For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

### Vapor Pressure & Changes of State

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Vapor Pressure and Changes of State
- Intermolecular Forces Overview
- Hydrogen Bonding
- Heat of Vaporization
- Vapor Pressure: Definition and Example
- Vapor Pressures is Mostly a Function of Intermolecular Forces
- Vapor Pressure Increases with Temperature
- Vapor Pressure vs. Temperature: Graph and Equation
- Clausius-Clapeyron Equation
- Example 1
- Heating Curve
- Heat of Fusion
- Example 2

- Intro 0:00
- Vapor Pressure and Changes of State 2:26
- Intermolecular Forces Overview
- Hydrogen Bonding
- Heat of Vaporization
- Vapor Pressure: Definition and Example
- Vapor Pressures is Mostly a Function of Intermolecular Forces
- Vapor Pressure Increases with Temperature
- Vapor Pressure vs. Temperature: Graph and Equation
- Clausius-Clapeyron Equation
- Example 1
- Heating Curve
- Heat of Fusion
- Example 2

### AP Chemistry Online Prep Course

### Transcription: Vapor Pressure & Changes of State

*Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.*0000

*We just finished discussing some areas of quantum mechanics, and today we are going to start talking about vapor pressure and changes of state.*0004

*Before I do that, though, I did want to mention one thing.*0012

*For those of you who have been sort of following along with your books and things like that, you are going to notice that one of the topics that is discussed towards the end of the quantum mechanics section, and in the beginning of the bonding section, in most books--one of the things that they discuss are things like periodic trends.*0017

*For example, what is the periodic trend as you move from left to right of the size of an atom?--or, let's say, the ionization energy?--or the size of an ion?--electronegativity?--things like that.*0035

*I have chosen, in this particular AP Chemistry course, not to necessarily (well, not "not to necessarily"--not to actually) discuss those things.*0047

*This is one of those things that I am actually going to leave to you as a student, and there are a couple of reasons why I am actually doing that.*0055

*One is: there may be a tendency, once you start talking about these things, to actually end up saying too much.*0062

*Too much of an explanation almost tends to make things more difficult.*0068

*Plus, the concepts themselves are not really...in fact, they are very, very basic and very, very simple.*0073

*I'm more concerned with some of the deeper issues of chemistry that come up in the problem-solving that is going to be sort of the greatest challenge that you face, both in your AP Chemistry exam, as well as the subsequent work that you do.*0080

*Things like periodic trends--yes, they are important, and yes, they do come up in questions in the multiple choice section of the exam; but again, they are sufficiently simple, in my estimation, that you can sort of read through the books.*0092

*The sections are very, very short, and they give you very quick, brief explanations; so I didn't want to spend too much time on it; I really wanted to concentrate on what I consider to be the most important thing.*0106

*Now, after we discuss vapor pressure and changes of state (a couple of lessons here), we are going to go on and discuss solutions--things like the colligative properties, vapor pressure of solutions...things like that.*0117

*And after that, I am going to discuss bonding; it's going to be the last topic I discuss before we actually jump into a full AP Chemistry exam.*0129

*Having said that, let's just go ahead and start with our vapor pressure and changes of state today.*0138

*These are very, very important topics--particularly the change of state.*0143

*OK, so let us, really quickly, mention something called intermolecular forces--again, another one of those topics that I am not really going to spend a lot of time talking about formally.*0147

*I'm just going to mention what they are, and if you need to fill in some of the gaps, you are welcome--by all means, do so; I would recommend reading the pertinent section in your books; it should take you no more than about 10 minutes, really--very, very simple--which is why I don't want to waste too much time on it.*0159

*Let me define some intermolecular forces (or, I should say, list some intermolecular forces).*0175

*The intermolecular forces are the forces between molecules--that is exactly what it is.*0185

*It is not intramolecular, but it is intermolecular.*0190

*They are the forces between the molecules--the forces between the individual molecules.*0194

*Now, it's a little bit strange to actually talk about something like intermolecular forces before we discuss something called bonding.*0205

*However, because it is not altogether necessary to discuss what these things are, it is not really a problem.*0212

*We are going to talk about these forces; we can just accept them as is, without worrying about necessarily where they come from, this and that; we are not going to be concerned with that.*0220

*When we discuss bonding, and we discuss something called the dipole moment, then this whole idea of an intermolecular force--everything will just fall into place for you.*0230

*It will be instantaneous--your understanding of what real intermolecular forces are.*0239

*I'm going to list three of the primary intermolecular forces that are the most important: they are hydrogen bonding, something called a London dispersion force, and dipole-dipole.*0245

*Essentially, what it is, is: basically, between two molecules of a substance, there are certain forces that exist.*0274

*They are of an electromagnetic nature; they are positive/negative charged; and the degree of these forces actually causes molecules to stick together with a certain strength, or not stick together.*0283

*That is it--that is really all you need to know, as far as intermolecular forces.*0295

*The greater the force between, let's say, a collection of molecules--well, it is going to be harder to break those molecules apart.*0298

*Your intuition tells you that, if you need to break those molecules (separate those molecules, in other words--just make them--from an aggregate, separate them out), you need to put more energy in.*0306

