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

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

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## Download Lecture Slides

## Table of Contents

## Transcription

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### Changes in Energy & State: Constant Volume

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
- Recall
- First Law
- Where Does (∆U = Q - W) or dU = dQ - dU Come from?
- Relations between Changes in Properties and Energy
- Relations between Changes in Properties and Energy
- Rate of Change of Energy per Unit Change in Temperature
- Rate of Change of Energy per Unit Change in Volume at Constant Temperature
- Total Differential Equation
- Constant Volume
- If Volume Remains Constant, then dV = 0
- Constant Volume Heat Capacity
- Constant Volume Integrated
- Increase & Decrease in Energy of the System
- Example 1: ∆U and Qv
- Important Equations

- Intro 0:00
- Recall 0:37
- State Function & Path Function
- First Law 2:11
- Exact & Inexact Differential
- Where Does (∆U = Q - W) or dU = dQ - dU Come from? 8:54
- Cyclic Integrals of Path and State Functions
- Our Empirical Experience of the First Law
- ∆U = Q - W
- Relations between Changes in Properties and Energy 22:24
- Relations between Changes in Properties and Energy
- Rate of Change of Energy per Unit Change in Temperature
- Rate of Change of Energy per Unit Change in Volume at Constant Temperature
- Total Differential Equation
- Constant Volume 41:08
- If Volume Remains Constant, then dV = 0
- Constant Volume Heat Capacity
- Constant Volume Integrated
- Increase & Decrease in Energy of the System
- Example 1: ∆U and Qv 57:43
- Important Equations 1:02:06

### Physical Chemistry Online Course

### Transcription: Changes in Energy & State: Constant Volume

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

*Today, we are going to start talking about, or to continue our discussion of the first law and energy and work and heat and things like that.*0004

*We are going to start talking about changes in energy and state.*0014

*In this particular lecture, we are going to be talking about constant volume processes.*0018

*We are going to start putting some constraints either to constant volume, constant pressure, constant temperature, things like that,*0022

*to see if we can learn some things about what is going on with the first law and with the energy.*0029

*Let us jump right on in.*0035

*Let us recall, this will go with black today.*0038

*U energy is a state function or a state property.*0045

*It does not depend on the path that you take.*0055

*It just depends on your beginning state and your ending state.*0058

*If I start at 0 feet sea level and ago up to 500 feet above sea level, the difference is 500 feet.*0063

*It does not matter whether I drop down to negative 300 and go up to 800, then come back to 500.*0072

*All that matters is the beginning and ending state.*0078

*Work and heat are not state functions, they are path functions.*0082

*W is a path function or path property.*0087

*Heat is also a path function.*0107

*A path function, its value, the value of work or heat depends on the path that we take.*0111

*I’m going from here to here, this symbols less work than this or this.*0117

*The amount of heat from here to here is going to be less than the amount of heat like that.*0124

*It depends on the particular path that we take.*0129

*The first law is this, it says that the change in energy, the total change in energy of the system is equal to*0132

*the heat gained by the system - the work loss by the system.*0145

*That is all that is.*0151

*We will often use this form.*0154

*This is the fully integrated form, we will often use the form, the differential form because*0156

*we will be leading much more on the mathematics and more sophisticated mathematics.*0161

*So du= dq – dw, the differential change in energy = the differential change in heat -the differential change in work.*0167

*A quick word about this, U is a state function.*0179

*W and Q are path functions.*0186

*The symbolism that I have used is a DDD and tends to imply that they are the same.*0189

*They are not the same.*0193

*Many books will differentiate notationally a state function and the path function.*0195

*They used this differential notation to standard, when you are used to the d for a state function.*0200

*They will use some variation of that for a path function.*0206

*You often see something like dw, that is Greek δ and dq or you will see d with a line through it.*0209

*For dw and dq, this let us know that these are path functions.*0219

*Most specifically, for mathematical point of view they are an exact differentials.*0224

*A state function is an exact differential and a path function is an inexact differential.*0228

*And we will be talking more about what we mean by exact and inexact and the properties of exact and inexact differentials.*0234

*I, myself, I do not like to differentiate because I believe, I mean physical chemistry and*0240

*Thermodynamics, there is already just an abundance of symbolism as it is.*0245

*To introduce new symbolism, it just confuses me.*0250

*To me, personally, it is just easier to remember that work and heat are path functions and energy is a state function.*0253

*Other than that, it should not cause you any problem.*0262

*I’m going to go and use the same, but they are not the same.*0265

*This is an exact differential, these are inexact differentials.*0267

*Let us go ahead and talk about integral from an initial state to a final state of this equals U2 – U1.*0272

