Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Physical Chemistry
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (10)

1 answer

Last reply by: Professor Hovasapian
Mon Aug 29, 2016 11:50 PM

Post by Mauro Souza on August 29 at 08:50:15 AM

Greetings Professor Hovasapian,

First, excellent explanations, thanks for the work!

Partial derivatives are taken by fixing variables and deriving to one variable at a time - like a slope in an axis. So if I have a x-y function one of the derivatives will be a slope with respect to x and another a slope with respect to y...so if I want the total rate of change, my logic would be to treat it as vector quantities and take the root of the sum of the squares of each rate, instead of just summing each rate. That analogy seems fair when I'm dealing with velocities (which are also rates) in x and y coordinates, for calculating things like total kinetic energy. Where is my logic wrong?

Thanks!

1 answer

Last reply by: Professor Hovasapian
Wed May 11, 2016 3:21 AM

Post by michael rise on May 7 at 05:40:42 PM

Hi Professor Hovasapian.

I would like to know why you have a negative (-nRT) as well as squaring the V denominator at around 20:50 and 31:40. Thanks and great lectures!

Best

your pupil

1 answer

Last reply by: Professor Hovasapian
Fri Mar 25, 2016 10:12 PM

Post by David Löfqvist on March 23 at 06:49:23 AM

Should one have taken multivariable caclulus before this course, or could you do them side by side?

1 answer

Last reply by: Professor Hovasapian
Fri Sep 4, 2015 10:04 PM

Post by Shukree AbdulRashed on September 4, 2015

Hello. Do you mind explaining how you got the expression for (partial derivative of P/ Partial derivative of molar volume) at 42:57? Thank you.

1 answer

Last reply by: Professor Hovasapian
Tue Sep 23, 2014 1:58 AM

Post by e b on September 22, 2014

At the clip 42:53min, why(dP/dT)_v is not equal to R/(v-b) - a/v^2

Math Lesson 1: Partial Differentiation

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 1: Partial Differentiation 0:38
    • Overview
  • Example I 3:00
  • Example II 6:33
  • Example III 9:52
  • Example IV 17:26
  • Differential & Derivative 21:44
    • What Does It Mean?
    • Total Differential (or Total Derivative)
    • Net Change in Pressure (P)
    • General Equation for Total Differential
  • Example 5: Total Differential 39:28

Transcription: Math Lesson 1: Partial Differentiation

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

Today, what we are going to do is a little bit of a math lesson.0005

We will be able to talk about partial differentiation.0010

I want to spend a fair amount of time on this and dedicate the entire lesson into this because of the idea of partial differentiation 0014

is going to be absolutely everywhere for the rest of the course.0021

It is important that we understand everything that is going on and get comfortable with this notion of partial derivative.0023

That is not very difficult but it just can be a little bit notational intensive, that is really about all that you have to worry about.0030

With that, let us get started.0036

We are accustomed to seeing things likes f(x)= sin x.0041

It only take the derivative of it we have f prime (x) = cos (x), this is a function of one variable.0051

In the words, x is your independent variable and f or y is your dependent variable.0060

We might have something a little bit more complicated, something like f(x) = sin x³.0069

In this case, f prime is a basic application, the chain rules, that something that you have done a million times.0079

Hopefully, it is going to be cos x³ × we take the derivative of what is in there.0085

We get x² and if we want to rewrite that it is not a problem but it is always a little bit more customary to write it as 3 x² cos x³.0092

Whatever you are comfortable with, it is your preference.0106

Many of the functions that we would be dealing with in Physical Chemistry, in thermal and quantum,0110

They are going to be functions to several variables.0116

Instead of just one independent variable x, you might have 2, 3, 4, or more independent variables.0119

To think the derivative of these functions, let us talk about how we do that and how we notate it.0128

Many of the functions we will deal with or functions of several variables and we have to deal with their derivatives.0138

Let us go ahead and do an example and that is the best way to proceed, I think.0181

The first example, I’m going to use F for our functions of several variables.0186

A function of x and y is equal to x² y³.0192

If I want to call this as z, it would be like z = x² y³.0200

Let us use this functional notation right here.0204

The derivative, when we take the derivative we take these things called partial derivatives.0207

Partial derivatives means, in this case we have a function of 2 variables, I’m going to differentiate with respect to x and I’m going to hold y constant.0211

