You see those lines that are between the phases? These are the boundary phases, right? The phase boundaries.

At those points, usually you have co-existent phases, like you have the liquid and the solid phase are at the boundary where the arrow is. The other part where you have somewhat of a curve, that's where you have the vapor-liquid phase boundary. Now what's so special about these boundaries? Well, when you start to heat up a solid, I'm just going to give you guys an example.

When you start to heat up a solid, when it's purely solid, let's say water here, because we have a specific temperature at which it melts. What happens is, you add heat, the temperature starts to rise. Then you reach zero degrees Celsius. Even though you're adding more heat, the temperature persists. It does not increase anymore.

Why? Because now you have an equilibrium between the solid phase and the liquid phase. What happens is, the solid starts to convert to a liquid and if you keep on adding heat, that's what's going to happen. If you keep the temperature constant and you stop heating, you keep it at zero degrees Celsius, equilibrium is just going to stay there.

Okay, so it's always going to be a solid and a liquid. So, you keep on heating, you reach somewhat of a steady curve for the temperature, right? Temperature remains constant until all the solid has converted to a liquid. After that, heating will cause the temperature to rise up again, okay?

So, you guys all know this, 273.15 Kelvin is the melting point for water, right? And 373.15 Kelvin is the boiling point. Now, just a tip for his exam, 273 and 273.15 are not exactly the same, okay?

So, he really stresses on these values, so make sure that the values that you use in your calculations are very accurate, okay? So, basically same with the solid and the liquid, same thing happens with the vapor and the liquid, right?

You reach a phase boundary where the arrow is pointing, that's where boiling starts to take place. 100 degrees Celsius, temperature stays constant as you're heating until all the liquid converts to a vapor. And at that point, after all the liquids have turned to vapor, temperature starts to rise again, okay?

So, now you look at this, you look at the first paragraph, you notice it says that you can actually keep water as a liquid at 285 degrees Celsius. That's 185 degrees more than the normal boiling point.

But then again, look at the pressure, 1000 PSI, right? So, does this make sense, that you can keep water as a liquid at such a high temperature? Yes, you're exerting a very strong pressure at it, right? You're keeping it nice and condensed.

Pressure plays a very important role in phases, right? The more pressurized something is, the more condensed it is. As in, between vapor and liquid, it's more liquidy. Between liquid and solid, it's more solid, alright? The less pressure you exert, the more the liquid or the water or the substance starts to expand and goes further more along the gas, towards the gas phase.

So, and I'm going to go back to the triple point. So, the definition of the triple point, as you can see, it's kind of like a point at which all three phases tend to intersect. And you can see it's kind of close towards zero degrees Celsius, right? If you can see from the plot, almost 273, 280, somewhere there.

Right? So, at this point, the substance exists as all three. Exists as a solid, exists as a vapor, and exists as a liquid. This is the definition of a triple point. What about the critical point? You notice that the critical point, the line kind of disappears.

The liquid and the vapor phase are rather indistinguishable. Okay? So, you kind of have, I wouldn't want to say a mixture, but you can't say if the substance is a gas or a liquid.

Okay. So, when you put water at one atmosphere, one atmosphere is kind of like the standard pressure, right?

It's the pressure that we're at. When the temperature is between zero degrees Celsius and 100 degrees Celsius, what do I have? I have a liquid, right? There is no solid. There is no vapor. Or let's just say, let's be more accurate. There's not enough solid or enough vapor that could account to the amount in there.

So, it's basically purely liquid. Below zero degrees, it's ice. Over 100 degrees, it's vapor. Exactly at these two points, and I've said this a number of times now, exactly at zero degrees Celsius, the liquid or the water can exist as both liquid and ice. And at 100 degrees Celsius, it can exist as both vapor and liquid, right?

Now, here we're looking at the phase diagram for carbon. Now, if you notice, and if you paid attention in the phase diagram for water as well,

you notice that ice was existing in like about, what, seven phases, right? So, you have seven different phases of solid ice or solid liquid, just different structures. Same thing with carbon here. You have two different structures, and I think you guys know this.

