 Will I get a round of applause at the end, too? I mean, that would be great. You're gonna know. Now? Now. Just based on this. That's awesome. Okay. So, something to add. I don't know if we'll get through the whole day because, apparently, my, uh, it's Ed's game. Mohamed went through a lot of slides, but we'll get through as much as we can. The test is next Tuesday, and it covers chapters 5 and 11 as well as any review chapters, like one through three. Take a look at the syllabus. If you're not sure, I would highly suggest to everyone that you read the text book. The text book has a whole lot of good information, obviously. That's why we buy it. But today we're gonna be covering a little bit further ahead in the course. So, guest lecture six. I'm that guest. Hello. Based on copies of solutions. So, when we think of chemistry, like you just picture what happens when we perform chemistry. You usually think of someone, are you raising your hand there? Yes. Oh, yeah, sure. Is that better? Yeah. Okay. So, typically, if I picture someone trying to perform chemistry, I don't usually picture them taking two solid blocks and rubbing them together. I picture them mixing some solutions. Solutions are very convenient. And much of chemistry occurs in solution. So, when we want a reaction to happen, we, in general, need two things to collide together. Yes? This isn't always the case, but we need two things to hit. And for that to happen, it's a lot easier in solution work. Everyone's rolling on top of each other. Yes? In a solid state, they're fixed. In a fixed spot, it's a lot harder to get them together to react. In the gas phase, yes, you have lots of collisions, but gases can be dangerous in general. We do lots of chemistry with gas phases. I'm just saying it's a lot easier to do it in solution. And so, understanding the characteristics of solutions is very important. So, for instance, it says here that there's a little chance of an uncontrolled escape of reactive chemicals. It's very true. In the gas phase, however, we can sometimes have some accidents. So, for instance, and I don't know if you guys have told us in the news recently, you don't have to look in the science magazines to hear about this. This was a giant accident back in 1984, where methyl, where methylate, excuse me, methylosocyanate was spilled as gas down in, where was it? Bopal India. Thousands and thousands of people were dead. It's because of things like this that we're always questioning, how can I do this chemistry in a safer way? And oftentimes we want to do it and look it forward. So, for instance, we know that the North America has a lot of natural gas, yes? We could use it for fuel. But it's very dangerous to transport natural gas. And so we always look for ways to convert that to a liquid fuel like methanol, convert methane to methanol, so that we can avoid accidents. So, solutions can, of course, be dangerous, too. I know I've spilled things on myself before. But while there are solutions of gases, uniform compositions throughout, in solids, sort of a frozen snapshot of a liquid solution, most solutions are liquids, with a liquid solvent and gaseous liquid or solid solute. So if you think about that for a second, that last line, liquid solvent and gaseous liquid or solid solute, hydrochloric acid, you guys usually use this liquid, right? HCl is actually a gas room temperature. And we dissolve it into a liquid, usually water. And that gives us the HCl acid we work with in the lab. So that's actually a solution. It's a gas mixed in. Same with a liquid-liquid solution. I could take something, any liquid or whatever, index and mix it with more water. That's a liquid-liquid solution. And a solid, a solid with liquid, that would be something like sodium chloride, which is solid room temperature in a liquid. So, but a solution is a single phase. If I take sodium chloride and mix it in with some water, I don't see two different phases present. That's because the sodium chloride is fully dissolved. So I have one phase, just the solution. So dissolving sodium chloride in water, it gives us a single phase, like I was saying, up to about 25% sodium chloride. Now that's by weight. So in your first chemistry class, you may remember, and I'm not sure if I do it here at UCI, but they may do a solubility test where you take some water and you put in sodium chloride until all of a sudden it will no longer dissolve the sodium chloride and it starts piling on the bottom. That's when it's reached its saturation point. And for water room temperature, that's around 25%, once no more salt can dissolve, it's saturated, adding further salt would give two phases. The solution, which is for sodium chloride, and the solid sodium chloride on the bottom. The saturation point depends on temperature. If we think about that just in general, we know that hot solvents tend to dissolve things faster and easier. They dissolve more material. This makes sense to us. A surprising note here, before this lecture, I thought sea water was a higher weight percent and just an interesting bit, you know, that if you drink sea water normally, you will die of dehydration. And all you have is just that little bit of 3.5 weight percent salt. I thought that was interesting. So a phase diagram. If a phase diagram is not familiar to you, then you should be reading the textbook more. Okay? This is a basic phase diagram, but instead of pressure versus temperature, like you may have seen, we have temperature versus the weight percent sodium chloride in water. I'm going to go through this slowly. So what we're taking is water at 0% here, no sodium chloride, and then we slowly change to 100%. So let's start over here on the left, where we have 0% sodium chloride. Now that's just water, right? So at 0 degrees Celsius, just water, well, that's going to be ice. It's going to be frozen. And above that, it'll be just liquid water until you get to 100 degrees Celsius and then it'll boil. So you probably have seen or heard about when roads freeze over, like in the mountain passes, it's very dangerous to drive over the ice. And so what they'll have is they'll come and salt the road. So you may have heard of that before. What they're doing is putting that salt into the water and there's a freezing point of depression. So you can see on this phase diagram, let's start at 0%, and then I put in a little bit of salt. All of a sudden, that freezing point starts to lower and lower and lower down here. And now I see that I can get all the way down to minus 20 degrees Celsius and the solution will freeze and this is safe for the roads. When they put that in there, the ice will melt in general, depending on the temperature. And