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### Math Lesson II

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• Intro 0:00
• Math Lesson II 0:46
• Let F(x,y) = x²y³
• Total Differential
• Total Differential Expression
• Example 1
• More on Math Expression 13:26
• Exact Total Differential Expression
• Exact Differentials
• Inexact Differentials
• The Cyclic Rule 21:06
• The Cyclic Rule
• Example 2

### Transcription: Math Lesson II

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

In the last lesson we started our discussion of entropy δS.0004

Before we actually talk about entropy and investigate how it behaves, I want to take a little bit of a break and discuss a little bit of mathematics.0010

The mathematics that I’m going to discuss here is a continuation of some of the mathematics that we started with, the discussion of partial differentiation.0019

This is going to be more technique in partial differentiation and it is a series of techniques that we are going to be using absolutely all the time,0027

not only for the rest of thermodynamics but once we get into quantum mechanics and spectroscopy and things like that.0034

Let us go ahead and get started.0040

Let F be a function of xy and we will let it equal x² y³.0047

Let us go ahead and form some partial derivatives here.0059

δF/δx is going to = 2xy³.0063

Over on this side, we will form δF/δy that is going to = 3x² y².0071

Let us go ahead and form the next of set of partials, the second partials.0083

This is a function of x and y. I can take δF/δx of this and I can take δF/δy of this.0087

I can take δF/δx of this and I can take δF/δy of this.0093

Let us go ahead and do that.0097

Let us start off with δx and I will notate it this way.0098

I will do δ/δx of δF/δx which is the same as δ² F δx² that is going to = 2y³.0102

I will form y, I will form δ/δy of the δF/δx which = δ² F δy/δx this notation that is going to = 6xy².0120

I come over here and I’m going to form δ/δx of the δF/δy which is notated as δ² F δx/δy that is going to = 6xy².0138

I will do δ/δy of the δF/δy which is δ ⁺2F δy² and that is going to equal 6x² y.0157

Notice, 6xy² 6xy² δ² F δy/δx δ² F δx/δy are mixed partials in different orders they actually equal each other.0172

δ² F δy/δx = δ² F δx/δy this is not an accident.0191

This is not a coincidence, this is generally true.0208

Let F again = RF of x and y begin from the total differential of this.0218

The total differential is δF = δF/δx holding y constant × δx + δF/δy holding x constant × δy.0231

What we saw above, in other word this thing here is true in general.0261

If F(xy) is a function and δF/δx δF/δy δ² F δy/δx and δ² F δx/δy exist, in other words if the derivatives of the function exist and0270

are continuous ... which in our case there always going to be continuous, then like we said δ² F δy/δx = δ² F δx/δy.0308

In other words, the cross partials or the mixed partials are always going to be equal.0333

If you are given a function of two variables, three variables, let us just stick with two variables for the time being.0337

If we are given a function F of two variables and it is well defined and continuous and if the derivatives, the first derivatives and the second derivatives,0342

if they actually exist or continuous then the mixed partials will always be equal to each other.0350

Mixed partials are equal.0360

For the total differential that we wrote which is δF = δF/δx holding y constant × δx + δF/δy holding x constant × δy, this means this is δF/δx.0377

If I take the derivative of this with respect to y and if I take the derivative of this with respect to x, they are going to be equal.0401

For the total differential, the derivative of this with respect to the other variable equals the derivative of this with respect to the other variable.0415

This means that δ/δy of δF/δx = δ/δx δF/δy.0426

Given a total differential expression derived from F = F(xy) from an actual function, given a function and you write down the total differential,0451

from that total differential the partial of the differential coefficients are this thing and this thing.0478

The top differential coefficients with respect to the other variable are equal.0509

It is up here, this blue, the derivative of this expression with respect to this variable = the derivative of this expression with respect to that variable.0526

Straight up, so outside and inside.0541

The derivative of this with respect to this variable = the derivative of this with respect to this variable or you can just look over here.0545

The derivative of this with respect to y = the derivative of this with respect to x, the other variable.0552

That is all that is going on here.0559

This is always true.0560

Let us go ahead and do an example here.0565

A thermodynamic example, we know from the first law that δU = δQ reversible - δW this is the definition of the first law of thermodynamics.0572

We also have the definition of entropy δS = δQ reversible/ T.0586

I will just multiply by T what I get is T δS = δQ reversible.0595

When we take this, we are just here and plug it into here I get the following.0602