*This is the fundamental property of a state function but it integrates the way you are used to according to the fundamental theory of calculus.*0285

*It is equal to δ U, this is an exact differential.*0293

*Exact differentials integrate like this.*0297

*You take the final state - the initial state of the integral of the particular function.*0299

*Exact differential, this is a state function.*0305

*When we integrated heat, dq, we do not get Q2 - Q1, we just get whatever the value of Q is.*0313

*This is not δ Q, this is an inexact differential.*0325

*Inexact differentials cannot integrate the same way but exact differential do.*0335

*You do not follow this final – initial.*0339

*It depends on the path so you just get a quantity.*0345

*dQ and dQ, it does not make any sense because a system does not possess heat.*0361

*Heat is something that shows up during a change of state.*0368

*Heat is actually a process, when there is a temperature differential of two of the system in the surroundings,*0371

*or two systems whenever it is, heat is the transfer.*0377

*Heat happens at the boundary.*0382

*That is what is going on, there is a system does not possess heat.*0384

*The system has a certain temperature.*0387

*Temperature and heat are not the same thing.*0389

*Heat is what happens during the change of state.*0391

*It makes no sense to say that there is a certain amount of heat at the beginning, a certain amount of heat in the end,*0394

*and change in heat is the difference between the two.*0400

*That does not make sense the system does not possess heat.*0404

*Granted, we tend to be little loose with our language usage when we talk about heat as if the system does possess it.*0407

*That is just part and parcel of the historical discussion of thermodynamics.*0414

*That sort of looseness in language that existed, signs in general, it is not a problem as long as you remember these things.*0417

*δQ makes no sense because a system does not possess heat.*0424

*And the same thing with work, when we integrate the differential of the work from one state to another,*0444

*all we are adding up all the work done along the particular path, we get the final work its equal to W.*0450

*It is not equal to dW because the system does not possess work.*0456

*It does not make any sense.*0462

*Work inexact path function.*0465

*The dW makes no sense for the same reason, a system does not possess work.*0471

*Here is what is interesting, the Q and W are path functions their difference is a state function.*0479

*That is extraordinary, that is truly amazing.*0488

*Q and W, they are path functions but their difference or sum in a different convention, the chemists convention usually write Q + W.*0492

*Our convention is Q – W but their difference, I will put sum here, is a state function, that is amazing.*0511

*State function, energy.*0525

*Where does this come from?*0534

*The question where does δ U = Q –W come from?*0536

*Or more often we use this with a differential = dq - du come from.*0545

*Let us talk about a cyclic process.*0557

*A cyclic processes is where you start with some initial state, you have a change of state,*0559

*you go to state 2 and then from state 2, you come back to state 1.*0563

*Exactly, it sounds like it is a cyclic process.*0567

*You go to some final state, you come back, and end up where you started.*0569

*The work of the cyclic process is equal to the integral of all the work done along the particular path that you take.*0575

*The integral dW, we will often put a circle around it to let us know that it is a cyclic process.*0586

*This is sort of an older notation, you probably do not see all that much anymore in modern books.*0590

*Cyclic just means cyclic, that is all it is, nothing strange about it.*0597

*In the Q, the heat lost or gained during a cyclic process that is equal to all the integrals of all of the dQ.*0602

*All of the differential heat elements along that.*0615

*Now these are not usually 0, in other words the cyclic work for a processes is not usually 0.*0618

*The cyclic heat is not usually 0.*0626

*These integrals are usually not equal to 0.*0633

*If they are, it is strictly a coincidence that is it.*0644

*In general, the cyclic integrals of path functions, the cyclic integrals of inexact differentials are not equal to 0.*0651

*In general, the cyclic integral of path functions / inexact differentials does not equal 0.*0667

*Now the cyclic integral of a state function does equal 0.*0686

*The cyclic integral of any state function does equal 0, as it must.*0696

*A state function depends on its initial and final state.*0713

*In a cyclic process, the final state and initial state are the same.*0716

*State - state you get 0.*0724

*In a cyclic involve any state function and the exact differential is equal to 0, as it must.*0726

*Initial, final, you go from initial to final and then you come back.*0735

*The cyclic integral of some dz = 0, where z is a state function.*0743

*Our empirical experience of the first law is the following.*0753

*This is what this says, this is the science, this is thermodynamics, it is all empirical.*0761

*Our empirical experience of the first law is this, if the system is subjected to a cyclic process or cyclic transformation,*0767

*either one, the work transferred to the surroundings, in other words,*0802

*the work gained by the surroundings to the surroundings = the heat transferred from the surroundings.*0817

*Now mathematically, this says this.*0846

*Our convention is we said that heat and work, these are effects that manifested in the surroundings.*0850

