true. Okay, I guess I didn't need to switch over. Now we're going to do some examples. So the best way to do these is really just tons and tons of examples. It's kind of the common theme for the majority of this lecture. Okay, so we have this. We have the contents of a natural gas tank, right? Because natural gas can be a whole variety of gases. It's not just one thing. And I give you the percentages of each one and I say, well, what's the partial pressure of each one? Given that the total pressure is one point two. So let's break each one up. We'll do CH4, C2H6 and C3H8. And we need to figure out the partial pressures of each one. So how are we going to do this? Well, this time I didn't give you a mole fraction, right? What did I give you? Percent. So are we going to take 88.6 and multiply by 1.2? No, what do we have to do first? We have to do the decimal, right? So we'll convert that and multiply by 1.2 to get our answer. So we get this. And just for sig fig purposes, we need to add that zero in there. So we need to round this to the three sig figs. And we have that. Yeah? Yeah? Would you ever ask us what's the partial pressure of carbon in CH4? You mean, OK. No, so I couldn't do that because the carbon itself wouldn't be a gas. I could ask you for the moles of carbon, which if I did that, you could take this. And how would you find moles of CH4? You'd have to fill into the ideal gas off. So I'd have to give you some more information, right? I'd have to give you some volume and temperature and things like that. And then you would know that however many moles of methane there is, how many moles of carbon would there be? One. So I can go to moles of the individual elements, but I can't go to a partial pressure of it. It doesn't make a lot of sense to talk about. OK, this one. So we convert that to a decimal. Careful with your know your single percent. Sometimes people miss the second moving over part of it. We multiply by the 1.20 ATM to get our answer. And we round to three figs, or excuse me, two figs this time. OK. And our last one. So once again, we convert to a fraction. We multiply by 1.2 atmospheres, and we write down our answer. And if I were to say, what is the mole fraction of methane, or CH4? What would you say? Do you have to calculate anything first of all? Let's say I ask that right from the beginning. I gave you this problem, and you know how I write the exams. Lots of times there's multiple parts of the question in one thing. And all I said is, I wrote all of this out, and I said, what is the mole fraction of each gas, and what is the partial pressure of each gas? What would your mole fractions be? Yeah, it would be these. In this one, it would be the 0.886. In this one, the 0.089. In this one, the 0.025. So if I give you the percent, you just turn it into a decimal point. And that's your mole fraction. That's why we multiplied this by this. We took our mole fraction, and we multiplied by our total. OK? All right. Ready for a much harder example now? You're supposed to be excited. No groans. Be happy. OK. So there's a little picture here, which I think probably got added after I had posted this. So let me zoom in a little bit, even though I probably ended up having to zoom out so you can see it. OK. This was because when this is a little bit of a confusing setup, but it's a very cool setup and something that you can do in lab, and you can do probably not at home, but definitely in lab. So we have a sample of oxygen gas collected over water. So what's happening here is some sort of reaction is going on here, some sort of reaction that makes oxygen gas. So in this case, there's a little beaker where something's going on. The other time this will happen is you can burn something and make CO2. So this is a way of measuring how much of a gas comes off a reaction, which is useful if you're trying to figure out something about how the reaction is happening. Maybe you want to know a percent yield, or you want to know something about a reaction, and you need to know what all your species are. You need to know how much product you get at each one. Well, this is the one way that you can test for oxygen, or that you can measure oxygen if it's the only gas that's evolved. So what happens is you have something going on here. You have a reaction happening. Oxygen gas comes through here. It's the only place for it to really go, and then it bubbles up. So you have a tube that originally was completely filled with water. You have a tube completely submerged with water. You stick it into a beaker, a flat thing of water here. And then the only air in here, or the only gas in here, comes from this reaction, which will all be oxygen. So this is the way that the setup works. And this is, I don't think you guys do this in lab unless it's been added in the last few years, but this is something that a lot of labs do. So what can you do with this if you have this, and let's say you put it in a graduated cylinder. You can measure that volume, right? Can you measure the temperature? Yeah, sure. So you know a lot of things about the system, and you can measure how many moles of gas are in here. And that's what we're going to do in this problem. We're going to try to figure out how many moles of oxygen came off this reaction. So when we say it's collected over water, this is how they actually go about doing it. They have a cylinder filled with water. They put a tube in. The gas bubbles through. And then you can, as long as this is like some sort of graduated system, you can see the volume of gas that you collect. So I say a wet gas. Well, how can a gas be wet? What that means is that it has water vapor in it. It's not pure. So if you take a tank of oxygen that you don't want any water in it all, they would have done something to take out the water vapor. In this case, it's sitting over top of water, right? We know that water evaporates. We know that water has a vapor pressure. That if you put some water down here, it's going to evaporate. And if you put a cap over it, it's going to condense and evaporate and do its thing. So there's water vapor in here. So that's going to change the volume a little bit. So we have two things, right? We have water vapor and we have the oxygen gas. So two things that we have to worry about. But those two combined have a volume of 7.28. So that's our total volume. And we have a pressure of 1.25 bar. And that's our total pressure. If all the water is removed, what volume will the dry oxygen gas occupy at this whole new set of conditions? So we're taking what we just learned with Alten's Law of Partial Pressures, and we're going to combine it with what we did last class, where we went through and did final and initial conditions. So a lot going on here. This is a tough question. So just to make life a little bit more fun, I also gave you two and the wrong unit so that we could practice playing around with units. So this right here, that's water vapor, right? That's the pressure of the water. So we know that this pressure is going to be what the water is. So then how do we go about finding the initial pressure of the oxygen? So let's write this out a different way. We know our pressure total has to be the addition of all the gases that are in there. So what is in this beaker right here for gases? Oxygen. So we know the pressure of oxygen plus? Yeah. OK. Now we're getting somewhere, right? Because we have this, we know our pressure total is 1.25 bar, and we know our pressure of our water is 23.8 millimeters of mercury. So we can find our pressure of oxygen. So we're getting closer. So first we have to do some unit issues. So you can really switch to pretty much any unit you want here. I happen to switch everything to millimeters of mercury. Sorry about that. That should be 1.25. So we're going to take the 1.25 bar, and I'm going to convert that into millimeters of mercury. You could convert it the other way if you wanted. And of course, all of these conversion factors are given to you on an exam, so don't spend any time memorizing them. OK. So we have this, and we can write down that value. And I'm going to have to zoom out a little bit now that you guys have had some time to look at the picture. OK. So that, once again, is going to be our total pressure. OK. So now we can start filling in some values. We need to know the pressure of oxygen. And just for the cleanliness of work purposes, I'm going to draw a line here. So we need to know pressure of oxygen. And we can get that through this equation, right? We know that our total pressure is only equal to the pressure of those two gases. So how do we get our pressure of oxygen? Pt, and then what? Do we add our pressure of water? Good. We subtract it. And so we get that. Now since we've switched our units around for this purpose, we can go ahead and fill in. Now, something that's worth kind of doing up here, because we're going to be doing a lot of calculations with things that we calculated, is to watch your sig figs start using your underline method. Where you just underline the last significant one so you can keep track of what's happening here. It's way too much work to try to trace it through otherwise. So we'll go through. We'll fill in our pressure of each one, because it's a bad idea to round every single time you fill in. To be honest, I tend to just save everything in my calculator. And then for the sake of units, put that. And so we can get our answer. Now we have to pay attention to sig figs a little bit here. What are we going to round to? So I'm hearing three, and I'm hearing tenths. So if this was multiplication and division, you're right with three. Because our lowest number is three sig figs. But this isn't multiplication and division. It's subtraction. So we round to our lowest digit, so our tenths place. So we round to that. So that is our pressure of oxygen. Are we done? No, right? This is our pressure of oxygen under what conditions? Place, right? This is our condition at 25 degrees. And this pressure, excuse me, the 1.25 pressure total in that volume. That's not what I asked you. That would have been too nice, right? I said, now we're going to change everything. Because in reality, if you were going to do this, probably you would change the situation. Probably you would end up, in order to dry the gas, you would have to do something to it. And so that's going to change your conditions around a little bit. So this is what we're going to do. We're going to do the exact same thing that we did last class. So this is going back to the material then. And we always started from this equation, right? Because we figured, well, we start with everything changing, and then we just cross off whatever doesn't change. So we start from this one more time. Now the nice part about this, do we have to worry about the fact that I'm in millimeters of mercury? No, right? Because whatever I plug in here, that's just what's going to come out from my final pressure. So that's not a big deal. OK. Or excuse me, we do actually have to worry a little bit. We have to convert from one or the other. So it doesn't matter what unit we put it in. It doesn't have to be in atmospheres. It doesn't have to be in millimeters of mercury. But it's got to be in the same thing. Now our volume. So let's go through and convert all of this. We're solving for what variable? V2. V2. So we'll fill everything else in. OK. So our P1 comes from here, right? Now I think in this case I converted everything to millimeters of mercury again. I was apparently in a millimeters of mercury day that day. For the sake of room, I'm going to be a little bit lazy about writing units. Perhaps, we'll see. So this is in millimeters of mercury. Now our V1, what is our V1? That's that, right? So we go ahead and rewrite that in. I've decided in hindsight not to be lazy about units. It's too complicated of a problem. So that's our V1. Now our N1. Do we have to worry about N here? No. That gets canceled out. So we don't have N, but we don't need to worry about N because we're not changing it. So that goes away. So our T1. Or is our T changing? Definitely. So I'm just going to fill in 25 here then, right? No, right? What do you have to do? Calvin, don't forget that. It's really the most painful mistake to have to grade on an exam because we know that you know the difference. But if you forget, it's wrong. So excuse me, I'm going to hold it for a moment. OK, next part then. P2. Do we know P2? Sure, it's given. So I'm just going to fill in 1.07 and go about my merry way, right? No. So we need to go ahead and we need to convert this. So that's an ATM. So we fill down ATM here. We fill in our conversion factor up here. You can always do this on a different sheet if you want to and then fill it in whichever way you