Raffi Hovasapian

Energy & the First Law III

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 Energy & the First Law III

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Section 2: Energy: Lecture 3 | 1:02:17 min

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2 answers

Last reply by: Joyce Ferreira
Wed Oct 25, 2017 12:42 PM

Post by Joyce Ferreira on October 22, 2017

Dear prof. Hovasapian,

I was asked to contrast a reversible and irreversible process in both physical and thermodynamics terms. I did not not how to explain it in physical terms. What would be the difference, please?

Thank You In Advance.

Best Regards.

1 answer

Last reply by: Professor Hovasapian
Mon Sep 7, 2015 6:52 AM

Post by Shukree AbdulRashed on September 6, 2015

Just to be clear, with all of these graphs, we're evaluating the work done from the perspective of the system (the gas under the piston)? Thank you.

1 answer

Last reply by: Professor Hovasapian
Tue Jun 23, 2015 1:39 AM

Post by Jinhai Zhang on June 22, 2015

Hello, Professor! In the text you referenced, it has the example of van der waals derivation for reversible isothermal work, and the book it use a - sign in front of the integral, and the answer it derived was -nRTIn(...)...
why it has an negative sign in front, and our discussion has no negative sign in front. is that the textbook you suggested made a mistake?

Energy & the First Law III

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
Compression 0:20

Compression Overview
Single-stage compression vs. 2-stage Compression
Multi-stage Compression

Example I: Compression 14:47

Example 1: Single-stage Compression
Example 1: 2-stage Compression
Example 1: Absolute Minimum

More on Compression 32:55

Isothermal Expansion & Compression
External & Internal Pressure of the System

Reversible & Irreversible Processes 37:32

Process 1: Overview
Process 2: Overview
Process 1: Analysis
Process 2: Analysis
Reversible Process
Isothermal Expansion and Compression

Example II: Reversible Isothermal Compression of a Van der Waals Gas 58:10

Example 2: Reversible Isothermal Compression of a Van der Waals Gas

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: Energy & the First Law III

Hello, welcome back to www.educator.com, welcome back to Physical Chemistry.0000

Today, we are going to continue our discussion of energy and the first law.0005

In the lesson, we have talked about the expansion of the gas.0008

Today, we are going to talk about the contraction of a gas + some other elements of work and things like that.0011

Let us go ahead and jump right on in.0018

Compression, as it turns out is going to be exactly the same as expansion in terms of the mathematics, the same equations apply.0022

Let us go ahead and recall what these equations are.0028

Let me go to blue and see here.0031

Compression, the same equations apply.0035

If we are expanding a gas under a constant external pressure, the amount of work or the numerical value is the external pressure × the change in volume.0042

If we have lots of differential elements, if in fact the pressure is not exactly constant throughout the whole stage,0055

if the pressure changes that is constant over a certain differential elements, we have this one which is the general expression v1 to v2, P external dv.0066

When we replace, when we take the differential limit as we go to more stages, we get to a point0080

where we are actually doing the expansion isothermally right along the isotherm.0087

We are not just doing it in 1 stage, 2 stages, or 50 stages.0093

We are actually following the isotherm when we do the expansion.0097

In that case, the work is going to be v1 to v2, the pressure dv, where we have actually replaced the external pressure0101

with the pressure of the system, the internal pressure of the system.0110

At that point, we are following the isotherm, the external pressure and the pressure of the system differ only by infinitesimal amount.0114

They are essentially equilibrium which is why we justified replacing this with this.0121

This is not the same and this is different.0126

This is external pressure, this pressure is the internal pressure of the system.0128

Let us go ahead and take a look at what a single stage compression looks like, the single stage compression.0136

We are taking a gas at a certain volume and we are squeezing it and making the volume smaller.0149

The final volume is going to be less than the initial volume.0155

A single stage compression and here I’m going to go ahead and draw the Pv diagram for a 2 stage compression so we can actually compare them.0159

We already have the background of expansion, the compression is exactly the same.0171

It is just reversed.0175

The paths says, underneath the isotherm that we are going to be above the isotherm.0177

It is going to look something like this.0182

We have an isotherm, in this particular case, because we are doing the compression this is our initial state down here.0186

It is a higher volume v2 v1.0194

We are going to be going from this initial state I’m going to call s1 and s2.0198