*We never measured something directly in the system.*0857

*We know what is happening in the system based on what we observed in the surroundings.*0859

*And we can take the systems point of view but what we are directly measuring is, is what is happening in the surroundings.*0863

*It is not a big deal as long as you think of the surroundings.*0870

*And if you want to talk about the system, you just switch your point of view.*0873

*When we talk about heat gain by work gained by the surroundings, we are talking about work done by the system.*0876

*Work that is lost by the system.*0882

*You just have to switch around.*0884

*Mathematically, it says this, a cyclic process dQ = dW.*0887

*This is our empirical experience of the first law.*0898

*Now let us see what happens here.*0902

*This is our mathematical or empirical experience of the first law.*0905

*What is going to happen is the following.*0908

*This is the same as, let me just move one of the integrals over to the other side which you end up with is this.*0911

*The cyclic integral of dQ - the cyclic interval of dW is equal to 0.*0919

*This is just the same as cyclic integral of dQ -dW = 0 because integration is a linear operator.*0929

*It is linear so I can just pull out the linear and the sort distributes.*0941

*We will talk about that little bit in calculus.*0946

*Now, I have a cyclic integral of something right here, this in the grand underneath the integral sign, the cyclic integral of it is equal to 0.*0949

*I know that the cyclic integral of any state function equal 0.*0961

*The cyclic integral of any exact differential = 0.*0965

*Therefore, dQ - dW must be some state function.*0969

*We call the state function energy, that is where this comes from.*0973

*We drop the definition on you but this is where it comes from becomes it comes from the empirical experience.*0978

*Empirically, this is what happens.*0982

*We want to give this a name we call it the energy of the system.*0984

*Any cyclic integral that = 0 is some state function.*0991

*We need to get the state function a name.*1006

*We call this state function which is dQ - dW or Q – W, we call this state function energy, the energy of the system.*1014

*This is where it comes from.*1031

*dU= dQ – dW, the differential change in energy of the system comes from*1036

*the differential change in heat - the differential change in work or the integrated form δ U= Q – W.*1045

*Remember, δ Q and δ W they make no sense.*1055

*Differential form and integrated form, we will be called the finite form.*1062

*Only δ U is defined.*1073

*This is the differential form, when we integrate this, when we integrate both sides, we end up getting is δ U = Q- W.*1079

*We do not get U, we get Q and W because they are path functions but du is a state function, we get the δ.*1087

*Only the δ U is defined, we do not measure energies of the system.*1096

*We measure changes in energy of the system, that is what we do.*1101

*Science best measures changes in things.*1104

*Only dU is defined, not U.*1107

*Let us see here, we have that δ U = Q – W.*1117

*If we measure the heat loss by the surroundings then subtract from it the work gained by the surroundings, both of which are easily measurable.*1135

*We get δ U, the energy change of the system.*1186

*If we measure the heat lost by the surroundings, the heat of the surroundings loses.*1202

*Once it loses something that goes into the system.*1209

*Then subtract from it the work gained by the surroundings, we get the change in energy of the system.*1212

*Basically, it always says that if you have a system and if you have the surroundings, if heat goes into the system,*1218

*if energy goes into the system as heat and a certain amount of energy to leave the system as work,*1225

*what you get is just the net change in energy.*1231

*It is just simple arithmetic is what it is.*1234

*I give you $100, take back $25, you are left with $75.*1237

*That is all this is, this is a simple accounting of energy in terms of work and heat, in terms of paper or coin money.*1242

*The change in state means changes in properties of the state of the system such as temperature, pressure, and volume.*1254

*The things that are easily measurable and they are properties of the state.*1286

*Also, temperature, pressure, and volume, these things are state functions also just like energy.*1290

*They do not depend on the particular path that you take to get there.*1297

*If you start with a system at 2 atm and then you increase it to 10 atm, and I take it down to 0.2 atm,*1300

*and take it back to 5 atm, the difference is from 2 to 5.*1308

*The change in pressure is 3 atm.*1311

*It does not matter how you got there, same with volume and the same as temperature.*1314

*These are state functions.*1317

*What we want to do, we are going to find relations between the changes in state,*1320

*the changes in these properties, and the change in energy.*1325

*The first law was work and heat, now we want to get a little bit more direct.*1328

*We want to express it in terms of things that the properties of the system itself, the temperature, the pressured, the volume.*1333

*What the changes in does, the quantities and say about the change in energy.*1339

*Let us find relations between changes in these properties and changes in energy.*1353

*Again, that temperature no b a P, temperature, pressure, and volume are state functions.*1382

*They are exact differentials, dT dP dV.*1398

*These are exact differentials.*1401

*We can start by assuming that energy is a function of temperature and pressure, temperature and volume, pressure and volume.*1406

