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### Energy & the First Law II

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
• The First Law of Thermodynamics 0:53
• The First Law of Thermodynamics
• Example 1: What is the Change in Energy of the System & Surroundings? 8:53
• Energy and The First Law II, cont. 11:55
• The Energy of a System Changes in Two Ways
• Systems Possess Energy, Not Heat or Work
• Scenario 1
• Scenario 2
• State Property, Path Properties, and Path Functions
• Pressure-Volume Work 22:36
• When a System Changes
• Gas Expands
• Gas is Compressed
• Pressure Volume Diagram: Analyzing Expansion
• What if We do the Same Expansion in Two Stages?
• Multistage Expansion
• General Expression for the Pressure-Volume Work
• Upper Limit of Isothermal Expansion
• Expression for the Work Done in an Isothermal Expansion
• Example 2: Find an Expression for the Maximum Work Done by an Ideal Gas upon Isothermal Expansion 56:18
• Example 3: Calculate the External Pressure and Work Done 58:50

### Transcription: Energy & the First Law II

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

We are going to continue our discussion of energy and the first law of gravity.0004

We are going to be spending several lessons on this topic, profoundly important because we are at the beginning of thermodynamics.0007

We definitely want to spend as much time as possible and make sure we understand what is happening at the basic level.0016

Later on, we will tend to move a little bit more quickly but for this topic it is important that you understand everything.0022

we are going to do a ton of example problems.0029

Not necessarily in the individual lessons as I have said before, due to the nature of the material we have to go through a fair amount of theory.0032

There are example problems in these lessons but I do not worry, we are going to do a ton of example problems in blocks,0040

in separate lessons themselves at the end of this particular unit.0048

Let us get started.0051

I’m going to state it, the first law says,0060

I think I’m going to go to my favorite color which is blue.0070

There are many statement of the first law, I’m going to give another one later.0085

I’m going to give this one here, it is probably the one that you are familiar with most.0089

It says that the energy of the universe is constant.0100

Basically, you know that energy cannot be created or destroyed, which means that the energy of the universe is constant, it does not change.0115

It moves around from system to surroundings, from surroundings to system, but it states as a whole, energy does not change.0124

We can express it like this, the energy of the universe is equal to the energy of the system + the energy of the surroundings.0133

The energy the system possesses, the energy of surroundings possesses, that is the total energy that stays a constant.0144

It moves but does not change, the total amount.0150

Let us go ahead and see what we can do here.0154

Let us say that energy 1 of the system + energy 1 of the surroundings, or initial or final I’m going to be using initial and final, 1 and 2 interchangeably.0157

The energy of the universe is a constant.0174

Some of the energies is some constant, let us call it C.0176

Let us undergo a change of state.0181

Upon a change of state, now the system has a new energy, the surroundings have a new energy.0186

Energy 2 of the system + energy 2 of the surroundings, it said it is constant so that is also equal to a constant.0194

I’m going to go ahead and subtract the first equation from the second equation and get something like this.0204

E2 of the system – e1 of the system + e2 of the surroundings – e1 of the surroundings is equal to C –C which is equal to 0.0212

Basically, it says e2 – e1 of the system that is a change in energy of the system.0244

E2 –e1 of the surroundings that is a change in energy of the surroundings that is equal to 0.0252

This is a statement of the first law of thermodynamics.0259

The change in energy that the system undergoes is equal to negative the change in energy that the surroundings undergo.0265

It is very simple.0271

If the surroundings loses 30 units of energy, the system has gained 30 units of energy.0272

If the system has lost 50 units of energy, the surroundings have gained 15.0277

Energy transfers at the total amount of energy that is contained is a constant value.0282

This is a state of the first law of thermodynamics.0290

We generally like the law like this, this is the one that you know of from general chemistry.0295

Generally, write the first law from one point of view.0305

Generally, writing from the systems point of view, in chemistry this is how you are accustomed to seeing it.0319

You are accustomed to seeing it like this, the change in energy of the system is equal to q – w.0323

You have probably seen as, the change in energy of the system is equal to q + w.0331

It has to do with conventions.0335

Again, the convention that we are taking is that when the system gains heat, heat is positive.0337

