Raffi Hovasapian

1st Law Example Problems I

Slide Duration:

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 1st Law Example Problems I

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Section 3: Energy Example Problems: Lecture 1 | 42:40 min

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Lecture Comments (15)

	1 answer Last reply by: Professor Hovasapian Mon Nov 26, 2018 5:22 AM Post by Mir afzal Khan on November 25, 2018 hello , sir i have question about 3rd example . its given in the example 110 kPa in the solution you wrote it down 100 kPa could you explain me these steps in little bit more detail , i will be very thankful to you sir
	1 answer Last reply by: Professor Hovasapian Wed May 31, 2017 6:31 PM Post by Vanessa Ralph on May 30, 2017 Hello, Thank you for your materials and help. I have a question regarding Example 4. I originally did my calculation for Pext in the expression dw=Pext dV as P2 = Pext and thus P2 = nRT/V2 = 1.148 atm. Thus w = (1.148 atm)(40L)(101.325 J/L*atm) = 4653.43 J. Why wouldn't this give the correct answer for an ideal gas moving along the isotherm? Also, I calculated w using your method and received 7,221.207 J as the answer for w = (3)(8.314 J/molK)(303K)ln(65/25)... ? What am I doing wrong? Thank you for your time!
	1 answer Last reply by: Professor Hovasapian Mon Sep 21, 2015 12:50 AM Post by Shukree AbdulRashed on September 19, 2015 Hello sir. Just a few questions. I don't see how you arrived at the conslusion that p1v1=p2v2? Any situation where it wont = 0? Does the "P" in PV=nRT only refer to the external pressure? Just to clarify, work would be negative in example 4 from perspective of the system? Is there anyway to solve for pressure in example 4? for these isothermal problems, if U= Q+ W, if one of Q comes out positive, then W should be negative and vice versa correct? Thank you.
	1 answer Last reply by: Professor Hovasapian Mon Sep 21, 2015 12:34 AM Post by Shukree AbdulRashed on September 19, 2015 Why is it appropriate to use kpA and decimeters for example 3?
	1 answer Last reply by: Professor Hovasapian Mon Sep 21, 2015 12:24 AM Post by Shukree AbdulRashed on September 19, 2015 For the equation U= Q - W, can't you just write U = Q + W, then figure out the appropriate signs based on the Point of view of the system?
	0 answers Post by Stuart Nystrom on September 17, 2014 ^for example 3
	3 answers Last reply by: Professor Hovasapian Wed Sep 17, 2014 10:03 PM Post by Stuart Nystrom on September 17, 2014 external pressure is 110kPa right? Not 100kPa... or am i wrong?

1st Law Example Problems I

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
Fundamental Equations 0:56

Work
Energy (1st Law)
Definition of Enthalpy
Heat capacity Definitions
The Mathematics

Fundamental Concepts 8:13

Isothermal
Adiabatic
Isobaric
Isometric
Ideal Gases

Example I 12:08

Example I: Conventions
Example I: Part A
Example I: Part B
Example I: Part C

Example II: What is the Heat Capacity of the System? 21:49
Example III: Find Q, W, ∆U & ∆H for this Change of State 24:15
Example IV: Find Q, W, ∆U & ∆H 31:37
Example V: Find Q, W, ∆U & ∆H 38:20

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: 1st Law Example Problems I

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

We have come to the very important parts in these lessons.0006

We have come to the beginning of the example problems.0010

We have gone through a lot of theory.0014

We have done a fair number of derivations that is a lot of mathematics.0016

As I promised you in the previous lessons, we re going to start on the example problems and we are going to do a lot of them.0020

When I say a lot of them, I mean a lot.0027

It is very important that we have good graphs of what is happening particularly with the first law.0031

With energy, with the work, and heat, that is going to lay the foundations.0038

We do not just handle them mathematically, that is true.0042

Of course, we want to because we have to do the problems but we want to understand what is going on.0046

That will carry us forward.0051

With that, let us go ahead and just jump right on in.0053

A lot of you are going to approach this particular class with memorizing a bunch of equations.0059