*We have to start anywhere we like and this is certain how it appears.*1415

*What you are seeing us do here on a page, here is where mathematics starts.*1418

*You are seeing the final result of this.*1426

*You are not seeing the process the sort of let up to it.*1428

*When a scientist sits down on a theoretical, he starts playing with mathematics.*1431

*He does not necessarily know where he is going.*1437

*He just starts playing with derivatives.*1440

*He starts playing with equations and by starting to take derivatives and taking other derivatives, integrating and moving this here.*1441

*You will see something that looks like it is important.*1451

*Using the final result of that, you are not seeing the process.*1456

*If it seems like we are pulling things out of the air, when we are doing these mathematics, we are not pulling things out of the air.*1459

*What you are seeing is the final result of all the work that is been done.*1467

*The truth is when somebody has done it for the first time, it is all over the place.*1471

*You go this way, and this way.*1475

*That is the real nature of science.*1477

*You do not see that in your experience and from your books.*1479

*It just seems like one day you are a scientist, woke up, and goes like that.*1482

*It did not happen like that.*1485

*Do not worry about it, it does not entirely make sense to you all at once.*1489

*Let us start with this.*1494

*We are going to let energy equal, we are going to take two variables, temperature and volume.*1497

*We are going to say that energy is a function of temperature and volume.*1504

*In other words, if I change the temperature, if I change the volume, or if I do both, the energy of the system changes.*1507

*If I can express it this way, you get the following.*1514

*I can express the differential change in energy this way, du = du dt vdt.*1518

*Do not worry, I would explain everything here.*1530

*DU dv, these are partial derivatives under constant V.*1534

*This is constant temperature DV.*1540

*This is called a total differential of this variable.*1547

*You have a function which is a function of two variables.*1552

*It is differential can be written like this, the total change in energy is equal, du = partial du dt × dt + du dv under constant temperature × dv.*1555

*Let us talk about what all this actually means.*1569

*You might want to go ahead and go back to the second lesson of the series, one that discusses partial differentiation*1572

*and actually introduces this thing called the total differential.*1579

*This is called the total differential for du.*1582

*It is based on the presumption that is a function of two variables.*1588

*First of all, we want you to notice that the differential of any state property which is an exact differential.*1595

*Sorry if I keep repeating myself, it is important though.*1618

*The differential can be written like this, it can be written in this form.*1623

*Any state property that you deal with, if you know that state property is a function of two variables,*1632

*you can write that state property the differential like this, always.*1637

*This is a proofing, any state property, any exact differential can be written like this on the right hand side of the equality sign.*1642

*The question is what does this mean?*1651

*Let us examine what this means.*1653

*What does this mean?*1658

*What if you are faced with something like this, what if you are faced with some mathematical equation*1663

*and you definitely want to give it a habit of stopping and examining what every single term means*1668

*and what every single term says and you want to get physical meaning.*1673

*In the case of du dt, this is the rate of change of energy with respect to temperature.*1678

*In other words, as the temperature changes by 1 unit, how much does the energy change?*1684

*This is a rate of change.*1689

*This is a derivative, is it like a derivative like any other derivative.*1690

*This happens to be partial derivative because we are dealing with a function of two variables instead of 1.*1693

*This really is just the same as du dt.*1698

*When you take that rate and you multiply it by the change in temperature, you get the total change in energy for the increment.*1700

*The same thing here, this is the rate of change of energy per unit change of volume.*1707

*Time and change in volume of the system experiences using a total change in energy.*1713

*You knew you want to get in the habit of doing this, a lot of work that we do specially in thermodynamics.*1718

*It is going to be strictly mathematical.*1724

*Actually, throughout all of physical chemistry, quantum mechanics, spectroscopy, you need to be able to assign physical meaning.*1727

*After you have done this a couple of times, pretty soon you are not intimidated by the mathematics anymore.*1732

*It makes sense what is going on.*1738

*There is nothing here that is counterintuitive.*1741

*It seems that way simply because you just unaccustomed of the mathematics.*1745

*You can get used to it like anything else.*1749

*This is something new that we are going to be using with some rather sophisticated mathematics.*1751

*But it is nothing that you do not understand, that you have not really been exposed to.*1755

*What you have exposed to, we will go over but do not just pull away from it, assign physical meaning to it.*1759

*Eventually, you will get them very quickly of knowing exactly what is going on and be able to relate this math*1766

*to relate what is happening physically in a system, which way temperature is moving, which way a system as the surrounding.*1773

*It says this way and that way, you will be able the see what is happening physically.*1780

*When you see what is happening physically, that is when you understand what is happening mathematically.*1783