This is going to be P1 this is going to be P2.0211

We start at a certain volume and pressure.0218

The external pressure has to be at least as large as the final pressure in order for the compression to take place.0224

We know the external pressure has to be bigger than the pressure of the system.0232

Otherwise, we would not be able to compress it.0234

The external pressure is bigger than the internal pressure, we have to compress it.0237

It has to be at least as big as this.0241

The external pressure that way, the smallest value of external pressure is going to be that.0243

What happens is this, we compress the gas, the volume, and the volume gets smaller.0249

As it gets smaller, the pressure rises and rises until it gets to a point where the pressure exactly matches the external pressure.0259

The pressure inside the system matches the external pressure.0270

The total work done is that right there, work =the external pressure × the change in volume or0275

the external pressure is here with this value, that is the height.0284

The change in volume is this value, so the work done during this compression, the amount of work that I have to do to compress this gas isothermally0291

under constant pressure is the area underneath that graph.0300

The same compression but let us do it in 2 stages.0306

This was our first path single state compression.0315

We have the isotherm, we have state 1, we have state 2, this volume 1, this is volume 2, this is pressure 1, this is pressure 2.0321

This time I'm going to compress it, I’m going to use a pressure that is going to be bigger than P1 but less than P2.0332

My first external pressure is going to be somewhere around there.0341

I'm going to go along this path and I'm going to compress the gas until I hit the volume decreases.0346

As the volume decreases, we know the pressure of the system is going to rise until the pressure hits that external value of pressure that is going to stop there.0357

I’m going to put more pressure and going to raise the external pressure to P2.0365

I’m going to squeeze it some more until the external pressure now matches or the internal pressure matches the external pressure.0372

The work that I have done is going to equal this first stage.0385

In this 2 stage compression, the amount of work that I have done is the area underneath here, the shaded area.0408

Clearly, the work done here is more than the work done here by that amount.0416

That is the difference.0427

The total work here is the work done in stage 1 + work done in stage 2.0431

This is the work in stage 1, this is a work in stage 2.0438

These are compressions under constant external pressures.0443

In this case, it was 1 external pressure.0461

In this case, it was 2 external pressures.0462

First here, and then here.0466

Clearly, these 2 stage compression requires less work.0471

We do not need the numbers, we can see it just from the area.0489

This has a lower area than this so it requires less work if I do this compression in 2 stages I do not have to work as hard to actually compress the gas.0492

Less work than 1 stage.0506

You can see where this is going, a multistage compression 50, 100, 150.0521

If I keep going it is going to require less and less and less work.0528

A multistage compression would require yet less work.0540

A multistage just turning into a calculus course.0566

A multistage compression from a higher volume to a lower volume.0575

This is volume, this is pressure, it would require less work that we do to compress the gas.0580

As we pass to the differential limit, smaller and smaller changes in volume, we find there is a minimum amount of work that0596

the surroundings must do to compress the system from an initial state to a final state.0632

This is the initial state, this is the final state, there is a minimum amount of work.0653

It is work, less work, the more stages I do the compression in, the less work I have to do.0658

At some point, the work that I do is the area underneath the graph.0664

Underneath all of these, it pass to the differential limit that work is going to achieve a minimum value.0670

It is going to be some value below which I cannot go.0678

This minimum work is achieved if we compress the gas isothermally.0683

If we actually go along the isotherm, we actually go this way.0715

Again, if you keep taking, you know this already from calculus along the isotherm.0721

We are not going to take that path or that path or this path.0732

We are going to take straight, we are going to pass right along the isotherm.0739

In this particular case, what you get is work.0745

Volume 1 volume 2, Pdv, Ihave replaced external pressure with the internal pressure of the system because0751

as we pass along the isotherm, the difference between the external pressure and the internal pressure of the system is infinitesimal.0760

It is so small that it really does not matter.0769

Therefore, I can replace the external pressure with the internal pressure.0775

They are essentially in equilibrium at all times along this path where we have replaced external pressure with the pressure of the system.0778

The pressure, internal pressure of the system itself, we do not put the subscript here but this is essentially where it is,0800

With the pressure of the system at any given moment along the compression.0806

This is the exact same thing as along the expansion.0818

When we do the isothermal expansion actually along the isotherm and that is the path we take, we get the maximum amount of work that a gas can do as it expands.0822