When the surroundings gain work, work is positive.0344

Under those conventions, the statement is actually written like this.0349

We are not going to be using E and we are going to be using e for the most part, we would it be using u for energy.0353

Δ u = q – w, these two ways, this is how you see that you are familiar with the first law of thermodynamics from your general chemistry course.0359

This is what this basically says that, the change in energy that the system undergoes or the surroundings0373

either one is equal to the heat that it gains or loses and the work that it gains or loses.0380

That is all this is saying.0386

This says, when the system undergoes a change of state, the initial energy of the system now there is some final energy of the system,0396

the net change is equal to the energy lost or gained as heat + the energy lost or gained as work.0427

Energy transfers in two ways.0461

It transfers as heat and it transfers as work.0463

When a particular system loses energy, it can lose it either as heat, gain or lose as heat, gain or lose as work.0466

Some of those are going to be the total change in energy of the system.0474

The change of energy of the system is equal to negative change in energy of the surrounding.0478

It is all about perspective and point of view.0483

In general chemistry, the point was not exactly hammered.0486

Basically, presented with this equation and we use to a bunch of problems.0492

A system gains this much energy as heat, loses this much energy as work.0498

What is the change in energy of a system?0502

We are going to do a problem like that in just a minute.0503

It is very important to take a step back, we want to think about things thermodynamically not just chemically.0505

There is a system, there is a surrounding, there is a very deep relationship between the two.0512

You have to make sure you do not forget about the surroundings or you do not forget that there is a system.0515

With that, let us go ahead and do a nice simple example problem just to get a sense of how this is dealt with numerically.0525

Our first example is going to be, a system does 100 J of work on its surroundings.0534

In the process, the system gains 125 J of heat, what is the change in energy of the system?0541

What is the change in energy of the surroundings?0548

Let us go ahead and take a look at what this looks like a diagrammatically.0553

Thermodynamics is like any other science, you want to draw as many pictures as possible, if you can draw a picture.0557

If you have schematics and pictures to fall back on, it will certainly help.0563

Let us see what we have got, I will go ahead and draw it here.0568

Let us go ahead and draw that as our system.0572

I will just go ahead and make a bigger box around it that is going to be our surroundings.0576

This is what the system does 100 J of work on its surroundings.0580

Work is actually leaving the system into the surroundings.0585

It is going to be positive because again that is our convention.0590

Once the work is done on the surroundings, work is positive.0593

Now in the process the system gains 125 J of heat.0600

Heat, the q, that is going to be positive because that is from the systems point of view.0605

When the system gains heat, the heat is positive.0611

Therefore, our q is equal to +125 J and our work is equal to +100 J.0615

We just go ahead and fall back on our equation.0630

The change in energy of the system is equal to q – w.0634

A change in energy of the system is 125 J - 100 J.0639

The change in energy of the system is equal to +25 J.0647

Why are we using this anyway?0656

You just have to look at, you do not have to do this equation.0658

If it gains 125 and it loses 100, it gains 125 is heat and lose 100 of work, there is 25 J left in the system.0661

That is the energy of the system.0671

We know that the δ u of the system is equal to –δ u of the surroundings.0674

Therefore, δ u of the surroundings is equal to -25 J.0679

The system has gained 25 J of energy net, the surroundings has lost 25 J of energy net.0687

When we say that the system has gain 125 J of heat, it has gained it from somewhere.0694

It has gained from the surroundings, that is was going on.0698

That is the transfer of energy.0701

Let us go ahead and then talk a little bit more about heat, work, and energy.0708

We have δ u - q – w.0717

The energy of a system changes in two ways, as a transfer of heat and or a transfer of work.0731

It is very important.0766

In thermodynamics, when we start to getting the problems, when we just start to speak normally about what is going on a particular process,0770

that has to be a little bit of an abusive language.0776

We speak of heat, we speak of work, almost in the same way that we speak of energy.0779

They are not.0785

A system possesses energy, it does not possess heat or work.0787

Heat or work quantities that are transferred only during a change of state.0792

A system in a given state does not have a certain amount heat.0798

It does not have a certain amount of work.0800

When it transfers its energy, either taking in some or giving out some, that transfer manifests as a heat or work.0803

It is very important to keep that in mind.0812

Systems possessed energy, energy not heat or work.0816

To label the point but it tends to fall by the wayside as we precede and this sort of abusive language has an effect on how we think about problems.0829