I get δU = T δS – δW.0607

You remember δW = P × δV so what I get is δU = T δS – P δV.0616

You have a total differential expression right there. δU = something × the differential S - something × that.0630

U is a function of S and V. So I have a differential expression which means that a partial of this with respect to this variable =0645

the partial of this with respect to that variable.0656

Now I can go ahead and write that.0659

Let me write this again on top of the same page, I have δU = T δS – P δV.0664

Therefore, a derivative of this with respect to V… So δT δV holding that constant = the derivative this with respect to that... δP δS,0679

holding that constant, this is negative.0699

That is that. This is a very interesting relation just from the fact that mixed partials are equal.0703

I have the total differential expression, I will automatically know that the differential this with respect to this variable = the differential of this0712

with respect to this variable holding the other variable constant.0719

So I have something like this.0723

This relationship this is an example, this relationship is one of a very important set of other relations, they are called Maxwell's relations.0725

This is the same Maxwell from electromagnetism.0757

This is not Maxwell’s equations, these are Maxwell thermodynamic relations, he did lot of work in a lot of fields.0761

These relations are one of very important set of relations called Maxwell's relations...among the variables of state of the system.0770

In this particular case, the variables involved are temperature, volume, pressure, and entropy.0789

We will come back to these but for right now this is just an example to show you that this applies to thermodynamics,0795

this mathematical theorem, but we will come back to Maxwell's relations later on when we discuss the free energy.0801

A function F(xy)…0819

So a state function F(xy) generates an exact total differential expression, δF = (δF/δx)y δx + (δF/δy)x δy0871

which guarantees that the mix partials are equal.0926

That takes care of the first part.0944

What is interesting is the converse is also true.0948

The converse is also true, converse is also true.0954

In other words, if we are given the total differential expression such as let us say δF = P δx + Q δy then if δP δy = δQ δx0965

then δF is exact and there exists an actual function F which is a function of the variables x and y.1012

If we are given the function, we write the total differential and the mix partials are equal.1036

If we are given a total differential expression and if we happen to take the mix partials, if the mixed partials are equal1043

then the actual function that gave rise to the differential expression that we wrote actually does exist.1049

In other words, first we are going from function we are differentiating down, now we are actually given the differential can we integrate backup.1055

That is what this is saying.1062

The only thing that I need to check is if the mix partials are equal, the δP δy= δQ δx,1063

then theoretically I can actually integrate this function and recover some function of x and y.1071

It goes in both ways.1076

There exist an actual F.1081

δQ and δQ for example, are examples of,1097

I will have to say a little bit more here.1109

If this δP/ δy if it does not equal the δQ/ δx, in other words if I'm given a total differential expression, I take the mixed partials,1115

I take the partials and they do not equal each other then there is no guarantee that such a function F(xy) exists.1129

In this particular case, δF is inexact.1157

δQ and δW are examples of inexact differentials.1165

For exact, we have the following.1194

In going from S1 to S2, U is exact δU=δ U which is U2 – U1.1198

If I go from S1 to S2 and back to S1, the cyclic of an exact differential = 0.1212

for inexact, if I go from S1 state 1 to state 2, the integral for example work it just equal the work.1222

It does not = δ work just the work.1232

If I do a cyclic S1 to S2 and if I come back to S1, the cyclic integral of an inexact differential does not = 0.1240

Let us see what else we can say, now let us talk about something called the cyclic rules.1256

Mixed partial is our first technique, now we are going to talk about another thing called the cyclic rule.1261

Let me move to black here.1267

There is another nice relation among the partial derivatives of a given function.1277

This one was actually quite beautiful and it is called the cyclic rule.1305

Let us start off, we will let z = z (xy).1314

z is a function of x and y, let us go over here.1322

For various values of x and y, I simply calculate z just the function of two variables.1330

z is basically like the third variable, it is the dependent variable.1352

Generally, we do things like F(xy) but we can say z(xy) because when I put an x and y and I did a particular calculation1355

it is going to spit out some number, that number we call z.1363

I have x and y which are independent variables and z is the dependent variable.1366

I have three variables.1371

z is a function of x and y so I have a total differential expression.1375

I can write δ(z) = δz/δx holding y constant × δx + δz/δy holding x constant × δy.1380

