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

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

### Math Lesson II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- 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

### Physical Chemistry Online Course

### 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 us just go ahead and start with an example here.*0044

*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 and*0270

*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 this*0712

*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 δy*0871

*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 δx*0965

*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 equal*1043

*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 calculation*1355

*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 y*1612

*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 and*1643

*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 temperature*1773

*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

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