If we do an isothermal compression along the actual isotherm itself, we get the minimum of amount of work0834

that we have to do that the system of the surroundings have to do on the system in order to compress that gas.0844

You can do a lot of work to compress a gas or you can do the minimum lot of work to compress a gas.0853

You can do a certain amount of work.0859

When the gas expands, it can do a certain amount of work but the maximum amount of work that I can do is done when it expands along the isotherm itself.0862

That is what is going on here.0873

Let us see here, let us go ahead and take a look at an example and see if we can make sense of some of this numerically.0881

A little long but it is very important, a lot of things are going to be dealt with here.0892

An ideal gas occupies 4 L at 2 atm and 25° C.0897

It is compressed to a volume of 1 L, calculate the minimum work needed to compress the gas isothermally by a single stage compression at constant pressure.0906

Here is the crucial, then calculate a minimum work required to compress it isothermally in a two stage compression.0918

First to a volume of 2 L, then from 2 L to 1 L, the initial status 4 L, the final state is 1 L.0927

It is a compression.0936

Single stage we are just going from 4 to 1 directly under constant pressure.0937

In a 2 stage, we are going to go from 4 to 2 and 2 to 1, under 2 different constant pressures.0941

Finally, calculate the absolute minimum.0948

The absolute minimum work required to affect the same change of state.0952

Let us go ahead and draw the Pv diagram to help us out here.0959

We are going to go like this, we have an initial state, we have a final state.0967

The initial volume is going to be 4 L and the final volume is going to be 1 L.0974

We said that the initial pressure is 2.0 atm and this is going to be our final pressure which we do not know yet.0982

Notice that they did not give it to us here.0993

I would have to look calculate that, that is not a problem, p1 and v1, p2 and v2.0995

We already know how to do that.0999

Let us see what we can do, a single stage compression.1004

Isothermally just means that the temperature stays the same, the minimum amount of work needed to compress the gas isothermally by a single stage compression.1010

In order to compress this gas from this state to this state, from 4 L to 1 L.1021

The external pressure has to be bigger than 2 to be able to stop the compression.1028

In order to reach the final pressure that was suppose to reach, a volume of 1 L whatever that pressure happens to be, it has to be at least as big as the pressure.1035

A single stage, the external pressure has to be at least as big as a final pressure.1046

For a single stage, we have the following.1062

The minimum work needed means that the external pressure has to be at least as large as the final pressure.1071

Let us go ahead and find out what the final pressure has to be.1104

P1v1 =P2v2, we are looking for P2.1109

Therefore, P2 = P1v1 / v2.1114

The initial pressure is 2 atm, the initial volume is 4 L, the final volume is 1 L.1122

Therefore, our final pressure has to be 8 atm.1137

That means the external pressure has to be at least 8 atm in order to affect that single stage compression.1142

The external pressure has to be there in order to do this compression.1157

Now that we know what the external pressure is 8 atm, let us go ahead and calculate the work for that.1162

Therefore, our work is equal to our external pressure × change in volume.1170

Our external pressure is 8 atm and our change in volume final – initial, 1 L -4 L=8 atm × -3 L =-24 L/ atm.1177

I’m not going to go ahead and convert it to joules, it is the numerical value that matters.1199

We have -24 L atm.1203

Let us go ahead and do this in 2 stages.1208

A 2 stage compression, for the 2 stage compression let us go ahead and draw that over here.1214

The 2 stage compression is going to look like this.1225

We have this and we said this is 4 and this is 2 atm, this is 4.0 L, this is going to be a Pf.1229

That is our final pressure, it is not a problem.1244

A 2 stage it is going to go from 4 L to 2 L and that is going to go from 2 L to 1 L.1250

Let us go ahead and put the 1 L here and let us put the 2 L here.1256

It is going to look like this.1262

That path it is going to take is going that way.1271

It is going to compress until it reaches a certain pressure inside and then we are going to have a new pressure.1273

It is going to be up here and it is going to compress some more.1281

The work done is going to be that right there.1285

The work total =work 1 + work 2.1288

The area underneath this rectangle, that is going to be work 1, the area underneath this rectangle that is going to be work 2.1298

This is going to equal the external pressure × the change in volume for the first stage and the external pressure and the change in volume for the second stage.1306