*So again, there is nothing that is not completely intuitive; it is just a force between molecules.*0317

*OK, now, I will talk about hydrogen bonding in this particular context.*0323

*Hydrogen bonding is the intermolecular force responsible for water being a liquid at room temperature.*0329

*Water is a very, very small molecule (H _{2}O); just based on mass alone, there is no reason in the world why it should be a liquid at room temperature, when every other molecule of that size is a gas at room temperature.*0348

*But, because of the hydrogen bonding--the forces that exist between the molecules--they actually stick together so much that it is actually in a liquid state at room temperature.*0363

*We actually have to put heat into it (we have to heat it up to 100 degrees Celsius) before we actually break the hydrogen bonds between the molecules to make them fly off and become gas particles.*0373

*It's responsible for water being a liquid at room temperature, and why it requires so much heat to boil off.*0386

*100 degrees Celsius is not a joke; anybody who has ever had a steam burn can tell you that; OK.*0410

*I'm just going to draw a network of what, basically, water looks like, as far as the hydrogen bonding.*0415

*We have oxygen, hydrogen, hydrogen; I'm going to put another oxygen; I'm going to put a hydrogen here; I'm going to put a hydrogen here; I'm going to put an oxygen there, put a hydrogen here; I'm going to put a hydrogen here and hydrogen here.*0422

*So, basically, what happens: as it turns out, the oxygen side of the molecule has a little bit of a negative charge; the hydrogen side of the molecule has a little bit of a positive charge.*0437

*Well, positive and negative attract; the distribution of charge is actually so great--it is not a complete positive and negative (I mean, this is a covalent molecule--in other words, it is not ionic--nothing has lost an electron here), but it is still so great that these hydrogen bonds, which are represented as dots, are really quite strong.*0447

*They actually hold the molecules together, which is why, at room temperature, water is a liquid.*0470

*They stick together so tightly; they don't want to let each other go.*0477

*I have to put in a certain amount of heat--I have to raise the temperature to 100 degrees Celsius--before I actually break these bonds and make the water molecules fly off as gas particles.*0481

*That is all that is going on--these hydrogen bonds.*0494

*This is what we mean by "intermolecular forces"; they are forces that exist between the molecules of a substance, and they control things like boiling point.*0497

*The higher the intermolecular forces of a particular substance, the higher the boiling point is going to be.*0505

*And again, there is nothing that is not intuitive about that.*0510

*OK, so breaking these bonds requires energy, as does the breaking of any kind of bond.*0514

*Thus, vaporization is an endothermic process.*0530

*OK, that is important to know: vaporization is an endothermic process.*0541

*Melting is an endothermic process: when you take something from a solid to a liquid, you need to put heat into it to melt it--it's endothermic: it requires heat to make it go forward.*0545

*The reverse: condensation from gas to liquid is an exothermic process.*0554

*Steam has to release heat whenever it actually cools down, and when it releases heat and cools down, it turns into water.*0560

*These forces become stronger than the amount of energy left in the water.*0567

*It is an endothermic process.*0574

*OK, in other words, H _{2}O liquid to H_{2}O gas--the ΔH for this is going to be a positive number.*0577

*Remember, a positive ΔH means endothermic; negative means exothermic (gives off heat).*0590

*OK, so let us define something called the ΔH _{vap}, the ΔH of vaporization--enthalpy of vaporization--heat of vaporization.*0596

*OK, so this is called the heat of vaporization, and it is the heat required (and remember--Joules--heat is energy) to vaporize 1 mole of something (of a liquid) at 1 atmosphere pressure.*0610

*In other words, standard pressure--normal atmospheric pressure is 1 atmosphere.*0642

*So, it's the amount of heat that I need, if I have one mole of a substance--of a liquid--if I want to turn all of that liquid into a gas...how much heat do I have to put into it?*0645

*That is the heat of vaporization.*0655

*It is often expressed in Joules per mole or kilojoules per mole, because by definition, it is for 1 mole of a substance.*0658

*Now, let's define vapor pressure.*0666

*Be very, very careful--do not confuse these two: one of them is a pressure--it is in atmospheres or torr; one of them is a heat--it is in kilojoules per mole.*0668

*OK, the vapor pressure is the pressure of the vapor above a liquid at equilibrium--I will explain what that means right now.*0676

*OK, if I have a flask, and if I all of a sudden inject some liquid into it, well, something is going to happen.*0701

*At a given temperature (25 degrees Celsius, 35 degrees Celsius, whatever it is), a certain fraction of the molecules at the surface actually have enough energy to jump up and vaporize.*0711

*There are going to be some molecules of this liquid that are going to all of a sudden start collecting above the liquid.*0723

*You are going to have this liquid in this flask, and there are going to be some molecules in the vapor state, bouncing around up here.*0731

*Well, some of these molecules up here are actually going to lose some energy, and they are going to come back down, and they are going to turn into liquid again.*0738

*At some point, this dynamic equilibrium--the number of molecules that is jumping out of the liquid state into the gas state and the number of molecules that are jumping--or dropping--from the gas state to the liquid state--is equal.*0748

