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

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

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## Table of Contents

## Transcription

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### Entropy

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
- Entropy, Part 1
- Coefficient of Thermal Expansion (Isobaric)
- Coefficient of Compressibility (Isothermal)
- Relative Increase & Relative Decrease
- More on α
- More on κ
- Entropy, Part 2
- Definition of Entropy
- Differential Change in Entropy & the Reversible Path
- State Property of the System
- Entropy Changes Under Isothermal Conditions
- Recall: Heating Curve
- Some Phase Changes Take Place Under Constant Pressure
- Example I: Finding ∆S for a Phase Change

- Intro 0:00
- Entropy, Part 1 0:16
- Coefficient of Thermal Expansion (Isobaric)
- Coefficient of Compressibility (Isothermal)
- Relative Increase & Relative Decrease
- More on α
- More on κ
- Entropy, Part 2 11:04
- Definition of Entropy
- Differential Change in Entropy & the Reversible Path
- State Property of the System
- Entropy Changes Under Isothermal Conditions
- Recall: Heating Curve
- Some Phase Changes Take Place Under Constant Pressure
- Example I: Finding ∆S for a Phase Change 46:05

### Physical Chemistry Online Course

### Transcription: Entropy

*Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.*0000

*We have been talking about energy and we did a bunch of problems concerning energy.*0005

*Today, we are going to talk about our next most important state property which is entropy.*0010

*Let us jump right on in.*0015

*Before we actually begin our discussion of entropy, there are a couple of quantities that I wanted to introduce mathematically.*0019

*And then from there we will begin our discussion of entropy.*0025

*These quantities tend to come up on a regular basis so I just want to introduce them here.*0028

*The first one is something called the coefficient of thermal expansion.*0035

*Let us go ahead and stick with black, the coefficient of thermal expansion.*0039

*And I'm going to put in here, in parentheses isobaric because this is done under constant pressure so*0055

*we designate it with a letter α = 1 / V × the partial derivative of V with respect to T at constant pressure.*0068

*I’m going to introduce them and I will go ahead and discuss what each one is individually.*0079

*The other one is called the coefficient of compressibility.*0083

*The coefficient of compressibility and this one is isothermal.*0092

*You will sometimes hear this we refer to as the coefficient of isothermal compressibility, it does not really matter.*0107

*The coefficient of the thermal expansion or coefficient of compressibility and this one we designate with the Greek letter kappa K.*0113

*Let us make this look a little bit more like a K here.*0122

*K = V the partial derivative of the volume with respect to a change in pressure under conditions of constant temperature.*0125

*Let us go ahead and define what these are.*0137

*Α the coefficient of thermal expansion.*0139

*Α is a relative increase in volume per unit increase in temperature.*0146

*That is what the partial derivative is.*0176

*It just says the rate of change in volume with respect to temperature.*0177

*If I increase the temperature, how is the volume going to increase?*0180

*That is exactly what this is.*0184

*This relative part, I will discuss it a little bit more in second but real quickly.*0185

*Relative means I’m dividing it by the initial volume I started off with.*0189

*I know that if I heat something up in general, it expands like cool it down it contracts.*0192

*This DV DT, if I keep the pressure constant it is just the rate of change of volume with respect to temperature.*0200

*It is a relative increase in volume per unit increase in temperature.*0208

*Kappa is the relative decrease in volume per unit increase in temperature.*0212

*If I increase the temperature, I’m sorry unit increase in pressure, the denominator is actually pressure.*0241

*Now this says, if I keep the change in volume with respect to pressure.*0251

*As I change the pressure, how is the volume changed?*0259

*If I keep the temperature constant of the system, if I increase the pressure of the system, what has to happen to volume has to decrease.*0262

*That is what this negative sign comes from, that is what this is saying.*0271

*kappa is a measure of the relative decrease in volume per unit increase in pressure.*0273

*Let us go ahead and talk a little bit more about what these are.*0281

*For α, the partial derivative part DV DT at constant P is a change in volume of the relative.*0283