Carbon can exist as graphite, can exist as diamond. Now, this kind of shows you why diamond can be so expensive, right? I mean, look at how much pressure you need to exert on graphite to turn it into diamond. This is a very expensive procedure. But nonetheless, people still tend to do it.

And it's a very slow process. It requires a catalyst. And so there are just a number of reasons as to why this process is very expensive and why diamonds are so damn expensive. Okay. So, again, as I said, when you have two phases,

when you're at a phase boundary, when you have your substance as a liquid and a vapor or a liquid and a solid, basically what happens is these two phases are at equilibrium. The atoms or the molecules start to jump from one phase to the other at random.

So basically what I'm trying to say is at the zero degree boundary, not all the solid remains a solid, not all the liquid remains a liquid. No. These atoms tend to jump around between the solid and the liquid phases.

So, yeah. There's an experiment that you can use in order to validate that. To validate that not everything that's a solid remains a solid, not everything that's a liquid remains a liquid. Usually what you do is you put, for example, you have a certain container that contains liquid water and gas or vapor water, right?

And you somehow introduce D2O. So you guys know what D is, right? D is deuterium. It's an isotope of hydrogen, right? And what you do is when you introduce that vapor into the container, at some point some of the D2O vapor will convert into a liquid.

Now you can use mass spectrometry or you can use certain lab techniques and you will see that after some time the liquid will contain some D2O. Okay, so really what I want you to get out of this slide over here is just that nothing remains the same, right? Everything is jumping between the two phases, all the atoms.

I guess the last sentence says it all, right? Now, so for some of you this might come off as a little odd. I know it was odd to me the first time I learned this, evaporating cooling.

So do you know when you're sweating, for example, when you're running on a hot day? Sweating is a very, sweating is a cooling process, right? That's why some animals can't stand the heat. Some animals don't sweat. Dry weather, they just die out. They don't cool down. So what happens basically is some molecules, when you have a substance and you start to heat it up,

let's say we want to vaporize some water, okay? What happens is the molecules in the water, the water molecules that have high enough energy to break through the barrier and kind of go out into the atmosphere, they just leave the solution. So what happens then? What happens is all the higher energy molecules within the liquid go out.

They leave. So whatever is left in the liquid is actually cooler than what it originally was, right? So, and you know, if it's an open container, the vapor completely goes out, right?

So there's no chance of the higher energy molecules to come back in again, so the water will be cooler. And I guess that's all I can say about this. That it's just, what's it called? It's just a cooling process.

Okay, this kind of explains it even better. So, what you notice is at lower temperatures, most molecules do exist at a lower temperature. But as you increase the temperature, you get somewhat of a more distributed number of atoms, right?

The curve kind of flattens out because now you have more molecules that are towards the higher energy end. Now, the shaded region here, the red shaded region in both graphs, are basically the molecules that have enough energy to escape the solution.

So imagine, if these higher energy molecules escape the solution, the solution is just going to get cooled, right? Now, hmm, where is it? Okay. Now you notice the last sentence, or the second statement in this slide says molecules may escape from a solid surface to like dry ice.

So, this is something I talked about last, not last week, yesterday, right? When we were talking about the dry, how that guy tried to dry his clothes under negative 15 degrees Celsius weather. Well, so you guys probably know this, phase transitions do not necessarily have to go from solid to liquid to vapor.

Transition can go directly from solid to gas. This is something we call sublimation, right? Sublimation. And what is a compound that is known for doing that under normal circumstances?

Carbon dioxide. So, carbon dioxide usually for, we call it dry ice. We call it dry ice because the solid directly goes to the vapor phase. It doesn't even, it doesn't, it doesn't liquefy. It doesn't get wet. So, these are all the different kinds of phase transitions that could occur.

You notice all of them are practically reversible. So, for sublimation, where you go from solid to gas, you have the opposite called deposition. And if you think about it, the name deposition makes sense. All the gas molecules are getting deposited on a surface, right? And they kind of solidify. Melting and freezing, you guys already know this. Vaporization and condensation also, assuming you guys already know this.