Let us go ahead and find out what these are, put it in, and see if we can come up with some numerical value for that.1319

For stage 1, it looks like this.1326

Stage 1, I need to find a certain pressure that is going to be bigger than 2 atm but is going to be less than the actual final pressure that we want,1337

The pressure that accompanies this 1 L.1345

It has to be somewhere in between here, in order to bring it here.1348

It has to be that value, the value of the pressure for the first stage has to match the pressure along the isotherm that takes us to 2 L.1351

Stage 1, we have P1v1 =P2v2, therefore, P2 =P1v1/ v2.1363

The initial is 2.0 atm, I’m going to go ahead and use my units here.1385

The initial pressure is 2 atm, the initial volume is 4 L, the final volume in the first stage is going to be 2 L.1393

I'm left with 4.0 atm, this is going to be my external pressure for stage 1.1407

The change in volume = 2 L -4 L =- 2 L.1419

I have the first stage, I have an external of 4 atm, I have a change in volume of 2 L.1444

I will do the actual multiplication of the n.1453

Let us do stage 2, for stage 2 my external pressure that is equal the external pressure from the first part, the single stage expansion.1455

We are still going from an initial state to a final state.1472

The final state is 1 L, that 1 L that is just the final state.1475

We already calculated what the pressure is going to be, that is going to be 8 atm.1479

Because now we need the pressure to be up here, in order to affect this compression, to go from 2 L to 1 L.1491

As we go from 2 L to 1 L, the pressure is going to rise until the pressure reaches whatever pressure this is for 1 L, which happens to be 8 atm.1501

The δ v of the second stage is equal to 1.0 -2.0 =s-1.0 L so we have that and we have that.1513

We can go ahead and calculate our work.1529

so the total work for this 2 stage is going to be work 1 + work 2.1533

Work 1 = the external pressure for stage 1 is 4 atm and the change in volume is -2.0 L.1544

This is going to be 8.0 atm × -1.0 L =-8 + -8 =-16.0 L atm.1556

Notice this is less than the single stage.1575

The single stage is -24 L atm.1579

In other words, the surroundings did 24 L atm of work on the system.1584

If we do it in 2 stages, the surroundings does 16 L atm.1590

It is dropping, it is confirming everything.1595

Let us go ahead and calculate the absolute minimum.1598

The absolute minimum is achieved, this is expansion along the isotherm itself, not this way or this way but along the isotherm itself.1603

This is expansion along the isotherm and that work is equal to the integral from the initial volume of Pdv.1623

This is an ideal gas.1644

We will put that in there, work =v1v2 nRT / vdv = when I do this integration and our T is a constant, I pull it out.1657

I have to do the integration in a previous lesson or we end up with the following nRT × log of v2 / v1.1675

Let us see what we have here, how can we solve this?1694

A couple of ways that we can do this, I think I’m actually going to do both ways.1700

We know what R is, we know what T is, we need to find what n is.1712

We already know what the final volume is 1, we know what the initial volume is 4 L.1717

We need to worry about the moles.1722

Let us go ahead and find n, Pv =nRT.1725

If we rearrange this to solve for n, I get n =Pv/RT.1728

The initial pressure is 2.0 atm, the initial volume is.1754

This is fine, it is not a problem.1769

We have the work equal to this, we do the integration, this is what we want to solve.1771

We have v1, v2.1774

We have T which is 25℃, the 298 K, we have R which is the 0.08206 if we are dealing with L atm, we need to find the number of moles.1775

For Pv =nRT, n =Pv/ RT.1789

In terms of Pv, we can either use the initial pressure, the initial volume or the final pressure and the final volume.1795

It actually does not matter what it is.1802

If you happen to notice, you do not have to go through this to find what n is.1805

Let us go back to Pv =nRT, nRT = the pressure × the volume of the system.1811

The pressure × the volume P1v1 =P2v2.1823

It does not matter, it is equal to P1v1 or I can use P2v2, it does not matter.1827

This nRT, this value right here, I can just take P1 × v1 if I need to.1834

If I want to do P2v2, that is not a problem.1841

In this particular case, nRT = let us use P1v1.1846

The initial pressure was 2.0 atm and the initial volume was 4.0 L.1851

Therefore, nRT = 8 L atm.1862

The work is equal to 8 L atm × nat log.1872

The final volume was 1 L, the initial volume was 4 L, when I do this calculation I get -11.1 L atm.1888