prefer. We fill in V2. And we fill in our final temperature. OK. So once we go through all of that and we do the algebra, which I'm not going to write out the algebra because quite honestly, it's the easiest way to just do it inside your calculator. We have V2 is equal to 8.50 liters. Now let's check our sig figs on this last part. So we know that this, let's just write in 290. So I'm going to actually just, it's not wrong, but I'm going to cross it off and write 298 for the sake of going through and doing the sig figs. Because we know we're going to be at our one's digits here. Now what's our sig figs here? We took 37 and we added it to 273. So is this really two sig figs? No, it's really three sig figs. So watch out for sig figs in those types of situations. OK. So now we have how many sig figs here? Four. Three sig figs here. Three, this is actually three by definition. It's three and three. So we need to have three down here. So I'm going to put a little underline there too. All right. OK. So questions on that very long complicated problem. This is one of those problems you probably are going to have to really get, go through and do looking at the notes. Do not looking at the notes, but constantly referring to them off and on for help. And then maybe do again on your own three or four times before you really get it. It's long and complicated. But it's really just combining a bunch of things that we know with one new thing. I gave you this new experimental setup. But we know that whatever that oxygen is, whatever that little volume is, is just our two pressures added together. We know that then if we want to find our pressure of oxygen after last class, last five minutes, we know that we can just subtract it from the total because there's only two components, right? It's only water and oxygen. So we subtract that to find our pressure of oxygen. Now we're back to what we did for the whole first class or the whole first half of class last time. And we said, OK, well, we have initial set of conditions and we have a final set of conditions. Let's solve for the unknown. All right. Any questions on that before I move on? You may be a little too lost for questions yet. That's fine. OK. We can revisit that one at a review session or something. All right. Next one. All right. So we have two containers connected by a valve that can be opened. Initially the pressure in volumes are as marked. Find the final pressures after the valve of art is opened. OK. So this could theoretically be a really complicated problem, but luckily Dalton's law makes it simple for us, which is really nice. So we can treat this as two completely separate problems. What we're going to do is we're going to pretend that we opened this valve and that there's nothing here, that there's no gas here, that it's just a vacuum. And we let the pink come into here and equilibrate. We're going to pretend that on one half of the problem. Then for the second half of the problem, we're going to go ahead and pretend that this goes into here, that this goes into a completely empty container over here. And that's going to be our second half of the problem. What that's going to wind up doing is giving us the partial pressures. So if we have the partial pressure of each gas, how do we find the final one? Yeah, we just add it together. So let's go ahead and do that. So let's start with the pink side. So what all is changing here? Did I say anything about the temperature? No. We assume it's constant. Is the moles changing? What do we think? Do everyone agree? Mostly because this is a tricky question. So they're right, but corals answers are good. Let's me know if you're understanding things. So why isn't the moles going to change? We're combining the pink and the blue, right? Yeah, but we're not talking about the blue right now. Exactly. We're not talking about the blue, so we don't care. All we care about is the pink going to change, and it's not. We're not going to take any way. We're not going to add in. So let's take away those R's. Those aren't necessary. So what do we have to worry about changing then? Just P and V. So it becomes P1, V1. It equals P2, V2. All right. So now we just fill in some numbers. So our P1 and our V1 are right here. Now P2 is what we're solving for. What is our V2? Is it 4 liters? No, what is it? It's 5 liters, right? Because it's total. We're adding it together. We're saying this pink one can just spread out over the whole space and nothing in its way. That's kind of what we're pretending. And so in that case, it's going to be 1 plus 4. So we solve that. And we get our first pressure. Now we're going to do the exact same thing for the blue side. So take a minute or so and do it. See if you can get the right answer. And then if you can figure it out, find the total pressure as well. Oh, yeah. So these are the initial values. So these are the initial values of the pink and the initial values of the blue. All right. I'll give you 20 more seconds. You can talk to your neighbor. It's fine. I have like 10 more seconds to see if you have the same thing. OK. What did you get for a pressure for the blue one? Do you guys agree? Two? Yeah. So I'm going to do this one solving for variables first. This one's final is 5 liters as well. Don't forget to add them up. OK. So that gives us our partial pressures. So in this case, I just told you to find the final pressure. But the easiest way to do it is to go through and solve for the partial pressures. There are actually other ways to do it as well. So let's go ahead and add these up. OK. So we get 2.20 atmospheres. All right. Questions on that one? Yeah. I know I did the hard one first and the easy one, but keep you on your toes. The next one, however, we're going to go back to hard. Really I just needed to give you guys a little bit of a breather between the two hard ones. That was the real reason for doing it here. OK. But I think you'll all be really happy that I'm doing this hard one. And as soon as I put it up and you see the format, you'll see why. And guess where I got it from? Yeah. Not Wiley. Sapling. OK. For those of you who don't know, Wiley is the other homework program that some of the classes use. That's where that came from. OK. Yeah. This is from your homework. So you have the worksheet in front of you, hopefully, because this is going to, you know, it's a long question with a lot of room, so I'm going to have to kind of zoom out and move around a little bit. But this is the same setup. And that's why we kind of started with the easier version of this. And this is the more intense version of it. So we have NO and O2. This is also my public service announcement for remember all that fundamental stuff that you may have, like, said, well, I'm going to just remember this stuff. Yeah. Make sure you remember this. Make sure you go back and do this, because this is a lot of the fundamental section that we sort of talked about a lot in lecture. Because as I kept warning you, it's going to come back in chapter four. Well, it's here. So we have two valves or two bulbs. They're connected with NO and O2. In the last situation, they didn't react, right? We just put them together, let them mix around, and found a new pressure. But in this case, they're reacting. They're forming a new substance. OK? And so it gets a lot more complicated now. Now we have to deal with a bunch of stoichiometry as well. So we need to find the pressures of each one. But to do that, we have to find the moles of each one first. And to find the moles of each one, what do we have to do? Are they going to react and completely form 2NO2, and we're going to have none of this and none of this? No. What's probably going to happen? I'd say that there's going to be a limiting reagent. So that's our first step. OK. So we're going to find the limiting reagent. So we need to find the moles of each one of the starting. We need to find moles of NO and moles of O2. So that's our first step. So to find our moles of O2, actually let's do moles of NO first. What equation are we going to use to find moles? What do we have? We have some volumes, and we have some pressures, and we have some temperatures. So what does that sound like? Do we have to worry about our units in this situation? Do we have R there? Does R have particular units? Yeah. So it's got to be atmosphere's leaders, right? So we're set there, so not a big deal. So we fill everything in. So this is all taken directly from the question. I'm skipping the units on R because honestly, we're really short on room here. OK. So we're going to have that. That's just calculator button pushing. And we get this. OK. So that's our moles of NO. Now what do we have to solve for? Well, if we're finding one that we're agent, we need moles of the other one, right? So exact same thing. And sapling was nice and only, it gave you everything in the right units on this very, very long, hard problem. I decided I didn't want to be nice to you on the previous one where I did that. So OK. So we plug this in. OK. So now we come to limiting reactant stage. There's a bunch of different ways to do limiting reactant, and everyone has their favorite way. I think in this one doing logic is the best way. You can choose to do it mathematically if you want. The two kind of are hand in hand. So if we look at this, if we have one oxygen, how many NOs do we need? Two. If we have five oxygens, how many NOs do we need? 10. If we have 20, 40. So we've got the pattern, right? How many do we need? Double. Now do we have that? So we look at oxygen. Do we have double the amount of NO that we need? No. So what one's going to be our limiting reagent? OK. All right. So how much NO are we going to have at the end of this reaction? Zero, right? So no moles, no pressure. OK. Next part, then. We need to figure out how much NO2 we're going to have, and how much what else are we going to have? Are we going to have any oxygen left? Let me reword the question. So which one was our limiting reagent again? NO. So that means that we have more oxygen than what we need, right? Yes. So does that mean we're going to have any oxygen left? Yes. So we're going to have to figure out how much NO2 we make, and how much oxygen we have left. So that's our next goal. So it doesn't really matter which order you do it in. I think I found moles of O2 first. So we're going to find moles O2 left, and find moles of NO2 made. OK. Let's do NO2 first. We'll see because it's the easiest. So if we used up all of our moles of NO. How much NO2 did we make? Let's break it down a little bit more. Let's say I make one mole, or I have one mole of NO. How many moles of NO2 would I make? One. If I had five, I would make five, right? They're the same, and that's because the coefficients are the same. So what we're going to do here is if we know how much NO2 is made, or we know how much NO we started with, which we do, we know how much NO2 is made. It's the same amount. So we just write it down, basically. We just take our amount of NO we started with, and we write it down. OK. Now oxygen's a little bit more complicated, right? Because we know how much we had to begin with, and what can we calculate that happened in this reaction? How much we made, how much we used up, how much is left over? What can we do with one calculation using this reaction? How much was used up, right? We know that if this has two, that one of them is going to get used up. If we have five, or 10, that five of them are going to get used up. So what we're going to need to do then is to find our final amount of oxygen. We're going to take our initial amount, and we're going to subtract away the thing that we used up. So let's do that. And let's just put a little f underneath moles for final, just to make it super clear. So obviously, if I were to give this on an exam, you wouldn't need to write down all of these sentences. I'm just doing that for you, so when you go back and look at this, you know what's going on. So moles of O2 left over. How much did we start with? We started with this, right? And we're going to subtract what was used. OK, how do we figure out what was used? Well, we know, I'm going to kind of do this as a sidebar, like I do sometimes. We know how much NO we started with, right? Now how is the moles of NO in the moles of oxygen related? By the mole ratio. So writing this out really big so that you can see it, and then I'll plug it into the smaller area, just want to make sure everyone can see. Our moles of NO go here, our moles of oxygen will go here. And what is the number that goes in front of the oxygen? 