So, and basically the statement up here, when it says that both solids and liquids have non-zero vapor pressure, this is more of an indication that sublimation can actually occur. Right? Telling you that the solid phase also can have a vapor pressure.

So, this is a very close, a problem very close to one of your problems in the syllabus. The one that we did yesterday. So, it tells you that this guy in Wyoming decides to hang his wet blue jeans out on a line in a dry winter day at temperatures zero degrees Fahrenheit.

So, that's way below the freezing point of water, right? I think. I don't know Fahrenheit very well, but I'm assuming this is lower than the freezing point of water. The jeans first freeze, but then they dry out. So, how can we explain this? I'm just looking for one word.

Sublimation, right? What happens is the temperature is too low for those wet clothes, or I don't want to say wet clothes. Let me rephrase that. So, the wet clothes, you put them out at zero degrees Fahrenheit. They're going to freeze, right? All the water that was in the liquid phase is going to turn into a solid phase. The problem is the temperature is too low for the solid or for the ice to go into melting, to go into a liquid, and then go to a vapor.

What happens is it directly goes into a vapor. Now, honestly, they won't be comfortable to wear. They're going to be very stiff and rigid. You guys can give it a shot if you want. It's a very slow process. You can't ensure that all of it would dry out. It would just be like wearing cardboards for pants, basically.

So, the same happens with the freeze-dried food. Also, what happens is water is taken off at low temperatures so that the food doesn't get cooked up until you want to actually eat it.

Right? But the texture is quite different. Same with the pants. Now, relative humidity. So, let's see.

If you're on a summer day and the relative humidity is rather high, evaporative cooling is slowed down. Why? Because it's all about equilibrium. If you have a lot of vapor in the air, chances are liquid is not going to go

into the vapor phase. The vapor phase is kind of saturated, especially when you have above 100% humidity. So, you don't cool off. Why? Because it's not favorable. When you have dry weather, evaporative cooling is more likely to occur. Why? Because you don't have a lot of water in the vapor phase. There is no humidity.

So, I guess this is just a fun fact. Why you feel hotter when there's higher humidity? Well, because you don't sweat as much. So, even when relative humidity is greater than 100%, it could almost rain. Because you want to think of that vapor and the atmosphere is just dying to turn into a liquid.

So, you get immediate rain. Now, we went over the fun facts. I'm going to go into more math here. And this is a very important equation. I suggest you guys know it very well. I can tell you

this. I'm going to give you a sneak preview. Something like this is going to come on the exam. So, basically, I don't know if you guys have heard of the Gibbs free energy, but the Gibbs free energy is basically a function, a very simple function of enthalpy and entropy. So, it's not only a function of energy, it's also a function of disorder. Right?

So, and this Gibbs free energy basically, this function kind of determines whether or not the process is going to be spontaneous. How? It's called the G by the way. Delta G is negative. Spontaneous. Positive. Not spontaneous. Negative means you're going from high energy to low energy. Positive. Low energy to high energy.

So, what happens when delta G is equal to zero? Equilibrium. Both processes are favorable. Right? So, let's say you have certain equilibrium between a liquid and a vapor. Right?

You can use this equation to be able to determine, for example, which is a function of the vapor pressure and the temperature, to be able to determine your heat or your energy of vaporization. And you'll see why this is important. So, before I go further with the slides, if you look at

this equation right here, you notice it looks like an equation of a straight line with a negative slope. Right? Your Y is your natural log of P. Your X is one over T. C is your intercept. Negative delta H over R is your slope. From simple math, you know, if you're given a certain number or a certain table of values for Y and X, you can determine the slope.

R is a constant. You can determine your delta H. Right? So, let's just see how this can be done. So, to make sure that what you're writing out is correct or to make sure that

you're getting the right result, you should always see a straight line with a negative slope, hopefully. Yeah. So, as I said before, you let your Y equal to the natural log of P. Your X is one over T. Your slope is negative delta H over R.

And finally, your intercept is C. Right? So, you plot Y versus X. You get a certain slope, which is negative delta H over R.

