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

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

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

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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
- The First Law of Thermodynamics
- Example 1: What is the Change in Energy of the System & Surroundings?
- Energy and The First Law II, cont.
- 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
- 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
- Example 3: Calculate the External Pressure and Work Done

- 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

### Physical Chemistry Online Course

### 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

*We are going to start with the first law of thermodynamics.*0056

*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 surroundings*0373

*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

*Of course, we will have more to say about this in subsequent lessons.*0389

*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

*The value of the state property, think about this example.*1260

*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 surroundings*1605

*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 energy*1617

*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

*These diagrams will help you solve your problems.*1662

*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

*I’m going to go ahead and shade this in.*1963

*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 pressure*3189

*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 have*3967

*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

1 answer

Last reply by: Professor Hovasapian

Thu Sep 14, 2017 4:14 AM

Post by peter alabi on September 12, 2017

Hi professor,

suppose instead of integration as follow [p(v)-Pext] like we did in this lecture, we integrate as follow {Pext-P(v)}? is the area between this function still work? and if so, what does that value means?

Thank for the great lecture, you've helped me passed my general chemistry, biochemistry, and AB calculus, now you've started saving my grade in physical chemistry. thanks a lot!

1 answer

Last reply by: Professor Hovasapian

Mon Feb 8, 2016 1:35 AM

Post by Van Anh Do on January 22, 2016

Hello. I have a question about what you said around 34:30. Why isn't the path for the gas expansion be along the isotherm? I thought as the gas expands in volume, the pressure and volume would follow that curve? Thank you.

1 answer

Last reply by: Professor Hovasapian

Tue Jul 28, 2015 11:24 PM

Post by Ayanna Hogan on July 24, 2015

What would be the are under the entire cure of the P-V diagram ?