1. We got that from where? The equation. Here is 2. So I just wanted to write it out big, and then I'm going to fill it into the equation on the notes we've been taking as well, but just because it's getting tough to see. OK. So we took our total that we started with, our initial amount, and we subtracted what we used. And when we get this then, we end up with 0.081. So now we have our final moles of oxygen. We have our final moles of NO2. And do we have our final moles of NO? It's 0. And so we're good there. Now what equations do we use to go ahead and solve? Now we're back to the exact problem that we had before. The problem where the question of, that's it, got asked. Now we're at the P1v1 over N1t1 question. And we have to look at what's changing. So if you go back to the question and we can look at everything that is and isn't changing, we can fill everything in. OK. So we're still not OK for time. OK. So when we go to do this, we put in our final amount of moles. Oh, thank you. Can we do this for each one? Is that too small? No, that's OK. You guys are all right. And we fill everything in. So now this becomes a scavenger hunt, right? Now you have to go through and figure out where everything is coming from. Make sure you grab all your final values and not your in-between values. So this is just R. Make sure you remember after all of this work that you've done to convert your temperature to Kelvin, right? Because it's easy to forget things like this after we've gone through everything. We're going to walk through it a little bit here once I finish it as well. And so we can talk back over everything that we did. So now I'm just filling in numbers. We've done all the hard parts, and we're filling everything in. So when you fill those into the calculator, I ran out of room, so I can't write it beside it. Now, of course, you know that your homework values are going to be different than this, so don't go filling those in right away, right? You know something changes your numbers on you. So that's your total pressure. Let's go back through and talk over what we just did in the last minute and a half. OK. So we started by saying, well, in order for me to find the final pressure, I need to find the initial pressures. So I know all of this detail. I can use pv equals nrt, and I can solve for my moles. Once I find my moles, I can use all of my knowledge that I remember from the first week of class, and I can use stoichiometry to solve for, one, the limiting reagent, so I know what's not going to be there. Two, how much of the non-limiting reagent is left over, and then how much of the product I made, and then after we have that, we have our moles of everything. So we still know that we have r and t and v. We can find that from the problem, but now we found our final moles, and we can fill in our final moles into the pv equals nrt equation again. So really at its heart, what this is doing is it's taking, and it's using, you know, origin factors just like we did before, and you have to go through moles. So you convert everything to moles. You do your stoichiometry, and then you convert everything back into pressure. And I'll leave this up as you guys are packing up and stuff so that you can see it if you can find the position of the other�
 Okay, so today we're gonna pretty much see.  All right, I think that's better.  Okay, so today if you guys can be quiet, we will.  Okay, we're gonna kind of start back and talk a little bit more about Dalton's partial  pressures.  I think really we'll spend most of today on doing the various versions of this in the  various examples because it is a little bit long.  We may make it into a fusion.  So I'm hoping to finish up the gas laws on next Monday and then be able to spend Wednesday  and Friday on review for you guys.  I think two review lectures would be best, but it really just depends on if we can make  it through or not.  I'm not 100% sure if we will or not.  So without further ado, we'll get going.  So review a little bit from the last five minutes last time.  So remember that partial pressure, the way that this works is that you can take the  pressures of each individual gas.  So let's say you have a mixture of three or four gases.  You can take the pressure of each one of them and then add them together and that'll give  you your total pressure.  So when we did this sort of with an equation form, this is it written out.  I'm sure you'll be having some computer issues.  So this is it written out in just pressure of each gas equals pressure of the total.  So using the fact that we know that pressure is equal to nRT over V, ideal gas law, we  can take and we can solve the pressure for each individual gas and then we can add them  all up together.  And we know that if we have a container with a mixture of gases in it, that the temperature  and the volume are going to be the same.  So we can go ahead and we can pull that out.  We can factor the moles out and then we can get this.  So what this really says is that you add up all of the moles and you can add that up and  then you get your total pressure.  So your moles of all your gas, you can fill it right into your ideal gas law as well and  solve for your total pressure that way.  So two different ways of doing it.  Now there's a very important concept here that we need to talk about and that's called  the mole faction.  So this X, this is actually the Greek letter chi, so that's how you say that.  And this says that if you have the, all this really is a fraction just like any other fraction,  right?  It's whatever you're talking about over top of the total.  So if you want to know the mole fraction of A, you take the moles of A and you put it  over top of the moles total.  It's no different really than saying the percent of A, like the mole percent of A, except  that you don't have the times 100% there.  It's just the decimal point or the fraction.  So that's what mole fraction is.  Now what this says is it kind of takes this concept, combines it with this one, and says  well we can take that chi, we can take that mole fraction, right, our part over our whole,  and we can just times that by the total pressure.  This isn't any different than the sort of how you work with percents your entire life,  right?  If you want to know what percent something is of a total, you take the fraction and you  multiply it by the total, okay?  So this is your total pressure.  So if I tell you that something has a total pressure of five atmospheres, but the mole  percent or the mole fraction of it is only 0.2, take 0.2 and times it by five, okay?  So don't let this seem like a new equation to you.  This is just using percents like you've always used them, except what's the first thing you  do if I say you have 75% of something?  You turn it into a decimal, right?  Well that's all this is.  It's just the decimal of the part over top of the whole.  So that's all the mole fraction is.  It's like a percent, except not with the percent.  I'm not sure why they go through the, you know, special aspect of naming it and different  and calling it chi and all of that, but you know, that's fine.  So don't let this be new to you, just kind of a different way of naming it.  Okay.  So before we do partial pressures with actual numbers, let's just go through and do it with  a quick picture.  So if I have this and I have this mixture of molecules, so I have some purples and some  greens of aqua or cyan and some pinks.  If I tell you that this has a pressure of 10, it doesn't matter what unit, I say 10.  And we want to know the partial pressure of each gas.  What do we do?  You have to find the mole fraction, right?  So let's start with the purple ones.  So first of all, let's figure out how many, what's our total number of molecules?  Our total mole of molecules.  So our total number of molecules we just count.  One, two, three, four, five, six, seven, eight, nine, ten.  Now if we want to know our mole fraction of the purple ones, what do we do?  Well how many purple ones do we have?  And I apologize for the colorblind people.  Statistics say there's at least a few of you and I do apologize.  One, two, three, four.  I won't do that on an exam obviously.  So there's four out of ten, right?  So your mole fraction is four out of ten.  So what would be our mole fraction of the pink ones?  Would it be six out of ten or three out of ten?  Good, three.  It doesn't matter that it's a diatomic, right?  We're treating all of these molecules and all of these atoms and all of these gases  as just like they're just spherical balls that are all the exact same.  They have no volume, they have no interaction, they have nothing.  They're just ideal gases.  So it doesn't matter if I drew these as diatomics and these as single.  So it would be three out of ten, right?  Now so what does that mean the pressure of the purple ones would be?  Four out of ten.  So just four, right?  So and you can go through, you can do that with each one.  So the blue would just be one out of ten.  That would be your fraction, it would be point one.  The greens would be two out of ten, point two.  So all mole fraction is is a count, right?  It's how many moles of something you have, or in this case how many atoms works the same  either way?  Over top of the total.  So now if we were to go through and say well what if the total pressure of the box is thirty  units then what would it be?  So that changes things a little bit, right?  But not too much.  So I'm going to have to flip over to the document screen for this so we can write.  So let's just do this for a couple of them.  So we talked about the pink and the purple ones.  And we talked about chi.  So let's write it down.  I would suggest it's doing this right on the slide.  That's why I didn't put it on the worksheet.  So we have our pink, we have our purple.  What is our chi for each one?  So you're looking at your picture.  And for the pink we had said, well there's three out of ten.  And for the purple we had said four out of ten.  Alright so let's just do it for those two.  I'll say that you can do the green and the blue at home.  So now I said well what if the total pressure is thirty?  Thirty whatever.  Because we're talking about molecules and not moles, it's kind of odd to use normal  units.  So we're going to call them thirty of man units.  What is going to be our pressure, our partial pressures?  How do we figure this out?  Well it's just a percent, right?  You're just treating this like a percent.  So if we want to know our partial pressure of each one, we take our mole fraction, we  multiply it by the total to get our answer.  If we took our purple we would do four out of ten.  In times if I had thirty to get our answer.  It doesn't matter that I drew one as sort of kind of your ideal gases, right?  Healing of neon, whatever, because it's just mono atomic and your other is diatomic.  Just take the mole fraction, multiply it by the total and you're done.  So part, Dolan's law is really nice.  It makes life kind of easy for us compared to if all of this wasn't true.  Okay, I guess I didn't need to switch over.  Now we're going to do some examples.  So the best way to do these is really just tons and tons of examples.  It's kind of the common theme for the majority of this lecture.  Okay, so we have this.  We have the contents of a natural gas tank, right?  Because natural gas can be a whole variety of gases.  It's not just one thing.  And I give you the percentages of each one and I say, well, what's the partial pressure  of each one?  Given that the total pressure is one point two.  So let's break each one up.  We'll do CH4, C2H6 and C3H8.  And we need to figure out the partial pressures of each one.  So how are we going to do this?  Well, this time I didn't give you a mole fraction, right?  What did I give you?  Percent.  So are we going to take 88.6 and multiply by 1.2?  No, what do we have to do first?  We have to do the decimal, right?  So we'll convert that and multiply by 1.2 to get our answer.  So we get this.  And just for sig fig purposes, we need to add that zero in there.  So we need to round this to the three sig figs.  And we have that.  Yeah?  Yeah?  Would you ever ask us what's the partial pressure of carbon in CH4?  You mean, OK.  No, so I couldn't do that because the carbon itself wouldn't be a gas.  I could ask you for the moles of carbon, which if I did that, you could take this.  And how would you find moles of CH4?  You'd have to fill into the ideal gas off.  So I'd have to give you some more information, right?  I'd have to give you some volume and temperature and things like that.  And then you would know that however many moles of methane there is, how many moles of carbon  would there be?  One.  So I can go to moles of the individual elements, but I can't go to a partial pressure of it.  It doesn't make a lot of sense to talk about.  OK, this one.  So we convert that to a decimal.  