Now, okay, this is not supposed to be hard for you guys, but now we want to understand where this Clausius-Clapeyron equation came from. So, basically the most important assumption you use for this equation to satisfy is treating your substance or your fluid as an ideal gas.

So, when you say ideal gas, what does that mean? Two things. Volume of your particles is negligible compared to the volume of the container. Right? And another thing you should take into account is that you assume that the vapor expands much faster or

expands at a higher rate than the liquid, and therefore will occupy a much larger volume than the liquid. So, you consider the liquid volume almost negligible compared to it. And you consider that the molar enthalpy of vaporization is independent of temperature. Otherwise, delta H is no longer a constant. You can't include it in your slope.

So, this should be a good hint. It tells you. This is an exam problem that can come up. Right? It gives you two pressures, two temperatures, right? So, two points on that

straight line, and he can ask you to determine the delta heat of vaporization. Actually, you just want to think of it as a plug-and-chug kind of question, right? He gives you a set of values or a set of variables and asks you to solve for one of the constants. Hell, he might ask you guys to solve for the rate constant, R. Sorry, for the gas constant, R. Right?

It will give you a certain rate of vaporization, two pressures at two different temperatures, solve for it. Now, so this is a practice problem. Let's say a compound is unstable at its normal boiling point. Right?

If it's unstable, that means it's reactive. So, your compound is not what it is. You want to purify it by distilling the material at reduced pressures. So, let's say we want to purify H2O2. The normal boiling point for H2O2 is 150.2 degrees Celsius.

But this is a very high temperature, and if you guys don't know, H2O2 is, let's just say a very scary compound. So, it has a lot of explosive risk at very high temperatures.

So, we want to try to purify H2O2 by letting it boil at a lower temperature. How do we do that? Let's try to think about it logically. Do we need a higher pressure or a lower pressure? Do we want to condense it more or do we want to vaporize it more?

Well, we'll just see in a sec. So, we want to determine what pressure should we distill H2O2 if we want to operate at 30 degrees Celsius. So, we want to boil this at a very low temperature. And they give you the delta heat of vaporization. And it's in calories per grams. Now, I'm assuming since he told you guys you're not responsible for numbers that much, he's probably going to give you the conversion.

So, you're going to know how to convert from calories to joules. And I'm assuming you guys already know how to convert from grams to moles if you're given the molar mass. So, this is what we do. A, convert your delta heat into joules.

And since you have the molar mass and you have the grams, you can determine the moles. Then you plug that into your CC equation. Now, notice how he put it. This is how he wants you to answer your questions on the exam. He doesn't want you to write out the equation like so and then rearrange it for your value while keeping the other ones as non-numeric terms.

No, he just wants you to directly set up your problem directly on the paper and directly plug in your numbers just as he did here. It saves a lot of time.

So, notice the pressure that we have to deal with. So, you notice that your pressure is almost 0.01 atm. Very low compared to what it usually is at high temperatures. And that makes sense because if you're exerting a very small pressure, your substance can boil at a very low temperature.

So, I'm just going to leave this here for a sec so you guys take note of it. So, never solve the equation symbolically when you can directly substitute your numbers.

Okay. So, by the way, before I continue, does anybody have any questions so far? We can go back. As far as you guys want, we can go back.

Again, I said it first. I'm going to say it again. If you guys want me to slow down, you guys want me to stay on one slide so you can take enough notes, just let me know, okay? We have a lot of time. I can slow down. Is there any slide you guys wish to go back to and kind of go over again?

All of them? Really? Are we going that fast? Okay. I'm sorry about that. Let's see.

Ah, no. Okay. So, yes. Don't call me professor, please. Just call me Mohammed. Okay.

That's very true. Okay. Okay, I'm kind of feeling bad now. I'm going too fast.

Alright. So, guys, really, I want you guys to try to tell me what- let me go through the whole- it's almost done. But anyway, we have a lot of time and what I can do is I can go back, we can kind of like go over certain things that I went over too fast, okay?

So, think of it like two lectures in a row. How does that sound like? Terrible, right? It's terrible. Okay, anyway. So, now these are a couple of notes that- so this is something I said in the very beginning of the lecture, right?