*Let us see what this means.*1792

*The first term du dt, du dt with this little subscript V on it, this is the rate of change of energy per unit change in temperature.*1796

*That is it, it is just a rate, it is just a derivative per unit change in temperature.*1818

*Look at the units, that is where I start.*1829

*I look at the units to help it make physical sense for me when I see a derivative like this.*1833

*We are talking about physical things, this is not just theoretical.*1842

*Well energy is in joules, temperature is in Kelvin, and this is J/K.*1846

*The V subscript that tells us that something is happening at a constant volume.*1854

*In thermodynamics, any subscript let us know that it is happening as we keep that variable fixed.*1859

*The rate of change of energy per unit change in temperature, just look at the units.*1867

*Therefore, if T changes by an amount, if the temperature changes by some differential amount dt,*1873

*and this du dt sub V × dt, it gives the energy change for the particular incremental change in temperature.*1889

*It gives the energy change, holding V constant.*1904

*Again just look at the units, du dt that is J/K.*1917

*If you multiply that by a change in temperature, a temperature is in Kelvin, we are left with joules.*1924

*You already notice this is heat capacity and you will see in a minute that it is.*1930

*We are just interpreting what this is right now.*1937

*We are just getting a sense of the mathematics.*1940

*Attaching meaning to the partial derivatives, what they mean, we do not just want them to be scrolls and scribbles on a paper, that mean nothing to us.*1944

*We know how to manipulate this so we have done calculus.*1952

*We are very good at calculus.*1955

*This is a simple calculus.*1956

*For the next one, du dv sub T well this is exactly what you think it is.*1962

*It is the rate of change of energy per unit change in volume at a constant temperature.*1972

*Therefore, if the volume of the system changes by some differential amount,*2006

*dv changes by an amount dv then this du/ dv sub T × dv.*2017

*It gives the total energy change for the differential energy change, du.*2042

*It gives the incremental energy change.*2047

*Again look at the units, energy is in joules, volume always take deci³, dv deci³.*2057

*Volume cancels volume, leaving you just energy, that is what is happening.*2068

*Therefore, this expression du=du dt sub V dt + du dv sub T dv says the following.*2076

*It is absolutely imperative that you do this, that you assign physical meaning, otherwise this stuff is going to get away from you very quickly.*2099

*Because from this point on, it is going to be essentially mathematical.*2108

*It is physical, we assign physical meaning to it but we are expressing these physical changes mathematically with partial derivatives.*2112

*This expression says the following.*2120

*If T changes by an amount dt and volume changes by an amount dv simultaneously, the amount dv*2127

*then the energy change which is du as the sum of the two.*2161

*This is totally intuitive.*2170

*It is just as sum of the two, if I change the temperature, this amount, and the rate at*2172

*which the energy changes per unit change in temperature, that is going to give me the total energy change for temperature movement.*2183

*And if the volume changes also, that change in volume × the rate of change of the energy of the volume gives the energy change for the volume.*2190

*If I add those two together because the total energy change, that is it.*2198

*That is all this is saying, you know this already.*2201

*When you know this intuitively, you know this since you were a kid.*2204

*There is nothing here that is strange.*2206

*It is just the symbolism that can look a little intimidating.*2210

*Let us see here, if we happen to know what these partial derivatives are, in other words*2216

*if we happen to know what du dt and du dv sub T are, we just integrate the expression.*2231

*We just put into the differential because we just integrate the expression.*2245

*We can just integrate this expression to get the total energy change.*2251

*We have the du, we have the expression on the right.*2278

*If we know what these partial derivatives are, we just put them in and integrate that function that gives us the total energy change.*2280

*Nice and simple.*2287

*Let us see if we can express these partial derivatives in terms of things that we know, in terms of things that we can measure.*2292

*We know what they mean.*2312

*Let us see if we can actually measure them somehow.*2313

*Let us see if we can express these partial derivatives, this and this, in terms of things we can measure.*2317

*Let us see, we have the du is equal to du dt sub v dt + du dv sub T dv.*2342

*We also have the first law which says du=dq – dw which is equal to dq and dw is equal*2362

*to the external pressure × the differential change in volume.*2374

*Right, that is the definition of work, pressure × volume.*2379

*I’m going to go ahead and put this expression over here on the left side so we get dq - P external dv*2382

*is equal to the right side du dt sub v dt + du dv sub T dv.*2394

*I'm going to ask that you actually confirm this.*2407

*There are a lot of symbols that I’m writing on the page and all of the subscripts and letters so it does get to be a little confusing.*2409

*Please make sure I’m actually writing this correctly.*2417

*In any case, this is where we start.*2423

*Let us see what we can do with this.*2425

*We have the first law, we have the expression of the change in energy in terms of the properties temperature, volume,*2427

