Raffi Hovasapian

Entropy

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Table of Contents

Section 1: Classical Thermodynamics Preliminaries

The Ideal Gas Law

46m 5s

Intro

0:00

Course Overview

0:16

Thermodynamics & Classical Thermodynamics

0:17

Structure of the Course

1:30

The Ideal Gas Law

3:06

Ideal Gas Law: PV=nRT

3:07

Units of Pressure

4:51

Manipulating Units

5:52

Atmosphere : atm

8:15

Millimeter of Mercury: mm Hg

8:48

SI Unit of Volume

9:32

SI Unit of Temperature

10:32

Value of R (Gas Constant): Pv = nRT

10:51

Extensive and Intensive Variables (Properties)

15:23

Intensive Property

15:52

Extensive Property

16:30

Example: Extensive and Intensive Variables

18:20

Ideal Gas Law

19:24

Ideal Gas Law with Intensive Variables

19:25

Graphing Equations

23:51

Hold T Constant & Graph P vs. V

23:52

Hold P Constant & Graph V vs. T

31:08

Hold V Constant & Graph P vs. T

34:38

Isochores or Isometrics

37:08

Physical Chemistry Entropy

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Section 4: Entropy: Lecture 1 | 49:16 min

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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?

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

Entropy, Part 2

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

Section 1: Classical Thermodynamics Preliminaries
	The Ideal Gas Law	46:05
	Math Lesson 1: Partial Differentiation	46:02
Section 2: Energy
	Energy & the First Law I	1:06:45
	Energy & the First Law II	1:06:33
	Energy & the First Law III	1:02:17
	Changes in Energy & State: Constant Volume	1:04:39
	Joule's Experiment	16:50
	Changes in Energy & State: Constant Pressure	43:40
	The Relationship Between Cp & Cv	32:23
	The Joule Thompson Experiment	39:15
	Adiabatic Changes of State	35:52
Section 3: Energy Example Problems
	1st Law Example Problems I	42:40
	1st Law Example Problems II	1:00:23
	1st Law Example Problems III	44:34
	Thermochemistry Example Problems	59:07
Section 4: Entropy
	Entropy	49:16
	Math Lesson II	33:59
	Entropy As a Function of Temperature & Volume	54:37
	Entropy as a Function of Temperature & Pressure	31:18
	Summary of Entropy So Far	23:06
	Entropy Changes for an Ideal Gas	25:42
Section 5: Entropy Example Problems
	Entropy Example Problems I	43:39
	Entropy Example Problems II	56:44
	Entropy Example Problems III	57:06
Section 6: Entropy and Probability
	Entropy & Probability I	54:35
	Entropy & Probability II	35:05
Section 7: Spontaneity, Equilibrium, and the Fundamental Equations
	Spontaneity & Equilibrium I	28:42
	Spontaneity & Equilibrium II	34:38
	The Fundamental Equations of Thermodynamics	30:50
	The General Thermodynamic Equations of State	34:06
	Properties of the Helmholtz & Gibbs Energies	39:18
	The Entropy of the Universe & the Surroundings	19:40
Section 8: Free Energy Example Problems
	Free Energy Example Problems I	54:16
	Free Energy Example Problems II	31:17
	Free Energy Example Problems III	45:00
	Three Miscellaneous Example Problems	58:05
Section 9: Equation Review for Thermodynamics
	Looking Back Over Everything: All the Equations in One Place	25:20
Section 10: Quantum Mechanics Preliminaries
	Complex Numbers	34:25
	Probability & Statistics	59:57
	SchrÓ§dinger Equation & Operators	42:05
	SchrÓ§dinger Equation as an Eigenvalue Problem	30:26
	The Plausibility of the SchrÓ§dinger Equation	21:34
Section 11: The Particle in a Box
	The Particle in a Box Part I	56:22
	The Particle in a Box Part II	45:24
	The Particle in a Box Part III	48:43
Section 12: Postulates and Principles of Quantum Mechanics
	The Postulates & Principles of Quantum Mechanics, Part I	46:18
	The Postulates & Principles of Quantum Mechanics, Part II	39:28
	The Postulates & Principles of Quantum Mechanics, Part III	35:32
	The Postulates & Principles of Quantum Mechanics, Part IV	29:55
Section 13: Postulates and Principles Example Problems, Including Particle in a Box
	Example Problems I	54:25
	Example Problems II	46:58
	Example Problems III	44:11
Section 14: The Harmonic Oscillator
	The Harmonic Oscillator I	35:33
	The Harmonic Oscillator II	43:04
	The Harmonic Oscillator III	26:30
Section 15: The Rigid Rotator
	The Rigid Rotator I	41:10
	The Rigid Rotator II	30:32
	The Rigid Rotator III	35:19
Section 16: Oscillator and Rotator Example Problems
	Example Problems I	33:48
	Example Problems II	46:43
Section 17: The Hydrogen Atom
	The Hydrogen Atom I	40:00
	The Hydrogen Atom II	35:58
	The Hydrogen Atom III	36:18
	The Hydrogen Atom IV	33:55
	The Hydrogen Atom V: Where We Are	51:53
	The Hydrogen Atom VI	51:53
	The Hydrogen Atom VII	34:29
Section 18: Hydrogen Atom Example Problems
	Hydrogen Atom Example Problems I	43:49
	The Hydrogen Atom Example Problems II	1:01:57
	The Hydrogen Atom Example Problems III	48:33
	The Hydrogen Atom Example Problems IV	48:33
	The Hydrogen Atom Example Problems V	48:33
Section 19: Spin Quantum Number and Atomic Term Symbols
	Spin Quantum Number: Term Symbols I	59:18
	Spin Quantum Number: Term Symbols II	34:54
	Spin Quantum Number: Term Symbols III	38:03
	Term Symbols & Atomic Spectra	57:49
Section 20: Term Symbols Example Problems
	Example Problems I	1:01:20
	Example Problems II	56:34
Section 21: Equation Review for Quantum Mechanics
	Quantum Mechanics: All the Equations in One Place	18:24
Section 22: Molecular Spectroscopy
	Spectroscopic Overview: Which Equation Do I Use & Why	50:02
	Vibration-Rotation	59:47
	Vibration-Rotation Interaction	46:22
	The Non-Rigid Rotator	29:24
	The Anharmonic Oscillator	30:53
	Electronic Transitions	1:01:33
Section 23: Molecular Spectroscopy Example Problems
	Example Problems I	33:47
	Example Problems II	1:01:05
	Example Problems III	33:31
Section 24: Statistical Thermodynamics
	Statistical Thermodynamics: The Big Picture	1:01:15
	Statistical Thermodynamics: The Various Partition Functions I	47:23
	Statistical Thermodynamics: The Various Partition Functions II	54:09
Section 25: Statistical Thermodynamics Example Problems
	Example Problems I	48:32
	Example Problems II	57:30

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 so0055

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 change0499

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 that0707

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 beginning0856

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 and1227

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 surroundings1290

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 heat1427

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 and1540

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 and1562

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 heat1749

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 and1840

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 pressure2648

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 H₂O.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

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