Careful with your know your single percent.  Sometimes people miss the second moving over part of it.  We multiply by the 1.20 ATM to get our answer.  And we round to three figs, or excuse me, two figs this time.  OK.  And our last one.  So once again, we convert to a fraction.  We multiply by 1.2 atmospheres, and we write down our answer.  And if I were to say, what is the mole fraction of methane, or CH4?  What would you say?  Do you have to calculate anything first of all?  Let's say I ask that right from the beginning.  I gave you this problem, and you know how I write the exams.  Lots of times there's multiple parts of the question in one thing.  And all I said is, I wrote all of this out, and I said, what is the mole fraction of each  gas, and what is the partial pressure of each gas?  What would your mole fractions be?  Yeah, it would be these.  In this one, it would be the 0.886.  In this one, the 0.089.  In this one, the 0.025.  So if I give you the percent, you just turn it into a decimal point.  And that's your mole fraction.  That's why we multiplied this by this.  We took our mole fraction, and we multiplied by our total.  OK?  All right.  Ready for a much harder example now?  You're supposed to be excited.  No groans.  Be happy.  OK.  So there's a little picture here, which I think probably got added after I had posted  this.  So let me zoom in a little bit, even though I probably ended up having to zoom out so  you can see it.  OK.  This was because when this is a little bit of a confusing setup, but it's a very cool  setup and something that you can do in lab, and you can do probably not at home, but  definitely in lab.  So we have a sample of oxygen gas collected over water.  So what's happening here is some sort of reaction is going on here, some sort of reaction that  makes oxygen gas.  So in this case, there's a little beaker where something's going on.  The other time this will happen is you can burn something and make CO2.  So this is a way of measuring how much of a gas comes off a reaction, which is useful  if you're trying to figure out something about how the reaction is happening.  Maybe you want to know a percent yield, or you want to know something about a reaction,  and you need to know what all your species are.  You need to know how much product you get at each one.  Well, this is the one way that you can test for oxygen, or that you can measure oxygen  if it's the only gas that's evolved.  So what happens is you have something going on here.  You have a reaction happening.  Oxygen gas comes through here.  It's the only place for it to really go, and then it bubbles up.  So you have a tube that originally was completely filled with water.  You have a tube completely submerged with water.  You stick it into a beaker, a flat thing of water here.  And then the only air in here, or the only gas in here, comes from this reaction, which  will all be oxygen.  So this is the way that the setup works.  And this is, I don't think you guys do this in lab unless it's been added in the last  few years, but this is something that a lot of labs do.  So what can you do with this if you have this, and let's say you put it in a graduated cylinder.  You can measure that volume, right?  Can you measure the temperature?  Yeah, sure.  So you know a lot of things about the system, and you can measure how many moles of gas  are in here.  And that's what we're going to do in this problem.  We're going to try to figure out how many moles of oxygen came off this reaction.  So when we say it's collected over water, this is how they actually go about doing it.  They have a cylinder filled with water.  They put a tube in.  The gas bubbles through.  And then you can, as long as this is like some sort of graduated system, you can see  the volume of gas that you collect.  So I say a wet gas.  Well, how can a gas be wet?  What that means is that it has water vapor in it.  It's not pure.  So if you take a tank of oxygen that you don't want any water in it all, they would have  done something to take out the water vapor.  In this case, it's sitting over top of water, right?  We know that water evaporates.  We know that water has a vapor pressure.  That if you put some water down here, it's going to evaporate.  And if you put a cap over it, it's going to condense and evaporate and do its thing.  So there's water vapor in here.  So that's going to change the volume a little bit.  So we have two things, right?  We have water vapor and we have the oxygen gas.  So two things that we have to worry about.  But those two combined have a volume of 7.28.  So that's our total volume.  And we have a pressure of 1.25 bar.  And that's our total pressure.  If all the water is removed, what volume will the dry oxygen gas occupy at this whole new  set of conditions?  So we're taking what we just learned with Alten's Law of Partial Pressures, and we're  going to combine it with what we did last class, where we went through and did final  and initial conditions.  So a lot going on here.  This is a tough question.  So just to make life a little bit more fun, I also gave you two and the wrong unit so  that we could practice playing around with units.  So this right here, that's water vapor, right?  That's the pressure of the water.  So we know that this pressure is going to be what the water is.  So then how do we go about finding the initial pressure of the oxygen?  So let's write this out a different way.  We know our pressure total has to be the addition of all the gases that are in there.  So what is in this beaker right here for gases?  Oxygen.  So we know the pressure of oxygen plus?  Yeah.  OK.  Now we're getting somewhere, right?  Because we have this, we know our pressure total is 1.25 bar, and we know our pressure  of our water is 23.8 millimeters of mercury.  So we can find our pressure of oxygen.  So we're getting closer.  So first we have to do some unit issues.  So you can really switch to pretty much any unit you want here.  I happen to switch everything to millimeters of mercury.  Sorry about that.  That should be 1.25.  So we're going to take the 1.25 bar, and I'm going to convert that into millimeters  of mercury.  You could convert it the other way if you wanted.  And of course, all of these conversion factors are given to you on an exam, so don't spend  any time memorizing them.  OK.  So we have this, and we can write down that value.  And I'm going to have to zoom out a little bit now that you guys have had some time to  look at the picture.  OK.  So that, once again, is going to be our total pressure.  OK.  So now we can start filling in some values.  We need to know the pressure of oxygen.  And just for the cleanliness of work purposes, I'm going to draw a line here.  So we need to know pressure of oxygen.  And we can get that through this equation, right?  We know that our total pressure is only equal to the pressure of those two gases.  So how do we get our pressure of oxygen?  Pt, and then what?  Do we add our pressure of water?  Good.  We subtract it.  And so we get that.  Now since we've switched our units around for this purpose, we can go ahead and fill  in.  Now, something that's worth kind of doing up here, because we're going to be doing a  lot of calculations with things that we calculated, is to watch your sig figs start using your  underline method.  Where you just underline the last significant one so you can keep track of what's happening  here.  It's way too much work to try to trace it through otherwise.  So we'll go through.  We'll fill in our pressure of each one, because it's a bad idea to round every single time  you fill in.  To be honest, I tend to just save everything in my calculator.  And then for the sake of units, put that.  And so we can get our answer.  Now we have to pay attention to sig figs a little bit here.  What are we going to round to?  So I'm hearing three, and I'm hearing tenths.  So if this was multiplication and division, you're right with three.  Because our lowest number is three sig figs.  But this isn't multiplication and division.  It's subtraction.  So we round to our lowest digit, so our tenths place.  So we round to that.  So that is our pressure of oxygen.  Are we done?  No, right?  This is our pressure of oxygen under what conditions?  Place, right?  This is our condition at 25 degrees.  And this pressure, excuse me, the 1.25 pressure total in that volume.  That's not what I asked you.  That would have been too nice, right?  I said, now we're going to change everything.  Because in reality, if you were going to do this, probably you would change the situation.  Probably you would end up, in order to dry the gas, you would have to do something to  it.  And so that's going to change your conditions around a little bit.  So this is what we're going to do.  We're going to do the exact same thing that we did last class.  So this is going back to the material then.  And we always started from this equation, right?  Because we figured, well, we start with everything changing, and then we just cross off whatever  doesn't change.  So we start from this one more time.  Now the nice part about this, do we have to worry about the fact that I'm in millimeters  of mercury?  No, right?  Because whatever I plug in here, that's just what's going to come out from my final pressure.  So that's not a big deal.  OK.  Or excuse me, we do actually have to worry a little bit.  We have to convert from one or the other.  So it doesn't matter what unit we put it in.  It doesn't have to be in atmospheres.  It doesn't have to be in millimeters of mercury.  But it's got to be in the same thing.  Now our volume.  So let's go through and convert all of this.  We're solving for what variable?  V2.  V2.  So we'll fill everything else in.  OK.  So our P1 comes from here, right?  Now I think in this case I converted everything to millimeters of mercury again.  I was apparently in a millimeters of mercury day that day.  For the sake of room, I'm going to be a little bit lazy about writing units.  Perhaps, we'll see.  So this is in millimeters of mercury.  Now our V1, what is our V1?  That's that, right?  So we go ahead and rewrite that in.  I've decided in hindsight not to be lazy about units.  It's too complicated of a problem.  So that's our V1.  Now our N1.  Do we have to worry about N here?  No.  That gets canceled out.  So we don't have N, but we don't need to worry about N because we're not changing it.  So that goes away.  So our T1.  Or is our T changing?  Definitely.  So I'm just going to fill in 25 here then, right?  No, right?  What do you have to do?  Calvin, don't forget that.  It's really the most painful mistake to have to grade on an exam because we know that you  know the difference.  But if you forget, it's wrong.  So excuse me, I'm going to hold it for a moment.  OK, next part then.  P2.  Do we know P2?  Sure, it's given.  So I'm just going to fill in 1.07 and go about my merry way, right?  No.  So we need to go ahead and we need to convert this.  So that's an ATM.  So we fill down ATM here.  We fill in our conversion factor up here.  You can always do this on a different sheet if you want to and then fill it in whichever  way you prefer.  We fill in V2.  And we fill in our final temperature.  OK.  So once we go through all of that and we do the algebra, which I'm not going to write  out the algebra because quite honestly, it's the easiest way to just do it inside your  calculator.  We have V2 is equal to 8.50 liters.  Now let's check our sig figs on this last part.  So we know that this, let's just write in 290.  So I'm going to actually just, it's not wrong, but I'm going to cross it off and write 298  for the sake of going through and doing the sig figs.  Because we know we're going to be at our one's digits here.  Now what's our sig figs here?  We took 37 and we added it to 273.  So is this really two sig figs?  No, it's really three sig figs.  So watch out for sig figs in those types of situations.  OK.  So now we have how many sig figs here?  Four.  Three sig figs here.  Three, this is actually three by definition.  It's three and three.  So we need to have three down here.  So I'm going to put a little underline there too.  All right.  OK.  So questions on that very long complicated problem.  This is one of those problems you probably are going to have to really get, go through  and do looking at the notes.  Do not looking at the notes, but constantly referring to them off and on for help.  And then maybe do again on your own three or four times before you really get it.  It's long and complicated.  But it's really just combining a bunch of things that we know with one new thing.  I gave you this new experimental setup.  But we know that whatever that oxygen is, whatever that little volume is, is just our  two pressures added together.  