it's a lot safer to drive on the road. It's a good application. So if we start up in here, just as a demonstration, let's say we're at room temperature 20, 25 degrees Celsius. We know that as we add salt, then it's going to dissolve, yes, until we reach this around 25%, 25% by weight, but in chloride. One thing to note here is the bending of this line. So you can see that it's bending slightly to the right, yes. As we increase in temperature, I can see that the weight percent, the total weight percent that I get for liquid solution, before I start getting liquid solution in chloride, it starts to increase. This is just a visual application to that thought where the hotter we hear that the higher temperature our solvent is at, the more we can solubilize that sodium chloride to get a higher percent. But depending on the temperature we're at, we'll pass into this phase, the liquid solution plus sodium chloride. Next is, so the notation A plus B is always means two phases. So in this instance, we would have liquid solution as one phase and the solid sodium chloride as another. If we took this liquid solution plus sodium chloride and we start to cool it, at one point we're going to get liquid solution plus sodium chloride that's hydrated. It's simple to think of cooling it down, more materials going to crash out, you're solubilizing less. But as you get down to really, really, really cold temperatures, you start putting that water on sodium chloride. So if we think of this philosophically, as you lower the temperature, it's a lot easier to get things to hit together and stick. So that sodium chloride tends to crash into one of those waters or rather it's water crashing into the same chloride and it'll stick. So we get this hydrated form. And as we increase the percentage of sodium chloride more and more and more, we see that we get the sodium chloride 2H2O plus just sodium chloride solid. And that's because if you think of this as a phase diagram, what we're saying in this region, I have almost no water left and it's almost all sodium chloride. So why do things dissolve? Same reason that gas spreads out to fill a container. It's statistics. It's hard to keep things in one spot. Things like to spread out. So dissolving increases the disorder of the solute. You can think of it just based on statistics and you don't have to take in a class to just think of it with common sense. So let's look at the statistics with just six atoms. We're just going to say krypton, but it doesn't matter. This is just six particles. It could be anything. So we're taking six atoms of krypton gas in a container. Starting with all six on one side in the stop clock flows. There's a picture in just a moment. I believe it's next. Yes, it's next here. So imagine we have this scenario where we have two containers and there's a stop clock here so it can effectively block it off. At one point we put six atoms which of course just mentally that's an obscurely small amount of atoms. This is just a demonstration. Six atoms of krypton on one side. And then we open the stop clock. We know just from common sense that that's going to equilibrate until they're on both sides. So why does this happen? Why does the pressure equalize? It equalizes because this is the most likely outcome. If I go back for a moment and take the moment when there are six atoms of krypton and then I open the stop clock what's going to happen? So I've got six atoms that are just crashing around all over the place and I have an opening. Naturally since there's nothing here it's a vacuum those are going to statistically start bouncing over onto the other side and then once I get onto their side they're also going to have a chance of bouncing back until of course I reach this final state where it's equal on both sides. So this is an example with only six. What are the chances of it not being equal on both sides? So here we've taken these atoms and we've covered them. Again, these are just krypton. They're just colors so you can see a difference in individual atoms. So we take these six different colors of krypton and this is a certain configuration. It's one possibility and when it talks about the weight this is a statistical weight. It's on mass. This is how much does this particular configuration contribute to the total possible outcomes and we'll go over the weight in one moment. So we have this scenario where I have six on one side that's a six to zero. So there is only one way to have all six on one side, right? That way right there. All six colors in one hole nothing anywhere else. So moving on, if I have if I change a little bit and bring over one a more even distribution has a higher weight. So before let me demonstrate we have a weight of one for this configuration. It has one possible way of happening. When I go to just five in one I see that I have six possible ways to do this. The configuration of five to one. Taking any one of these colors yellow, red, blue, green sort of blue magenta, I don't know taking one or three on either side there are six different ways to do this just statistical randomness. Crypto on one side Crypto on the other side So by the time we get to four and two there are a whole lot more possibilities. We can take, obviously I'm not going to go through each one of these, but you can take any combination of these colors with four on one side and two on the other and you find that there are 15 different ways. You can see how this grows exponentially and in fact it is exponential. Kind of configurations. So for the three to three configuration there are 20 different ways to color the atoms. The configurations, two to four well you know let me go back for just one second so you can imagine that with all these possibilities with four and two now we move one more over and with three and three there are 20 different ways to color those atoms 20 different possibilities and then we had two containers one on the left and one on the right and we were starting with the one on the left being more filled. If you go on the opposite side one on the right being more filled all the statistics are the same you're just doubling that possibility So the configurations two and four five and one and zero and six have those same weights as four and two, five and one just as much. So the total number of configurations that we have is given by one plus six plus 15 plus 20 plus 15 plus six plus one equals 64 You can see what we were doing starting with six and zero, one, right five and one and that contributes five and so forth until it's even and then you can go to the other side. That's how many possible orientations of the system there were and that's 64 total which as I mentioned earlier