I’m going to treat the other variables constant.0220

I want to take the derivative with respect to y, I'm going to hold the x constant.0223

Here is what it looks like notationally.0227

The derivative of f with respect to x is equal to, again, I’m taking with respect to x.0231

I differentiate the x while I treat this like a constant, I ignore this.0237

This is going to be 2 xy³ that is it.0243

If I take the derivative of the function with respect to y, I'm going to treat this as a constant.0250

We will take the derivative of this so this is going to be 3y² or 3x² y², if I want to write it in a more conventional form 0260

by putting that numbers first and variables afterward.0269

That is it, when we differentiate a multi variable function we are still differentiating 1 variable at a time.0274

Instead of writing df dx, the symbol that we use is this symbol that looks like a little curly d.0282

This tells me that I'm taking a partial derivative.0295

It tells me that I have a function of several variables and the particular variable with respect to which I’m differentiating is this variable right here, the x.0298

In this case, it is the y.0306

Everything you are doing is exactly the same.0308

There is thus nothing new that you have to learn.0310

The only thing you have to sort of get used to is holding the other variables constant, all the other variables whether their 1, in this case, or 2, or 3.0312

Just treat them like constants.0321

The only problem is going to be one of mechanics and keeping everything straight especially when you have to start dealing with more complicated functions.0322

Just go slowly and take it one at a time with all you are doing.0330

You are not doing anything different than what you have done before.0334

Do not let this notation actually intimidate you.0339

It is just df dx, df dx is a single variable, this is multi variable.0341

And in fact, a little later in the lesson you are going to see how we can actually treat this right here, as if it were a single variable.0347

In some sense, we are still taking the derivative into single variable function.0355

We are treating this function as if it is a function only of x.0360

That is why, but this symbol is used to represent the fact we are dealing with a function of several variables but still it is just a derivative.0364

It is still just a rate of change.0371

In other words, as I change x, how does f change?0373

The change is according to that rule right there.0376

There is nothing new here, this still represent a slope just like any derivative always represents a slope at a given point.0379

Everything is exactly the same.0386

Let us do another example here.0391

Let us try, I think I will stay with red.0396

This is example 2, so this time let us have a function of 3 variables, f (xyz) is going to be equal to e ⁺x × y × sin y × z.0401

Let us go ahead and start, we are going to take a partial with respect to x, a partial with respect to y, a partial with respect to z.0425

Our df dx, notice the x does not participate in here so the sin of yz is treated like a constant.0431

It is only this, the derivative of this with respect to x is going to equal y × e ⁺xy × sin yz.0442

Let us make z a little bit more clear.0455

That is the partial with respect to x.0459

Let us go ahead and take the partial with respect to y.0462

The y shows up here and the y shows up here so this is going to be a derivative of a product function.0469

We are going to take this × the derivative of that + that × the derivative of this.0474

This × the derivative of this + this × the derivative of that because the y shows up in both of these functions.0481

This × the derivative of that, we are going to get e ⁺xy × cos yz.0490

And again, the application of the chain rule, the derivative of those inside is going to be × z + this × the derivative of that.0499

Sin yz × the derivative of this with respect to y, which is going to be x × e ⁺xy.0508

And if I want to simplify it a little bit, I can bring this is z forward, I get z e ⁺xy × cos yz + x.0520

E ⁺xy × sin yz is a slightly simplified form.0535

Again, it is not necessary, just sort of conventional.0540

Let us go ahead and take the partial with respect to the third variable.0544

So df dz, I’m not holding the x and y constant.0548

In this particular case, z shows up here so this function does not participate.0557

This is not going to be a product rule.0562

It is just going to be e ⁺xy × cos yz × the derivative of what is in here which is going to be y.0564

That is = y e ⁺xy × cos yz and that is it.0576

A simple, straight differentiation, the only problem is going to be keeping it all straight.0585

That is the biggest problem that you are going to run across.0590

Let us do another example here.0594

Example 3, let us go ahead and return to example 1.0598

Let us return to example 1, we said that f(xy) =x² and y³.0606

Let me actually go to the next page.0621

We said that the derivative with respect to x was 2 xy³ and we said that the derivative with respect to y is equal to 3x² y², there we go.0627