*It is just a straight change in volume per unit change in the temperature.*0300

*However, we take this rate of change so we take this and we divide by the volume that we actually start off with.*0315

*We take this rate of change and divide by,*0336

*I will just go ahead and call it the volume.*0349

*We start with, before we make any change, before the change actually begins.*0356

*I’m just going to call the initial volume, just divide by the initial volume.*0361

*You start with a certain volume, you heat something up by the initial volume, I think it is better.*0366

*We do not have to write quite so much.*0372

*When we take the rate of change and when we divided by the initial volume that we start off with, this is what gives us.*0379

*What this does is tell us what percentage of the original volume, what percentage of the initial volume V does the change DV represent.*0388

*In other words, if I had 10 ml of something and if I heat that up and now the new volume is 11 ml.*0434

*The change is 1 ml, it went from 10 to 11 so that is 1.*0442

*If I divide that 1 by the initial amount that I started off with which was 10, I get 1/10.*0446

*So that gives me the percentage which means the volume increased by 10%.*0452

*What is nice about this, when we actually divide by the initial amount to get a relative increase,*0456

*instead of just a normal increase is it does not matter how much we start off with.*0462

*I start off with, in this first example that I gave I had 10 that went to 1, that is a 10% increase.*0469

*It is at 10 ml of the particular solution what if I had 100 ml of that solution?*0475

*That solution, it is the same solution so it is going to behave the same way.*0480

*That 100 ml, if I heat it up by the same amount, it is going to go up.*0485

*The change is going to go from 100 to 110 ml.*0490

*The 10 ml change divided by 100 that I start off with still gives me 10%.*0494

*This α, this relative increase, the change divided by the amount that I start off with gives me a percentage change*0499

*that eliminates the need to worry about how much is actually there.*0508

*That is why it is relative increase is more important, it is more valuable information that just the increase.*0513

*For K, this -DV DP at constant temperature this is just the normal change.*0522

*It is the change in volume per unit change in pressure.*0541

*Now by dividing this by the initial volume of the system, we recover the percentage of the initial volume that this – δ V represents.*0553

*When we take a particular change and we divide it by the initial amount that we started off with, it gives us a relative change.*0604

*That δ V represents how much of the original V, it gives us a percentage.*0612

*Notice, if we do so we have something like this.*0618

*In the case of kappa, we have this K = -1 / V DV DP this is a unit of volume.*0623

*The units of volume cancel which are left with is just a particular unit of pressure.*0632

*When you divide this, the change by the initial amount, you get a percentage, that is what it is saying.*0639

*It is saying a 100 ml of something will change by certain percentage.*0645

*It is giving it to you relative to the amount that you start off with.*0649

*That is really nice.*0653

*I will not go ahead and I will not say any more about that, I hope that actually makes sense.*0658

*Let us go ahead and now begin our discussion of entropy.*0662

*Entropy, I’m going to begin by just giving you the classical thermodynamic definition of entropy.*0671

*I'm not going to tell you what entropy is.*0680

*I’m not going to try to tell you what entropy is.*0682

*What we are going to do is we are going to give this mathematical definition and then we are going to star playing with this mathematical definition.*0685

*We are going to start investigating how entropy behaves and in the process of discovering how it behaves,*0692

*the hope is that it will give you a sense of what entropy is.*0701

*Entropy is a very elusive property.*0704

*To this day, I still think that the best way to think about entropy is sort of the way that*0707

*it was introduced you in General Chemistry, just qualitatively in general.*0711

*You want think about it as the disorder of the system.*0716

*How much general disorder is there in a system.*0719

*I still think that is the best qualitative way of actually looking at it.*0722

*Now later on, we will give a precise definition of what entropy is in terms of the distribution of energy and the distribution of particles within a given volume.*0726

*But the definition I’m about to give does not require that we actually think about something in terms of particles.*0735