*The rate at which the vaporization and the condensation happens is equal, and it reaches an equilibrium.*0762

*Well, at that equilibrium, on average, there is going to be a certain number of molecules that are hanging around above the liquid, in the gas phase.*0768

*Well, that is what pressure is: pressure is a certain amount of gas particles, in the gas phase, bouncing against the walls and creating this thing called pressure.*0778

*That is what vapor pressure is: if I set a particular temperature in a flask, if I inject some liquid into it and allow that liquid to just come to equilibrium with the vapor on top of it, I can actually measure the vapor pressure at that temperature.*0789

*And of course, at a higher temperature, the vapor pressure is higher--which we will get to in just a minute.*0804

*That is what vapor pressure is: it is the pressure of the vapor above a liquid at equilibrium.*0809

*Now, we can calculate the vapor pressure (actually, let me write this over here on the side; I want a little bit more room so I can draw) of a liquid with a simple barometer.*0817

*It is actually really, really cool if you do this sometime in your lab--at a given temperature, of course.*0851

*Let me write that in parentheses: at a given temperature.*0860

*So, I am going to have a little barometer here; there is mercury in this little dish, and there is an inverted tube that we fill with mercury; and then we take this tube, and we drop it into this, and then the mercury level drops in the tube.*0867

*So, up in here, there is a vacuum, and there is a certain height at which it just stops.*0884

*The pressure that the (this is mercury in here, and it's mercury in the tube) mercury--the weight of this amount of mercury here is pushing down this way, but the atmospheric pressure is pushing down this way, and the height of this mercury is going to be 760 millimeters.*0891

*That is 760 torr; that is equal to 1 atmosphere.*0911

*Atmospheric pressure: that is what a barometer does--it measures atmospheric pressure.*0914

*Well, now, if I do this same thing--if I have the same apparatus--and if I take this, something very, very interesting happens.*0918

*If I take a syringe full of liquid water with no gas bubbles in it, and if I get the syringe so that I can actually bend the syringe so it's like that (OK, so here is the syringe, and the little stopper), I actually can pump some water into this.*0928

*If I pump some water into this, well, water is less dense than mercury; so water will actually go to the top.*0948

*Once it reaches the top of this, some of the vapor of the water will start to...the water will start to vaporize.*0954

*And, what will end up happening is: you will end up with a little bit of water vapor at a given temperature, like that.*0961

*What you will end up with--there will be a little bit of water here on the surface; it will be in equilibrium with this; but what happens, when you measure this: you are actually going to end up (oh, by the way, I should say: this is going to be at 25 degrees Celsius)...this is 760 with a vacuum; when you put water in there and allow it to come to equilibrium with the vapor, you end up measuring 736 millimeters.*0971

*Well, 760 millimeters, minus 736 millimeters, equals 24 millimeters, which is equal to 24 torricelli.*0998

*This is the vapor pressure of water at 25 degrees Celsius.*1010

*This is how we measure it: the vapor that collects above the mercury column from the water that has been injected into this tube actually pushes down this mercury from 760 down to 750, down to 740, to 736.*1016

*And, as it pushes it down, that is the pressure at that temperature; this is exactly how they measure it.*1030

*That is it; so, that is how we measure vapor pressure; now, let's write a couple of other things.*1038

*Vapor pressure varies widely among liquids, and vapor pressure is mostly a function of the intermolecular forces of the particular substance we are talking about.*1046

*So, vapor pressure is mostly a function of intermolecular forces; in other words, the greater the intermolecular forces in a molecule (in a substance, I should say--a substance)...not the greater...the lower the vapor pressure.*1071

*All that means is that, if I have a certain liquid that has really, really high intermolecular forces, that means that those molecules want to really just stick together.*1130

*So, when you put those molecules at the top of that mercury column, or in a flask that has been evacuated--well, a very small number of those molecules are going to actually have enough energy to jump out of the liquid--to break away from those intermolecular forces and jump into the vapor phase.*1141

*There is not going to be a lot of those vapor molecules hanging around above the liquid, so the vapor pressure is going to be low.*1159

*In other words, there is not going to be a lot of vapor on top of the liquid.*1168

*For a substance that has very weak intermolecular forces, it is going to actually vaporize very, very easily, because a greater percentage of the molecules are going to have the low energy, naturally, to just jump into the vapor phase*1171

*You can think of something like gasoline, or something like diethylene...well, actually, gasoline is probably the one that is in most of your experience.*1185

*Gasoline, at room temperature, will vaporize very, very easily--it has a very low vapor pressure, whereas something like water does not.*1193

*You have to raise it to 100 degrees Celsius before it actually starts to seriously vaporize.*1201

*Yes, at 25 degrees Celsius, you are going to have some water vapor (24 torr); well, 24 torr is really not that much, compared to, say, 760 torr (which is room pressure--atmospheric pressure at sea level).*1207

*OK, let's see: the lower the vapor pressure, because more energy is required to break the bonds...*1223