We want to be very clear which is why I keep we are reiterating it.0840

Work and heat are quantities that are transferred during a change of state as the only time.0849

Remember from the previous lesson, they show up only during a change of state.0871

Consider the following analogy.0884

Let us say, energy is money, energy is dollars.0887

Heat is paper, and work is coin.0904

When you go to the bank, you can deposit or withdraw money.0913

You could deposit or withdraw it as paper money or coin money, that is the thing.0915

Energy is a money, energy is the dollars, and that is what you have, that is what you possess, a certain amount.0920

The coin is like work, you could deposit money as coin, you could deposit it as paper.0930

Heat and work are the paper and the coin, respectively.0939

There are different ways of actually transferring money, of moving money.0943

They are not actually things that the bank account possesses.0948

The bank account has a certain amount of money.0952

The energy is the money, the heat is the paper, the work is the coin, that is generally how it works.0955

Let us go ahead and take a look at quick scenario here.0960

So in the first scenario, I have an initial state, I have \$100 in the account.0963

Let us say I deposit \$150 as paper and let us say I withdraw \$50.00 as coin, then I withdraw another \$25.00 as paper.0975

My final state equals \$175.1000

The second scenario, my initial state, I’m going to start off with \$100 in the account.1006

I’m going to deposit \$1000 as coin.1016

I’m going to withdraw \$900 as paper.1023

I’m going to withdraw \$50, again as paper.1033

I’m going to go ahead and deposit \$25 as coin.1040

My final state again \$175, initial state \$100, final state is \$175.1049

Second scenario, initial state \$100 and final state \$175, this is energy.1057

Energy went from 100-175, energy went from 100 to 175.1063

It did not do it in the same way, different amounts of heat and work were transferred.1069

Clearly, the initial state and final states are the same but that paths taken to get from the initial state to final state are not the same.1076

Energy is what we call the state property.1091

Heat and work, I’m just going to go ahead and call them q and w.1101

They are called path properties.1107

Let me go over here, energy is a state property q and w are called path properties.1108

Most of the time they are also called path functions.1126

We are using the word function and properties interchangeable.1130

I, personally, do not like the use of the word function, path function or state function, you can call it state function as well.1133

The reason I do not is because we already have a specific idea what a function is.1138

Somehow when you think about a state function or path function, for me I'm looking for some kind of function.1143

When I say state property or path property, it is telling me that at some property, that is something is happening.1151

A state property is energy, it is something that has a specific numerical value, q and w.1159

Although they are not properties of the system, they are properties quantities that actually show up during a change of state.1165

They have actual numerical value, algebraic they are positive or they are negative.1172

Path function that does not really matter.1177

You are going to see them refer to as both.1179

Energy is a state property, q and w are path properties.1181

Now here is the difference, the value of state property does not depend on the path taken to get there and the path taken.1188

It depends only on the initial state and the final state.1222

The value of the path function, the value of a path property absolutely does depend on the path.1233

We will have more to say about paths in just a minute.1254

We are going to be talking about paths a lot.1256

If I’m here at Los Angeles and I want to go to Denver Colorado which is roughly a mile above sea level.1263

Here is my initial state and here is my final state.1271

The only thing that matters is the fact that I’m going from here to here is the height, let us say 1 mile.1274

I have gone up above sea level 1 mile.1280

There is a thousands of paths that I can take to get there.1284

When I talk about the change in state, the change in state is 1 mile.1287

When I talk about the path that I have taken to get there, I can go from California to Miami.1292

I can go from Miami to New York.1298

I can go from New York at a space station and at the space station back down to Colorado.1300

I have done a lot of work to get there and I have spent a lot of heat to get there.1305

Or I can just take a single trip straight shot from Los Angeles to Denver.1309

I have not expanded as much work.1317

I have not expanded as much heat.1318

The initial and final states are the same but the paths are very different.1321

The quantities of work and heat expanded are different.1324

This is why.1328

Height is a state property in this particular case.1329

Elevation, the difference in elevation but the paths that I take to get there, the amount of gas that I spend,1333

the amount of work that I do, the height of for which I go and come.1340

All of that is actually different so it depends on the path.1346

Mathematically, it is just going to become very significant.1349

We will have more to say about when we cross that.1353

Let us talk about work, let us talk about pressure, volume, work.1360

When a system changes volume, consider a gas in a cylinder gas and a container.1374