*If I just have a block of steel, it does not matter what that block of steel is made of, it is still going to behave a certain way.*0742

*That is what our experiences are, our empirical experience of the thermodynamic behavior of things.*0749

*The definition I'm going to give is a purely empirical, purely thermodynamic definition.*0755

*We are going to use this definition and later we will define what entropy actually is.*0761

*We want to get a sense of how it behaves so if you come a little bit more comfortable with it.*0766

*I’m going to be writing all of these down so no worries.*0772

*The definition of entropy is this, DS = DQ reversible/ T.*0775

*Just take that as your basic definition.*0788

*Do not worry about it, do not clutter up your mind around that just yet.*0792

*It is absolutely fine, reversible T.*0796

*Let us go ahead and say some things about it.*0802

*Pretty much what I just said a moment ago.*0806

*We will not discuss what entropy which is designated with the letter S.*0811

*We will not discuss what entropy is right now.*0827

*For now, we will treat it mathematically which really is the best approach when dealing with entropy.*0843

*Entropy is one of those things that you can end up actually saying too much about in the beginning*0856

*and it ends up making it much more difficult to deal with.*0860

*If you just deal with the mathematically first, it actually makes it easier to understand.*0863

*For now, we will treat it mathematically and investigate how entropy behaves under various circumstances.*0872

*How entropy behaves, if we know how something behaves, we are going to get more comfortable with it.*0887

*It is going to give us a better sense of what it is and how entropy behaves under various circumstances.*0893

*And what I mean by various circumstances, under conditions of temperature pressure and volume.*0902

*What happens if I raise the temperature and raise the pressure but keep the volume constant, things like that.*0907

*How it behaves under different circumstances and a given system.*0912

*Knowing how this behaves will help us eventually understand what it is.*0924

*I apologize for all this writing, this is just the initial phase, we want to get a couple of things out of the way.*0951

*We will relate entropy to the spatial and energy distributions of the particles that actually make up the system.*0964

*We will do the later on.*1009

*Now, these particles which make up the system, they comprise the structural model.*1011

*They comprise a structural model, what I mean by that is we are actually telling you what the structural of particular piece of iron is or a particular gas in a flask is.*1035

*We are telling you what it is made of.*1049

*We are giving you what its structure is.*1051

*These particle make it comprises structural model.*1055

*The above definition the DS = DQ/ T.*1058

*I will write that out again.*1064

*The definition DS = DQ reversible/ T, it does not require a structural model, that is what is nice about it.*1067

*It does not require a structural model.*1079

*You do not need to know what a system is comprised of.*1084

*You do not need to know how it is constructed and the behavior is the same, this represents a behavior.*1087

*This is very convenient.*1097

*Entropy is an extensive state property like energy which we designated as U.*1104

*Extensive means it depends on how much is there.*1129

*If 2 mol of a particular gas has a change in entropy of 10, 4 mol of that is going to have a change in entropy of 20.*1132

*It just depends on how much is there.*1140

*Remember what state property is, a state property does not depend on the path that you take in order to get from one state to another.*1143

*Heat and work are not state properties.*1152

*How much heat and work is involved in a particular transformation depends on the path that you take.*1155

*Energy, it does not, all that matters is where you begin and where you end.*1159

*The path that you take absolutely does not matter.*1165

*The only thing that matters is the ending and the beginning.*1167

*That is a state property.*1169

*Entropy is a state property, volume is a state property, pressure is a state property, temperature is a state property.*1171

*Heat and work are not state properties.*1177

*As a state property or state function, DS is an exact differential.*1181

*It is very important, it is as profound consequences for its mathematics.*1196

*Let us talk about what the definition actually says.*1205

*I will write the definition again, I will write it up here for convenience DS = DQ reversible/ T.*1208

*It means exactly what it says.*1217

*The definitions says, if I make the differential change to a system in going from state 1 to state 2 and*1227

*going from state 1 to state 2 and I conduct this change along a reversible path.*1256