I said that when you need to- when you want to solve out a numeric equation or a numeric problem, I told you 273.50 and 273 actually make a difference. So, what he usually wants you to do is don't round your numbers until you get to the very end of the- or the very last step of your calculations, okay?

So, for ex- like- he even said that specifically, like if you can put all the numbers in the calculator that appear on the calculator, you can put all those numbers in, you go ahead and do that, okay? And leave the rounding step to the very, very last one. Now, you guys have to pay attention to this equation here, the Claus

-Clapeyron equation, because we only talked about vaporization, delta H of vaporization, right? Now, let's say we're converting from solid to liquid. Fair enough. Or just solid or something. Okay.

Let's say I want to deal with a different process. Let's say I want to- I want to discuss melting or freezing. I don't use the same delta heat, right? I use a different delta heat. Which one? Fusion or melting? Basically the same, just opposite sides, right?

So, another thing that you guys should probably know, because he might give you a problem where you have a sublimation experiment going on, but instead of giving you the delta heat of sublimation, he gives you the delta heat of fusion and the delta heat of vaporization.

Now, logically speaking, sublimation is kind of like taking a shortcut to a two-step process, right? Because the solid first, or conventionally, should first melt, then after it melts, it should vaporize.

If you're doing that in one step, it's just going to be the sum of the two, right? Makes sense. Okay. Now, if you remember, I talked about the Gibbs free energy. I said that

the Gibbs free energy is basically a measure of how spontaneous a reaction can be. And as you can see, it's a function of delta H, which is your enthalpy, and a function of delta S, which is your entropy. Now, we said at equilibrium, delta G is equal to zero. Why? Because

both processes have the same relative disorder or have the same relative enthalpy. So, when do you have equilibrium in terms of phases? Well, when you have a phase transition. When sublimation is occurring, when melting is occurring, when fusion is occurring, vaporization or condensation, right?

So, at these specific phase transition points, and let's generalize more, in the case where you have an equilibrium, what you can do is you can actually solve, you can set up the equation as such, where you can find the temperature at equilibrium, you can find the delta heat at equilibrium, or the delta S at equilibrium, right?

So, really simple- really simple concept, and it just goes on or just comes from the fact that delta G is equal to zero, right? Yes. The numbers.

So, as I told you- so, can I go back to the previous slide, or somebody slipped my throat? Oh, thank you. Okay. Let's see. Which slide? You mean the one with the problem? Here?

Okay, so remember, he said that in his first lecture, I think, or the second one, he said that you guys don't have to know the numbers. Numbers will be given. Now, they gave you here- what's it called? They told you normal boiling

point. What is a normal boiling point? When does a normal boiling point occur? What's the pressure? One atmosphere, right? It's just a standard pressure. So, you guys should know this. It makes sense. He gave you a temperature. What did he do with that temperature? He converted it to Kelvin, right? He said always convert our temperatures to Kelvin. What else? He gave you the other temperature that you need to find, right? In this case, it's 30 degrees Celsius.

So, these numbers are given. He even gave you delta H. He's probably, most likely, going to give you the conversion, okay? So, that's why I always say pay attention to the units. Because even if you have no idea what the question is asking,

even if you have no idea what these numbers mean, you can do some unit analysis and know what you need to get. Like, he can tell you. You have calories. You have grams. You have joules per mole. I want you to give me kilojoules. How do I do that? Here's a set of conversions. You look at it. You reason it out. You see

what's supposed to cancel with what, and you should be able to get your answer. Yes? Okay. So, just to be on the safe side, any time you're dealing with temperature, always convert to Kelvin, unless otherwise stated.

Pressure? See, this is the thing with pressure. He's, on the exam, he's going to let you know. But if he doesn't tell you, like, I want this in these units or this in these units, you don't need to worry about it.