When the system changes volume, we better think of a balloon.1383

Either by being compressed by an external pressure or by expanding against an external pressure, work is done.1393

Work is done, in other words work is transfer.1429

Let us see here, we say that when gas expands, the volume of the system increases.1444

The gas expanding, the volume is increasing.1458

So volume of the system increases.1461

This is the system doing work on the surroundings.1478

Or work is flowing from the system to the surroundings.1490

Work is positive.1507

When gas is compressed, you know that the volume of the system decreases.1514

System this is the same as the surroundings is doing work on the system, which is the same as work flowing from the surroundings to the system.1549

In this particular case, work is negative because it is flowing from the surroundings.1581

Work or taking the point of view of the surroundings.1584

The magnitude of the work that is done in expansion or compression is equal to that external pressure × the change in volume.1590

If you want numerical values for the work that is done in gas expanding, the work that a gas does on the surroundings1605

is equal to the external pressure × the change in volume that the gas experiences.1612

If I compress the system, the work done by the surroundings on the system, the magnitude for numerical value of the amount of energy1617

that transferred is going to be that external pressure × change in volume that the system experiences.1626

Let us look at what an extension looks like on a pressure volume diagram.1638

Analyzing the pressure volume diagram, analyzing all these diagrams is going to be very important.1645

You probably did maybe a little bit of it in general chemistry, I’m guessing not altogether too much.1651

In thermodynamics it is going to be profoundly important.1655

Definitely, take your time and make sure you understand everything is happening on these diagrams.1658

They will allow you to follow certain paths to say this is happening, this is not happening.1665

It is very important.1670

Let us see what an expansion looks like on a PV diagram.1672

Let us go ahead and do this.1678

I’m going to go ahead and draw the diagram here.1681

I will go ahead and draw this like that.1687

I will go ahead and mark my things.1691

This is going to be the initial state.1694

This is going to be some initial pressure P1.1697

Some initial volume v1 and T.1700

We are going to actually keep T constant.1703

This right here, this is an isotherm.1706

For our purposes, you want to deal with only a couple variables at a time.1712

Again, we have pressure up here and we have volume on this axis.1716

We want to be able to make things easier on us.1721

We are going to do all of our things around an isotherm.1724

It is all that it means, is that we are keeping the temperature constant when we do this expansion or this compression.1727

This is the initial state, P1 v1 T.1733

This is going to be our final state.1737

This is going to be P2 V2 and again T, temperature is constant.1739

What we are doing is following.1743

Let us say we have some cylinder with a piston, I will move this a little bit further down here.1745

We have some cylinder and we have some piston arrangement.1753

On top of that piston, we can go ahead and put a mass.1758

We have little pegs here, there is a gas in here.1761

This is a P1 v1 initial pressure, initial volume.1765

There is a certain pressure, there is a certain mass.1770

That mass accounts for the external force, external pressure.1773

What is going to happen, this is the initial state.1777

What we are going to do is remove those kegs and if the internal pressure is actually going to be bigger than the pressure outside, the gas is going to expand.1779

It is going to push the piston up.1789

That is what is going to happen.1791

It is going to go to end up looking like this.1795

It is going to end up rising until the internal pressure which is now P2, with a new volume V2, is equal to the external pressure that comes from the mass.1803

When the external and internal pressures are the same, the piston is going to stop.1812

That is what is going to happen.1817

This is what is happening physically.1818

I have a gas in the cylinder, it is under pressure, I go ahead and I release the kegs, it undergoes an expansion.1821

When it undergoes an expansion, we are going to look at what this looks like on the pressure volume diagram.1828

Here I’m going to mark P1 and here I’m going to mark P2.1833

Here I’m going to mark volume 1, volume 2.1838

We are starting up here.1842

Here is what happens, in order for there to be an expansion, the pressure on the inside has to be bigger than the external pressure.1844

Here is what happens, if this mass is a single mass, the external pressure does not change.1852