no coincidence as being exponential it's an exponential statistical analysis two to the sixth the number of containers two raised to the power of the number of atoms you'll mention in a moment and I'll just say it now you can imagine that with 64 and only six atoms just think of how many atoms are in a balloon right if those were just if those were two metal containers and you had an atmosphere and one and opened it up it's exponential like that the number of possibilities that include atoms mostly on one side they don't contribute anything at all it's statistically insignificant it doesn't matter and so by the time you get to any sort of significant amount of atoms there's no there's no point in even considering a situation where the atoms are on mostly on one side they're equal they will equal about 50 for 64th of the time moving randomly we'll look and see a 4 to 2 or a 3 to 3 or a 2 to 4 and that's just adding up those possibilities equals up to 50 to 64 and about 14 to 64 times we'll see that 6 to 0 5 to 1 or 1 to 5 or 0 to 6 so like I was saying if we took that 64 and made it 10 billion right 10 to the point 3 that's way beyond billion you're talking about huge numbers now so when we view it this way the gases spread out just because it's more likely when the first scientists started looking at these systems they didn't have to have the chemicals they could just envision what's mathematically correct to first approximations when you don't consider chemical interaction math describes almost everything it goes into physics, it goes into chemistry it describes the world it's only when you start taking exact scenarios where you have to start considering how atoms interact together etc so it's more likely for them to be found equally distributed and jammed into a smaller volume likewise and again it's just going to statistics if we roll a paradise the most likely outcome is to get a 7 so we have large numbers of atoms and nearly even distribution is overwhelmingly likely considering the solvent to be like a vacuum and the solute to be like a gas so consider for a moment if I had a beaker full of water with solid sodium chloride on the bottom before it's dissolved it's just the instant before we stir it and it dissolves that's like a concentrated one balloon that's concentrated with gas atoms it doesn't like to just sit there when there's room in the solution so considering the solvent to be like a vacuum and the solute to be like a gas it is overwhelmingly likely that a solute will dissolve if there is no energy of interaction between the particles like the ideal gas case so they're saying in the event that sodium chloride had no interaction with another sodium chloride well it would easily disperse just like a gas molecule but we know that this isn't entirely true just like we know that the ideal gas equation isn't totally true if we interact together we know that a solute is going to have interactions together so in sodium chloride we know that sodium and chloride have that ionic force together yes while in the crystal there are lots of ionic forces one sodium cation is not interacting with just one chlorine ion it's interacting with lots of chlorine ions so there are lots of forces pulling them together it takes energy to separate one sodium chloride so there are several energies that we have to consider if I take water in sodium chloride how do I get that sodium chloride to dissolve well there are solute-solute forces so the solute being sodium chloride to sodium chloride solute-solvent forces so sodium chloride interacting with water and they're solvent-solvent water interacting with water for me to get a sodium chloride from another sodium chloride and a water from another water together I have to break a water-water interaction a sodium chloride-sodium chloride interaction and then put them together so you can imagine, let's say perhaps for a moment that sodium chloride and water weren't missable they didn't lag each other it would be hard to break sodium chloride which has a strong bond and water which has a strong bond together and put them together to get an interaction these are sort of things we need to consider so if all of the forces were the same if water had an equal energy of interaction with another water and sodium chloride has an equal energy of interaction with another sodium chloride and they had an equal energy of interaction with each other then they just dissolve because entropy is to increase that is statistics of the three interactions we must first separate the solute molecules so that sodium chloride sodium chloride and the solvent molecules that water and water and then mix the two together so water has a network of hydrogen bonds we should know this by now that keep forming and breaking randomly in the liquid so that's something I have to consider when I break that water-water interaction it's not just dipole-dipole there's also the hydrogen bonds so I have to break that first and that costs energy other molecules that can form hydrogen bonds are then quite soluble in water so yes can you slow that a little bit? oh sure, am I going too fast for you guys? you want me to back up? two slides? just tell me when when? when I heard it? okay, alright, this is my first time be gentle so let's see where I enter the interaction let me go back to the slides and go through it again alright what's funny is I heard Mahalak went through quickly and I was like, not me not me, are you going to have me? alright this way do you want me to just... what time is it? I got plenty of time, right? 11.24, wow 24 minutes have passed and I know that's most of the slides well ah, keep it down so Mahalak went through most of my slides that's my excuse but we do have some sample problems at the end some review questions for the exam and we're going to go over those as well now to take up some time or we can just have a light conversation I'm down for anything so huh? I haven't been keeping track of the website has he gotten the other ones up yet? what was that? so the slides for up to lecture four are on and we're at lecture like six right? that's true, so he hasn't been here all week so I anticipate he'll have the slides up I'll make sure to ask him about it too yeah, good I'm glad we had to talk okay so we have, it's only 11.24 alright let's go through this again um okay too far? you guys are never happy for anything okay we're good? I'm getting mixed signals from everyone that's more? okay we're going to start here anyone has any problems? get out and leave okay we're going to start here I'll go through it again a little more slowly alright um so county configurations remember we're talking about these two containers