*OK, now, let's start dealing with some equations here.*1252

*Vapor pressure increases with temperature, which makes sense: as you raise the temperature of a substance, on average, you are raising the number of molecules that have enough energy to jump into the vapor phase--to break their bonds--to break their chains, if you will--to break free from the intermolecular forces.*1259

*That makes sense: at a higher temperature, there is going to be a higher amount of vapor molecules in equilibrium with the liquid phase, which means the vapor pressure is going to be higher at that temperature.*1285

*OK, so the temperature at which a liquid's vapor pressure reaches 760 torr is called the boiling point.*1297

*At this temperature, 100% of the molecules have enough energy to escape into the gas phase.*1331

*That is the whole idea: 100% of the molecules have enough energy to overcome their intermolecular forces.*1345

*The boiling point--that is at which every single molecule, as enough time passes, will go into the gas phase.*1368

*OK, so now, when we plot vapor pressure versus temperature, y versus x, we get a non-linear graph.*1375

*We know that, when we deal with non-linear graphs, more often than not, we are going to probably end up taking a logarithm and trying to see if we can linear-ize it, if it's linear.*1406

*Well, guess what--it is.*1414

*So, let me go ahead and draw the non-linear version first: so we have something like this, where here we have the temperature; here we have the vapor pressure; so let's go ahead and just do water.*1416

*If we start at 0 here, and let's say this is 100 degrees Celsius, and the pressure, let's say, is 1, 2, 3, 4, 5, 6, 7, 8, 760: that is 1 atmosphere--an atmosphere, as it turns out, goes something like that.*1431

*As you raise the temperature higher and higher and higher and higher, the actual vapor pressure rises.*1456

*Well, at some point--as it turns out, at 100 degrees Celsius for water--the vapor pressure reaches 760 torr at 100 degrees Celsius.*1463

*We call that the boiling point; that is the whole idea behind the boiling point--it is when the vapor pressure of a liquid reaches 1 atmosphere pressure or 760 torricelli.*1476

*Well, as it turns out, it turns out that the logarithm of the vapor pressure, plotted against the reciprocal of temperature, is logarithmic--or is a straight line.*1486

*It gives a straight line.*1507

*And how did they find this out?--very simple: they just kept trying it until something worked; it's that simple.*1509

*That is how science works.*1514

*A lot of times, you will be given all of these equations that seem to fall out of the sky.*1516

*It's true, some of them we can derive--but for all practical purposes, when we are faced with a graph like this, literally what they did is: they probably ended up taking the logarithm of the vapor pressure and trying it against this; the reciprocal of the logarithm--they tried it against this; the reciprocal of this against this.*1520

*At some point, they hit upon the logarithm of these values--the y-values--versus 1 over these values; it ended up giving a straight line.*1535

*That is it--just trial and error: that is how you do it.*1543

*Don't think that scientists just sort of look at this stuff and automatically know, "Oh, yes, that makes sense; I'll just do this"; it doesn't work like that.*1551

*They make it look like it works like that, but that is misleading--it doesn't work like that.*1557

*OK, so what we end up with is this equation here: the logarithm of vapor pressure (I'm going to go ahead and just call it V/P, vapor pressure: you can call it P _{vap}--yes, that is fine) equals -ΔH, over R (which is the gas constant), 1/T, plus C.*1563

*What you end up getting, when you actually plot...this is going to be 1 over temperature, and this is going to be the logarithm of vapor pressure; you end up with some straight line, as it turns out--something like that.*1595

*Well, here is what is great: so this is your y-value; this is your x; this is your b (your y-intercept); this is your m.*1611

*As it turns out, your m is the (oops, I think I should probably write the subscript here; so I'm going to erase this so I can make a little more room for myself) ΔH of vaporization--the heat of vaporization.*1624

*1/T, plus C: this equation right here gives you a straight line; so when you plot, take 1 over the temperature and the logarithm of the vapor pressure that you get; the slope of that line happens to be the heat of vaporization, divided by the gas constant.*1642

*This C is specific to the particular substance, so water has a specific C; dichloromethane has a specific C; benzene has a specific C--things like that.*1663

*So, c is characteristic of the given liquid.*1674

*Now, that is that; so let's do a little bit of math here.*1677

*I don't know where I should...you know what, that is fine; I'll go ahead and do it right here.*1689

*I don't want to turn the page just yet.*1694

*All right, note that C is independent of temperature; so, if I solve this equation for C--if I said C equals...well, let me do it...so, I have: the logarithm of vapor pressure at one temperature, plus ΔH _{vap}/R, times 1/T; that equals C.*1697

*All I have done is taken this particular thing and moved it over to that side, so C equals this thing.*1741

*Well, C also equals the logarithm of the vapor pressure at a second temperature, T _{2}, plus ΔH of vaporization over R, times 1/T_{2} (this is T_{1}) equals C.*1748

*Well, C equals both of these things: the vapor pressure at one temperature and the vapor pressure at another temperature--this is the same; 1/T _{1} and 1/T_{2}.*1767

*Well, C is C; it's a constant; so I can set these two equations equal to each other, and then sort of rearrange them, and here is what I end up getting.*1776