Let me go ahead and tell you right off the bat that is not going to work.0068

Memorizing equations will work for a specific set of problems or for a specific type of problem, this is higher science,0072

this is more sophisticated science or sophisticated mathematics.0080

The idea is to have a reasonable degree of understanding to have a handful of equations at your disposal and let the problem at hand,0084

However, it is worded because it is going to be worded and the same thing is going to be worded in 10 different ways.0092

You need to be able to extract that information, find out what the problem is asking,0098

and use the handful equations that you have at your disposal to derive what you need.0102

You are not going to be asked to do a complex derivation that is just for your scientific literacy, the derivations that we went through.0108

But as long as you understand what is happening, keep the equation that you learn to a minimum,0116

just the fundamental set of equations and that is what I'm going to give you.0121

In order to solve all these problems, I'm going to give you the basics of equations that you need to know in order to do all these problems.0124

Our fundamental equations, the ones that you want to know are the following.0133

Fundamental equations and you can use these for every single problem.0140

You can use a subset of them for every single problem but these are the ones that you have to know.0145

Fundamental equations.0156

Work, we have to know what work is and that is the following.0163

I'm going to give almost all of these equations the differential form, the finite version is just a δ in front of it for the state functions, for the path functions.0170

Do not stick any δ in front of it.0177

The work is equal to the external pressure × change in volume that is the definition of work.0182

Now what about energy in the first law?0191

Energy in the first law says that change in energy of the system is equal to the heat withdrawn from the surroundings - the work produced in the surrounding.0195

That is DQ =- DW.0220

The definition of enthalpy is very important, it is equal the energy of the system + the pressure of the system × the volume of the system.0225

This is not external pressure, the pressure of the system.0242

The definition of the heat capacities is very important.0247

The heat capacity definitions we have two of them.0268

We have a constant volume heat capacity, the constant volume heat capacity is defined as0271

the heat withdrawn from the surroundings divided by the change in temperature of the system.0278

When I said heat withdrawn from the surroundings under constant volume, under constant volume0284

the heat withdrawn from the surroundings is that heat that goes into the system.0290

It is just a questionable point of view.0293

It is really important be able to switch back point of views because that means you understand what is happening thermodynamically.0296

Qualitatively what is going on is which moving and in which direction.0302

That was identified with this partial derivative DU DT.0307

If you have a function, if you have the energy and it is a function of temperature and volume,0314

the partial derivative of that function with respect the temperature is that heat capacity under constant volume.0321

And of course, we have the constant pressure heat capacity.0327

The constant pressure heat capacity = DQ DU DT,0330

the heat withdrawn from the surroundings under conditions of constant pressure ÷ the change in temperature.0337

Again, this is the same that definition of the heat capacity is the same.0343

It is heat ÷ temperature.0347

This happens to be idea, these three lines for definition.0350

This happens to be identified with the enthalpy.0354

This is the change in enthalpy per change in temperature.0358

Under constant pressure conditions that is how best defined.0360

The fundamental equations, you have this one.0365

It would do this in red.0369

This one you have to know, this one you have to know, and you have to know these two.0373

When I say you have to know, you have them as a piece of paper right next to you.0381

Eventually, you are going to do enough problems so we are going to memorize them.0384

These are the ones that you will memorize not the others.0387

Let us go back to blue which you wan to know as far as the mathematics.0394

In mathematics we have two relations of a mathematics that you have to know.0402

DU = CV DT + DU DV DV this is the general equation.0410

The change in energy of the system, if the energy is a function of temperature and volume,0422

the differential total change if I change temperature and volume, if I change both it is this.0430

And I also have an analogous one for the enthalpy, the DH.0437

DH = CP DT + DU DP, constant temperature DP.0442

If the enthalpy is a function of temperature and pressure, the change in enthalpy of the system is this + that, if I change both temperature and pressure.0454

If they say this is a constant pressure process this goes to 0 because DP = 0.0463