Let us go ahead and take the second derivatives.0650

Notice that we have 2 derivatives, we have a df dx.0652

This is a function of x and y, it is still a function of 2 variables.0657

I can take the derivative of this with respect to x, a derivative with respect to y, and I have this derivative which is a function of x and y.0661

I can take the derivative of x with the derivative with respect to x and a derivative of this with respect to y.0670

I can have 4 second partial derivatives.0676

For this, it could be true for this.0680

What I have is the following, I have ddx of the df dx, and I have the ddy of the df dx.0683

Over here, I have the partial with respect to x of the df dy and I have the partial with respect to y of the df dy.0705

This one I have ddx with ddy of it.0722

I have 4 partials altogether, let us go ahead and evaluate what each of these is.0725

When I go ahead and, this is a nice notation because it tells me I'm taking the derivative with respect to x of this function which is already a derivative.0731

You can put this together the way that you did it with a single variable calculus.0742

We often notate this as d² f dx².0746

I will go ahead and use that notation.0751

D² f dx² that is that one right there.0755

I take the derivative of this with respect to x, I’m going to get 2y³.0761

Let us go ahead and take the derivative of this one.0770

I can it as d² f dy dx, I will put the dy dx separate.0773

D² f, that comes from the top and it is going to be dy dx.0779

The order is important, this tells me that I already have the df dx, I’m taking the derivative with respect to y.0786

I’m moving from right to left.0792

I’m going to differentiate this function with respect to y.0796

With respect to y, I get 3 × 2 xy² = 6 xy², that is this one.0800

Let us go ahead and do this one.0810

This is going to be df² dx dy.0812

Notice that the order is reversed in these 2.0821

This is dy, it is dx first and then dy, this is dy first and then dx of that.0824

When I take the derivative of this with respect to x, I will get 6xy².0835

Maybe the last one which is d² f, dy², I end up with 6² y.0844

Let us take a look at these, 2y³ 6xy² 6x² y.0857

Notice something with these 2, they are the same.0864

Notice that these 2 are equal.0873

This is not a coincidence, this is generally true, not always but it is generally true for multi variable functions.0877

Functions was first and a second derivatives exist and are continuous over their domain.0900

You do not necessarily have to worry so much about this but just a quick word, again, we are talking about a function of 2 variables.0925

Another 2 independent variables are x and y, which means that now the whole plane is the domain.0931

If you take a function of the first derivative exists, if the second exists, and if the derivatives are continuous, which they are, is a continuous.0938

Every one of these is continuous.0945

It will always be the case that the mixed partials, the second derivative dy dx, dx² df dy, the reverse order will always be equal.0947

That will always be the case.0963

If the first and second derivatives exist in a continuous over their domain.0965

For our purposes, that is always going to be the case.0970

It is not going to be a problem, I will deal with any exceptions individually but this is really important.0972

We will be using this particle property a lot in subsequent lessons.0979

These are called mixed partials.0985

Mixed partials or cross partials, mixed partial derivatives.0987

That and that would generally be equal and does not matter any order.1003

This also is actually true for functions of 3 variables or more.1009

Let us say you have 3 variables xyz, let us say you take df dx, df dy, and df dz, like we did in example 2 where we have a function of 3 variables.1014

If you are to take df and then dx dy or if you are to take the df dy and dx dz, you can switch the order of the partial differentiation.1027

All of those will actually be equal, profoundly, the property of functions of several variables.1039

It is a very important property.1045

Let us go back to blue, in thermodynamics.1050

Thermodynamics is what we are going to be covering in the first part of this Physical Chemistry course, the classical thermodynamics.1066

Thermodynamics is customary to show explicitly which variable or variables are being held constant in a notation.1075

Example 4, let us go ahead and take the ideal gas law Pv = nrt.1120

In most of this case, I wrote it in terms of n.1131

Pv= nrt, let me rearrange this and write it as P = nrt / v.1136

Now what I have done is I have expressed p as a function of 3 variables, as a function of n, as a function of t, and as a function of v.1144

P is a function of these 3 variables, it is a multi variable function.1156

I can form 3 partial derivatives, I can take dp dn, dp dt, dp dv.1160

Let us go ahead and do that.1166

Dp dt, notationally what we do is we write n and v, I did a partial derivatives like normal but notice I put this partial derivative sign in parentheses and I have written nv.1171

I have explicitly stated that I’m holding the n positive and I’m holding the v positive.1184