*I end up getting: the logarithm of the vapor pressure at temperature 1, minus the logarithm of the vapor pressure at temperature 2...*1785

*You know what, I'm not going to use this; I'm sorry--this V/P--I think it's going to end up confusing you guys; I'm just going to use one variable here.*1799

*So, here I have been saying vapor pressure, vapor pressure; I'm going to use this symbol, OK?--P _{vap} at T_{1}; I think that is a little bit better.*1807

*Otherwise, you are going to think it's V/P, and it's not volume over pressure; it's not that.*1818

*...minus the logarithm of the vapor pressure at the second temperature (T _{2}) is equal to -ΔH_{vap}, over R, times 1/T_{1} minus 1/T_{2}.*1823

*This is our equation of interest, and let's see what this equation tells us.*1844

*It gives us a relationship between the vapor pressure at one temperature and the vapor pressure at a second temperature, and so here are the two T's, and they are related through the gas constant, and they are related through their ΔH of vaporization.*1851

*So, if I have my ΔH of vaporization, and if I have one temperature...again, this is just an equation with a bunch of different variables.*1872

*I might have 2 of them; 3 of them; I can find the third or fourth, depending on what is necessary.*1879

*OK, you are going to see this equation sort of written in several different forms.*1884

*Notice, I have a logarithm minus a logarithm; I can write it as the logarithm of this divided by this.*1891

*I can actually take this negative sign, distribute over this, and write 1/T _{2} minus 1/T_{1}; this is my personal preference--I like it this way.*1897

*I don't like equations to be too heavily manipulated; otherwise, you sort of don't see what is going on.*1905

*But that is just me personally; but this is just how I like to write it.*1911

*This is called the Clausius-Clapeyron equation--not that the name is altogether that important, but I suppose it's nice to give credit where credit is due.*1916

*And again, you will see it in different guises.*1931

*Let's go ahead and do an example.*1933

*Example: OK, the vapor pressure of H _{2}O at 25 degrees Celsius is 23.8 torricelli, and the heat of vaporization of water is equal to 40.7 kilojoules per mole.*1936

*In other words, if I have 1 mole of liquid water--18 grams--I have to hit it with 40.7 kilojoules, 40,700 Joules, to vaporize that water; that is a lot of energy.*1967

*That is a very, very high heat of vaporization.*1980

*OK, the question is: What is the vapor pressure at 60 degrees Celsius?*1983

*OK, so I have the heat of vaporization at one temperature--I have the 23.8 torr; I have the heat of vaporization, so I can use the Clausius-Clapeyron equation to go ahead and calculate the vapor pressure at another temperature.*1995

*That is it: so let's go ahead and do it.*2006

*Let us rearrange our...so I'm going to write it as: The logarithm of the pressure--the vapor pressure--at 60 degrees Celsius is equal to the logarithm of (you know, should I put...you know what, let me put the variables in, and then I'll put the numbers in afterward--not a problem) the vapor pressure at 25 degrees Celsius, plus ΔH _{vap} over R, times 1/T_{1}, minus 1/T_{2}.*2008

*And now, let's substitute some values in here: so we get--this is equal to: the logarithm of 23.8 torricelli, plus 40,700 Joules (OK, so it gave it to you in kilojoules per mole, but because R is in Joules per mole-Kelvin, I have to change the unit to Joules--they have to match, so again, you always want to be careful with the units), 8.3145 Joules per mole-Kelvin, times 1/298, minus 1/333.*2055

*And when we solve this, we end up getting: the vapor pressure at 60 degrees Celsius equals 133.8 torricelli.*2100

*And again, this is a logarithmic thing here; so when you actually take the logarithm number, you are going to get this; you are going to have to exponentiate; so don't forget that.*2115

*OK, this is a logarithmic equation; you have to exponentiate to get your final answer.*2122

*There you go: 133.8 torr--given one vapor pressure and the enthalpy of the vaporization, we can calculate another vapor pressure.*2127

*OK, so now, I want to talk about something called a heating curve, which is profoundly important.*2140

*A heating curve: it is a plot of temperature versus time, as energy is added to a system at a constant rate.*2151

*Generally, what we do is: we start with a solid, and then we start raising the temperature--we just add energy; we add...*2183

*I'm sorry--we don't raise the temperature; we add energy.*2189

*We add energy; we add energy; we measure the temperature--that is what we do.*2191

*So, I'm going to do the heating curve for water: it's profoundly, profoundly important--not very intuitive--I promise you, when you see what this thing looks like, it's not going to be very intuitive at all.*2197

*You are probably thinking, if you just keep adding energy, well, the temperature is just going to keep going up.*2209

*It is true--the temperature does keep going up--but there are two points where it actually plateaus, and it stays constant--very, very counterintuitive.*2213

*Let's mark this one as -20 degrees Celsius; let's put 0 over here; and let's do 2, 4...let's do 20, 40, 60, 80...this will be 100 degrees Celsius.*2220

*And then, this, of course, will be...time is 0; so this is time, and this is temperature.*2236

*So again, what we have here is time; as we go, in certain increments of time, we are adding energy to the system; we are adding energy to the system.*2245