yeah thank you this is all in video you're making me look bad anyway okay so we have two containers right? and what did we do? we took one container that had six atoms in it and we opened it up to another container that had zero just by statistics we're going to become equal as I mentioned before when we open up this container at first and we had six on one side they're crashing around randomly and they're just going to by statistics eventually find this hole and make their way to the other side this is going to keep happening until we have an equal amount going from one side to the other and an equal amount coming from that second side first it's an equilibrium we have three on each one switches to the other another switches back and so forth but this is only with six atoms so this is only with six atoms so in this distribution where we have five to one five on one side and one on the other this configuration label is five to one five in the first container and one on the other and its statistical weight is six there are six ways that I can put five on one side and one on the other right and I want to do a one to five so one on one side and five on the other that also is going to have a weight of six but it will be five on this side and one on the other so we have to consider what are all the possible ways for me to arrange these atoms so with four and two continuing that line of thought we have fifteen different ways where four are on the side and two here so green blue blue purple right yellow red you just switch one of them keep the red on here but switch the yellow with that light blue one that's a whole different configuration a whole other possibility for these atoms when we consider their randomness continued on there are fifteen different ways to arrange four excuse me four and two those have four that one has one, two, three, four four, five, six, seven there I'm glad we redid this I'm sure this is why you were all confused so in any case ways to put four and two fifteen different ways and then for the three and three there are twenty you can give this yourself when the numbers are small you can just write out all the combinations and add them up if you like that so for that three for three there are twenty different ways to color the atoms but then those configurations of two to four, five and one and zero and six as I mentioned they have the same weights as four to five to one there's another typo I didn't write this so two to four and five to one first container has two second container has four that has the same as the first container having four and the second container having two this one five and one has the same as one and five and then zero and six six and zero why wouldn't each one have the same weight which let's go back so two and four, five and one why don't each have two and four five and one zero and six have the same weight so let's go back to the images so compare compare six to zero if the atoms are moving around totally randomly there's only one possible way for me to have six and zero with six and the first and zero and the second if I go just to five and one and we'll go back and forth between this there are six different ways to do this so if I just take a snapshot of this gas totally random if I took a million snapshots I'm going to get a statistical distribution between what my snapshot sees zero and six are five and one it's totally statistical based on what the chances have been seen six and zero are five and one I know that six times more often I'm going to see five and one and that is more likely there are more configurations for this than for six and zero weight as a traditional sense it's more like yeah I must have said this weight is not a mass yeah so I was blazing through before this weight is not a mass at all this is a statistical weight okay it's saying distributed to all the possible configurations this is just a number that represents what chance do I have of seeing this situation so going back by one this has a weight of one there's only one chance of having six and zero one possibility going on to the next I have five possibilities of having five in the first and one in the second and so this has a weight of six it's six times more likely to see this than six and zero okay does that answer your question okay thank you for bringing that up I didn't realize that people had missed that the weight thing so four and two so you can see there's possibly about five possible configurations take that a step further go to four and two here are all the possible ways for me to write four and one and two in the second there are 15 different combinations so if the atoms are totally at random going to have some configuration I have 15 different ways that show me this exact arrangement remember these atoms with all the different colors there's still just one atom they're all just krypton so if I took a picture of this right now left from the bottom from any one from the right it would all look exactly the same to me and so I would count that as another one of my four and two if I keep doing that snapshot, snapshot, snapshot over and over again I'm going to have 15 different possible ways but only one possible way to see six and zero and that's where this comes from that's where the statistical reasons come from so for that three and three which is taking the same thing to the next configuration there are 20 different ways to color those atoms and then you're asking why they don't have the same weight so the configurations so the configurations two in one four in the other five in one one in the other and zero in one six as flipping your diagram so six and zero has one possible configuration and zero and six has one possible configuration together I have two possibilities of ever taking that image and seeing six in either with zero in the other so the total possible the total configurations that are possible are 64 so let's think about that six and zero I take a snapshot there are two and 64 chances that I will take a snapshot with six in one and zero in the other very small and if I have I'll go to the next page there you go and 50 out of 64 those times if the atoms are moving randomly we'll look and see that four in one and two in the other or three in one and three in the other because what I'm doing is adding the 20 so remember 20 corresponded to three and three so three and three plus the chances to see four and two so four and two that was 15 so four and two 15 three and three that's 20 and two and four which is another 15 that's how many that's the sum four and two 15 15 30 50 50 out of 64 times I'm going to see one of these and 14 out of 64 times we'll see 6 0 5 1 1 5 or 0 6 so