*That is why this RV is here, along a reversible path.*1275

*DQ which is a heat and that is gained or lost in that transformation, I will call it the heat withdrawn from the surroundings*1290

*because we generally view things from the point of view of the surroundings.*1306

*The heat withdrawn from the surroundings, it is not big deals it is the negative of we are withdrawing something from the surroundings.*1310

*We are putting it into the system.*1321

*It is just a question of perspective.*1322

*The heat withdrawn from the surroundings for the transformation divided by T the temperature at which you are conducting this transformation,*1325

*the temperature at which this differential change is taking place,*1362

*It gives me a numerical measure for the differential change and this so called state property as change in entropy.*1387

*Here is the definition.*1416

*This basically says, if I take a system from state 1 to state 2 and right now we are just worried about the differential change.*1419

*If I make a differential change, if going from state 1 to state 2, and if I conduct this change along a reversible path the amount of heat*1427

*that is gained or lost in this transformation depending on your perspective of the surroundings or system.*1436

*If I take that amount of heat and if I divide it by the temperature at which the transformation is taking place,*1446

*I get the change in entropy of the system or the surroundings depending on your particular point of view.*1453

*It is just a straight definition, this is really no different than the definition that was given for energy.*1458

*You remember the definition for energy was DU = DQ – DW.*1464

*Again, energy was expressed in terms of the heat and work that transpires during a transformation.*1471

*In the case of the state property, entropy it is only has nothing to do with the work, it is only related to the heat that transpires during this transformation.*1478

*It is really no different than what came before.*1489

*DU was expressed in terms of heat and work, DS change in entropy is expressed in terms of the heat.*1493

*The only difference is we decide as far as the definition is concern that this heat has to be, that has transpired during a reversible path.*1499

*You remember what a reversible path was.*1508

*Remember, let us say we are expanding a gas, let us say that this was pressure, this was the volume axis with a PV diagram,*1510

*this was pressure 1 and this was pressure 2, volume 1 and volume 2.*1521

*We can go from here to here, that is one path.*1525

*We can go from here to here, it is another path.*1528

*If we go from here to here, or if we follow this path, if we follow the isotherm that actually made a reversible path.*1531

*That is all this is saying.*1539

*Now instead of working with energy, if I go from here to here and if I calculate the heat that is gained or lost and*1540

*if I divide the heat by the temperature during each increment of that step, I get the change in this property called entropy.*1547

*We have defined what entropy is in terms of what it really is.*1556

*We just defined it mathematically.*1560

*There is some number that is changing to some property of the system that is changing and*1562

*we can assign a numerical measure for it, based on some things that we can measure.*1567

*We can measure the heat gained or lost and we can measure the temperature at any step of that change.*1571

*We have DS = DQ along a reversible path divided by T.*1582

*If we integrate over the entire path, this is a differential.*1590

*If we integrate along the entire path, the whole thing not just the differential change we get the following.*1599

*We get the integral DS = the integral from state 1 to state 2 of DQ reversible/ T.*1612

*This is an exact differential so the integral of an exact differential is just δ S if you find that change state 1 to state 2.*1624

*And that is going to = the integral from 1 to 2 DQ reversible/ T, whatever this happens to be in our particular measurement.*1637

*It is very important to be very clear about what this definition is.*1650

*In fact, what any definition is.*1672

*Let us go over here.*1684

*Let us be very clear about what we give a definition of something, what is actually it is saying.*1687

*What does it mean? What does the left side of the equality sign mean?*1693

*What does the right side of the equality sign mean?*1696

*Let me rewrite the definition again up here so we have a page DQ reversible/ T.*1699

*S is a state property of the system, there is some property that is measurable.*1708

*However, we do not measure S directly, the way we measure a length or a volume.*1725

*We do not measure S or DS directly.*1738

*However, we have discovered that many years of experimentation discovered that if we measure the heat*1749

*that transpires along reversible path and divide by T, then add the sum of all of these along the entire path.*1770