The external pressure is constant.1859

This external pressure here is what happens.1863

We started at volume 1, I pull up the pegs.1866

Now the internal pressure is going to cause the thing to expand so therefore, the volume is going to change.1871

The volume is going to increase.1881

The volume is going to change, the volume is going to change, it is going to keep increasing as the volume changes.1882

You know that the pressure on the volume have an inverse relationship.1889

As the volume of the system increases, the pressure inside the system decreases.1893

It decreases along this line here, the isotherm.1897

We are keeping the temperature constant.1902

As the volume increases, the pressure is going to decrease.1905

It is going to go down.1910

It is going to come to a point when the volume has reached such a point, that the external pressure is now the same as,1914

This external pressure is a constant pressure that comes from the mass.1924

When it actually reaches the value of P2, P2 the piston is going to stop.1928

That is what happens.1936

The pressure inside is bigger than external pressure.1938

Well it is going to expand.1941

As it expands, the pressure starts to drop when the pressure on the inside P2 reaches the external pressure which comes from the mass is going to stop.1944

The pressure and this pressure is the same so we get this.1954

We said that the work done during expansion, the numerical value is equal to the external pressure × that change in volume,1966

The external pressure happens to be this.1973

It happens to be the same as P2 because that is when the piston will actually come to a stop.1975

That is this value right here, external pressure.1984

The change in volume is v2 – v1, that is this length right here.1988

External pressure × change in volume, the area underneath that rectangle is the work done during this expansion against the constant pressure.1992

That is what is happening here.2003

The path taken in this process is this one, we start with a given pressure.2006

We release depends it starts to expand until it reaches this point.2015

We went from an initial state to a final state via this path right here, that is the path that we took.2022

The work done during this expansion is the area underneath that rectangle.2028

Let us go ahead and see.2035

Work is equal to P external × change in volume which is P external × v2 – v1.2040

Again, this is just the area underneath that.2053

This is a single stage or one state expansion.2059

All that means is that you have a single constant external pressure in a single volume expansion.2073

We get our expansion in one stage.2093

We went from initial state to final state, initial state to final state in a single state which we allowed to go this way.2095

That is the path that we took.2105

When we talk about path, this is actually where the word path comes from.2108

It comes from these pressure volume diagrams and the particular path that the system is following in order to get from an initial state to a final state.2111

What if we do the same expansion, in other words, going from P1 v1 to P2 v2.2129

Same initial state, the same final state, this time what if we do it in two states?2139

What if we do the same expansion in two stages?2144

Here is what it looks like physically.2151

We have a cylinder, that is there, this one is a little higher.2154

This one is going to be up here.2170

Basically, what we do in this case is we put a certain mass on there.2171

When we put a total mass on there, this is going to be P1 and this is v1.2175

This is going to be P prime, this is going to be v prime, and this is going to be P2, and v2.2179

We put a certain mass there.2186

The pressure in here is bigger than the external pressure, it is going to expand a little.2188

It is going to expand a little until it reaches a certain height and it is going to stop.2192

What we do is we take this mass off and we replace it with another mass.2197

The mass that is a little bit lighter.2206

It is going to expand some more until it reaches another height.2208

Let us see what this looks like in the pressure volume diagram.2214

These PV diagrams there are very useful and very important.2219

Again, we have the same initial state and the same final state.2224

This is the initial state which is P1 v1 and this is the final state which is P2 v2.2228

Our initial pressure, our final pressure, our initial volume, and our final volume.2238

Here is what the expansion looks like, this two stage of expansion.2246

Here is what happens, I'm going to put a mass on there such that the external pressure is actually less than P1 but is bigger than P2.2250

I’m going to go ahead and put it right there.2260

I’m going to call it P external prime, here is what happens.2263

The minute I pull out these pegs, the expansion is going to start.2272

I pull out the pegs, it is going to start at volume 1.2277

What is going to happen is it is going to expand until it reaches that point.2281

This point is S prime, this is P prime, V prime.2288

As the volume expands, the pressure inside the system decreases because as volume expenands, pressure decreases, that is the relationship.2292

Doing this along the isotherm and so the pressure goes down.2301

Again, the pressure goes down until the pressure inside the system is the same as the external pressure that comes from the mass and then stops.2306