The single stage compression took 24 L atm of work.1906

A 2 stage compression took 16 L atm of work.1912

I can take more stages 3, 10, 50, 100 this number is going to keep dropping.1917

At a certain point, it is going to hit a minimum.1922

That minimum is the area underneath the isotherm from volume 1 to volume 2.1924

The minimum amount of work that I have to do in order to compress this gas is 11.1 L atm.1932

Notice, because the final volume is less than the initial volume, the sign automatically takes care of itself.1939

This -11.1, the amount of work that is done is 11.1.1947

It is negative, it is a work done by the surroundings on the system.1953

That means energy is transferring as a work from the surroundings to the system.1957

This is the minimum amount of work that I have to do with the surroundings has to do in order to compress this gas.1963

Let us see what we have got here.1973

A gas can expand or be compressed isothermally along many paths.1978

When we see that they are doing expansion or a compression isothermally, that does not mean we are doing it along the isotherm.2004

We can do it along any path.2011

Isothermal just means that we are keeping the temperature the same as it moves from one state to the next, initial to final or any intermediate stage.2013

An isothermal expansion or compression is not the same as an isothermal expansion or compression along the isotherm.2024

When the isotherm itself is the path, that represents a maximum or minimum quantity, that is the difference.2032

When we say isothermal, isothermal just means that it means they are keeping the temperature the same.2039

You can follow any path you want.2044

When we say isothermally along the isotherm, that represents the absolute minimum and absolute maximum quantities.2046

Gas can expand or compressed isothermally along many paths.2058

The path that follows the isotherm is the one that gives the maximum work done during expansion.2065

It also gives the minimum work required for compression.2105

In both cases, the numerical value of the work done is = to the integral from the initial volume to the final volume of the pressure × the differential volume element,2125

Where P is the pressure of the system at any point along the transition.2146

Along the isotherm, if the path that we actually take is along the isotherm, the external pressure and the internal pressure of the system may differ infinitesimally.2164

We are justified in replacing P external by P itself.2212

It is very important.2227

Let us go ahead and talk about reversible and irreversible processes.2233

You are going to hear the word reversible used all the time throughout thermodynamics.2238

Let us have a change of pace, let us go ahead and try red.2248

Reversible and irreversible processes.2257

We are going to subject gas and we are going to subject it to a cyclic process.2270

We are going to start at an initial state.2276

We are going to take it to a final state.2278

We are going to do something else would bring it back to the initial state.2280

We are going to do it in two different ways.2283

We are going to have some initial state s sub I, which is going to be pressure 1 volume 1.2288

It is going to be isothermal so the temperature is going to stay the same.2297

We are going to have s2 or s final and this is going to be a pressure 2 volume 2 and T.2300

What we are going to do is we are going to get 2 stages.2311

We are going to start at stage 1 and go ahead and expand that gas.2313

We are going to expand it to a final state.2319

What we are going to do is we are going to compress it.2321

Under compression, we are going to bring it back to its initial state.2326

We are going to do it in 2 different ways.2329

Process 1, we are going to do it as a single stage expansion with the external pressure = P2 and we are going to do a single stage compression.2339

We are going to bring it back with the external pressure = P1.2369

The second process, I will do the same thing.2376

I’m going to start with state 1, we are going to take it to state 2.2381

We are going to take it back to state 1 but this time the path it would take would follow the isotherm itself.2383

The process 2, this is going to be the infinite multistage expansion.2392

We know from calculus or from what we just saw, the infinite multistage expansion up and down, that differential infinitesimal amount,2406

that is the actual path along the isotherm.2413

An infinite multistage expansion and compression then compression along the isotherm.2415

Let us see what we got here.2440

Let us go ahead and look at process 1.2443

Process 1 looks as follows.2447

Let me draw it here because I want some room.2456

This is state 1, this is the initial state, and this is the final state.2467

It is P1 P2 v1 v2, this is v1 v2 P1 P2.2479

A single stage expansion going from here to here in a single stage expansion would look like this.2487

This is going to be the work done by the expansion.2501

A single stage compression, now we go backwards.2505

Single stage compression has to follow this path.2508

The work of the compression =the work under this whole rectangle.2516

The work of the expansion is a work under this rectangle.2521

Notice that the work of the compression is a lot bigger than the work of the expansion.2526