*we sent them equal to each other and we are left with this.*2435

*Let us see if we can make sense of this.*2437

*This is the equation we start.*2440

*I apply this equation to our changes of state.*2442

*We apply this equation to various changes of state.*2447

*Here is where it begins, various changes of state.*2461

*The first change in state we are going to do is we are going to change of state under conditions of constant volume.*2469

*A system goes from state 1 to state 2 but in that process, the volume stays constant.*2484

*We have dealt with the volume change.*2488

*If V remains constant and then this dv equal 0, V final V initial, the volume stays the same.*2498

*There is no differential change in volume so dv = 0.*2513

*Let me write this expression again, we have dq - P external dv = du dt sub V × dt + du dv sub T × dv.*2517

*If dv = 0 then this goes to 0, that goes to 0, what we are left with is the following.*2540

*Dq, we are going to go ahead and put that V there because we are under constant volume process, = du dt sub V dt.*2548

*Let us see, it gets to U = dqv and we write the other equation also, du = dq - the external dv.*2569

*This goes to 0 so we are also left with du = dqv.*2589

*What we have here is the following, under constant volume, the change in energy of the system = the change in heat.*2599

*It is equal the heat that is a lost by the surroundings or that heat gained by the system, if you want to take the systems point of view.*2607

*Again, we are taking the surroundings point of view for the most part.*2615

*For a constant volume processes, the heat that the surroundings loses happens to equal the change in energy.*2619

*If I want to know what the change in energy is, I just have to measure how much heat the surroundings loses.*2627

*Over here, change in heat, the amount of heat that the surroundings loses which is equal to the change in energy*2632

*is equal to the rate of change of the energy of the temperature × the temperature increment.*2641

*This equation which relates dqv which is the heat withdrawn from the surroundings, the heat lost by the surroundings.*2651

*The heat withdrawn from the surroundings to an increase in temperature of the system which is the dt.*2669

*Well, both of these are easily measurable, the change in temperature and the change in heat.*2697

*We can measure the change in heat lost by this, we can measure the heat lost by the surroundings.*2711

*We can measure the change in temperature of the system.*2717

*The ratio which is dqv/ dt it is called heat capacity.*2722

*The change and heat over a change in temperature, that is the definition of heat capacity.*2733

*In this particular case, because it is a constant volume process, this ratio of the heat of the surroundings loses*2738

*divided by the temperature by which the system, the temperature increase of the system,*2745

*this is defined as the constant volume heat capacity.*2754

*Constant volume heat capacity is defined as heat capacity in this particular case because*2760

*the process is happening under constant volume it is called the constant volume incapacity.*2766

*Dq sub V dt which is defined as a constant volume heat capacity C sub V.*2776

*That happens to be associated, identified, with this partial derivative.*2784

*A partial derivative which is the change in energy or change in temperature at constant volume.*2792

*We were able to associate some partial derivative with something that we can easily measure,*2799

*this P capacity which happens to be under constant volume.*2807

*The heat capacity just happens to be the heat withdrawn from the surroundings divided by the change in temperature of the system.*2811

*You are in chemistry, you are accustomed to thinking about it as a heat gain by the system divided by the temperature increase of the system.*2822

*That heat change of the system divided by the temperature change of the system, that is fine, it is the same thing.*2831

*Again, we are taking the surroundings point of view.*2836

*This is the heat lost by the surroundings.*2838

*It happens to be the heat gained by the system.*2841

*It is just a question of point of view, as long as you know what is happening.*2844

*That is the very definition of heat capacity, that is how we define it.*2848

*Let us just go ahead and do what we do with differentials.*2853

*We can just move this over here so dq sub V = cvdt.*2856

*We just move that there.*2869

*This is the differential form and we know this already.*2872

*The change in heat is equal to the heat capacity × the change in temperature.*2879

*We know this from general chemistry, constant volume.*2883

*Let us go ahead and write the integrated form here.*2888

*Our integrated form is, let me integrate that function, we get the energy = the integral from temperature 1, temperature 2, and CVDT.*2895

*We said that dqv=cvdt, dqv is equal to du.*2923

*That is what we get as the first law tells us when there is no change in volume.*2932

*We just put this over here, we will get this thing and we integrate it.*2936

*In other words, du is equal to cvdt and we integrate both sides.*2941

*The integral of du is δ U, the integral is that integral from temperature 1 to temperature 2.*2947

*From du= dqv, we also get this integrated version which is δ U= QV.*2958

*Q is a path function, so there is no δ Q, the change in energy of the system happens to equal at a constant volume processes,*2974

*the change in energy of the system happens to equal the heat withdrawn from the surroundings.*2982