Now what I do, is I'm going to go and say this is our V prime.2318

I would take that mass off and replace that with a lighter mass.2325

If I go with a lighter mass, now the external pressure is less than the P2.2329

It is going to expand some more.2335

It is going to expand some more.2338

It is going to keep expanding until now the external the pressure inside the system matches the external pressure at which point it will stop.2340

Again, we have gone from initial system to final system as P1 v1.2350

The P2 v2, the same weight but now we have taken a different path.2356

We have done it as a two stage expansion.2362

Well the first expansion, the amount of work done,2367

I’m going to have these lined up properly, sorry about.2374

Let us go ahead and change this form.2377

This is v1, the work done during the first stage, this expansion that is that area.2384

The work done during the second expansion, that is that area.2395

This is the isotherm, I think I have everything here.2403

I will say this is work 1, first stage, and this is work 2 that is the second stage.2410

The total work done in this two stage expansion is equal to work done during the first stage + the work done during the second stage.2418

Nothing new there, let us compare the two diagrams.2426

In the single stage, this was the work that is done.2444

This is the one stage expansion, this is the two stage expansion.2470

This area right here, that is the same as this area right here.2477

The two stage expansion did more work.2482

It did more work by this amount right here.2485

This is what is important.2490

We see that the two stage expansion has done more work.2492

The area is larger, the initial state and the final state are exactly the same but the paths taken are different.2512

Because the paths taken are different, the work is going to be different.2542

Now this path vs. this path.2548

They are very different paths.2557

This is why we call work a path function, this right here.2562

It is based on the Pv diagram.2575

Schematically, it is following a different path to get from an initial state to a final state, an initial state to a final state.2578

It can take any path it wants but the fact that a different path gives me a different value of the amount of work that it did.2585

This is why we call that a U path function.2591

Its value depends on the path we take.2599

As a reminder, heat is also a path function.2611

Q was also a path function.2617

Not just work but heat and work are path functions.2619

Let us take a look at a multistage expansion.2631

A multistage expansion, which I will go ahead and draw it here, I’m not going to do much analysis on it.2635

It is going to look exactly like what you think it looks.2647

A multistage expansion is going to be, if this is our initial state and this is our final state, one stage, two stage, multistage looks like this.2650

That is the total work done is going to be the area underneath everything.2663

That is what a multistage expansion looks like.2674

Exactly what you think.2678

You can see where this is going.2681

This looks a lot like a calculus course, one more stages and one more stage, tinier rectangles this way.2683

More and more area.2689

You can see where this is going.2695

If we keep increasing, the number of stages, we get this.2707

We can keep increasing the number of stages until the steps that we take, the paths that we take, in other words,2729

the change in volume until the changes in volume now become a differential length.2741

You know where this is going.2752

Instead of writing work equals P external, change in volume which is true.2754

This is valid and this is actually valid equation for constant pressure process, this is how we get the work.2761

If we do in multiple stages, it is working differentials instead of in large quantities like this P external × the differential volume element.2768

This little increments differential volume increases.2783

The differential work that is done in going from let us say this point to this point,2788

is going to equal the external pressure from here to here × the differential volume element.2794

We are just breaking this up into a bunch of little rectangles and adding up all of that.2800

Now so this is going to be a very important relation.2809

We are often to be working with differentials as opposed to large scale stuff like that.2814

If we assume and it is a pretty fair assumption to make that our external pressure remains constant over the differential volume change.2820

Since, DW is the differential amount of work in going from here to here, if I want the total work done in going from here to here.2852

I integrate all of those, I add them up, I add up all the differential work elements.2860

Our total work is equal to the integral from state 1 to state 2 of the differential work element.2871

The differential work element is equal to this, v1 v2, P external dv.2882

That is it, we just gone from δ v to dv.2889

We are just working smaller and smaller, this is it.2893

This is the most general expression for the pressure volume work done by a gas as it expands isothermally.2906

This is an isothermal expansion, we are keeping the temperature the same.2938

If we do not keep the temperature the same, the amount of work that is actually different.2943

This is nice, this is an integral.2949

If we happen to know how the pressure changes with volume, we can go ahead and solve the integral in order to get the work done along that particular path.2952

If we happen to know how the external pressure changes as a function of volume then we just evaluate the integral.2965