In other words, the partial with respect to t, a partial with respect to temperature, a constant number of moles and constant volume.1187

The things that I hold constant, I will write down in explicitly notation nr/ v, dp dn.1197

This time I’m holding constant is going to be t and v.1211

I’m writing down there and I will get RT /v.1213

I hope you actually confirmed that I'm getting the correct partials.1218

Dp dv, this time we are holding n and t constant.1225

If you are holding only one of the constant then only 1 of these shows up.1231

If you are holding 2 of them constant, 2 of them show up, that is all it is.1234

It is just a notational convention in Physical Chemistry, = -nrt /v².1242

Here is where I want to stop and I want talk a little bit about what these derivatives mean.1257

Again, there is going to be a lot of mathematical notation on the page and often time, it is going to feel overwhelming.1262

It is always nice to stop and not just to get lost in the mathematics but to stop and understand that these things have physical meaning.1271

You always want to invest your mathematical equations with physical meaning because it means something physically, something is going on.1279

There is a lot of sophisticated mathematics here and a lot of it is theoretical but there is nothing theoretical about the Physical Chemistry.1285

We are talking about what is going on with chemical system and we are using mathematics to describe it.1293

Every bit of mathematics has a physical meaning and we to invest it with that to maintain that connection to it.1299

Let so let us talk about what derivatives mean.1306

What does dp dt and v = n/Rv, what does it mean?1315

What exactly what you think it means?1330

What is a derivative?1332

You know that derivative as a slope, you know that is the rate of change.1334

That is what is important, you always want to interpret that partial derivative or derivative as a rate of change.1338

If I have some function, in this case P = nrt/ v, my ideal gas law.1344

For p, with this dp/ dt is, it is telling me that if I make some small change in the temperature, how is the pressure going to change?1349

If I hold the number of moles and the volume constant, I do not worry about those, it is a measure of how much the pressure changes with a change in temperature.1361

I already know this from single variable calculus.1371

Whenever I see something like dy dx, what I’m saying is, if I change x a little bit how does y change?1374

That is all the derivative it is at that particular point.1388

Again, it means the exact same thing and you always want think about it that way and you always want to assign units to it.1391

And in this particular case, the pressure is atmosphere, the temperature will be Kelvin.1399

What the derivative is, it is telling you how many atmospheres per Kelvin, per unit Kelvin.1405

At any time you do a division, a derivative, a rate of change, the denominator is always going to be 1 miles / hr, atm/K, atm/mol, L/K.1412

Whatever happens to be that is what it is, it is the rate, a derivative is a rate.1428

That is all that is happening here.1435

It is a measure of how much P changes when I change T.1440

In this case, T only because we are doing partial differentiation.1462

Again, examine the units.1468

I have P on top, dp dt, this is going to be atm if I happen to be dealing with atm/ K, atm/ °K, a derivative is a rate.1479

If I have the rate, in this particular case, atm/ K and if I multiply by a particular temperature which is in Kelvin, the K and K cancels leaving me with atm.1496

That is all what is going on, it is just a rate.1510

It is the same thing that you have been doing.1512

Again, keep track of the units, assigned them, remember that it is a rate, and all this will start to make sense.1515

In fact, if you want you can do this.1523

Let us go ahead and write again, dp/ dt = nr/ v.1527

You remember from single variable calculus, you get something like this.1537

I will do this in red.1539

Remember, you did dy dx = 2x, the derivative of x².1541

If y = x², dy dx = 2x, if you actually want to measure the specific change in this particular variable and this variable you treat it like a fraction 1549

and you move this over, what you got was this.1562

At a particular value of x, if you were to change x by this much, your change in y would be this much.1567

We call this the derivative, we call this the differential, you can do the same sort of thing here.1576

A single variable calculus dy/ dx does actually represent a fraction.1581

This notation, this partial differentiation notation, this is a symbol but you can still treat it formally as if it were a division of some sort.1587

You can write it like this and you go back to blue.1597

I will stick with red.1599

In some sense, you can go ahead and move that over so that you can get something that looks like this dp = nr/ vdt.1603

This is the same thing, if I make some small change in temperature, how is the pressure going to be affected?1613

This is the differential, we tend not to write it like this.1619

This partial differentiation notation is telling us that we are dealing with the multi variable function.1624

However, because we actually held every other variable constant this is still just a function of a single variable in some sense.1630