*If we are measuring temperature, the change in energy of the system happens to equal the integral of the change in temperature,*2989

*the integral of the heat capacity of the system × the change in temperature from one temperature to the other.*2997

*These are the equations that are important.*3003

*If this is heat capacity happens to be, if T is constant over the range of temperature increase then we have δ U = CV δ T.*3010

*In other words, over a certain temperature increase, let us say 25 to 50, if the heat capacity does not change.*3047

*We do not need to integrate, we can just take the change in temperature.*3052

*This is how we seen it before in general chemistry.*3055

*Our assumption was that heat capacity is constant over a range of temperatures.*3057

*As it turns out, heat capacities temperature dependent.*3063

*The hotter something gets, the heat capacity changes.*3066

*This is the wheel equation, that right there.*3070

*These two equations, in other words, δ U= the integral from temperature 1 to temperature 2 of the constant volume heat capacity × that.*3078

*The change in energy = the heat withdrawn from the surroundings, these two equations,*3094

*they express the energy change of the system, in terms of measurable quantities.*3104

*That is what we wanted.*3122

*In terms of measurable quantities, that is good, that is what we want.*3125

*In science, we measure things.*3135

*We need to be able to measure things.*3137

*The theory that we got from manipulating to get a partial derivatives, that tells us one thing*3140

*but we need to associate these partial derivatives as something we can measure.*3146

*We have something that dudt V this, we can associate it with the heat capacity.*3150

*Heat capacity is easily measurable.*3160

*We measure the heat lost by the surroundings or the heat gained by the system and we divide by the temperature change in the system.*3162

*We have something mathematical that is associated with something physical, something we could measure.*3168

*This is beautiful, it is what we want.*3174

*Mathematically, δ U = QV.*3179

*Δu QV they have the same sign.*3198

*They have the same sign.*3206

*Now given our particular convention, a positive value of heat means that heat is flowing from the surroundings to the system.*3209

*Heat flows from the surroundings, in other words through the system.*3231

*If heat is positive and change in energy is positive.*3255

*A positive QV implies that δ U of the system, this is what all these means.*3263

*I’m just explicitly writing at this time in general.*3272

*It means that δ U is positive.*3278

*That means the system has increase in the energy of the system.*3284

*A negative QV means that heat is flowing to the surroundings.*3295

*In other words, the heat is flowing from the system.*3302

*The system is losing heat, therefore, the energy of the system decreases.*3305

*Δ U is negative.*3311

*If δ U is Q, Q is positive W is positive.*3314

*If Q is negative δ U is negative.*3317

*This is a decrease in energy.*3321

*The constant of volume heat capacity is always greater than 0.*3330

*A positive temperature change implies that δ U is positive.*3339

*It is an increase in energy, a drop in temperature implies that the δ U is negative.*3348

*If the temperature of the system increases, the energy of the system increases.*3358

*If the temperature of the system decreases, the energy of the system decreases.*3361

*That is what is going on here.*3365

*This is actually really important.*3367

*Therefore, at a constant volume, at constant V the temperature is a direct representation of the energy of the system.*3369

*Let us see what temperature actually is.*3398

*Temperature is a measure of the energy, the average energy of the system.*3400

*OK let us go ahead and do an example.*3407

*It is a very simple example.*3410

*A lot of these lectures I’m only going to have a lot of these preliminary lectures.*3412

*I already have 1 or 2 examples, do not worry about that.*3417

*In subsequent lessons and for several lessons, after we get these preliminary theoretical discussions out of the way,*3423

*that is where we are going to do the bulk of our examples.*3429

*It is not going to be a lesson where we have a little bit of discussion and a handful of examples.*3432

*I’m going to set aside complete lessons, several of them at the end of this unit, to do all of the example problems that we need.*3437

*Do not worry there is only 1 or 2 showing up in these particular lectures.*3446

*We are going to do a lot and when I say a lot I mean a lot.*3449

*We definitely need to get familiar with this material.*3454

*Handling the first law, we need to set a good foundation.*3456

*We are going to do a lot of problems, I promise you.*3459

*Let us take a look.*3464

*The first example we have, 2 mol helium gas, they are taken from a temperature of 25° C to a temperature of 55° C.*3466

*The molar heat capacity happens to be 3 ½ R, notice molar heat capacity.*3477

*This is the amount of heat J/ K/ mol.*3485

*Molar heat capacity, if I just said heat capacity it would be J/ K/ C.*3491

*If I say specific heat capacity, we are accustomed to in general chemistry it would be J/ K/g or J/℃/ g,*3497

*because specific heat does talk about mass, lower heat capacity, J/ mol/ K/ mol.*3506