Let us go ahead and redraw.3003

As we one more stages, one more stages, we get higher and higher work.3009

There is only so many stages you can go.3015

There is a maximum amount of stages that you can have.3019

Basically, when you take the differential and when you take the differential volume element to 0, what you can end up getting is the integral under this.3022

Clearly, there is a maximum, there is an upper limit on the amount of work that a gas can do upon isothermal expansion.3031

There is an upper limit because there is an upper limit on the area.3052

Basically, that is it.3059

You are going to get more and more until you basically have the area underneath the curve.3062

There is an upper limit on the area under the isotherm.3069

There is a maximum amount of work a gas can do upon isothermal expansion.3085

This max is achieved when the path we take is directly along the isotherm.3114

Instead of going to little differential elements when we take the limit of the differential, now what we do is we just move right along that isotherm.3138

That is our path, not this, but right along the isotherm.3151

Straight down this way and straight up that way when we do our compression.3159

Let us see what we got here, in order for a gas to expand, the pressure inside the system at any given moment has to be bigger than the external pressure.3167

That is how expansion takes place.3183

If a pressure the same, there is no expansion.3185

It has to be greater that.3187

For this differential changes in volume, for the small changes in volume, the pressure inside has to be slightly bigger than the external pressure3189

or the external pressure is has to equal the internal pressure - from differential amount in pressure.3202

This says that the external pressure is going to be just slightly less, slightly by differential amount, an infinitesimal amount less than the pressure inside.3213

If this is the case, the work equals change in volume 1 to volume 2 of the external pressure × dv, that is equal to the external, that is equal to the pressure of the system.3223

The internal pressure - dp dv that equals the interval for v1 v2 of Pdv - the interval from v1 to v2 of dp dv.3240

This is just mass and will actually goes to 0, this is the second order differential.3260

A differential × a differential and it goes to 0.3265

Do not worry, what you are left is with is this.3268

The work is actually equal to the integral from v1 to v2, the pressure dv.3276

The pressure here, this is the pressure of the system.3284

Before, under constant pressure conditions, the initial equation we said is that it is the external pressure that defines the work that is done.3287

In this case, if we are traveling along the isotherm, with only really infinitesimal amounts we no longer have to use the external pressure.3296

We can use the internal pressure, the pressure of the system.3305

Because the pressure of the system and the external pressure, in order for there to be small infinitesimal changes, they are almost the same.3308

When we are moving along this isotherm, the pressure outside and the pressure inside are essentially equilibrium, that is what is going on here.3316

We have managed to eliminate the external pressure term and deal only with the pressure of the system, the pressure inside.3323

P external has been replaced by P which is a pressure of the system at any given moment along the isothermal expansion.3333

Let us go ahead and do an example here.3373

Example 2, we want to find an expression for the maximum work done by an ideal gas upon isothermal expansion from an initial volume to a final volume.3380

Ideal gas, we already know what an ideal gas is.3394

We know the equation state for the ideal gas, it is Pv=n RT.3397

We also know this, we also know that the work is equal to the integral from v initial to v final of Pdv.3405

We need to find an expression for P as a function of v.3415

We have an ideal gas right here, let us just rearranged it.3419

The pressure of an ideal gas is equal to nrt/v.3422

I’m just going to go ahead and put that in there and what I get is the following.3426

Work equals the integral from the v1 to v2.3430

I will just use v1 and v2.3434

Dp dv which is equal to now nrt/ vdv.3437

Nrt is a constant so I’m going to pull that out.3445

Nrt v1 to v2 of dv/ v, 1/ vdv.3448

I know what the integral of dv or v is, it is the natural logarithm of v.3457

What we get is work is equal to nrt × log of v2 – log v1.3463

I will go ahead and rearrange that using the properties of logarithms.3477

The log of v2/ v1, there you go.3481

For an ideal gas expanding isothermally, this is the expression for the work done by that gas on its surroundings.3488

This is the maximum amount of work in an ideal gas can do as it expands, n × R× T × log of the final volume divided by the initial volume.3496

This is a very important relation to understand.3510

This is the isothermal expansion, the maximum amount of work that an ideal gas can do upon isothermal expansion.3513