We know that then if we want to find our pressure of oxygen after last class, last five minutes,  we know that we can just subtract it from the total because there's only two components,  right?  It's only water and oxygen.  So we subtract that to find our pressure of oxygen.  Now we're back to what we did for the whole first class or the whole first half of class  last time.  And we said, OK, well, we have initial set of conditions and we have a final set of conditions.  Let's solve for the unknown.  All right.  Any questions on that before I move on?  You may be a little too lost for questions yet.  That's fine.  OK.  We can revisit that one at a review session or something.  All right.  Next one.  All right.  So we have two containers connected by a valve that can be opened.  Initially the pressure in volumes are as marked.  Find the final pressures after the valve of art is opened.  OK.  So this could theoretically be a really complicated problem, but luckily Dalton's law makes it  simple for us, which is really nice.  So we can treat this as two completely separate problems.  What we're going to do is we're going to pretend that we opened this valve and that there's  nothing here, that there's no gas here, that it's just a vacuum.  And we let the pink come into here and equilibrate.  We're going to pretend that on one half of the problem.  Then for the second half of the problem, we're going to go ahead and pretend that this goes  into here, that this goes into a completely empty container over here.  And that's going to be our second half of the problem.  What that's going to wind up doing is giving us the partial pressures.  So if we have the partial pressure of each gas, how do we find the final one?  Yeah, we just add it together.  So let's go ahead and do that.  So let's start with the pink side.  So what all is changing here?  Did I say anything about the temperature?  No.  We assume it's constant.  Is the moles changing?  What do we think?  Do everyone agree?  Mostly because this is a tricky question.  So they're right, but corals answers are good.  Let's me know if you're understanding things.  So why isn't the moles going to change?  We're combining the pink and the blue, right?  Yeah, but we're not talking about the blue right now.  Exactly.  We're not talking about the blue, so we don't care.  All we care about is the pink going to change, and it's not.  We're not going to take any way.  We're not going to add in.  So let's take away those R's.  Those aren't necessary.  So what do we have to worry about changing then?  Just P and V. So it becomes P1, V1.  It equals P2, V2.  All right.  So now we just fill in some numbers.  So our P1 and our V1 are right here.  Now P2 is what we're solving for.  What is our V2?  Is it 4 liters?  No, what is it?  It's 5 liters, right?  Because it's total.  We're adding it together.  We're saying this pink one can just spread out over the whole space and nothing in its  way.  That's kind of what we're pretending.  And so in that case, it's going to be 1 plus 4.  So we solve that.  And we get our first pressure.  Now we're going to do the exact same thing for the blue side.  So take a minute or so and do it.  See if you can get the right answer.  And then if you can figure it out, find the total pressure as well.  Oh, yeah.  So these are the initial values.  So these are the initial values of the pink and the initial values of the blue.  All right.  I'll give you 20 more seconds.  You can talk to your neighbor.  It's fine.  I have like 10 more seconds to see if you have the same thing.  OK.  What did you get for a pressure for the blue one?  Do you guys agree?  Two?  Yeah.  So I'm going to do this one solving for variables first.  This one's final is 5 liters as well.  Don't forget to add them up.  OK.  So that gives us our partial pressures.  So in this case, I just told you to find the final pressure.  But the easiest way to do it is to go through and solve for the partial pressures.  There are actually other ways to do it as well.  So let's go ahead and add these up.  OK.  So we get 2.20 atmospheres.  All right.  Questions on that one?  Yeah.  I know I did the hard one first and the easy one, but keep you on your toes.  The next one, however, we're going to go back to hard.  Really I just needed to give you guys a little bit of a breather between the two hard ones.  That was the real reason for doing it here.  OK.  But I think you'll all be really happy that I'm doing this hard one.  And as soon as I put it up and you see the format, you'll see why.  And guess where I got it from?  Yeah.  Not Wiley.  Sapling.  OK.  For those of you who don't know, Wiley is the other homework program that some of the  classes use.  That's where that came from.  OK.  Yeah.  This is from your homework.  So you have the worksheet in front of you, hopefully, because this is going to, you know,  it's a long question with a lot of room, so I'm going to have to kind of zoom out and  move around a little bit.  But this is the same setup.  And that's why we kind of started with the easier version of this.  And this is the more intense version of it.  So we have NO and O2.  This is also my public service announcement for remember all that fundamental stuff that  you may have, like, said, well, I'm going to just remember this stuff.  Yeah.  Make sure you remember this.  Make sure you go back and do this, because this is a lot of the fundamental section that  we sort of talked about a lot in lecture.  Because as I kept warning you, it's going to come back in chapter four.  Well, it's here.  So we have two valves or two bulbs.  They're connected with NO and O2.  In the last situation, they didn't react, right?  We just put them together, let them mix around, and found a new pressure.  But in this case, they're reacting.  They're forming a new substance.  OK?  And so it gets a lot more complicated now.  Now we have to deal with a bunch of stoichiometry as well.  So we need to find the pressures of each one.  But to do that, we have to find the moles of each one first.  And to find the moles of each one, what do we have to do?  Are they going to react and completely form 2NO2, and we're going to have none of this  and none of this?  No.  What's probably going to happen?  I'd say that there's going to be a limiting reagent.  So that's our first step.  OK.  So we're going to find the limiting reagent.  So we need to find the moles of each one of the starting.  We need to find moles of NO and moles of O2.  So that's our first step.  So to find our moles of O2, actually let's do moles of NO first.  What equation are we going to use to find moles?  What do we have?  We have some volumes, and we have some pressures, and we have some temperatures.  So what does that sound like?  Do we have to worry about our units in this situation?  Do we have R there?  Does R have particular units?  Yeah.  So it's got to be atmosphere's leaders, right?  So we're set there, so not a big deal.  So we fill everything in.  So this is all taken directly from the question.  I'm skipping the units on R because honestly, we're really short on room here.  OK.  So we're going to have that.  That's just calculator button pushing.  And we get this.  OK.  So that's our moles of NO.  Now what do we have to solve for?  Well, if we're finding one that we're agent, we need moles of the other one, right?  So exact same thing.  And sapling was nice and only, it gave you everything in the right units on this very,  very long, hard problem.  I decided I didn't want to be nice to you on the previous one where I did that.  So OK.  So we plug this in.  OK.  So now we come to limiting reactant stage.  There's a bunch of different ways to do limiting reactant, and everyone has their favorite way.  I think in this one doing logic is the best way.  You can choose to do it mathematically if you want.  The two kind of are hand in hand.  So if we look at this, if we have one oxygen, how many NOs do we need?  Two.  If we have five oxygens, how many NOs do we need?  10.  If we have 20, 40.  So we've got the pattern, right?  How many do we need?  Double.  Now do we have that?  So we look at oxygen.  Do we have double the amount of NO that we need?  No.  So what one's going to be our limiting reagent?  OK.  All right.  So how much NO are we going to have at the end of this reaction?  Zero, right?  So no moles, no pressure.  OK.  Next part, then.  We need to figure out how much NO2 we're going to have, and how much what else are we going  to have?  Are we going to have any oxygen left?  Let me reword the question.  So which one was our limiting reagent again?  NO.  So that means that we have more oxygen than what we need, right?  Yes.  So does that mean we're going to have any oxygen left?  Yes.  So we're going to have to figure out how much NO2 we make, and how much oxygen we have left.  So that's our next goal.  So it doesn't really matter which order you do it in.  I think I found moles of O2 first.  So we're going to find moles O2 left, and find moles of NO2 made.  OK.  Let's do NO2 first.  We'll see because it's the easiest.  So if we used up all of our moles of NO.  How much NO2 did we make?  Let's break it down a little bit more.  Let's say I make one mole, or I have one mole of NO.  How many moles of NO2 would I make?  One.  If I had five, I would make five, right?  They're the same, and that's because the coefficients are the same.  So what we're going to do here is if we know how much NO2 is made, or we know how much  NO we started with, which we do, we know how much NO2 is made.  It's the same amount.  So we just write it down, basically.  We just take our amount of NO we started with, and we write it down.  OK.  Now oxygen's a little bit more complicated, right?  Because we know how much we had to begin with, and what can we calculate that happened  in this reaction?  How much we made, how much we used up, how much is left over?  What can we do with one calculation using this reaction?  How much was used up, right?  We know that if this has two, that one of them is going to get used up.  If we have five, or 10, that five of them are going to get used up.  So what we're going to need to do then is to find our final amount of oxygen.  We're going to take our initial amount, and we're going to subtract away the thing that  we used up.  So let's do that.  And let's just put a little f underneath moles for final, just to make it super clear.  So obviously, if I were to give this on an exam, you wouldn't need to write down all  of these sentences.  I'm just doing that for you, so when you go back and look at this, you know what's going  on.  So moles of O2 left over.  How much did we start with?  We started with this, right?  And we're going to subtract what was used.  OK, how do we figure out what was used?  Well, we know, I'm going to kind of do this as a sidebar, like I do sometimes.  We know how much NO we started with, right?  Now how is the moles of NO in the moles of oxygen related?  By the mole ratio.  So writing this out really big so that you can see it, and then I'll plug it into the  smaller area, just want to make sure everyone can see.  Our moles of NO go here, our moles of oxygen will go here.  And what is the number that goes in front of the oxygen?  1.  We got that from where?  The equation.  Here is 2.  So I just wanted to write it out big, and then I'm going to fill it into the equation  on the notes we've been taking as well, but just because it's getting tough to see.  OK.  So we took our total that we started with, our initial amount, and we subtracted what  we used.  And when we get this then, we end up with 0.081.  So now we have our final moles of oxygen.  We have our final moles of NO2.  And do we have our final moles of NO?  It's 0.  And so we're good there.  Now what equations do we use to go ahead and solve?  Now we're back to the exact problem that we had before.  The problem where the question of, that's it, got asked.  Now we're at the P1v1 over N1t1 question.  And we have to look at what's changing.  So if you go back to the question and we can look at everything that is and isn't changing,  we can fill everything in.  OK.  So we're still not OK for time.  OK.  So when we go to do this, we put in our final amount of moles.  Oh, thank you.  Can we do this for each one?  Is that too small?  No, that's OK.  You guys are all right.  And we fill everything in.  So now this becomes a scavenger hunt, right?  Now you have to go through and figure out where everything is coming from.  Make sure you grab all your final values and not your in-between values.  So this is just R. Make sure you remember after all of this work that you've done to  convert your temperature to Kelvin, right?  Because it's easy to forget things like this after we've gone through everything.  We're going to walk through it a little bit here once I finish it as well.  And so we can talk back over everything that we did.  So now I'm just filling in numbers.  We've done all the hard parts, and we're filling everything in.  So when you fill those into the calculator, I ran out of room, so I can't write it beside  it.  Now, of course, you know that your homework values are going to be different than this,  so don't go filling those in right away, right?  You know something changes your numbers on you.  So that's your total pressure.  Let's go back through and talk over what we just did in the last minute and a half.  OK.  So we started by saying, well, in order for me to find the final pressure, I need to  find the initial pressures.  So I know all of this detail.  I can use pv equals nrt, and I can solve for my moles.  Once I find my moles, I can use all of my knowledge that I remember from the first week  of class, and I can use stoichiometry to solve for, one, the limiting reagent, so I know what's  not going to be there.  Two, how much of the non-limiting reagent is left over, and then how much of the product  I made, and then after we have that, we have our moles of everything.  So we still know that we have r and t and v. We can find that from the problem, but now  we found our final moles, and we can fill in our final moles into the pv equals nrt equation  again.  So really at its heart, what this is doing is it's taking, and it's using, you know,  origin factors just like we did before, and you have to go through moles.  So you convert everything to moles.  You do your stoichiometry, and then you convert everything back into pressure.  And I'll leave this up as you guys are packing up and stuff so that you can see it if you  can find the position of the other�
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Okay, so today we're gonna pretty much see. All right, I think that's better. Okay, so today if you guys can be quiet, we will. Okay, we're gonna kind of start back and talk a little bit more about Dalton's partial pressures. I think really we'll spend most of today on doing the various versions of this in the various examples because it is a little bit long. We may make it into a fusion. So I'm hoping to finish up the gas laws on next Monday and then be able to spend Wednesday and Friday on review for you guys. I think two review lectures would be best, but it really just depends on if we can make it through or not. I'm not 100% sure if we will or not. So without further ado, we'll get going. So review a little bit from the last five minutes last time. So remember that partial pressure, the way that this works is that you can take the pressures of each individual gas. So let's say you have a mixture of three or four gases. You can take the pressure of each one of them and then add them together and that'll give you your total pressure. So when we did this sort of with an equation form, this is it written out. I'm sure you'll be having some computer issues. So this is it written out in just pressure of each gas equals pressure of the total. So using the fact that we know that pressure is equal to nRT over V, ideal gas law, we can take and we can solve the pressure for each individual gas and then we can add them all up together. And we know that if we have a container with a mixture of gases in it, that the temperature and the volume are going to be the same. So we can go ahead and we can pull that out. We can factor the moles out and then we can get this. So what this really says is that you add up all of the moles and you can add that up and then you get your total pressure. So your moles of all your gas, you can fill it right into your ideal gas law as well and solve for your total pressure that way. So two different ways of doing it. Now there's a very important concept here that we need to talk about and that's called the mole faction. So this X, this is actually the Greek letter chi, so that's how you say that. And this says that if you have the, all this really is a fraction just like any other fraction, right? It's whatever you're talking about over top of the total. So if you want to know the mole fraction of A, you take the moles of A and you put it over top of the moles total. It's no different really than saying the percent of A, like the mole percent of A, except that you don't have the times 100% there. It's just the decimal point or the fraction. So that's what mole fraction is. Now what this says is it kind of takes this concept, combines it with this one, and says well we can take that chi, we can take that mole fraction, right, our part over our whole, and we can just times that by the total pressure. This isn't any different than the sort of how you work with percents your entire life, right? If you want to know what percent something is of a total, you take the fraction and you multiply it by the total, okay? So this is your total pressure. So if I tell you that something has a total pressure of five atmospheres, but the mole percent or the mole fraction of it is only 0.2, take 0.2 and times it by five, okay? So don't let this seem like a new equation to you. This is just using percents like you've always used them, except what's the first thing you do if I say you have 75% of something? You turn it into a decimal, right? Well that's all this is. It's just the decimal of the part over top of the whole. So that's all the mole fraction is. It's like a percent, except not with the percent. I'm not sure why they go through the, you know, special aspect of naming it and different and calling it chi and all of that, but you know, that's fine. So don't let this be new to you, just kind of a different way of naming it. Okay. So before we do partial pressures with actual numbers, let's just go through and do it with a quick picture. So if I have this and I have this mixture of molecules, so I have some purples and some greens of aqua or cyan and some pinks. If I tell you that this has a pressure of 10, it doesn't matter what unit, I say 10. And we want to know the partial pressure of each gas. What do we do? You have to find the mole fraction, right? So let's start with the purple ones. So first of all, let's figure out how many, what's our total number of molecules? Our total mole of molecules. So our total number of molecules we just count. One, two, three, four, five, six, seven, eight, nine, ten. Now if we want to know our mole fraction of the purple ones, what do we do? Well how many purple ones do we have? And I apologize for the colorblind people. Statistics say there's at least a few of you and I do apologize. One, two, three, four. I won't do that on an exam obviously. So there's four out of ten, right? So your mole fraction is four out of ten. So what would be our mole fraction of the pink ones? Would it be six out of ten or three out of ten? Good, three. It doesn't matter that it's a diatomic, right? We're treating all of these molecules and all of these atoms and all of these gases as just like they're just spherical balls that are all the exact same. They have no volume, they have no interaction, they have nothing. They're just ideal gases. So it doesn't matter if I drew these as diatomics and these as single. So it would be three out of ten, right? Now so what does that mean the pressure of the purple ones would be? Four out of ten. So just four, right? So and you can go through, you can do that with each one. So the blue would just be one out of ten. That would be your fraction, it would be point one. The greens would be two out of ten, point two. So all mole fraction is is a count, right? It's how many moles of something you have, or in this case how many atoms works the same either way? Over top of the total. So now if we were to go through and say well what if the total pressure of the box is thirty units then what would it be? So that changes things a little bit, right? But not too much. So I'm going to have to flip over to the document screen for this so we can write. So let's just do this for a couple of them. So we talked about the pink and the purple ones. And we talked about chi. So let's write it down. I would suggest it's doing this right on the slide. That's why I didn't put it on the worksheet. So we have our pink, we have our purple. What is our chi for each one? So you're looking at your picture. And for the pink we had said, well there's three out of ten. And for the purple we had said four out of ten. Alright so let's just do it for those two. I'll say that you can do the green and the blue at home. So now I said well what if the total pressure is thirty? Thirty whatever. Because we're talking about molecules and not moles, it's kind of odd to use normal units. So we're going to call them thirty of man units. What is going to be our pressure, our partial pressures? How do we figure this out? Well it's just a percent, right? You're just treating this like a percent. So if we want to know our partial pressure of each one, we take our mole fraction, we multiply it by the total to get our answer. If we took our purple we would do four out of ten. In times if I had thirty to get our answer. It doesn't matter that I drew one as sort of kind of your ideal gases, right? Healing of neon, whatever, because it's just mono atomic and your other is diatomic. Just take the mole fraction, multiply it by the total and you're done. So part, Dolan's law is really nice. It makes life kind of easy for us compared to if all of this wasn't true. Okay, I guess I didn't need to switch over. Now we're going to do some examples. So the best way to do these is really just tons and tons of examples. It's kind of the common theme for the majority of this lecture. Okay, so we have this. We have the contents of a natural gas tank, right? Because natural gas can be a whole variety of gases. It's not just one thing. And I give you the percentages of each one and I say, well, what's the partial pressure of each one? Given that the total pressure is one point two. So let's break each one up. We'll do CH4, C2H6 and C3H8. And we need to figure out the partial pressures of each one. So how are we going to do this? Well, this time I didn't give you a mole fraction, right? What did I give you? Percent. So are we going to take 88.6 and multiply by 1.2? No, what do we have to do first? We have to do the decimal, right? So we'll convert that and multiply by 1.2 to get our answer. So we get this. And just for sig fig purposes, we need to add that zero in there. So we need to round this to the three sig figs. And we have that. Yeah? Yeah? Would you ever ask us what's the partial pressure of carbon in CH4? You mean, OK. No, so I couldn't do that because the carbon itself wouldn't be a gas. I could ask you for the moles of carbon, which if I did that, you could take this. And how would you find moles of CH4? You'd have to fill into the ideal gas off. So I'd have to give you some more information, right? I'd have to give you some volume and temperature and things like that. And then you would know that however many moles of methane there is, how many moles of carbon would there be? One. So I can go to moles of the individual elements, but I can't go to a partial pressure of it. It doesn't make a lot of sense to talk about. OK, this one. So we convert that to a decimal. Careful with your know your single percent. Sometimes people miss the second moving over part of it. We multiply by the 1.20 ATM to get our answer. And we round to three figs, or excuse me, two figs this time. OK. And our last one. So once again, we convert to a fraction. We multiply by 1.2 atmospheres, and we write down our answer. And if I were to say, what is the mole fraction of methane, or CH4? What would you say? Do you have to calculate anything first of all? Let's say I ask that right from the beginning. I gave you this problem, and you know how I write the exams. Lots of times there's multiple parts of the question in one thing. And all I said is, I wrote all of this out, and I said, what is the mole fraction of each gas, and what is the partial pressure of each gas? What would your mole fractions be? Yeah, it would be these. In this one, it would be the 0.886. In this one, the 0.089. In this one, the 0.025. So if I give you the percent, you just turn it into a decimal point. And that's your mole fraction. That's why we multiplied this by this. We took our mole fraction, and we multiplied by our total. OK? All right. Ready for a much harder example now? You're supposed to be excited. No groans. Be happy. OK. So there's a little picture here, which I think probably got added after I had posted this. So let me zoom in a little bit, even though I probably