*We are going to start at -20 degrees Celsius, so we are going to start with a solid block of ice.*2254

*As we add energy to this system, well, the temperature of that block of ice starts to rise.*2259

*All of a sudden, when it hits 0 degrees Celsius, it gets flat--interesting.*2265

*And then, once all of that ice is melted and turns to water, then the temperature starts rising again.*2272

*It starts rising again, rising again, rising again, rising again; and all of a sudden, when it hits 100 degrees Celsius, it levels off; I keep adding energy, but the temperature doesn't change.*2279

*And then, at some point, when all of the water has been converted to gas, it starts rising again.*2289

*This is called a heating curve; it follows the path from one phase (solid) to another phase (liquid) to another phase (gas) as you keep adding energy, and depending on that particular temperature.*2296

*So, here we have ice; here, we have water; and here, we have steam.*2309

*Now, let's talk again about what it is that is exactly happening here.*2320

*We start off with a solid block of ice; we start adding energy to it--we heat it up, basically--and we have a thermometer attached to it--a digital thermometer (or whatever--it doesn't need to be digital); we are measuring its temperature.*2324

*Well, the temperature keeps rising from -20 degrees, all the way to 0; but something happens.*2339

*Once it hits 0 temperature, we keep adding energy--we are adding energy, but we notice that the temperature of the block of ice is not actually changing--that is very, very odd.*2344

*Well, the reason it is not changing is because, at that temperature, all of the energy that is going into that block of ice is being used to break up the bonds that are holding that ice together, to convert everything to a liquid.*2354

*Up until that point, nothing is really melting; it is actually just absorbing heat, absorbing heat, absorbing heat, according to a certain heat capacity.*2369

*It is absorbing heat, and then at 0--at that point, all of the excess heat doesn't go towards a temperature rise; it goes towards converting all of the solid to liquid.*2379

*Once everything is converted to liquid, then it starts absorbing heat and rising in temperature, rising in temperature.*2389

*It is still a liquid, still a liquid, still a liquid; all of a sudden, when it gets to 100 degrees Celsius, it stops again.*2395

*At that point, all of the excess energy that I am adding to it is not making the temperature rise; it is actually using all of that energy to break all of the intermolecular bonds to send off all of the molecules, every single molecule--give it enough energy to completely go into the gas phase.*2401

*At this point is when we are completely in the gas phase; now, once again, we add more energy, more energy; the temperature of the steam keeps rising.*2419

*Steam doesn't stay at 100 degrees Celsius; you could actually keep increasing the temperature of the steam--you can put energy into steam; it will get hotter and hotter and hotter and hotter.*2429

*That is what is going on; so this is a heating curve.*2441

*Now, the way I have drawn it, it looks as though the slopes of these things are actually the same; they are not, as it turns out.*2445

*These have different slopes, because ice, liquid water, and steam have different heat capacities.*2451

*And remember what heat capacity is: it is the amount of heat necessary to raise the temperature of a gram of that substance 1 degree Celsius, or the molar heat capacity--one mole of that substance by 1 degree Celsius.*2459

*OK, so that is it: when you see something like this (which you will see), this is a heating curve: it describes the behavior at different temperatures, and the temperatures at which it stabilizes are precisely the melting point and the boiling point.*2473

*OK, so now, let's define something else.*2491

*Earlier, we defined the heat of vaporization; now, we are going to define something called the heat of fusion, ΔH _{fus}.*2494

*It is the heat required to melt one mole of a liquid at 1 atmosphere pressure.*2503

*I'll just put "1 atmosphere"; not a problem--OK.*2526

*So, again, the greater the intermolecular forces, the harder it is to actually separate those molecules--so the greater the heat of fusion.*2528

*Here, at this point, we are talking about the ΔH of fusion.*2539

*OK, all of the excess energy that I am putting into this--the temperature is not rising; I actually need to put in a certain amount of energy (which is different for each substance) to melt the solid to a liquid.*2546

*Once everything is melted, then the temperature starts rising again.*2560

*At this point (well, actually, the whole thing), this is the ΔH of vaporization.*2564

*The amount of energy that I am putting into it equals the heat of vaporization; that is how much heat per mole that I need to put in to convert all of the liquid completely to gas.*2572

*Once it is converted to gas, at that point, then the temperature will start rising again.*2584

*This is very counterintuitive--you don't expect this to happen.*2589

*You think that, if you just keep adding heat to something, the temperature is going to rise.*2592

*It doesn't; as a solid, it rises, but the conversion from solid to liquid--it stabilizes, and it stays constant at 0.*2595

*And then it starts rising again, as a liquid; once it makes the transition from liquid to gas, it needs to make the transition fully; there needs to be no liquid present.*2605

*Any liquid that is present--that temperature will not rise; I promise you, it will not rise.*2615

*Only when it is completely turned into gas--then any excess energy will actually push the temperature higher.*2620

*So, OK, now let's do an example.*2626

*Example: How much energy (this is going to be a bit of a marathon problem) does it take to convert 0.50 kilograms of ice at -25 degrees Celsius (so we're starting with a block of ice at -25 degrees Celsius) to steam at 220 degrees Celsius?*2632

