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Lecture Comments (13)

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


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

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 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

Transcription: Vapor Pressure & Changes of State

Hello, and welcome back to, 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 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 (H2O); 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, H2O liquid to H2O 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 ΔHvap, 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 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 phase1171

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 Pvap--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 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, I have: the logarithm of vapor pressure at one temperature, plus ΔHvap/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, T2, plus ΔH of vaporization over R, times 1/T2 (this is T1) 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/T1 and 1/T2.1767

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?--Pvap at T1; 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 (T2) is equal to -ΔHvap, over R, times 1/T1 minus 1/T2.1823

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/T2 minus 1/T1; 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 H2O 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 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 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 ΔHvap over R, times 1/T1, minus 1/T2.2008

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, ΔHfus.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 know what, I'll just say C of water; how is that?--C of H2O) 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, 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 1.6x106 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 to discuss changes of state.3149

We'll see you next time; goodbye.3153