*We want to find the change in energy of the system and we want to find the heat that is lost by the surroundings for the transformation.*3512

*We know that δ U is equal to the integral from temperature 1 to temperature 2 of CVDT, we know that already.*3524

*We also happen to know that δ U is equal to QV.*3537

*We find this and we find that.*3541

*Let us go ahead and work this one out, it is really simple.*3544

*Δ U is equal to the integral of T1 to T2, the CV or constant volume heat capacity is 3 ½ rdt.*3548

*The 3 ½ R is not a function of T, it is a constant so we can pull it out, = 3 ½ R × the integral of T1T2 dt.*3561

*That = 3 ½ R × δ T, notice I have not put the values in.*3579

*I may have to change this to 298 and change the 55 to whatever it is, 55 goes to 73.*3584

*In this particular case, because this is constant I do not to have to evaluate the integral.*3593

*This right here becomes R δ T, temperature is a state function.*3598

*The integral of the state function is δ of the T.*3605

*This is really simple, let us just go ahead and put the values in.*3610

*We get 3 ½ × R, the R value we are going to take is 8.314 J/ mol/ K.*3614

*The change in temperature is going to be the 55 - to 25, so 55° - 25°.*3624

*The δ T, in terms of Kelvin and Celsius is the same.*3634

*A difference of 1° C is the same as 1° K.*3639

*I do not actually have to convert to it to K when I’m doing δ.*3642

*This is going to be 30 K, K and K cancels and I end up with is, if I did my arithmetic correctly which more often that I do not, 374 J/ mol.*3646

*We are left with J/ mol.*3663

*Since we have 2 mol of helium, we multiply that and we get a total of 748 J.*3667

*748 J that is the change in energy.*3680

*So δ U= 748 J.*3683

*QV happens to equal that, QV = 748 J.*3688

*Under these circumstances of constant volume, this 2 mol of helium gas that goes for 25° to 55° C, the energy change was 748 J.*3697

*The energy of the system, the heat lost by the surroundings is 748 J.*3707

*The 748 J went from the surroundings to the system, that is what happened.*3715

*Let us go ahead and close this out.*3726

*The important equations for this particular lesson, we have the heat capacity which is defined as dq/ dt,*3731

*the change in heat over the change in temperature.*3749

*The heat lost by the surroundings divided by the temperature increase of the system or*3751

*the heat gain by the system divided by the temperature increase of the system.*3756

*Totally your choice, as long as you are consistent.*3760

*This happens to equal, it is associated with the derivative, this partial derivative,*3763

*the change in energy or the change in temperature at constant volume.*3770

*Du = CVDT, du = dq, δ U + integral from T1 to T2 of CVDT, this is just the differential form, this is the integrated form.*3776

*Δ U = Q not dq.*3805

*At constant volume, the heat that is lost by the surroundings = the change in energy of the system.*3814

*If I want to know what the change in energy of the system is, all I have to do is measure how much heat is lost by the surroundings.*3827

*And that is what I do, I do not measure directly into the system.*3835

*I measure what is happening in the surroundings which is why we keep saying surroundings.*3838

*I know that in chemistry we are accustomed to thinking about the system but where we can take our measurements*3843

*in order to find what was happening in the system is in the surroundings.*3848

*The change in energy of the system is equal to at under constant volume heat capacity ×*3854

*the differential change in temperature for the particular increment.*3861

*If I integrate all those increments over the temperature change, I get the change in energy of the system.*3865

*Thank you so much for joining us here at www.educator.com.*3873

*We will see you next time for a continuation of discussion, bye.*3876

0 answers

Post by Joyce Ferreira on October 22 at 09:12:49 PM

I am really enjoying your lessons. Your teaching skills are extraordinary. Thank you very much and God bless you!

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 1:23 AM

Post by Van Anh Do on February 14, 2016

For the example, I thought Cv is the heat capacity at constant volume and since the problem doesn't say that the system has a constant volume the whole time, how do we know to plug 3/2R in for Cv? Thank you.

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 1:14 AM

Post by Van Anh Do on February 14, 2016

For this lecture, are we assuming that pressure is also constant? I'm not sure why we're able to leave out pressure. Thank you.

1 answer

Last reply by: Professor Hovasapian

Wed Jun 3, 2015 7:41 PM

Post by Joseph Carroll on June 3, 2015

Exemplary teaching methods Professor Raffi! Now, I can honestly profess that all the calculus courses I took are being utilized in a quite a profoundly and complementary way in relation to the physical world. Thanks for the excellent break down of the total differential of U(T,V) :-). I am using the Atkins 10th edition of physical chemistry, and although I read the section 2D before your lesson and understood the physical relation behind the mathematics roughly, you made the ideas come together effortlessly for me.