*We are going to take a .5-kilogram block of ice; we are going to keep adding heat to it; we want to know how much energy will convert all of that, at -25 degrees Celsius to start, all the way to steam that is carrying a temperature of 220 degrees Celsius.*2672

*How much energy do I have to put into it?*2687

*We have to do this in 5 steps.*2689

*We have to account for this step--the rise in temperature of the ice; the melting of the ice; the rise in temperature of the liquid; the vaporization of the liquid; the rise in temperature of the vapor.*2692

*We have 5 calculations to make--so let's get going.*2704

*OK, now, we have some information here: we have the heat capacity of ice--it is given to us: that is equal to 2.1 Joules per gram per degree Celsius.*2710

*We have the heat capacity of liquid water (I'll just say C of liquid...you know what, I'll just say C of water; how is that?--C of H _{2}O) is equal to 4.18 Joules per gram per degree Celsius--you know that from our work in thermochemistry.*2723

*And, as it turns out, the heat capacity of steam is 1.8 Joules per gram per degree Celsius.*2739

*The heat of vaporization of water is equal to 40.7 kilojoules per mole, and our last bit of information: the heat of fusion of water is equal to 6.0 kilojoules per mole.*2752

*This was actually kind of a surprise to me: 6 kilojoules doesn't seem very much, but 6,000 Joules is still kind of a lot.*2767

*OK, so let's quickly draw our picture again, so we have something to reference.*2773

*I'll actually make it kind of small, just so we have a pictorial; so I'm going to draw my picture over here.*2779

*So again, we are looking at something like this: 1, 2, 3, 4, 5: we have 5 calculations to make.*2785

*Let's start: from -25 degrees Celsius, all the way to 0 degrees Celsius, we are going to use Q=mcΔT.*2793

*We are putting heat into something; it is going to equal...that heat capacity equation, Q=mcΔT...the mass, times the heat capacity, times the change in temperature.*2803

*OK, because they are asking for how much energy: so Q=mcΔT.*2814

*Well, that is equal to...the mass is 500 grams (right? .5 kilograms), times...well, this is ice, so it is 2.1 Joules per gram per degree Celsius; times the change in temperature--the change in temperature: -25 degrees to 0--this is 25 degrees Celsius.*2821

*When we calculate this, we get 26,250 Joules; that is our first calculation.*2840

*It takes 26,250 Joules to convert .5 kilograms of ice...actually, to raise the temperature to 0 degrees Celsius.*2848

*We haven't actually converted it yet.*2859

*Now, at 0 degrees Celsius (0 degrees--ice), we want to convert that to water at 0 degrees; so here is where we are actually going to convert it--we are going to melt it.*2861

*This is where we do...well, we have 500 grams, times 1 moles equals 18 grams (right?--1 mole of water is 18 grams), and we have: the heat of fusion of water is 6 kilojoules per mole, so 6,000 Joules per mole.*2875

*Gram, mole, mole, gram, gram; when I do that, I get 166,667 Joules; that is the second part.*2896

*This is the first part; this is the second part.*2907

*So, it takes 26,250 Joules to raise the temperature of ice from -25 to 0.*2910

*At 0, it takes an additional 166,667 Joules to melt that ice, to convert it into liquid water.*2916

*Now, we need to raise the temperature of water from 0 to 100.*2924

*So, that calculation looks like this: so we have 0 degrees Celsius to 100 degrees Celsius; again, we go back to the Q=mcΔT, except now, we are going to use the heat capacity 4.18, because now, we are talking about a liquid.*2929

*It equals, again, 500 grams of water (that hasn't changed); 4.18 Joules per gram per degree Celsius; and this time, our temperature change--0 to 100--is 100 degrees Celsius.*2946

*We end up with 209,000 Joules.*2960

*Now, it takes 209,000 Joules to raise the temperature of water from 0 degrees Celsius to 100 degrees Celsius; that was the third phase.*2965

*Now, we need to add enough heat to actually change all of that liquid water to gas; this is going to be the fourth phase.*2975

*At 100 degrees Celsius, we have water, and we need to convert that to steam at 100 degrees Celsius.*2981

*Notice: the temperature stays the same until everything is converted.*2992

*Once again, we have 500 grams; well, 1 mole happens to be 18 grams, so that is the mole conversion; and the heat of vaporization is 40,700 Joules per mole.*2996

*That means 1 million, 130 thousand, 555 Joules are necessary to actually change 500 grams of water to 500 grams of steam; that is a lot of energy.*3012

*I'm telling you--hydrogen bonding: profoundly strong.*3027

*Now, we need to take steam from 100 degrees Celsius all the way to 220 degrees Celsius.*3031

*We go back to Q=mcΔT.*3038

*We have: well, mass--again, we still have 500 grams; this time it is of steam; it's 2...no, 1.8; I'm sorry; the heat capacity of steam is 1.8 Joule per gram per degree Celsius.*3042

*And this time, we are raising it 120 degrees (right?--100 to 220 is 120); and we end up with 108,000 Joules.*3058