Therefore, we tend to write it like this, dp = nr/ vdt, this is more in line with what we did with single variable calculus and treating it like a fraction.1639

This is the differential, a small change in temperature × this partial derivative which is this is equivalent to writing dp is equal to rdp/ dt × dt.1655

This is atm/ K and this differential change in temperature is K.1675

K and K will cancel giving me this thing in atm.1683

It gives me the differential change in pressure upon a differential change in temperature.1688

If you do not like the differential part, if you prefer Δ, δ P = the derivative × δ T.1691

It is the same thing that we did in a single variable calculus, the derivative notation this is the differential notation.1701

This is the derivative notation, the differential notation.1707

However, because this is technically not a fraction, we tend to write it this way.1711

Again, we are talking about we are changing just 1 variable.1718

The v is being held constant, the n is being held constant, for all practical purposes this is still just a single variable derivative.1721

This is the differential, that is all you are doing here.1731

Let us go ahead and just double check this.1742

I have nr/ vdt, n is mol, R is L, atm/ mol K, I will go ahead and use that particular unit.1745

Volume is L and dt is in K.1765

Mol cancels mol, L cancels L, K cancels K, we are left with atm.1771

Sure enough, this gives us some value pressure in atm.1777

It matches, this is a common analysis that you want to do if you feel like you are getting lost, treat it like any other unit problem when you are canceling units.1783

Let us go ahead and introduce something very important, something called the total differential with a total derivative.1796

I will go ahead and I think I will go back to black here.1803

The total differential or total derivative, you often here it called both.1811

Total differential is actually more appropriate and here is why.1830

Let us start with our completely intensive form of the ideal gas law Pv = RT.1839

Let us go ahead and solve for pressure, RT/ v.1847

In this particular case, we are expressing pressure as a function of 2 variables, the T and v.1854

We have a function of 2 variables.1864

We already know that we can take dp dt holding v constant that = R/ v.1868

We know we can take the partial of the pressure with respect to the molar volume.1882

Dv because I’m holding temperature constant, we want to specify the notation, this is going to be – RT/ the molar volume².1887

Again, if you want to rearrange this, what we did before, we rearranged this, we move this over here so expressed it as dp =R/ v × dt.1902

And then we express these as just regular derivatives Rv dt, same thing here dp = - RT/ v² dv.1918

I will go ahead and convert these, I will write it as dp = -RT/ v².1935

We will write this as a regular derivative instead of a partial derivative, something like that.1943

Let me go ahead and we put this back into notational form.1950

This R/ v is equal to this, I’m going to substitute this expression into here and this one right here it is equal to this notationally.1954

Therefore, I’m going to take this and put back in here.1966

What we have is following, dp= dp dt that constant volume × dt.1969

This make sense, this is exactly what we did before.1981

The partial, the differential, the change in pressure comes from the change in temperature × the derivative.1985

That is what we have always done, dt dt, this is a K and K, K and K cancels leaving me just pressure.1994

The same thing over here, I have dp is equal to, I’m going to go ahead and put this notation here, dp/ dv, holding T constant × dv.2002

This is a volume, this is a volume, they cancel.2017

The differential change in pressure is equal to the differential change in volume × the derivative at that particular point.2020

If I want to change 1 variable as the temperature, I’m going to get a corresponding change in the pressure.2028

If I hold the temperature constant and if I change the volume, I’m going to get a corresponding change in pressure.2033

Another question is what if I want change both?2039

What if I want to not hold any 1 of them constant?2042

What if I want to change both the temperature and the volume, I want to see what the net change in pressure is.2045

We have a way of doing that, we just add the individual changes.2052

Here is what it looks like, let me write this out.2056

Each of these, that and that expresses how P changes if I change 1 variable.2062

Now the question is, but what if I want to know what happens to P when I change both the variables simultaneously?2094

I will just add these 2 up.2100

I will add this to that and express it to something called the total differential and here is what it looks like.2102

Dp= dp dt, the constant volume × dt + dp dt, that constant temperature × dv.2113

Here is what it says, if I make a small change in temperature and I make a small change in volume, I take the small change in temperature 2128

and multiply by the rate of change of the pressure with respect to temperature from the equation.2136

What my equation happens to be?2143

I’m going to get the change that comes from the change in pressure that comes from change in temperature.2145