But I know him. He's very efficient. So, he would probably tell you in what units he wants you to get. Like, how many milliliters of this? How many kilojoules per mole will you get? You know? So, he'll give you the units that you need to get, eventually. If he doesn't, this is on me, on the video camera, so, on my responsibility, if you're not required or he doesn't ask or it's not

obvious that you need to express your answers in a certain unit, you express it in any other unit you want. Okay? But Kelvin, always. Temperature. Always in Kelvin. Right? Yeah. So, anyway, to go back to your question, so, all this given over

here, along with the conversion that they did up here, for the calories to joules, that's all you really needed to get that answer. Now, I'm assuming we have done enough exercises for you guys to know the value of the gas constant, even though I'm pretty sure it's probably going to be given on the exam. Yes?

Let me let you in on a little secret. Okay. It probably doesn't really matter. So, let's see. He says the normal boiling point is 150.2 degrees Celsius, right?

And it has a pressure of one atmosphere, so that's your P1 and your T1. For example, 30 degrees Celsius tells you you have- okay, this is your T2, right? And you want to find your P2. So, what is happening to the temperature? The temperature is going down. Right? So, we said that T1 is 150 and T2 is 30. Right?

Why? How do I know this? Because he's telling you, this is the initial temperature, 150, and it's going down to 30. You want to take it down to 30. Okay? So, basically whatever is stated- you should be able to tell from the wording of the problem. I

don't know if you can tell from here, but you see it doesn't matter. Remember, boiling is going to happen. This is the delta H of vaporization, right? Boiling point is going to happen for a

certain pressure and a temperature, right? A point. It's a straight line. Delta H is constant. Right? So, it doesn't matter which one is your P1, which one is your T- which one is your P2, as long as they're in this form. Right? So, you can't, like, for example, put your P1 as one atmosphere and your T1 as 30. No. Okay? But other than that, it doesn't really matter. Yes?

Oh, that's probably a typo because it's- he says here it's 30 degrees Celsius. Okay? Okay? Don't worry, I will go over the slides. Yes?

Delta S is your entropy. Okay? Basically a measure of disorder. No hands? Okay. Delta G? Delta G is your Gibbs free energy. Okay?

Now, question. Anybody- well, okay first. Delta what and what? Melting- you know what melting is, right? Fusion- you don't?

Oh, okay. Melting is when solid goes to a liquid. Fusion is going to be? Fusing together? Combining together? Just the opposite? So, liquid to solid, right? Now, here's- I think here there's another typo. So, boil and vaporize are basically the same thing, right?

So, I would go with boil and condense- and condensation or vaporization and condensation. Okay? So, one of these two pairs down here has to be different. It has to be condensation. Okay?

But, I mean, if you guys understand the concept, that's all that's important. Okay? And don't worry, again, as I said, I'm going to go back to previous slides. I'm going to make sure you guys get some good notes. Don't worry about it. Now, Trouton's law or Trouton's rule- and this usually applies to liquids who tend to- tend to usually be like- tend to act a little ideally, right?

For example, water doesn't apply here because you have a lot of hydrogen bonding. Let's put it this way. Trouton's rule applies to molecules or applies to compounds that are- technically don't do hydrogen bonding. Because they're molecules that don't relatively interact with each other very much.

Remember, that's one of the things about ideal gases or ideal fluids. Molecules don't interact with each other. So, a liquid and a gas are both basically disordered phases, right? The solid is the only ordered phase.

If you put a water- if I hit a water- a glass of water on the table, it's going to get irritated. If I do that for a solid or for ice, the ice is just going to stay the same. It's not going to budge, right? So, they're both disordered. The only difference is one is condensed, the other one is non-condensed, right?

So, the entropy of vaporization is basically the same for a lot of molecules that don't have a lot of interaction between each other. Because basically- remember, heat of vaporization is going from a liquid to a solid- from a liquid to a vapor.

So, when we're saying that they're both disordered, right? So, the only thing that's really changing is it being condensed to it being non-condensed, right? So, and this is usually very similar in the case of all molecules.

So, I guess what I'm trying to get at is that Truton's rule tells you is that the entropy of vaporization for most molecules or most types of compounds is relatively the same. Because the only thing that's changing is how condensed it is and how condensed it's not going to be, right? Which is a very similar phenomenon for most molecules.

