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Lecture Comments (6)

1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 1:23 AM

Post by Van Anh Do on February 14 at 10:00:23 AM

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 at 09:52:25 AM

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.

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 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

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 and0240

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