the six and the zero right that contributes one the zero and the six that contributes another one the five and the one that contributes six the one and the five that contributes six total 14 out of 64 so the chance of me seeing very few on one side and more on the other is very small only 14 and 64 when I only have six atoms since the number of possibilities increases exponentially the chance of me having most of a gas on one side and little gas on the other that becomes completely insignificant and we don't even consider the possibility so they spread out because it's more likely for them to be down equal if I take a million atoms since we've seen a half a million on one and a half a million on the other that's incredibly high of course give or take a couple atoms who's counting right? likewise, if you're a gambling person we roll a pair of dice most likely outcome is to get a 7 quick tangent went to Vegas for my first time a few weeks ago, for my anniversary spent $100 lost it all thank you it was roulette I was up 16 and then 10 minutes later I'm down 50 I don't know how it works when we have large numbers of atoms a nearly even distribution is overwhelmingly likely so considering that solvent to be like a vacuum as you were mentioning before and the solute to be like a gas overwhelmingly likely that a solute will dissolve if there is no energy of interaction between the particles like the ideal gas case so let's picture this again we have that beaker of water with a bunch of sodium chloride just sitting in a mass it's not likely for that sodium chloride to just sit in a mass as long as those interactions are the same like the ideal gas case no energy of interaction in fact there are forces between atoms in condensed phases as I was mentioning earlier we know that sodium chloride has an interaction with it and we know that in a sodium chloride crystal that sodium chloride is interacting with several other sodium chloride and it's hard to break them apart than strong ionic forces that require energy to break one sodium chloride from another sodium chloride solute, solute, that's the water oh excuse me, sodium chloride in this case so breaking apart sodium chloride from another sodium chloride that takes energy the solute and solvents when I take that sodium chloride and the water approaches where we have interacting forces and in the case of water and sodium chloride that's a positive energy it's favorable to those tracks and I have solvents so water and water so I have water it takes energy to break that apart so I have two unfavorable energies breaking water away from water breaking sodium chloride away from sodium chloride and I have a favorable energy taking sodium chloride and interacting with water and if that interacting energy is strong enough it can overcome the energies that I'm having to break so if all of the energies are exactly the same then breaking a water apart from a water and placing it with sodium chloride well that's no different energy wise if the interaction of water with water I break it apart but then that sodium chloride comes in and I get that energy back well then it doesn't matter and it's just going to be statistics and it's going to dissolve because we're increasing that entropy delta s is greater than zero if this were the first time going through it would that be a good pace? I'll keep that in mind so the enthalpy delta H of dissolving measures the sum of those three interactions we must first separate the solute molecules and the solvent molecules and then mix the two together and if that is a good energy it will happen naturally and if it takes more energy to do that than it will not do it's not going to happen water has that network of hydrogen bonds forming and breaking randomly in the liquid other molecules can also form those hydrogen bonds and then are quite soluble in water let's go back to what I was saying a moment ago I have to break water from the water and I have to break something else from something else let's say ammonia NH3 well I know NH3 has hydrogen bonds so when I take that ammonia and bring it closer to water I know they're going to form strong hydrogen bonds this increases that interaction between water and that solute so it could be more likely to dissolve in this case it would along with the soluble in water so ethanol, methanol, ethylene glycol they're miscible with water or they have ethanol has a hydrogen bond yeah it has an aliphatic region but it also has that OH group and that forms hydrogen bonds with water and so it can dissolve same thing with ethylene glycol and methanol forming these hydrogen bonds on the other hand a long chain alkane like hexane with H atoms only on carbon it's not going to dissolve very much we get two layers like salad dressing or if you're just in your lab you have oil and water so let's talk about this for a second then if we have oil we'll just say it's hexane and water we all know that why don't we mix what kind of forces does it have what kind of forces I mean just dispersion mostly it doesn't have any H atoms it doesn't have a dipole it doesn't have any ion or dispersion hexane doesn't hold together very well that's why it has such a low boiling point water however holds together very powerfully it has strong hydrogen bonds it has dipole-dipole forces it holds together so when I try to put the two together I have to break a hexane-hexane interaction that's easy because these dispersion forces aren't very strong but I have to break a strong water-water interaction now if the hexane-water interaction were powerful that would be okay but it's not hexane doesn't have any hydrogen bonds it can get into water it's not a dipole so it can interact with water's dipole it's just sort of like dispersion forces or an ion-induced dipole that sort of thing it's a weak interaction and so you're not going to break apart strong water with weak hexane because their interaction is also weak so they separate so we're on to some practice problems exam time is next week remember what Dr. Schock has said oh it's not time to leave we have some problems I mean you can leave if you want I'm not going to track but we've got time we've got 11.43 so we've got plenty of time there are about 5 questions here and the exam is Tuesday it's going to cover chapters 5 and 11 plus 3 view 1 through 3 but he'll probably focus on 5 through 11 I'm sure but you just in case you need to know 1 through 3 so we'll just go through some exam questions right now oh not exam questions some review questions and we'll see how it goes the boiling point and the melting point of a pure substance what influence does molecular shape have on the liquid range so I think what I'll