*In other words, integrate the entire path.*1795

*We get δ S for the transformation.*1814

*That is what this is saying, that is what the definition is.*1820

*There are something that we want to identify, this thing called S.*1824

*We are going to identify in terms of things that we already know DQ and T, that is what the definition is.*1834

*We do not measure S directly, what we do is we measure the heat that has given off or withdrawn in a process and*1840

*we divide by the temperature at which a process that takes place that gives a number.*1847

*That number we say is equal to the state property, that is what the definition is.*1851

*When you see definition of mathematics, what is on the left they are saying that what is on the left = what is on the right.*1856

*It is the thing on the right hand side of the equality that is what you are measuring.*1864

*That is what your experimental data is that stuff.*1867

*It is equal to this thing that we are defining on the left.*1870

*So definitions are very important.*1874

*It is very important since S is a state property δ S absolutely does not depend on the path taken to go from S1 to S2.*1882

*Since S is a state property, δ S does not depend on the path taken to go from S1 to S2.*1923

*That is the whole idea behind a state property.*1929

*All that matters is where you begin and where you end that is why we have a δ S.*1931

*Now the path can be reversible or irreversible.*1939

*The path, do not worry I will just contradict myself of what came before.*1945

*The path can be reversible or irreversible.*1949

*However, if we use the equation δ S = the integral from state 1 to state 2 of the heat withdrawn during the process divided by the temperature,*1960

*if we use this equation to actually calculate δ S by solving this integral, if we use the equation to calculate δ S then the path has to be reversible.*1982

*And the path has to be has to be irreversible path.*2004

*You are going to discover in mathematics and in science that we will give a definition of something.*2013

*Definition is there for the sake of having a definition, it is a starting point.*2021

*It gives us the starting point on which we can actually build but when we actually go to measure or calculate things like the DS,*2027

*we often do not use the definition because we find simpler ways of doing it.*2040

*We find other ways of actually doing it.*2043

*The definition is there as more of a formal structure but we do not necessarily use it.*2045

*In this particular case, to calculate the δ S.*2051

*This is the definition and it depends on a reversible path.*2055

*If we use this equation to calculate δ S then we have to use irreversible path but there other ways to calculate δ S.*2060

*And in that case, the path does not matter.*2067

*That is the difference because S is a state property, how you get from one state to another does not matter,*2070

*only if you can use this particular equation, the definition, and calculate δ S that is when you have to use a reversible path.*2076

*Fortunately, we do not have to do that.*2082

*Let us start investigating how S actually behaves.*2089

*The first thing we are going to discuss is entropy changes under isothermal conditions.*2093

*Let us go ahead and do that.*2100

*Entropy changes under isothermal conditions and again you know the isothermal means that the temperature is held constant.*2104

*We do the same thing with energy.*2120

*How does energy behave under isothermal conditions?*2122

*Now we are doing it with entropy.*2125

*We have δ S = the integral from state 1 to state 2 of DQ reversible/ T.*2129

*Isothermal means T is constant.*2140

*If T is constant we can pull it out from under the integral sign so what we have is δ S = 1/ T × the integral 1 to 2 of DQ reversible.*2142

*Δ S = 1/ T the integral of DQ is just Q, it is the entire heat for the entire path.*2160

*This is the differential for one piece of it.*2170

*If I follow the entire path, I get Q, I get the particular heat that is withdrawn from the surroundings.*2173

*Q reversible/ T that is our important equation.*2182

*Δ S = Q reversible/ T.*2188

*If a particular transformation takes place isothermally, if I keep the temperature constant during that transformation,*2193

*all I have to do is find out the heat that was withdrawn from the surroundings or the heat that went into the system,*2199

*But depending on your perspective and divide by the temperature at which that took place.*2205

*Once the temperature is constant, I just divide by the temperature and that gives me my change in entropy for that particular process.*2209