And usually the value for that molar entropy is 87.5 joules per mole per kelvin. For most molecules, add that specific pressure and add their normal boiling point, okay? So, really, all you really need to get out of this slide, you need to understand,

is that unless hydrogen bonding is involved- and here, of course, you're doing an approximation. Because they're not going to be exactly 87.5. But unless hydrogen bonding is involved, unless the interaction between these molecules is strong, the entropy is constant, okay?

So, I'll show you how this can apply. Yeah, exactly. So, the heat of vaporization is what's different, not the entropy. So, what we're saying, basically, is since both of them are disordered, the entropy is going to stay the same. Because the entropy is only a measure of the disorder, right?

It's the energy of vaporization that changes between one compound to the other. So, this is a good practice problem to understand the method. So, they're asking you to find the normal boiling point of hydrogen sulfide and of argon, knowing their enthalpies of vaporization are 18.6 and 6.4 kilojoules per mole, right?

Now, remember when we said delta G is equal to zero, which happens for most phase transitions or equilibria, you can relate the temperature to the enthalpy and to the entropy, right? So, like T boil, for example, is equal to delta H boil or over delta S boil.

Here, they're asking us to find T boil. They gave us delta H boil. What do we need? Well, we need the entropy, right? But since we said and we made that assumption, according to Trudeau's rule, that the change in disorder is basically the same,

so you consider delta S for these molecules to be, or delta S of vaporization, so I'm really stressing on this, it's just the delta S of vaporization, you consider it to be a constant. So, for both of them it's 87.5, just plug them into the equation, you make sure your temperature is in kelvin.

You plug it all the way and you should be able to find it. Does that make sense? Yes? I want to hear a unified yes. Okay, thank you. Alright, so, are we all good? No? Oh, one more point.

As you notice, they gave us the true values down here. You see that they're kind of different. So, they're very close, but they're not exactly the same. That's why it's an approximation, okay? They can't all exactly have 87.5. Okay. Now, so this is what I was talking about.

So, as far as you guys need to know, as long as it doesn't do hydrogen bonding, Trudeau's law applies. Why? Because water is a little bit, for example, water does hydrogen bonding.

Any molecule with HF in it is basically also, what's it called? HF molecules also do hydrogen bonding. And anything that has an NH bond also does hydrogen bonding, right? You guys should know this. So, we know that a hydrogen bond is a strong type of intermolecular force.

And so what it does is it makes the liquid a little more ordered than it's supposed to be, right? And therefore, Trudeau's law doesn't apply for them. So, this order is disrupted going to the vapor pressure where there's no hydrogen bonding.

But, with these liquids or with things that actually do hydrogen bonding, the order takes a while to get disrupted. So, notice how large the entropy is of vaporization for water. It's 100 kilo- joules per kelvin per mole as compared to 87.5. It's not very low.

So, it would likewise be higher for other liquids that show hydrogen bonding. Okay? So, this is an easy thing to understand. It shouldn't be too- too terrible or too difficult. So, are we all good on this slide? No. Yes. Can you repeat that?

Okay. Because they're considering it as the molar entropy of vaporization.

This is the entropy per mole, right? If that small m is not there, it's just going to be joules per kelvin. Okay? So, notice there's a difference in the units between- what's it called? Between entropy and enthalpy. Can somebody tell me why?

Why is there a difference between entropy and enthalpy? What's the difference? Hmm? Well, the units can help you determine what is different. And it's something I said in the previous slide- in a previous slide.

So, what's the unit of enthalpy? Joules per mole, right? Or joules. It's a unit of energy. What about entropy? Joules per kelvin. It's not exactly a unit of energy. It's an important thing for you guys to know. That's why we multiply it by the temperature in the delta G equation.

Because delta G, or G, gives free energy. It's an energy value, right? So, when you multiply your entropy- I'm just going to go back. I'm going to come back here. Here. Notice it's delta H minus T, delta S. So, let's do a little unit analysis. This is kind of like a practice for you guys. Delta S is joules per kelvin.

Multiplied by temperature, what does it turn into? Joules, right? So, it's an energy term now. Yes?