do for these problems especially because we have plenty of time is I'll just give maybe 30 seconds for you guys to think about a problem and then we'll go over it together so go ahead and just think about this for about 30 seconds I'll try to keep quiet it's very hard for me I like to talk but I'll try it it's very hard the factors influence the boiling point and the melting point of pure substance and how do you affect that with the range so we know that the boiling point is going to be influenced by molecular forces boiling point is how much energy do I have to put in to get those to break apart from each other put them into the gas phase so very strong interactions are going to increase my boiling point water has very strong interactions it's got a high boiling point hexane does not have very strong interactions no boiling point so we consider ion-ion ion-dipole-dipole-dipole London dispersion forces and hydrogen bond if you ever forget what those mean go look in the textbook I really encourage people to read the textbook I read the textbook, look where I'm at up here teaching you it works so the melting point is influenced by similar factors but packing is very important I don't recall if Dr. Stock had ever mentioned packing in his lectures but how well layers and layers of the club how it fits together influence how closely they can get together and how strong that interaction is so we know that if I just took an ion-ion like a positive and a minus charge the closer I bring them together the harder it is for me to pull them apart same thing with solids if I have things that pack really well together then when it comes time to try and break them apart to go from the melting point to the liquid phase well it's going to be harder to pull them apart into the liquid phase if they're packed really tightly together if something doesn't fit at all it's not as able to interact as well and you can break it apart a little bit easier and that's going to lower its melting point so a molecule that is round or symmetrical may have a high melting point because it fits together well so a symmetrical molecule has only dispersion forces it may also have a low boiling point and still have a small range of our witches of the liquid some molecules like CO2 have no liquid range at 180 N pressure they sublime again a phase diagram we should be getting very familiar with these pressure versus temperature we know that with dry ice with CO2 that's minus 78.5 degrees celsius that's at about 195 right in that range so it's unfortunate this graph doesn't show that place but here we're in the solid form dry ice right so as one bar is around 180 N at a temperature of around 195 kelvin we're in this region right here where we would be a solid and as it warms up we enter right into the gas phase instead of the liquid phase so we need almost 60 bar pressure to have CO2 be a liquid or room temperature so if I increase this pressure let's say I'm at 100 bar and then I increase the temperature then I'll enter into a liquid phase and eventually it's super critical yes what was that you would just like a question if you don't want to realize what do they mean by how long? it is so if somebody has a melting point of 0 degrees celsius and a boiling point of 100 degrees celsius that's a 100 degree celsius range okay I throw out a big range you can put it in liquid they're saying that when things have very very very a small amount of forces sometimes it can't even stay in the liquid very well once you melt it it goes into the gas phase pretty quickly so you would have a very small range I don't know what there's no range at all but they're saying there would be a much smaller range of 10 degrees celsius I can't think of any examples that have a really small range except for this one maybe so are there any more questions on that problem? alright I'll see you in the next one okay so a gas mixture has 90% argon bi-volume and 10% helium bi-volume to treat these as agiogases if low pressure effusion through a pinhole is used to try to separate the two gases what would the percentage of helium bi-volume be after the process now we're not going to sit here while people work out a whole problem take a minute to think about it write out whatever you want think about how you address this problem and then we'll go over together okay let's talk about it so we're using Graham's law it tells us the relative rates of effusion through a small orifice go like the inverse square root of the masses so we'll look at this for one moment that is r1 over r2 those are rates not radii the rate of the first thing going through a small hole divided by the rate of the second thing going through a small hole equals the mass of the second thing over the mass of the first take the root so let's talk about it for one second does this make sense I have argon and helium argon is very heavy helium not so much I put them into a container with a small hole if they're at the same temperature which they should be which one's moving faster helium when we have our own equations helium is moving very quickly it's bouncing around very very fast statistically it has a better chance of getting through that hole because it's going to hit that hole more often the size of the hole doesn't have anything to do with the difference in size between argon and helium argon is still infinitesimally small compared to the size of the hole so it's only based on the mass helium is bouncing around and hit that hole more often so if we take this and we plug in those masses and take the rates we can find out the ratio of the rates we can't solve r1 and r2 we're taking the ratio helium is 4 grams per mole and argon is 40 and so what this means is that the ratio that relative rate and remember that's r1 over r2 the ratio is 3.162 that's what we're saying in helium we'll go through that hole 3.162 times as fast as whatever argon does we haven't insinuated any kind of rate or a time we're not saying how long this happens for it we're saying that for one moment the rate is going to be 3.162 times the rate of argon and so if we set that rate of argon diffusion in any units we know that the rate of diffusion of helium is going to be 3.162 times whatever that rate is and hence after one stage of a fusion so we can have this in seconds or hours whatever it depends on how accurate you want to be after one stage of a fusion through an orifice we'll have relative numbers of atoms so argon at some rate going through I'm going to have 0.9 of whatever my units are going through helium there's only 0.1 worth of helium versus argon remember it was 90% argon and 10% helium and so the 10% of helium at 3 times faster rate I'm going to have 