*Notice the unit Q/ T J/ K.*2216

*Let me repeat that.*2225

*This says in going from S1 to S2, I simply take the heat for the entire process which is Q reversible and divide by T which happens to be constant happens.*2228

*Isothermal conditions are very easy to find the entropy change because it is really easy to measure how much heat is gained or lost in the process.*2277

*We just measure it and take the temperature.*2284

*It just tells you how much heat is gained or lost in a particular process.*2289

*You divide by the temperature that you run the experiment under and then you have a change in entropy.*2292

*That is pretty fantastic.*2297

*This gives me δ S for the process.*2301

*This equation, this is what used to calculate changes in entropy involving a change of phase, liquid to gas or gas to liquid, solid to liquid, liquid to solid, things like that.*2312

*This equation is used to calculate δ S values for changes of phase specifically the δ S of vaporization and δ S of fusion.*2328

*Δ S of vaporization is the change in entropy in going from liquid to gas.*2362

*The δ S of fusion is the change in phase in going from solid to liquid or liquid to solid vaporization, liquid to gas, gas to liquid, either direction is fine.*2367

*I said earlier that the best way to think about entropy qualitatively is still in terms of these orders.*2381

*In terms of the randomness of the system.*2387

*A solid is a very order thing, as it melts, as it becomes liquid it is becoming more disordered.*2390

*δ S is going to be positive.*2396

*As a liquid goes to gas, a gas is a much more disorder thing than a liquid is.*2399

*In going from liquid to gas vaporization the entropy is going to be positive.*2405

*In other words, the entropy of the gas is going to be higher than the entropy of the liquid.*2411

*Therefore, the final - the initial entropy of the gas - the initial of the entropy of liquid, you are going to get a positive number.*2416

*That is what is going on.*2424

*If you are going the other way, if you are condensing from gas to liquid we have a negative entropy, -δ S, negative change in entropy.*2425

*If you are going from liquid to solid you are becoming more ordered.*2433

*There is going to be less order in your final product, the solid and there was in the liquid the initial phase so you are δ S is going to be negative.*2437

*Qualitatively thinking of it in terms of disorder is very important.*2446

*This equation is used to calculate δ S values for changes in phase specifically the δ S of vaporization and the δ S of fusion.*2453

*Let us go ahead and recall what a heating curve looks like from general chemistry.*2460

*Recall the heating curve, what happens when I take a piece of something, solid piece I just keep heating up and keep putting more energy to it.*2466

*What happens to it?*2481

*Here is what it looks like it.*2482

*We are making too big here.*2485

*This axis is temperature and this axis is energy, we are going to just keep adding energy to something.*2487

*Solid phase, let us go ahead and draw it first and tell you what is going on here.*2494

*This is the solid phase, this is the liquid phase, and this is the gas phase.*2501

*There is a temperature at which it melts and there is a temperature at which the thing boils.*2506

*Let us take ice not water, if I have solid ice and it is below 0°C, if I keep heating up the temperature is going to rise.*2510

*I’m adding energy to it and the temperature is rising.*2525

*It is going to get to a particular temperature, in this particular case it is going to be 0°C.*2529

*At 0°C, that ice starts to melt.*2533

*The solid starts turning into water, the phase is changing.*2536

*As that phase change is taking place, as it is melting notice the temperature does not rise.*2540

*All of the energy that I put into it from this point to this point, it goes toward converting the solid to the liquid.*2546

*Here is our phase change, our phase change from solid to liquid.*2554

*If I go the other way it is liquid to solid.*2564

*Once it is actually all converted to liquid, as I keep adding energy to it, the temperature is going to rise.*2566

*It is going to rise and I put energy and heated it up.*2575

*At some point, it is going to reach the point at which the liquid, the water starts to boil.*2577

*Now from here to here, the temperature does not change any more, the temperature does not rise.*2582

*I’m still adding energy and still heating the thing up but all the energy that put into it is being used to convert the liquid water to water vapor.*2588