Therefore, the total work = the work of the compression + the work of the expansion = the work of the compression is going to be P1 × v1 - v2.2537

In the compression, this is the final state and this is the initial state.2557

It is going to be v1 - v2.2562

The work of the expansion is going to be P2 × v2 - v1 v2 - v1.2566

I’m going to switch this around, v2 - v1.2575

Therefore, this is going to become -P1 × v2 - v1.2578

Whenever I switch these, I factor out a negative.2583

The factored negative is going to be put there + P2 × v2 - v1.2586

I have v2 – v1 v2 – v1.2593

This is going to = P2 - P1 × v2 - v1.2596

Pressure 2 - pressure 1, pressure 2 - pressure 1, this is negative, v2 - v1 this is positive.2604

What I end up is a negative total work done.2612

The total work done in a cyclical process I start here, I expand the gas in a single stage.2621

The work is this much.2628

I compress the gas through the path, the work is this much.2631

The total difference in work, the total work done actually ends up being negative.2637

Notice, I end up losing that much work.2641

The surroundings ends up having to do that much more work in order to compress the gas,2646

it does not recover the work that was done along the expansion.2650

The total work is negative, mean work has been destroyed in the surroundings.2657

Work is negative that means work has flowed from the surroundings to the system.2677

It is negative, its float from the surroundings to the system.2684

This cyclical process we would expect it to be the same but it is not.2688

You end up doing more work in your compression than you get back in the expansion.2692

Or you end up doing less work in the expansion but it takes more work to actually compress the gas back to where it was before, back to its initial state.2697

the total work of the process is not the same.2707

Work has left the surroundings.2711

There is a weight in the surroundings that is actually lower than it was before.2715

The surroundings has lost work net.2719

Let us go ahead and look at process 2.2727

Now process 2 looks like this.2732

Here is the isotherm, here is the initial state, here is the final state, it is going to be P1, P2, v1, v2.2741

We are going to do this along the isotherm.2755

We are going to expand along the isotherm and we are going to compress back along the isotherm.2758

The work of expansion along the isotherm = the integral from v1 to v2 of Pdv.2768

That is this way.2782

We will go back to our initial state, the work of compression that is equal to v2 v1.2784

We went from v1 to v2, we are going from v2 to v1 of Pdv.2795

It is the same, the internal pressure of the system at any point along the compression of expansion so these are the same.2802

Therefore, the work total is equal to the work of the expansion + the work of the compression is equal to the integral from v1 to v2 of Pdv + the integral from v2 to v1 of Pdv.2810

I’m going to switch this integration and turn this into a negative sign.2832

Equals the integral from v1 to v2 of Pdv - the integral v1 v2 of Pdv =0.2838

A single stage expansion followed by a single stage compression to go from state 1 to state 2, then back to state 1.2854

We ended up with a total loss of work.2861

Work is negative it is 0, the surroundings lost work.2862

That is done isothermally, we kept the temperature the same.2868

In this case, instead of following this path and this path.2873

We follow this path during the expansion and this path during the compression.2881

We see that the work is actually 0.2885

There HAs been no net loss of work by the surroundings.2889

The work is 0, this is rather extraordinary.2893

No net work has been lost.2903

In other words, the system has been restored to its original state which is no different than the first process.2910

That process the system has return to its original state.2931

In its initial state, we find s1 and s2 and back to s1.2935

There is no difference here.2939

In both cases, the system has been restored to its original state and the surroundings have been restored to their original state.2940

The surroundings have been restored to their original state.2955

The work is 0, the surroundings have been restored to their original state.2959

We call this irreversible process.2974

Any process that restores the surroundings to their original state is irreversible process.2981

Any cyclical process that ends up bringing the system back to its original state but where something is not the same in the surroundings, that is an irreversible process.2988

Let us go ahead and write this down.3001

Reversible process, when a system undergoes a change of state by a specific process that is restored to its original state by the same sequence in reverse,3003

then if the surroundings are also restored to their original state, the process is called reversible in both directions.3007

When a system undergoes a change a state by a specific process, goes from an initial state to another state and is restored back to its initial state.3117

It goes from s1 to s2 and goes from s2 to s1 by following the same sequence of steps that it went from s1 and s2 but in reverse.3124