0.3162 amount of something coming through and so what happens is total I have the 0.9 plus the 0.3162 that's how much got through and this is how much helium there is that's now 26% so we increased from 10% helium to 26% helium I believe that Dr. Schachter used uranium as an example for this in the first lecture on ways to separate things it can be quite useful all they don't use uranium as a separation by method they're not going to use centrifuges so oh any questions? I didn't realize that was that kind of thing say again? yeah sure we're all just writing down or do we have questions? what? oh sure it owns an actual question on it so we have one chamber and another in that second chamber we let stuff transfer over that's now 20% or to the 26% of helium that's something to consider is that rate doesn't stay the same in that first stage we had 9% argon and 10% helium after we let that go through we know that my second container has 26% helium why is that important? that means that in the first container I now have even less helium by percentage than I did before I had 10% before now whatever the amount is if I keep letting that going the differencing rate which was 3.162 that's going to start decreasing because as I get closer and closer to there being almost no helium there's going to be more argon going through than helium your question asks what would the percentage of helium by volume be after the process? where are you looking at? because the percentage of volume is going to be 26% that's the new container whatever went through is now 26% helium yes so this makes sense right? let's say 1 is helium so the rate for helium is going to increase faster and faster relative to argon as the mass of argon or whatever my second one is it's heavier and heavier versus helium so let's pretend it's not argon for a second when this is argon we have 3.164 so m2 40 over m1 4 take the root 3.164 if I made m lead let's just say 200 grams per mole that now is 200 over 1 and take the root that's going to mean r1 over 2 is now reusing because helium is so much faster than whatever that really heavy gases that I put in there does that make sense? I couldn't hear that whole thing oh it just makes the mass easier if we had a real situation we would want to know exactly how much argon we have this is a demonstration of using ratios to compare relative rates we just said we say that r1 over r2 is 3.162 so I'm pretending that whatever the rate of r1 is I just call that 1 I'm saying relative to that so r2 is 1 if the ratio r1 over r2 is 3.162 that lets me say that the rate of helium transfer is 3.162 times whatever r1 over n is I think I'll have to limit that one we've had plenty of time right I wish there was a clock 11.58 class is out at 12.20 we've got a few more questions if you have more questions come down and ask me afterwards so we'll move on did anyone want just a moment to write something down a little bit you got it? so another practice problem alright recall that a Van der Waals gas follows the equation of state as follows I'm not going to read through that remember that the ideal gas equation is just pv equals nrt and here we've said pv equals nrt is great but it's not fantastic so we've added in these correction factors a n squared over v squared so where the gas specific parameters a and v depend on a gas if for xenon we have a equals 4.192 liters squared atm per mole and b equals 0.05156 liters per mole then what is the volume of one mole of xenon at 5 degrees celsius and 2 atm pressure to 3 significant digits now I'd like to take a moment here to point out something about units so if you were just looking at this you think what are the units of a I know what the units of n are n squared it's moles that's going to be moles squared and the v that's liters squared for me to add two things together they have to be the same units and so automatically I know the units of a because for a n squared over v squared to equal atm a has to be liters squared so atm so that you have atm per mole squared you can't sell out the mole squared there that's just a quick way to address one of the units in a situation similar to nb for volume equals liters nb also will have to be liters and so v is going to be liters per mole because n is mole so mole times liters per mole gives you liters so go ahead and take a moment to try and find the where was it and what is the volume try and find the volume of one mole of xenon at 5 degrees Celsius with 218 pressure with three significant digits tip don't try and solve for v in that equation you're going to make a quick guess using the equation that you have that's not quite as accurate as that one but it will let you almost get the answer so earlier in the quarter remember he has the get really close and then punch in the calculator a couple of guesses until you're spot on right so don't try and solve for v alright so let's go ahead and flip on over so first thing as Dr. Shaka is fond of telling you remember to convert Celsius to Kelvin keep everything in the same units his advice is to use pv equals nrt guess almost the right answer and then you're going to nudge your way to the correct answer so figure out that nrt and do not round r because we need the accuracy so something he keeps on harping on in the class is don't round too much if he asks for the accuracy don't think at the end just make sure at the end I have three digits if you cut out the digits earlier in the problem it's going to ruin your accuracy down in the end so you have to keep that many digits all the time so in this case we're using as many digits of r as is practical going out to four is plenty good that's going to give us all the accuracy we need and more to get our three digits in the end so figure out nrt don't round r and that equals 2, 4, 3 liter atm from here so we started with the ideal gas estimate of the volume when we begin guessing so or using the works view you can graph it or whatever but make sure you get three digits correct verify your answer by substituting it back in the end of the equation so he's saying well we know nrt if we go back to the to the full equation we're not making any assumption by just solving for nrt so we can do that right off the bat it's this side with the more complex equation that causes us grief so we can start with just pv as a quick guess which doesn't have the a and b in it and then we can in his words skoosh our way over to the correct answer so if we let our function of volume equal to atm plus 4.192 liter atm volume squared times volume minus 0.05156l so we'll go back for a second what we're doing is we're taking oh we have this equation right we're trying to figure out the volume so we can