*This is the other phase change.*2597

*The two important temperatures are the melting temperature and the boiling temperature.*2602

*This phase change is from liquid to gas.*2605

*It behave like this solid, liquid, gas, this is what a heating curve looks like.*2610

*Once everything is gas, I had more energy to it, now of course the temperature just keeps rising.*2614

*Notice, the temperature does not change during changes of phase, the process is isothermal, the temperature stays the same.*2621

*That is why we can use what we just did with entropy.*2631

*During changes of phase, the change is isothermal, the temperature does not change so we can use this.*2633

*Now since phase changes take place not only isothermally but they also take place under constant pressure*2648

*we usually do not pressurize to watch the phase change.*2661

*We can in certain circumstances but when we are watching ice melt and then vaporize, it is just happening just under normal atmosphere pressure.*2664

*It is just a constant pressure process.*2671

*Since phase changes take place under constant pressure, we remember that Q is actually = to the enthalpy under conditions of constant pressure,*2674

*the heat of a particular transformation is actually = to the enthalpy of the transformation.*2690

*This δ S a vaporization which = the Q of vaporization/ T = δ H of vaporization/ T.*2696

*In order to find the entropy of the vaporization process, as it goes from liquid to solid, all I have to do is calculate the δ.*2714

*If I look it up or I calculate it, it is the same in the heat and that δ H because you are under constant pressure conditions, the heat and δ H are the same thing.*2724

*I just divide it by the particular temperature.*2735

*In this particular case, it is going to be a boiling temperature.*2737

*Similarly, if I want the δ S of fusion it is just the heat of fusion / T which in chemistry we call the δ H of fusion.*2741

*And we divide in this particular case the melting temperature.*2754

*That is very important.*2759

*For any phase change, that δ S of that process of that phase change = enthalpy of the process divided by the particular temperature at which that phase change takes place.*2762

*Let us go ahead and do example problem nice and simple.*2780

*What is the δ S, what is the change in entropy for the transformation of 150 ml of water from the liquid to gas phase as boiling point?*2789

*For water, the δ H of vaporization is 40.7 kl J /mol .*2798

*What that means is that for every mol of water I have to put 40.7 kl J of heat into it to convert liquid water to gas water, that is all δ H means.*2802

*Again, because this is an extensive property, it actually matters how much is there.*2813

*Let us see what we can do, 150 ml of water is about = 250 g because the density of water is 1 g /ml g/ cm³.*2819

*Let us find how many mol is this.*2831

*150 g × 1 mol of water is 18 g so what we have is 8.33 mol of H _{2}O.*2833

*The δ S of vaporization = the δ H of vaporization divided by the boiling temperature.*2847

*The δ H of vaporization is 40.7 so we have 40.7, it is kl J/ mol and the boiling point is 100°C but we do not use Celsius temperature, we use K.*2855

*This is going to be 373 K.*2868

*When I do this division, I end up with 0.109 kl J /mol K.*2872

*I have a 8.33 mol, mol cancels mol, so I end up with 0.909 kl J/ K or if I want 909 J/ K.*2892

*This is the change in entropy.*2909

*If I have 1 mol of water, the change in entropy as I take it from a liquid to a gas phase at 100°C is going to be 0.109 kl J/ mol K.*2912

*For the 150 ml the change in entropy is 909 J/ K that is it.*2929

*Isothermal process, δ S of vaporization just use the δ H of vaporization divide by the boiling temperature.*2938

*Δ S of fusion just take the δ H of fusion divide by the melting temperature.*2947

*Thank you so much for joining us here at www.educator.com.*2954

*We will see you next time, bye.*2955

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 2:03 AM

Post by Jinhai Zhang on December 16, 2015

heating curve flat region is that the critical point?

1 answer

Last reply by: Professor Hovasapian

Wed Nov 11, 2015 4:32 AM

Post by Manish Shinde on November 10, 2015

under what condition would a substance have a melting point that is independent of pressure?