If the surroundings are also restored to their original state, the process is called reversible.3136

If during the change, in going from s1 and s2 then back to s1, if the surroundings are not restored to their original state, the process is irreversible.3144

Reversibility represents an idealization.3155

In this particular case, it represents the expansion along the isotherm.3159

It represents the compression along the isotherm.3164

When that is the case when we actually perform a process on expansion and a compression, when we do it isothermally along the isotherm,3168

there is no net change, there is no loss of work, that is what is important.3178

A reversible processes are important.3183

You saw that we can actually do the expansion and compression isothermally along any path.3186

There are specific paths, mainly one specific path along the isotherm both this way and that way.3192

That makes the process reversible where everything is restored exactly to where it was before.3201

It is as if nothing happened.3206

Reversible processes are important because they represent maximum and minimum quantities or values.3208

We already know before we talk about reversible processes, we talked about how if you travel along the isotherm,3238

you are going to end up getting the maximum amount of work in the expansion of the gas isothermally3245

or the minimum amount of work required for you to actually compress that gas isothermally.3251

We are giving it a name, when you are doing this isothermal expansion reversibly, in this case along the isotherm that is when you achieve your maximum and minimum values.3257

We will go back to blue, to make it look a lot better.3274

The isothermal expansion or compression of a gas along the actual isotherm is the reversible isothermal expansion or compression.3286

As opposed to just an isothermal expansion or compression which can happen along any path in a single stage or in multiple stages.3329

The maximum work obtained by an expanding gas is irreversible isothermal expansion.3367

The minimum work required to compress the gas is the reversible isothermal compression.3402

Again, when we say that something is happening isothermally, all we are saying is that the temperature is being kept constant for the process.3433

We are not saying that you are actually moving along the isotherm.3441

If we mean we want to move along the isotherm, we will say you are explicitly along the isotherm.3445

We will just use the term reversible that is what is going on here.3451

Isothermal can be any path but reversible, isothermal is one path.3455

It is along the isotherm itself.3460

It is what gives you maximum and minimum quantities of work.3462

Of course, net work again is equal to the integral from an initial volume to a final volume of Pdv.3467

Let us go ahead and finish up with an example here.3485

One more page, example number 2, the van der Waal’s equation states is a correction to the ideal gas law to account for deviation from ideal behavior in real gases.3490

The ideal gas law you know is Pv =nRT or something which is a little more instructive nRT/ v.3502

The van der Waal’s gas says that the pressure does equal to nRT =v.3514

There is a pressures that have to be make, P = nRT/ v - nb – an²/ v².3518

Or a and b are constants that have to do with a particular gas in question.3527

We want you to find an expression for the work done in the reversible isothermal compression of a van der Waal’s gas from an initial volume to a final volume.3533

Let us go ahead and see what we can do.3545

We already know the work is equal to initial v final Pdv.3547

We have an expression for P, it is right here.3563

Let us go ahead and put that in there.3566

It is equal to the integral from vi to to vf of nRT/ v - P – an²/ v² dv.3570

Let us go ahead and solve this integral now.3591

We will go ahead and separate that out.3593

The work is equal the, I’m going to integrate this first term right here and our T is a constant.3596

That is v – P, that has to stay in there.3606

It is going to = nRT × the integral of v1 to v2 of dv/ v - b – nb.3609

V – nb – n² ×the integral from v1 to v2 of dv/ v².3624

Dv/ v – nb, this is a logarithm, dv/ v² is v² is the same as v -2 dv.3637

You can do that the way you integrate any other thing, what you get is the following.3646

I will go to the next page here, work =nRT × log of v2 - nb/ v1 - nb – an² -1 / v.3652

v1 to v2 which we will end up with the final expression of nRT × log of a2 - nb/ v1 – nb.3680

It comes out here so you will get +an² 1/ v2 -1/ v1.3695

There you have that.3707

If we are dealing with van der Waal’s gas in a particular situation, we decide to use the van der Waal’s equation instead of the ideal gas equation.3708

If we want to find out what the amount of work done by the isothermal reversible expansion or compression of the van der Waal’s gas is, this is the expression for it.3716

Thank you so much for joining us here at www.educator.com.3731

We will see you next time for further discussion of energy in the first law, bye.3733

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