If they say it is a constant temperature process DT = 0.0467

This goes away.0471

This is the general equation.0473

You are letting the constraints of the system tell you what is happening, what you need to do these terms.0474

Let us go ahead and I will write is the fundamental concepts.0482

I will right this on this page.0488

Fundamental concept, isothermal, when I say something is isothermal that means it is a constant temperature.0495

Mathematically, it means the following.0515

It means that the DU is= 0 or finite δ U = 0.0517

Isothermal means that the temperature does not change.0522

If the temperature does not change, the energy of the system does not change, DU = 0.0525

In your equations, you can set DU= 0.0529

You can out 0 in for DU.0533

Adiabatic means DQ = 0, it means no heat is actually allowed to flow.0535

If they say the word insulated that means it is adiabatic DQ = 0 or the finite form δ Q.0548

Because heat is a path function, it is not a state function.0560

Only state functions have the δ.0563

Isobaric, if you happen to see the work that just means constant pressure,0569

When I say constant pressure, they will use the word isobaric.0575

Mathematically, that means D P = 0.0583

If you see isochoric or isometric and the usual you would not use these terms.0590

The usually terms for pressure and volume, they usually say constant volume and constant pressure.0598

This is constant volume.0603

For constant volume, DV = 0, that is the mathematics there.0604

Final set, for ideal gases, these are just the basic things that we need to know in order to solve the problems.0614

This is what we want to have at our disposal.0622

For ideal gases, we know that PV = nr, that is our equation of state from ideal gas.0625

We also have that relationship between the constant pressure and constant volume heat capacities that is CP - CV = nr.0635

Or in terms of molar, CP mol - CV mol= R.0645

Molar just means that you divide everything by n that is all it means.0653

From ideal gas, go back to the general equations DU DV under constant temperature = 0 for ideal gas.0657

If it is not ideal, it is not going to be 0.0671

Actually, in this particular case probably will be 0.0674

We want to write a general equation for ideal gas is definitely = 0.0678

And for ideal gas is DH DP.0682

The change in enthalpy, the change in pressure per unit change in pressure, if I change the pressure by one unit, 1 atm how is the enthalpy going to change?0686

0 from ideal gas the enthalpy is not a change of pressure.0697

It is not a function of pressure.0703

The energy is not a function of volume.0705

Energy is a function of temperature.0708

The enthalpy is a function of temperature from ideal gas.0710

With that, let us just go ahead and jump into this very long process of example problems.0716

We are going to do a lot of them like it is very important.0721

I want to understand this very well.0725

Let us see what we got.0725

Our first example is going to be 2 mol of an ideal gas undergo the following changes in state a, b, and c.0730

What is the change in temperature for each change in state?0737

Molar constant volume heat capacity = 12.5 J/ mol/ K.0742

In this case they have given us a molar value.0748

They give us that, if we want to keep an eye on units as much as anything else.0751

2 mol of ideal gas undergo the following changes in state.0760

Let us go ahead and deal with part of a.0763

Let us go ahead and recall our conventions before we actually get to the problem.0766

Our conventions are the following.0773

Heat is positive when heat flows into the system.0777

Work is positive when work is produced.0800

When you see the work, work is produced that means work is positive.0811

Work is produced in the surroundings.0814

Work we are looking at something from the point of view.0817

In the case of work, we are looking at the surroundings point of view.0819

When you see the work produced, it means work is positive in the surroundings.0823

If you see work is destroyed that means it is negative.0827

That means work is leaving the surroundings.0830

Work is going into the system.0833

Heat is positive when heat flows into the system.0835

That means it is flowing out of the surroundings.0838

It is just a question of point of view but this is our convention.0840

Let us go ahead and see what we have got.0845

We want to know what the temperature changes.0848

The equations that we have at our disposal, we are talking about heat and work.0851

They want to know change in temperature.0855

It looks like we would do something like this.0858

We have an ideal gas.0862

We know that if we are dealing with an ideal gas we are going to use these equations right here DU =DQ – DW.0866