Here I'm going to get the total change that takes place when I change the volume a little bit.2151

If I add each of those changes that gives me the total change in the pressure when I change both of these.2156

This is called the total differential and is profoundly important.2162

It will show up a lot in our subsequent work.2166

This is called the total differential.2173

Once again, this part right here, it represents the small change in pressure when I make a small change in temperature.2185

It is just the change in temperature × that derivative with respect to temperature.2196

You already know this.2201

This term represents the change in pressure that accompanies a small change in volume.2204

It is the small change in volume δ v × the derivative of the function with respect to v.2212

This is the same as write dy dx = z, dy = zdx, that is it.2219

This is a rate × some value, this is rate × value, temperature and temperature they cancel leaving the pressure.2230

Volume and volume, they cancel leaving the pressure.2240

Pressure + press pressure = the total change in pressure when I change both variables simultaneously.2243

That is what I'm doing here.2250

This is called a total differential or the total derivative.2253

If you want to see the analysis, dp dt this is atm/ K × K and this is atm/ L.2260

Let us say L/ mol × change in volume which is L/ mol, atm atm, my final answer is atm.2273

That is what it is telling me.2288

In general, given the function which is a function of 2 variables x and y, the total differential is a total differential f is equal to the partial of f 2295

with respect to the first variable, holding the second variable constant × the change in x + the partial of the function with respect to y, holding x × the differential change in y.2333

It is very important, this is the total differential.2351

In general, if you have a function of 2 variables, the total differential always looks like this.2356

Let us go ahead and do an example here.2362

Let us see, where are we?2369

This is going to be example 5.2370

Van der Waal’s equation, you remember van der Waal’s equation from general chemistry, it is a fix of the ideal gas law 2380

to bring it more in line with a description of the way gases behaved.2394

As the equation state because it gives you the state of particle system.2398

I’m going to write it in this form RT/ v - b – a/ v².2408

Find dp, the total differential.2425

Let us be formal about this, P is a function of the variables T and v.2439

It is very important to take out your variables and know what your variables are, what your independent variables are.2449

This is also constant, b is a constant, T and v are the only 2 variables.2455

P is a function of the 2 variables T and v.2462

Let us go ahead and take dp, let us write the individuals.2472

Let us go ahead and write the formula for the total differential, dp= dp with respect to the first variable, holding the second variable constant × 2483

the differential of the first variable dt + the partial of the function with respect to the second variable, holding the first variable constant × the differential of the second variable.2494

We can just plug these in.2510

We have a function right here, once we actually have a function, we will find a derivative of the function with respect to T.2512

We will find the derivative of the function with respect to v, and we will put it in there.2523

We will leave that alone, we will leave that alone, and we will leave that alone.2527

And we will have our actual total differential for a specific function.2532

In this case, the van der Waal’s equation.2535

Let us go ahead and do it.2540

Dp, when I go ahead and differentiate this with holding v constant, so v does not matter, I end up with the following.2546

I end up with R with respect to T, it ends up being R/ v –b, that is dp dt, that is that one.2559

Let us go ahead and take the derivative with respect to v holding T constant, the derivative of this holding, this time T constant, which we will end up is the following.2574

You end up with - RT / v- b² - a × 2 × v/ v⁴.2585

I will go ahead and cancel, I'm left with my final expression which is -RT/ v – b².2607

A – and - becomes + and it becomes +2a/ b³.2617

Now that I have this and this, I will plug them in to here and here.2624

I have my final expression.2639

The total differential of P= R/ v – P² × dt + -RT/ b - b² + 2a/ b³ × dv.2643

This is the total differential for the van der Waal’s equation.2687

It is all based on given the function of 2 variables, the total change in that function when you vary both variables simultaneously is df dx 2692

holding y constant × the change of x that you make + the derivative of the function with respect to y, the partial with respect to y × the change in y that you make.2713

That is it.2727

Hopefully, for those of you that have a background in multivariable calculus, it is not a problem.2732

A lot of these should be a review.2738

You may not have seen total differential, some multivariable class talk about it and some do not.2739

For those of you that have never seen partial derivatives or the total differential, I hope that this is actually make you feel a little bit more comfortable.2745

Starting with the next lesson, we are going to jump right into thermodynamics.2754

Until then, welcome back to www.educator.com and I will see you next time, bye.2757