Let us go ahead and do another example.3526

Example 3, an ideal gas occupies a volume of 1.5 deci³ at 2.0 atm.3532

If this gas is to expand isothermally to a new volume of 3.5 deci³ under a constant external pressure,3546

what is the largest possible value that the external pressure can have and how much work is done by the gas in this expansion under this value of external pressure?3555

We have an ideal gas, we know it occupies a volume of 1.5 in deci³ at 2 atm.3569

If this gas is to expand isothermally to a new volume under a constant external pressure, the expansion is constant external pressure here.3578

What is the largest possible value of the external pressure can have if it is going to actually affect this change?3589

Let us go ahead and see what this looks like.3596

Amount of volume of 1.5 deci³ at 2.0 atm.3612

I need to go to a volume of 3.5 deci³ and I’m going to do it isothermally under a constant external pressure.3622

Constant external pressure this means a single stage expansion, there is going to be some value.3633

Basically, it is like this, I know that this is some final pressure here, I need to get from this state to state.3638

I need to get from this pressure to whenever this pressure is.3646

I need to go from 1.5 to 3.5 this is done isothermally, we are going to move along here but this is done under constant pressure.3649

Because it is constant pressure, it is going to have to happen as a single stage expansion.3659

In order for the gas to expand, the external pressure has to be less than the initial pressure.3667

In order to expand, and now in order to reach this state, I could have any external pressure I want to up to the final pressure of the system.3677

In other words, I can have a external pressure here, external pressure here.3693

Any of those values will be just fine because the expansion will take place.3700

Remember, it will go this way, this is the amount of work done.3704

This way, this is the amount of work done.3708

Whatever the value is, they want to know the largest possible value that the external can have.3709

What the largest possible value that the external pressure can have is actually going to be P2.3721

In order to reach that state, I have to reach an internal pressure of P2.3733

In order to reach an internal pressure of P2, I have to be able to get to P2, that is the largest value.3736

So the external pressure it can be bigger than 0 but it has to be less than or equal to the final pressure.3740

Therefore, the largest value that the external pressure can have is the P2.3748

Let us go ahead and calculate that.3753

That is very easy.3755

P1 v1 =P2 v2, I have P1 v1 and I have v2, it was very easy for me to calculate.3759

My final pressure P2 is going to equal P1 v1/ v2.3769

It is going to equal, the initial pressure is 2 atm, the initial volume is 1.5 deci³, my final volume is going to be 3.5 deci³.3778

I do not have to make any conversions here because units are going to go ahead and cancel.3794

My P2, my final pressure is going to end up being 0.857 atm.3800

This is the largest value that the external pressure can have in order for this to happen.3808

It can be anything less than that.3813

If it is less than that, less work is going to be done.3815

If it is a little more than that, more work is going to be done.3817

At a certain point, when the external pressure is the same as the final pressure of the system that I want,3820

that is going to be the largest value that I can have in order to actually reach P2.3828

My external pressure is equal to P2 is equal to 0.857 atm.3837

In order to calculate the actual work done, the work is equal to the external pressure × the change in volume.3846

Here is where we have to make some conversions.3855

Let us just go ahead and do this.3860

We can go ahead and do it this way.3865

The external pressure we said is going to be 0.857 atm, the change in volume is going to be from 1.5 to 3.5.3868

It is going to be 3.5 - 1.5 which is going to be 2.0 L.3877

We get 1.714 L/ atm that is a perfectly valid number.3885

Let us go ahead and convert it to J just for the heck of it.3891

1.74 L/ atm × 8.314 J / 0.08206 L/ atm.3895

This is the relationship between a L/ atm and a joule.3912

When I do this calculation, I end up with 173.7 J.3915

If you are wondering where this relationship came from between L/ atm and joule, these are the two values of R.3922

R is the same.3930

I have 8.314 J/ mol K.3932

I have mol K 0.08206 L/ atm.3941

Mol K cancels mol K, I have myself a conversion factor, J L/ atm.3953

L/ atm is unit of energy.3959

Here is the conversion factor, 101.3 is what this number is but I like to write it that way.3961

There you go, that is the maximum amount of, this is the work done by the maximum value that the P external can have3967

in order to undergo this expansion under constant external pressure.3981

Thank you for joining us here at www.educator.com.3987

We will see you next time for our continuation of energy in the first law, bye.3989