Or we are going to use the finite form because they actually give us the values DU = Q – W.0880

We are also going to use the following.0885

We are going to use DU = CV DT.0887

The second term the DU has this one + the partial derivative, the DU DV this is an ideal gas.0892

DU DV = 0 because it = 0 we are left with this equation.0898

Or the finite form δ U = CV δ T.0903

Will we are looking for changes in temperature so we try to find δ T.0911

Therefore, I'm going to solve this for δ T.0916

Δ T = DU/ CV.0919

I have CV, I just need to find δ U and do the division.0924

Let us go ahead and do that.0929

Part A, they say that 500 J of heat flows out as heat, that means is flowing out of the system.0930

That means it is positive when heat flows into the system.0941

If it flows out that means Q is negative so this is -500 J.0944

Work they are telling me 150 J of work is destroyed which means work is negative.0952

Work = -158 J.0959

Δ U = Q - W = -500 -150.0966

Therefore, δ U = - 350 J.0977

Therefore, δ T is = δ U/ CV.0986

This is going to be - 350 J ÷ CV.0994

They gave us the molar heat capacity.1002

This is 12.5 J/ mol/ K.1005

They are 2 mol.1008

We do a quick calculation here.1010

CV the molar heat capacity it is going to be 12.5,1013

Let me just do this on the next page.1019

I have 12.5 J/ mol/ K × 2 mol that gives me 25 J/ K.1023

Now I can do by problem.1041

My δ T in this particular case = -350 J ÷ 25 J/ K.1043

J and J cancel and if I get my arithmetic correctly which I’m never quite sure that I do, -40 K.1054

That is your answer.1064

The system experience is -14°K drop in temperature.1066

Let us see here, 500 J of energy flows out of the system and 150 J of energy flows into the system.1078

The net loss of energy by the system is 350 J.1091

If the system loses energy, the energy of the system is going to drop.1096

Part B, part B says that the amount of heat is 400 J, 400 J flows in.1105

They also say that 400 J of work is produced, +400.1117

Δ U= Q - W = 400 -400 = 0.1129

δ T = 0/25 J/ K.1141

The change in temperature is 0.1146

What happened here is the following.1148

You have your system, the boundary, and you have your surroundings 400 J of heat flowed into the system.1151

400 J of work was done on the surroundings.1163

400 J went this way, 400 J went this way.1167

There is no change.1170

From the system's point of view 400 came in and 400 left.1172

From the surroundings point of view, 400 left and 400 came in.1177

Δ U is 0, that is what is happening.1180

If δ U is 0 and δ T is 0.1186

That is what we want here in this particular case, the temperature change.1189

Part C , they are telling me that 0 J flow as heat so Q = 0 and they are telling me that 113 J of work is destroyed.1194

Work = -130 J.1207

Δ U = Q - W = 0 --130 J.1212

Therefore, δ U= 130 J.1220

Therefore, δ T =130 J ÷ 2 mol or 12.5 J/ mol/ K gives us 25 J/ K.1226

J cancels J and what we are left with is 5.2°K increase.1234

System surroundings, no amount of heat flowed.1252

Work 113 J was destroyed that means this is negative.1257

That means it is flowing out of the surroundings into the system.1262

This much work was done.1269

If these much work as energy transfers to the system, the energy of the system rises.1270

The energy of the system rises, the temperature of the system rises.1276

5.2 K change in temperature of the system.1281

This is what we are looking at, we are looking at the system.1287

Our convention is that heat is positive when it flows into the system.1290

Work is positive when it is done on the surroundings or it is negative if it is done on the system.1296

It is just a question of perspective.1305

Let us look at our next example here.1310

Doing a particular change of state of 55 J of work or destroyed and the internal energy of the system increases by 200 J,1315

the temperature of the system rises by 12°K.1324

What is that heat capacity of the system?1327

The heat capacity of the system = DQV / DT or Q / δ T.1332

Well they give us the change in temperature that so we have the δ T.1347

All we have to do is find Q, this division that will give us that heat capacity of the system.1353