ended up having to zoom out so you can see it. OK. This was because when this is a little bit of a confusing setup, but it's a very cool setup and something that you can do in lab, and you can do probably not at home, but definitely in lab. So we have a sample of oxygen gas collected over water. So what's happening here is some sort of reaction is going on here, some sort of reaction that makes oxygen gas. So in this case, there's a little beaker where something's going on. The other time this will happen is you can burn something and make CO2. So this is a way of measuring how much of a gas comes off a reaction, which is useful if you're trying to figure out something about how the reaction is happening. Maybe you want to know a percent yield, or you want to know something about a reaction, and you need to know what all your species are. You need to know how much product you get at each one. Well, this is the one way that you can test for oxygen, or that you can measure oxygen if it's the only gas that's evolved. So what happens is you have something going on here. You have a reaction happening. Oxygen gas comes through here. It's the only place for it to really go, and then it bubbles up. So you have a tube that originally was completely filled with water. You have a tube completely submerged with water. You stick it into a beaker, a flat thing of water here. And then the only air in here, or the only gas in here, comes from this reaction, which will all be oxygen. So this is the way that the setup works. And this is, I don't think you guys do this in lab unless it's been added in the last few years, but this is something that a lot of labs do. So what can you do with this if you have this, and let's say you put it in a graduated cylinder. You can measure that volume, right? Can you measure the temperature? Yeah, sure. So you know a lot of things about the system, and you can measure how many moles of gas are in here. And that's what we're going to do in this problem. We're going to try to figure out how many moles of oxygen came off this reaction. So when we say it's collected over water, this is how they actually go about doing it. They have a cylinder filled with water. They put a tube in. The gas bubbles through. And then you can, as long as this is like some sort of graduated system, you can see the volume of gas that you collect. So I say a wet gas. Well, how can a gas be wet? What that means is that it has water vapor in it. It's not pure. So if you take a tank of oxygen that you don't want any water in it all, they would have done something to take out the water vapor. In this case, it's sitting over top of water, right? We know that water evaporates. We know that water has a vapor pressure. That if you put some water down here, it's going to evaporate. And if you put a cap over it, it's going to condense and evaporate and do its thing. So there's water vapor in here. So that's going to change the volume a little bit. So we have two things, right? We have water vapor and we have the oxygen gas. So two things that we have to worry about. But those two combined have a volume of 7.28. So that's our total volume. And we have a pressure of 1.25 bar. And that's our total pressure. If all the water is removed, what volume will the dry oxygen gas occupy at this whole new set of conditions? So we're taking what we just learned with Alten's Law of Partial Pressures, and we're going to combine it with what we did last class, where we went through and did final and initial conditions. So a lot going on here. This is a tough question. So just to make life a little bit more fun, I also gave you two and the wrong unit so that we could practice playing around with units. So this right here, that's water vapor, right? That's the pressure of the water. So we know that this pressure is going to be what the water is. So then how do we go about finding the initial pressure of the oxygen? So let's write this out a different way. We know our pressure total has to be the addition of all the gases that are in there. So what is in this beaker right here for gases? Oxygen. So we know the pressure of oxygen plus? Yeah. OK. Now we're getting somewhere, right? Because we have this, we know our pressure total is 1.25 bar, and we know our pressure of our water is 23.8 millimeters of mercury. So we can find our pressure of oxygen. So we're getting closer. So first we have to do some unit issues. So you can really switch to pretty much any unit you want here. I happen to switch everything to millimeters of mercury. Sorry about that. That should be 1.25. So we're going to take the 1.25 bar, and I'm going to convert that into millimeters of mercury. You could convert it the other way if you wanted. And of course, all of these conversion factors are given to you on an exam, so don't spend any time memorizing them. OK. So we have this, and we can write down that value. And I'm going to have to zoom out a little bit now that you guys have had some time to look at the picture. OK. So that, once again, is going to be our total pressure. OK. So now we can start filling in some values. We need to know the pressure of oxygen. And just for the cleanliness of work purposes, I'm going to draw a line here. So we need to know pressure of oxygen. And we can get that through this equation, right? We know that our total pressure is only equal to the pressure of those two gases. So how do we get our pressure of oxygen? Pt, and then what? Do we add our pressure of water? Good. We subtract it. And so we get that. Now since we've switched our units around for this purpose, we can go ahead and fill in. Now, something that's worth kind of doing up here, because we're going to be doing a lot of calculations with things that we calculated, is to watch your sig figs start using your underline method. Where you just underline the last significant one so you can keep track of what's happening here. It's way too much work to try to trace it through otherwise. So we'll go through. We'll fill in our pressure of each one, because it's a bad idea to round every single time you fill in. To be honest, I tend to just save everything in my calculator. And then for the sake of units, put that. And so we can get our answer. Now we have to pay attention to sig figs a little bit here. What are we going to round to? So I'm hearing three, and I'm hearing tenths. So if this was multiplication and division, you're right with three. Because our lowest number is three sig figs. But this isn't multiplication and division. It's subtraction. So we round to our lowest digit, so our tenths place. So we round to that. So that is our pressure of oxygen. Are we done? No, right? This is our pressure of oxygen under what conditions? Place, right? This is our condition at 25 degrees. And this pressure, excuse me, the 1.25 pressure total in that volume. That's not what I asked you. That would have been too nice, right? I said, now we're going to change everything. Because in reality, if you were going to do this, probably you would change the situation. Probably you would end up, in order to dry the gas, you would have to do something to it. And so that's going to change your conditions around a little bit. So this is what we're going to do. We're going to do the exact same thing that we did last class. So this is going back to the material then. And we always started from this equation, right? Because we figured, well, we start with everything changing, and then we just cross off whatever doesn't change. So we start from this one more time. Now the nice part about this, do we have to worry about the fact that I'm in millimeters of mercury? No, right? Because whatever I plug in here, that's just what's going to come out from my final pressure. So that's not a big deal. OK. Or excuse me, we do actually have to worry a little bit. We have to convert from one or the other. So it doesn't matter what unit we put it in. It doesn't have to be in atmospheres. It doesn't have to be in millimeters of mercury. But it's got to be in the same thing. Now our volume. So let's go through and convert all of this. We're solving for what variable? V2. V2. So we'll fill everything else in. OK. So our P1 comes from here, right? Now I think in this case I converted everything to millimeters of mercury again. I was apparently in a millimeters of mercury day that day. For the sake of room, I'm going to be a little bit lazy about writing units. Perhaps, we'll see. So this is in millimeters of mercury. Now our V1, what is our V1? That's that, right? So we go ahead and rewrite that in. I've decided in hindsight not to be lazy about units. It's too complicated of a problem. So that's our V1. Now our N1. Do we have to worry about N here? No. That gets canceled out. So we don't have N, but we don't need to worry about N because we're not changing it. So that goes away. So our T1. Or is our T changing? Definitely. So I'm just going to fill in 25 here then, right? No, right? What do you have to do? Calvin, don't forget that. It's really the most painful mistake to have to grade on an exam because we know that you know the difference. But if you forget, it's wrong. So excuse me, I'm going to hold it for a moment. OK, next part then. P2. Do we know P2? Sure, it's given. So I'm just going to fill in 1.07 and go about my merry way, right? No. So we need to go ahead and we need to convert this. So that's an ATM. So we fill down ATM here. We fill in our conversion factor up here. You can always do this on a different sheet if you want to and then fill it in whichever way you prefer. We fill in V2. And we fill in our final temperature. OK. So once we go through all of that and we do the algebra, which I'm not going to write out the algebra because quite honestly, it's the easiest way to just do it inside your calculator. We have V2 is equal to 8.50 liters. Now let's check our sig figs on this last part. So we know that this, let's just write in 290. So I'm going to actually just, it's not wrong, but I'm going to cross it off and write 298 for the sake of going through and doing the sig figs. Because we know we're going to be at our one's digits here. Now what's our sig figs here? We took 37 and we added it to 273. So is this really two sig figs? No, it's really three sig figs. So watch out for sig figs in those types of situations. OK. So now we have how many sig figs here? Four. Three sig figs here. Three, this is actually three by definition. It's three and three. So we need to have three down here. So I'm going to put a little underline there too. All right. OK. So questions on that very long complicated problem. This is one of those problems you probably are going to have to really get, go through and do looking at the notes. Do not looking at the notes, but constantly referring to them off and on for help. And then maybe do again on your own three or four times before you really get it. It's long and complicated. But it's really just combining a bunch of things that we know with one new thing. I gave you this new experimental setup. But we know that whatever that oxygen is, whatever that little volume is, is just our two pressures added together. We know that then if we want to find our pressure of oxygen after last class, last five minutes, we know that we can just subtract it from the total because there's only two components, right? It's only water and oxygen. So we subtract that to find our pressure of oxygen. Now we're back to what we did for the whole first class or the whole first half of class last time. And we said, OK, well, we have initial set of conditions and we have a final set of conditions. Let's solve for the unknown. All right. Any questions on that before I move on? You may be a little too lost for questions yet. That's fine. OK. We can revisit that one at a review session or something. All right. Next one. All right. So we have two containers connected by a valve that can be opened. Initially the pressure in volumes are as marked. Find the final pressures after the valve of art is opened. OK. So this could theoretically be a really complicated problem, but luckily Dalton's law makes it simple for us, which is really nice. So we can treat this as two completely separate problems. What we're going to do is we're going to pretend that we opened this valve and that there's nothing here, that there's no gas here, that it's just a vacuum. And we let the pink come into here and equilibrate. We're going to pretend that on one half of the problem. Then for the second half of the problem, we're going to go ahead and pretend that this goes into here, that this goes into a completely empty container over here. And that's going to be our second half of the problem. What that's going to wind up doing is giving us the partial