easily make this a function that depends on volume and that's just pressure plus a n squared over volume times volume minus nb so let's go back up to that slide so our function pressure plus the 4.192 liter squared atm over v squared so that's a n squared divided by v squared times v minus the nb so he just figured out these numbers and put them in and this is our function for v so if we plug in values of v and see what we get our starting guess is nRT over p we should all know that definitely and that is 11.4121 liter so we set up the table where this is the volume in liters this is the function of volume in liter atm our target value which should be 11.4121 and then our little comment on that value when we do the ideal gas law we have 11.4121 our starting value is 22.2 when we do volume based on ideal gas law that's 11.4121 and the function when we plug volume into our function is 23.0867 liters atm our target value which is nRT we're trying to get pv equals nRT ideal gas law wasn't good enough Van der Waals equation has a and b we're taking that first guess putting in the function we know that nRT is 22.8243 we're trying to see how accurate we are so volume 11.4121 gives us two bigger values so we go a little bit smaller in volume 11.2 we'll put it into our function and that's a little too small so 11.3 we put that in and we're just scooching his word is not mine I think scooching is a weird way to describe it I value is tiny so 11.25 that's a bit small so depending on the accuracy we're looking for we know that the answer is in between 11.25 and 11.3 so these are three digits like it asked that's 11.3 right big question? so so over okay so Van der Waals equation accurate not so much we can replace all of this with pv and that's the ideal gas equation so going forward your question was why does volume equal nRT over p that's just using the ideal gas law right did that answer your question? I feel like I missed something in your question is there something that I didn't get to? okay it did any questions with that? no? I think there's one more problem let me just go ahead we have an x-ray diffraction question and we have 12.08 alright we got 12 minutes I think this is the last problem an x-ray diffraction experiment using copper k-alpha x-rays so that's just describing what level of x-rays is at of wavelength 1.54 on a pure metal crystal gave a first order Bragg peak at an angle theta 14.17 degrees if you're not familiar with this go read the book they have a quick and simple explanation on diffraction of how this is working so we have a metallic crystal it's just layer upon layer a simple cubic probably of metal so the incident angle of radiation is 14.17 degrees and they tell us the wavelength is 1.54 angstroms assuming the crystal has a cubic lattice and that the peak arises what is the dimension of the unit cell if the structure is simple cubic what is the radius of the atom what further information would you need to identify the actual element so I'll go ahead and give you one minute to work on this a little bit excuse me and then we'll work over it it's funny to be down here and actually see people like sleeping with their heads something like that but obviously I can't because their eyes are closed real quick before we go over it when I ask you guys up there in the balcony is my camcorder still going it is this is my first time teaching a lecture with this many people if you all just give a shout to my mom and dad I'm going to send it to them so on three we're all going to say hi mom hi dad one, two, three hi mom, hi mom, hi dad love you okay alright oh my wife hi Daniela thank you okay let's talk about this alright so we have a first order diffraction and that's code for n equals one if we're thinking about this equation one, normally this would be n n equals one in the Bragg equation hence we have one times the wavelength which we already have because they gave us that equals two times d sin theta they gave us theta when we're solving towards d so let's go back one second here because there was an assumption that was made and we didn't talk about it when it says that assuming the crystal has a cubic lattice and the peak arises from the atoms in the corners of the cube what that's saying is d was a measure that's making my slit where the x-rays are being diffracted it's not always the case that one set of layers makes that diffraction we're making that assumption so if I just solve for d well maybe that comes down to 10 angstroms and that could correspond to 5 or 6 layers we made the assumption that it's from the first layer and so d is just the distance in between the two corners on our point okay so if we solve for d that's .3145 nanometers and for a simple cubic unit cell this is too odd remember we were trying to solve the radius of the atom if it's a cubic lattice I have atoms at all the cubes and they're touching each other if they're touching each other and I know the distance in between the middle point of those two atoms because they're right at the corners then I know the radius of each atom and the distance and the radius is just .157 nanometers so cut that in half so we're trying oh sorry we're just writing down do we have questions questions so far just writing down alright moving on I'm hearing a lot of murmurs so we were trying to find the let me read the end of the question we were trying to find the mass what information do you need to identify the actual element so what do we need to identify the element well we already have the volume we have the radius what's the last thing that we would need for a simple cubic cell there is one atom per unit cell that's because at each corner we have one eighth of that atom contributing to the unit cell at corners it's one atom so we can figure out how many moles there are in a cubic centimeter so if we then knew the density of the material in grams per centimeter squared then we would know how many grams there are per mole and that lets us of course identify the atom because we know the grams per mole but we just look over at the periodic table and that's all we need so and this is very very true re-electrical experiments are of course a lot more complex than this nobody takes crystal structures of things that are just one element in there they're incredibly incredibly complex so density yeah so we need the mass any questions? oh why don't you go ahead and start going you can come down and ask me do you know anything about the notes okay so notes for the exam they can only be notes that you've taken in class so you can't go outside of class write a book come bring it in you wouldn't want to do that anyway how are you going to have time to look at it write