We need Q, we know that δ U = Q – W, they tell us that a particular change of state 55 J of work are destroyed.1360

They tell us that the internal energy of the system increases by 200 J.1377

Δ U = 200 = Q – 55 J of work is destroyed -55.1381

Therefore, 200 = Q + 55.1395

Our Q is = 145.1401

Therefore, let us put that back in there.1412

Our constant volume heat capacity is going to be 145 J ÷ 12°K.1415

If I did my arithmetic correctly, let us check.1424

I never get my arithmetic correct.1427

12.1 J/ K that this is our heat capacity.1430

Our molar heat capacity, notice there is no mention of a particular moles or anything like that.1435

If I want the molar heat capacity I would divide by the number of moles and that would give me J/ K/ mol.1440

This is just J/ K.1445

Be very careful of the units.1448

That was pretty straightforward, nothing too strange here.1454

We are starting off easy working our way up.1457

3 mol of an ideal gas expand isothermally.1463

Isothermally tell us that the temperature is constant.1466

Isothermally against a constant pressure 110 kg Pascals for 25 dm³ to 65 dm³.1474

We want you to find Q,W, δ U and δ H for this change of state.1482

This is a pretty standard problem.1489

Find the heat, find the work, find the change in energy, find the change in enthalpy.1490

This is a really common problem.1494

Some of the problems last for one of the other but again it is nice to ask for all four because that is the relationship.1497

This is concerned with work, this and this.1502

There is a relationship among these.1504

Let us talk about, fundamental concept.1508

Isothermal that means the change in temperature = 0.1512

If the change in temperature = 0 that means there is no change in energy so that means that DU = 0 or DU = 0.1522

I automatically have my DU=0.1532

Isothermal means δ U= 0 ideal gas.1536

That was nice.1542

They gave us a constant pressure 110 km Pascal and the unit of change in volume 25 dm³ to 65 dm³.1545

A pressure × a change in volume =work.1554

Work = the external pressure × the change in volume, that = 100 × 10³ Pascal × 40 × 10³ m³.1559

This is in dm meters, this is km Pascal or Pascal.1580

It is a Pascal km that is a Joule.1585

We have to make sure working in the right units.1589

1 J = 1 Pascal m³.1591

We have to make those conversions.1596

In this case, 10³ the -3 cancel, 100 × 40 that leaves us with 4000 J.1598

The work is 4000 J.1609

It is positive that means 4000J of work is done on the surroundings.1612

4000 J of work is produced.1616

Δ U= Q – W, δ U is 0, isothermal.1623

0 = Q – W.1630

Therefore, Q = W.1634

W is 4000 so Q = 4000.1639

There we go we just found Q.1643

We have taken care of work, we have taken care of Q, and we have taken care of δ U.1645

Now we are almost done.1652

All we have to worry about is our δ H.1653

Δ H, our recommendation with δ H we will start with the definition of enthalpy.1655

Our definition of enthalpy is H = U + PV.1661

We want δ H.1667

The Δ operator that = δ U + PV and this operating on this.1669

The δ operator is linear operator, distributes just regular distribution, consider the symbols they just mean symbolic distribution.1677

That = δ U + δ PV.1685

We already know what δ U is, δ U is 0.1695

Therefore, δ H =δ PV.1698

This is not P δ V, this is δ PV.1704

It is pressure 2 × volume 2 - pressure 1 × volume 1, that is what δ PV is.1709

Δ PV =P2 V2 - P1 V1, this is not P δ V, this is not constant pressure.1725

This is the pressure of the system.1736

Δ PV is P2V2 – P1 V1.1742

This is my ideal gas P1 V1 =P2 V2.1744

P1 V1 = P2 V2 because P1/V1 = P2/ V2.1750

The temperature is constant so the temperature drops off.1756

You are left with this basic equation from general chemistry.1758

P2 V2 – P1 V1, if they are equal then δ PV = 0.1764

Therefore, δ H = 0.1771

Their work is 4000 J, your heat is 4000 J.1774

Your change in energy of the system is 0.1779

Your change in enthalpy of the system =0.1782

That is what is going on here.1786

Here is the system, here is the surrounding, 4000 J of work is being produced.1788