*Now, we have to add all of this together: the 108, the 1 million, the 209, the 167...the everything else.*3072

*When we add it up, we get 1,640,284 Joules, or 1.6x10 ^{6} Joules (mega-Joules!).*3079

*That is our final answer.*3096

*So, the moral of the story is: If you are changing phases, you have to account for the phases themselves.*3098

*If you are going from solid to liquid, you have to take care of the rise in temperature of the solid; you have to take care of the change of the solid to liquid, and then the rise in temperature of the liquid.*3106

*Or, if you are going the other way around--if you are taking liquid to solid--the drop in temperature of the liquid; the conversion of the liquid to solid; and the drop in temperature of the solid.*3117

*You have to consider each phase.*3128

*Think about the heating curve: every segment of the heating curve--that is what you have to account for.*3130

*On the places where temperature rises, you use Q=mcΔT, heat capacity of the particular phase; and on those places where the temperature is constant, you use the heat of vaporization or the heat of fusion.*3135

*I hope that helps.*3148

*OK, thank you for joining us here at Educator.com to discuss changes of state.*3149

*We'll see you next time; goodbye.*3153

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 1:06 AM

Post by Jason Smith on December 12, 2015

Hi professor! Hope you're having a good day.

Is the following reasoning correct?

The reason why sodium chloride dissolves in water is because the ion-dipole forces between H2O and the Na cation/Cl anion are stronger than the hydrogen bonds between the H and O atoms?

Am I conceptualizing this correctly?

Thank you in advance!

1 answer

Last reply by: Professor Hovasapian

Tue Mar 17, 2015 4:22 PM

Post by Jason Smith on March 14, 2015

I have a theoretical question that I would love your opinion on professor: if you continue to heat a substance once it has reached a gaseous phase, what will happen? Is there such a thing as "infinitely hot"? Will the gas particles themselves "disappear"? Or will they simply move more quickly? Is there ever a point where the gas particles get so hot that they move infinitely quick? Hope these questions make sense.

1 answer

Last reply by: Professor Hovasapian

Tue Mar 17, 2015 4:10 PM

Post by Jason Smith on March 14, 2015

For the flask example at the 13:00 minute mark or so (when you talked about vapor pressure), is the top of the flask closed off to the surrounding environment (have a top on it to prevent vapor from escaping). Also, vapor will vapor pressure always remain constant assuming that the temperature remains the same? Thank you professor.

1 answer

Last reply by: Professor Hovasapian

Tue Jul 23, 2013 5:20 AM

Post by KyungYeop Kim on July 20, 2013

I have two questions on solutions.. In a simple equation MgCl2 >>> Mg + 2Cl, I seem to confuse about what 2Cl means. I do know it means that 1 mole MgCl2 produces 2 mole Cl, but what about concentration? If I'm asked to calculate concentration of Cl-, then would I be correct in finding out the concentration of MgCl2 and dividing it by 2?

Sometimes Cl exists in gas form. Does it hold true in above case? also, if there's Cl2 instead of 2Cl, does it make any difference in calculating concentration and etc.?

Solution is not a very intuitive concept for me.. Thank you in advance.

1 answer

Last reply by: Professor Hovasapian

Sun Jun 2, 2013 2:55 PM

Post by KyungYeop Kim on May 31, 2013

A Hydrogen atom is said to be partially charged, thus being attracted to other differently charged atoms of molecules. (For instance, the hydrogen bonds in between H2O molecules) But what's special about hydrogen bonding? Aren't all partially charged atoms attracted to the oppositely charged atoms? Or is it something that only happens to hydrogen?

0 answers

Post by Professor Hovasapian on April 20, 2013

Hi Antie,

R is the Ideal Gas Constant. For Gases, in units of L*atm/mol*K the value is 0.08206. In units of J/mol*K it is 8.31. Sice we are discussing energy, we use the 8.31value.

For an element or compound in a given state (solid, Liquid or gas), this equation expresses how much energy is gained or lost by that compound upon increase or decrease of Temperature, respectively. The origin is based on Heat Capacity: the amount of energy necessary to raise the temp of a compound 1degree Celsius (or Kelvin).

Now, heat of fusion is a simple conversion. The heat of Fusion is the amount of energy a compound gains or loses as it transitions from Solid to Liquid and Liquid to solid, respectively.

So, if something has a Heat of fusion of 10 kJ/mol, and there are 3 mols of it, then it requires 30 kJ of energy to melt it...or it releases 30 kJ into the surroundings when it freezes.

Let me know if this does not make sense. And no worries -- please feel free to ask as many questions as you would like and need. I'm happy to help in any way I can.

Take good care

Raffi

1 answer

Last reply by: Professor Hovasapian

Sun Apr 21, 2013 3:01 AM

Post by Antie Chen on April 19, 2013

In Clausius-Clapeyron Equation, what's the R represent? And how I know it is 8.3145 J/mol*K in Example 1?

I am also confused the origin of the equation q=m*c*delta T and the equation to calculate heat of fusion in Example 2?

Sorry for so much question :P