pressures. So if we have the partial pressure of each gas, how do we find the final one? Yeah, we just add it together. So let's go ahead and do that. So let's start with the pink side. So what all is changing here? Did I say anything about the temperature? No. We assume it's constant. Is the moles changing? What do we think? Do everyone agree? Mostly because this is a tricky question. So they're right, but corals answers are good. Let's me know if you're understanding things. So why isn't the moles going to change? We're combining the pink and the blue, right? Yeah, but we're not talking about the blue right now. Exactly. We're not talking about the blue, so we don't care. All we care about is the pink going to change, and it's not. We're not going to take any way. We're not going to add in. So let's take away those R's. Those aren't necessary. So what do we have to worry about changing then? Just P and V. So it becomes P1, V1. It equals P2, V2. All right. So now we just fill in some numbers. So our P1 and our V1 are right here. Now P2 is what we're solving for. What is our V2? Is it 4 liters? No, what is it? It's 5 liters, right? Because it's total. We're adding it together. We're saying this pink one can just spread out over the whole space and nothing in its way. That's kind of what we're pretending. And so in that case, it's going to be 1 plus 4. So we solve that. And we get our first pressure. Now we're going to do the exact same thing for the blue side. So take a minute or so and do it. See if you can get the right answer. And then if you can figure it out, find the total pressure as well. Oh, yeah. So these are the initial values. So these are the initial values of the pink and the initial values of the blue. All right. I'll give you 20 more seconds. You can talk to your neighbor. It's fine. I have like 10 more seconds to see if you have the same thing. OK. What did you get for a pressure for the blue one? Do you guys agree? Two? Yeah. So I'm going to do this one solving for variables first. This one's final is 5 liters as well. Don't forget to add them up. OK. So that gives us our partial pressures. So in this case, I just told you to find the final pressure. But the easiest way to do it is to go through and solve for the partial pressures. There are actually other ways to do it as well. So let's go ahead and add these up. OK. So we get 2.20 atmospheres. All right. Questions on that one? Yeah. I know I did the hard one first and the easy one, but keep you on your toes. The next one, however, we're going to go back to hard. Really I just needed to give you guys a little bit of a breather between the two hard ones. That was the real reason for doing it here. OK. But I think you'll all be really happy that I'm doing this hard one. And as soon as I put it up and you see the format, you'll see why. And guess where I got it from? Yeah. Not Wiley. Sapling. OK. For those of you who don't know, Wiley is the other homework program that some of the classes use. That's where that came from. OK. Yeah. This is from your homework. So you have the worksheet in front of you, hopefully, because this is going to, you know, it's a long question with a lot of room, so I'm going to have to kind of zoom out and move around a little bit. But this is the same setup. And that's why we kind of started with the easier version of this. And this is the more intense version of it. So we have NO and O2. This is also my public service announcement for remember all that fundamental stuff that you may have, like, said, well, I'm going to just remember this stuff. Yeah. Make sure you remember this. Make sure you go back and do this, because this is a lot of the fundamental section that we sort of talked about a lot in lecture. Because as I kept warning you, it's going to come back in chapter four. Well, it's here. So we have two valves or two bulbs. They're connected with NO and O2. In the last situation, they didn't react, right? We just put them together, let them mix around, and found a new pressure. But in this case, they're reacting. They're forming a new substance. OK? And so it gets a lot more complicated now. Now we have to deal with a bunch of stoichiometry as well. So we need to find the pressures of each one. But to do that, we have to find the moles of each one first. And to find the moles of each one, what do we have to do? Are they going to react and completely form 2NO2, and we're going to have none of this and none of this? No. What's probably going to happen? I'd say that there's going to be a limiting reagent. So that's our first step. OK. So we're going to find the limiting reagent. So we need to find the moles of each one of the starting. We need to find moles of NO and moles of O2. So that's our first step. So to find our moles of O2, actually let's do moles of NO first. What equation are we going to use to find moles? What do we have? We have some volumes, and we have some pressures, and we have some temperatures. So what does that sound like? Do we have to worry about our units in this situation? Do we have R there? Does R have particular units? Yeah. So it's got to be atmosphere's leaders, right? So we're set there, so not a big deal. So we fill everything in. So this is all taken directly from the question. I'm skipping the units on R because honestly, we're really short on room here. OK. So we're going to have that. That's just calculator button pushing. And we get this. OK. So that's our moles of NO. Now what do we have to solve for? Well, if we're finding one that we're agent, we need moles of the other one, right? So exact same thing. And sapling was nice and only, it gave you everything in the right units on this very, very long, hard problem. I decided I didn't want to be nice to you on the previous one where I did that. So OK. So we plug this in. OK. So now we come to limiting reactant stage. There's a bunch of different ways to do limiting reactant, and everyone has their favorite way. I think in this one doing logic is the best way. You can choose to do it mathematically if you want. The two kind of are hand in hand. So if we look at this, if we have one oxygen, how many NOs do we need? Two. If we have five oxygens, how many NOs do we need? 10. If we have 20, 40. So we've got the pattern, right? How many do we need? Double. Now do we have that? So we look at oxygen. Do we have double the amount of NO that we need? No. So what one's going to be our limiting reagent? OK. All right. So how much NO are we going to have at the end of this reaction? Zero, right? So no moles, no pressure. OK. Next part, then. We need to figure out how much NO2 we're going to have, and how much what else are we going to have? Are we going to have any oxygen left? Let me reword the question. So which one was our limiting reagent again? NO. So that means that we have more oxygen than what we need, right? Yes. So does that mean we're going to have any oxygen left? Yes. So we're going to have to figure out how much NO2 we make, and how much oxygen we have left. So that's our next goal. So it doesn't really matter which order you do it in. I think I found moles of O2 first. So we're going to find moles O2 left, and find moles of NO2 made. OK. Let's do NO2 first. We'll see because it's the easiest. So if we used up all of our moles of NO. How much NO2 did we make? Let's break it down a little bit more. Let's say I make one mole, or I have one mole of NO. How many moles of NO2 would I make? One. If I had five, I would make five, right? They're the same, and that's because the coefficients are the same. So what we're going to do here is if we know how much NO2 is made, or we know how much NO we started with, which we do, we know how much NO2 is made. It's the same amount. So we just write it down, basically. We just take our amount of NO we started with, and we write it down. OK. Now oxygen's a little bit more complicated, right? Because we know how much we had to begin with, and what can we calculate that happened in this reaction? How much we made, how much we used up, how much is left over? What can we do with one calculation using this reaction? How much was used up, right? We know that if this has two, that one of them is going to get used up. If we have five, or 10, that five of them are going to get used up. So what we're going to need to do then is to find our final amount of oxygen. We're going to take our initial amount, and we're going to subtract away the thing that we used up. So let's do that. And let's just put a little f underneath moles for final, just to make it super clear. So obviously, if I were to give this on an exam, you wouldn't need to write down all of these sentences. I'm just doing that for you, so when you go back and look at this, you know what's going on. So moles of O2 left over. How much did we start with? We started with this, right? And we're going to subtract what was used. OK, how do we figure out what was used? Well, we know, I'm going to kind of do this as a sidebar, like I do sometimes. We know how much NO we started with, right? Now how is the moles of NO in the moles of oxygen related? By the mole ratio. So writing this out really big so that you can see it, and then I'll plug it into the smaller area, just want to make sure everyone can see. Our moles of NO go here, our moles of oxygen will go here. And what is the number that goes in front of the oxygen? 1. We got that from where? The equation. Here is 2. So I just wanted to write it out big, and then I'm going to fill it into the equation on the notes we've been taking as well, but just because it's getting tough to see. OK. So we took our total that we started with, our initial amount, and we subtracted what we used. And when we get this then, we end up with 0.081. So now we have our final moles of oxygen. We have our final moles of NO2. And do we have our final moles of NO? It's 0. And so we're good there. Now what equations do we use to go ahead and solve? Now we're back to the exact problem that we had before. The problem where the question of, that's it, got asked. Now we're at the P1v1 over N1t1 question. And we have to look at what's changing. So if you go back to the question and we can look at everything that is and isn't changing, we can fill everything in. OK. So we're still not OK for time. OK. So when we go to do this, we put in our final amount of moles. Oh, thank you. Can we do this for each one? Is that too small? No, that's OK. You guys are all right. And we fill everything in. So now this becomes a scavenger hunt, right? Now you have to go through and figure out where everything is coming from. Make sure you grab all your final values and not your in-between values. So this is just R. Make sure you remember after all of this work that you've done to convert your temperature to Kelvin, right? Because it's easy to forget things like this after we've gone through everything. We're going to walk through it a little bit here once I finish it as well. And so we can talk back over everything that we did. So now I'm just filling in numbers. We've done all the hard parts, and we're filling everything in. So when you fill those into the calculator, I ran out of room, so I can't write it beside it. Now, of course, you know that your homework values are going to be different than this, so don't go filling those in right away, right? You know something changes your numbers on you. So that's your total pressure. Let's go back through and talk over what we just did in the last minute and a half. OK. So we started by saying, well, in order for me to find the final pressure, I need to find the initial pressures. So I know all of this detail. I can use pv equals nrt, and I can solve for my moles. Once I find my moles, I can use all of my knowledge that I remember from the first week of class, and I can use stoichiometry to solve for, one, the limiting reagent, so I know what's not going to be there. Two, how much of the non-limiting reagent is left over, and then how much of the product I made, and then after we have that, we have our moles of everything. So we still know that we have r and t and v. We can find that from the problem, but now we found our final moles, and we can fill in our final moles into the pv equals nrt equation again. So really at its heart, what this is doing is it's taking, and it's using, you know, origin factors just like we did before, and you have to go through moles. So you convert everything to moles. You do your stoichiometry, and then you convert everything back into pressure. And I'll leave this up as you guys are packing up and stuff so that you can see it if you can find the position of the other�
Wiley is the other homework program that some of remember from the first week of class, and I can position of the other