If I restrict my choices of x and y such that z never changes and I can do that, I can choose x and y, z is a function of x and y.1400

I can choose x and y such as z stays the same, it never changes.1431

In other words, δz = 0 then 0 = δz/δx y δx + δz/δy (x δy).1436

I’m going to make a little bit of change here because I'm actually setting z as constant δz = 0.1461

I can write this as δx holding z constant + δz/δyx and I can write this as δy holding z constant.1472

I’m going to go ahead and divide by this term right here, divide by the δy sub z.1490

What I end up getting is 0 = δz/δxy δx/δy z + δz/δy x.1500

I'm going to multiply by the reciprocal of that term.1524

When I multiply by the reciprocal everything I get 0 = δz/δx sub y × δx/δy sub z × δy/δz sub x + 1.1537

I’m going to rearrange and I’m going to move this over to the other side and I have the following.1566

I have δz/δx holding y constant × δx/δy holding z constant × δy/δz holding x constant = -1.1574

This is the cyclic rule.1595

If I have some function which is a function of x and y, z is a function of x and y, there is a relationship that exists among the partials.1603

The partial of z with respect to x holding y constant × a partial of x with respect to y holding the z constant × a partial of y1612

with respect to z holding x constant is always going to = -1.1619

Notice the relationship among the variables z x y, x y z, y z x, all three are represented in each case.1626

The zx xy yz, that is the best way to think about it.1637

Basically, what you can do if you want, you can just write the first variable, the xyz and1643

you can write the other variables in any combinations underneath that, that just do not repeat it.1648

You can write it as, y z and x and just go ahead and do something like this.1652

δx/δy δy/δz δz/δx xyz is constant yz x is constant, zx y is constant and it is always = -1.1663

That is the relationship that exists between the x, the y, and z.1672

Let us take a look at an example.1678

Example 2, let be the three variables be temperature, pressure, and volume be our three variables.1685

It make some difference which will be expressed as a function of the other two because T is function of P and V,1703

P is a function of T and V, V is a function of P and T.1710

It is just we arranging the equation, it is not a big deal.1712

We immediately write down the relations so let us do it this way.1717

Let us write the T, P, R.1721

Let us write P and T, V and I will pick T here, I will pick V here, and I will pick P here, this is PT V, TV P, VP T.1726

What I have the following.1749

I have δP δT holding V constant × δT δV holding P constant × δV δP holding T constant = -1.1751

I have a beautiful relationship.1768

The rate of change of pressure with respect to temperature under constant volume × the rate of change of temperature1773

with respect to volume under constant pressure × the rate of change in volume with respect to pressure under the change in temperature = -1.1779

This relation is valid.1791

The cyclic rule tells me that it is valid.1799

I have a relationship between T P and V, PV = NRT or PV = RT there is a relationship that exists between these.1801

I can go ahead and just flat out just write down the relationship that exists between the partial derivatives because I know the cyclic rule is true.1813

Let us go ahead and rearrange to make it a little bit nicer.1825

Recall, we said α =1/ V × δV / δT under constant pressure.1830

I'm going to go ahead and move the V there so I get δV δT under constant pressure = α V.1849

I have kappa the coefficient of compressibility -1/ V × δV δP under constant temperature.1860

I’m going to move it there so I get δV/ δP under constant temperature = - kappa × volume.1869

This thing right here becomes this δP δT under constant volume.1886

δT / δV I have δV / δV δT = α × volume this is just the reciprocal.1902

δT this is going to be × 1/ α × volume.1913

δV δP= - KV that is = -1.1921

I just put this and this into here appropriately and the V cancel and I'm left with δP δT V × - kappa/ α = -1.1932

Therefore, δP/ δT the constant V = α/ kappa.1956

in other words, if I keep the volume constant for every unit increase in temperature or unit change in temperature,1966

the change in pressure is α/ kappa, the coefficient of thermal expansion divided by the coefficient of compressibility.2000

I have this amazing relationship and these are tabulated for a lot of things or if they are not, they are easily calculated.2011

I have this amazing relationship that exists simply by virtue of the cyclic rule.2017

This is just an example of one application of the cyclic rule.2022

We will see other examples and other applications.2026

Thank you so much for joining us here at www.educator.com.2031

We will see you next time for the discussion back to entropy.2033

Thank you, bye.2038