It is expanding the gas, the system is doing work on the surroundings.1798

If energy as work goes this way, if the system does work on the surroundings, the energy of the system is going to decrease.1806

The energy of the system decreasing implies that the temperature of the system is going to decrease.1815

But this is an isothermal process.1821

Isothermal means we are not allowing the temperature to decrease.1822

Where is that extra energy coming from in order to make sure that the temperature stays up?1826

It is coming from the surroundings.1832

Work is being done on the surroundings.1834

The surroundings is actually giving back heat to the system, heat is 4000 J.1836

Heat is coming into the system that is what is happening here.1842

This is what this is saying.1847

In this isothermal process, the system does work on the surroundings by pushing against it.1850

In the process of pushing against it and doing work its losing energy as work.1855

If it loses energy as work, the energy is going to drop but the temperature is going to drop.1858

If energy drops, the only way for the temperature to stay the same because we are making sure this happens isothermally,1863

it has to pull energy as heat from something else.1869

The only other source it has to do that is the surroundings.1872

It pulls heat this way so heat goes this way, the temperature stays the same.1875

That is what is going on here.1882

This is the mathematics and this is what is happening physically.1884

I hope that makes sense.1887

Let us do some more here.1893

We have 3 mol of ideal gas to 30°C expands isothermally and reversibly from 25 dm³ to 65 dm³.1900

Find Q, W, δ U, and δ H.1909

The expansion is the same, it is going from 25 to 65 dm³.1911

The only thing that is different is it is happening isothermally.1916

It is an ideal gas but now it is happening reversibly.1919

It is not just isothermal, we are not just keeping the temperature constant but we are also doing it reversibly.1922

We are moving along the isotherm.1927

We are not just going this way.1930

Let us go ahead and do this.1934

In this particular case, isothermally and reversibly we are going to use this one DW = P external × DV.1936

Reversible means that the external pressure = to the pressure of the system.1947

The P external = P itself, they are in equilibrium.1960

That is what reversible means, the external pressure is = to the pressure of the system.1963

Therefore, DW = PDV or we are going to do the derivation itself instead of memorizing equation.1968

We are going to do the derivation because we can, we have a mathematics just a simple calculus, PDV.1978

Therefore, work = to the integral from V1 to V2 of PDV.1984

It equals the integral from V1 to V2 of nrt / V DV because for ideal gas P = nrt / V.1996

I’m just substituting this value in for P.2008

Our T is a constant so I will pull it out so I get nrt × the integral from V1 to V2 of DV / V.2012

I end up with nrt LN of V2 / V1, that is my work.2022

And I just put my values in2031

I have 3 mol, R is 8.314 J/ mol K.2033

This is happening at 30°C.2045

It is isothermal because it stays at 30°C.2047

This is 303 K.2050

In this case I do not have to worry about changing this to meters because the unit cancels 65/25.2053

I think this is supposed to be, I think I did my numbers wrong here.2065

I think this is supposed to be 75.2070

If I’m not mistaken, I think this is supposed to be 75 but any way you can work out the actual arithmetic itself does not really matter.2075

All that matters is the process.2085

3 mol × log of, I think I’m going to ahead and keep it as written.2087

I’m going to do 65/ 25, the dm³ end up cancelling.2095

The ratio is what matters so I ended up with 8303 J.2102

I can recall this 8303 come from this being 65 of this come from being 75.2112

I may have actually been thinking of 75 when I did this problem and put it to my calculator but this is the correct process.2120