How? Of course there's an enthalpy of vaporization. Enthalpy of vaporization is different. How though? How is it different from what?

So, the Gibbs free energy combines two different factors for determining whether a process is going to be spontaneous or not spontaneous. There are two things that you should know about spontaneity, right?

Hey, something becomes more disordered, it's more spontaneous, right? If you put some salt and you put some pepper over it, right over it, you shake it up, you're not going to have a layer of salt and a layer of pepper, right? They're going to mix up together, right? Why? Because it's more spontaneous to have more disorder. And that's what delta S is. It's a measure of disorder. Now, enthalpy, on the other hand, is a measure of energy, right?

Energy is a different term. If something has higher energy, it's less stable, less spontaneous. Okay, if anybody wants to raise their hand, please raise it higher.

No? Okay. So, are we good here? So, this is a good way of demonstrating how hydrogen bonding disrupts the trend.

So, you notice that they have a really high boiling point, right? Molecules that tend to have high hydrogen bonding have a harder time breaking apart when you heat them up, right? Now, this is a question. You guys know, in terms of electronegativity,

fluorine is more electronegative than oxygen is more electronegative than nitrogen, right? Why is H2O at a higher boiling point than hydrogen fluoride?

So, yes, it has more opportunity to do hydrogen bonding, right? It has two hydrogens, two lone pairs, has more chance of doing hydrogen bonding, right? And so, it holds things more tightly. Exactly. And this is something I guess you guys already know as well, but decreasing the molar mass also leads to a decrease in the boiling point.

So, this is an important thing you guys should know as well.

Okay? Now, of course, this decreasing in the molar mass does not apply, does not apply to things that hydrogen bond. We're talking for ordinary molecules, right? Because you can see it kind of shoots up when you start doing hydrogen bonding.

Are we all okay? Thank you. All right. Okay. So, can somebody tell me why methane does not follow the H-bonding trend?

I mean, it's bonded to four hydrogens. I think it might do a little hydrogen bonding. Why? Anybody?

You? Exactly. Exactly. So, carbon is not electronegative enough to polarize the hydrogen enough for it to do hydrogen bonding. So, I guess these are a couple of things you guys should know before you go into the exam.

Condensed phases are far more complex than gases, right? Gases are easy to study, pv equal nRT. Condensed matter, condensed phases, they take, they're much more complicated.

So, you know that according to the phase diagram, two phases can coexist, but only at a certain point, right? So, for a given pressure, you need a given temperature for two phases to exist. And I'm going to go over that again.

Now, we do not yet have a good theory to be able to predict, for example, that water should freeze at 273.15 Kelvin at one atmosphere. How do we do that, or how do we get this? By experiment. Now, the fourth bullet point, it kind of tells you an important thing that you should know about x-ray diffraction.

It's used. Why do we do x-ray diffraction? What it simply does, it determines the structure of your solids, because we said that solids that tend to crystallize, tend to crystallize in certain structures.

And it's important for application purposes to understand why, or not understand why, but understand how the solid crystallizes. It's important to know why iron forms a body centered cubic structure, right? Why sodium chloride has somewhat of a phase centered cubic structure.

Again, last point, and I guess you guys should know this like, palm of your hand. The unit cell is the smallest unit in a crystal, and we said, stand in the bullet point above it. Solid is a regular periodic array, so it's basically a set of these repeating unit cells, right?

So the whole 3D lattice is basically just made up of copies of those, of that specific unit cell. So how is that helpful? Why would I want to know this? Well, because it means that you can isolate a single unit cell, and just knowing how that unit cell looks like, and just knowing all kinds of information and properties about that unit cell,

you can generalize it and determine a more general, get a more general idea about how the crystal looks like, or the properties about these crystals. Okay? So, I'm going to go back to some slides that I feel I need to go over more, all right?

And I'm very democratic, so if you guys choose or there's something in specific you want me to go over, please feel free to tell me. Okay? So, as long as we're all okay on this slide, I'd like to go back to the top.

Is everybody okay with that? No? You're a no or a yes? Yes. Okay.