The number itself is actually not that relevant.2127

This takes care of the work.2131

So work = 8303 J.2133

Isothermal implies that the temperature is constant.2139

If the temperature is constant that means there is no change in energy.2144

Therefore, isothermal means that δ U = 0.2148

If δ U = 0 that means 0 = Q – W, Q = W.2160

Therefore, Q = 8303 J.2178

Enthalpy is going to be the same as that problem that we just did.2187

Let me go over it, H = U + PV.2193

Δ H = δ U + PV = δ U + δ PV.2200

Δ U = 0 PV, this is an ideal gas so P1 V1 = P2 V2 so that also = 0.2211

Therefore, δ H for this process = 0.2222

And again this is the pressure of the system not the external pressure.2226

The previous example, in example 3, Q = W = 4000 J.2236

In this example Q = W = 8303 J.2245

This one was just a isothermal against an external pressure 110 km Pascal.2255

This one was not just isothermal but it was also reversible.2262

It was also reversible, this is the maximum amount of work that this gas can do in expanding from one volume to another.2274

The other path that was taken was 4000 J, 8303 J is the maximum amount of work that this gas can do in expanding.2286

That is was happening here.2298

Example 5, we have 4 mol of an ideal gas.2304

They are compressed this time isothermally from 100 L to 20 L by a constant pressure of 7 atm.2308

Find Q, W, δ U and δ H.2316

This is pretty straightforward.2319

We have to take on it by now.2319

We have done a couple of problems just like this.2322

We are just going to keep doing a lot of problems that are the same just to get a sense of it and to become comfortable with that.2324

That is what we want.2331

Let us go ahead and do work first.2333

Work = the external pressure × the change in volume.2335

They give us the change in volume that is just the 120.2342

It is final - initial so 20 -100 so it was actually -80.2346

The external pressure is 7 atm and -80 L so we end up with -560 L atm.2352

-560 let us go ahead and convert this to J.2370

-560 L atm × 8.314 J ÷ 0.08206 L atm, that cancels.2374

I’m going to go ahead and express in kJ, -56.72 kJ.2393

It is an isothermal process.2399

We already know that an isothermal process δ U = 0.2402

If δ U = 0 that means 0 = Q - W which means that Q = W so Q = -56.7 kJ.2408

That takes care of Q and we already work out δ H part.2421

Δ H = δ U + δ PV, we are dealing with an ideal gas and this is 0, this is 0.2427

Our δ H is 0.2439

In this particular case, what is happening is this.2442

Here is the system, here is the surroundings, notice in this case that the other work and heat are negative.2444

Work is -5.2454

Sorry I’m not writing it properly -56.7.2461

This is right, I have the next line.2467

Everything is good.2469

The work done is actually negative, that means work is destroyed.2478

It means work is leaving the surroundings 56.7 kJ is moving from the surroundings to the system as work.2481

Heat -56.7 kJ.2490

Heat is negative when it s leaving the system.2493

This is a compression.2500

The surroundings are doing work on the system 56.7 kJ of work.2502

56.7 kJ energy transfer as work from the surroundings to the system.2508

This is an isothermal process which means the temperature have to stay the same.2513

If we put energy in the system, the temperature is the one to rise but we want to keep the temperature down.2518

We have to take energy away from the system in order to keep the temperature at the same level.2524

Heat moves as energy or energy moves as heat from the system to the surroundings.2530

Therefore, you get a total of δ U = 0.2537

In other words, there is no energy change for the system.2541

That is was going on here.2546

Let us go ahead and we will stop it with these 5 example problems.2548

Certainly there are more example problems to come, quite a few more.2553

Thank you for joining us here at www.educator.com.2557

We will see you next time, bye.2559

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ISBN: 0935702997

Publisher: University Science Books

Year: 1997

"As the first modern physical chemistry textbook to cover quantum mechanics before thermodynamics and kinetics, this book provides a contemporary approach to the study of physical chemistry. By beginning with quantum chemistry, students will learn the fundamental principles upon which all modern physical chemistry is built. The text includes a special set of ""MathChapters"" to review and summarize the mathematical tools required to master the material Thermodynamics is simultaneously taught from a bulk and microscopic viewpoint that enables the student to understand how bulk properties of materials are related to the properties of individual constituent molecules. This new text includes a variety of modern research topics in physical chemistry as well as hundreds of worked problems and examples."

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