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 1 answerLast reply by: Professor HovasapianTue Sep 15, 2015 2:54 AMPost by Shukree AbdulRashed on September 14, 2015Hello. I'm not understanding how you assume that Cv is a constant? Isn't there some sort of interdependence between the two variables: temperature and heat capacity, thus the heat capacity should be integrated or carried along as well? Thank you. 1 answerLast reply by: Professor HovasapianSat Jun 20, 2015 5:59 PMPost by Joseph Carroll on June 11, 2015Say I have an isolated system, with an internal heat source, and there is a mole of an ideal diatomic gas in that system. If the temperature of that system increases, then consequently each individual gas particle will gain more kinetic energy, and since the isolated system has a fixed volume, the molecules will collide with each other at a faster rate, and increase the overall temperature and thus internal energy of the system.â€¨â€¨ As a result, by the first law of thermodynamics, we know that even though energy was neither created nor destroyed there was a transfer of heat energy into the system from the internal heat source, which caused the diatomic perfect gas particles, on the molecular level, to increase in each molecule's translational, rotational, and vibrational motion (and also the rate of collisions between molecules).â€¨â€¨ Now concerning entropy: For the case stated above would it be true that there was an increase in entropy due solely to these facts: There is an increase in the overall distance (faster translational motion) each individual gas particle can travel in a given second, due to a greater velocity, that thus the gas particle could occupy a larger magnitude of space/volume than before heating the gas [this results in a larger dispersal of energy since the heat energy is being distributed to the gas molecules' increase in translational, rotational, and vibrational motions, resulting in a further traveling apart in a given second than before heating]. â€¨â€¨Moreover, since the gas particles are confined and moving faster they are colliding more rapidly. This results in many more possible ways the molecules could arrange themselves due to an increased rate in collisions, and thus resulting in an increase in the many different number of ways we can look at this mole of gas in a given microstate, or snapshot, of all the molecules' possible orientations. Is this what entropy change really is on the molecular level Professor Raffi?

### Entropy As a Function of Temperature & 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
• Entropy As a Function of Temperature & Volume 0:14
• Fundamental Equation of Thermodynamics
• Things to Notice
• Entropy As a Function of Temperature & Volume
• Temperature-dependence of Entropy
• Example I 26:19
• Entropy As a Function of Temperature & Volume, Cont. 31:55
• Volume-dependence of Entropy at Constant Temperature
• Differentiate with Respect to Temperature, Holding Volume Constant
• Recall the Cyclic Rule
• Summary & Recap 46:47
• Fundamental Equation of Thermodynamics
• For Entropy as a Function of Temperature & Volume
• The Volume-dependence of Entropy for Liquids & Solids

### Transcription: Entropy As a Function of Temperature & Volume

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

Today, we are going to continue our discussion of entropy and we are going to talk about entropy as a function of temperature and volume.0005

Let us jump right on in.0013

I have to give you a little bit of a warning, for this particular lesson and the next lesson,0016

and several ones that follow there is going to be a lot of mathematics.0020

Most of the derivation, the mathematics in itself is really not difficult, it is mostly just taking derivatives0029

and you know equating one equation with another and see which coefficients are the same and identifying one thing with the other.0035

The only thing you have to worry about, I’m going to warn you about is there is going to be a lot of symbolism on the page.0042

It is really important to not lose the four from the three and things like this.0050

At the end of these lessons, I will stop and recap and I will tell you what is actually important as far as equations to remember0056

and what you should concentrate on.0064

I just wanted to give you fair warning that there is going to be a lot of symbolism on page, a lot of mathematics.0066

Let us see what we can do with it.0072

Nothing is altogether difficult though.0075

We defined entropy as follows, we said that DS = DQ reversible / T.0078

This relates a change in entropy to an effect in the surroundings or the system.0091

It is just a question of perspective in the surroundings, basically, the quantity of heat withdrawn from the surroundings divided by the temperature.0113

It is related to an effect in the surroundings.0138

Can we use the equations we have so far from our previous work.0141

This is the question, can we use our equations to relate entropy to changes in the state variables of the system which we can actually measure very easily.0149

In other words, temperature pressure and volume are easily measurable.0188

Not that heat, it is not that it is just that basically if I can find a way to use the equations that we had our disposal to be able0196

to express this δ S or the DS in terms of the pressure, the temperature, and the volume and maybe energy, it would be very convenient.0202

That is what we are going to set out to do, that is our goal, that is our big picture.0214

We want to express entropy in terms of temperature, pressure, and volume.0218

Let us see what we can do.0222

We know that in a reversible transformation, so I just go ahead and put it over here.0226

In a reversible transformation and from our discussion of energy we said that the external pressure is actually equal0232

to the pressure of the system because of something reversible the system and the surroundings are always essentially in equilibrium.0245

The external pressure we just said equal to the internal pressure.0251

Therefore, our definition of work because we are going to bring work back into this, definition was P external × DV.0257

When P external = P then DW is just equal to P DV.0268

First law expression, energy = DQ and in this case reversible, - DW.0281

DU = DQ reversible, DW is just P DV.0292

We have this DS = DQ reversible / T so if I move the T over here, I get the DQ reversible.0303

I get that DQ reversible = T DS.0312

I’m just going to go ahead and put this into here and I get the following I get DU = T DS – P DV.0317

I’m going to switch things around and move this over here and I'm going to put T DS = DU + P DV.0347

I’m going to divide by T, everything by T, in order to get DS alone, I’m going to be left with DS =1/ T DU + P / T DV.0359

This is a very important equation, this is called a fundamental equation of thermodynamics.0376

It is going to be the starting 0.4 much of the work that we do subsequently.0386

This is the fundamental equation of thermodynamics and is nothing more than a combination of the first law of thermodynamics which is the definition of energy.0391

Instead of the second law of thermodynamics which was just an expression of the definition of entropy.0402

First law is energy and second law is entropy, we put those together and we come up with a fundamental equation of thermodynamics.0409

You do not need to memorize this because the derivation as you see is very simple.0420

Just put the definition of entropy in terms of the T DS into here and just manipulate it.0428

That is it, it is very simple.0432

I will go ahead and actually go to the next page here and rewrite the equation again.0437

We have DS =1/ T DU + P/ T DV.0441

This expresses a change in entropy, P2 changes in energy and volume.0454

In other words, what this says is I can change the energy, I can change the volume, or I can change both.0478

Either one of those or combined that changes the entropy. In other words, I have two independent ways of changing the entropy.0485

I can change the energy of the system, I can change the volume of the system, either one of those changes0493

that I make in the system will change the entropy of the system, that is all this equation says.0497

This expresses a change in entropy that changes in energy and volume as well as the temperature and pressure.0504

Notice I have T and I have P, notice that every variable is represented, temperature, pressure, volume, energy, and entropy S U T PV is altogether.0516

That is as why it is called a fundamental equation of thermodynamics.0548

Some other things to notice, the first thing is the coefficients, the 1/ T and the P/ T.0553

I think I will go ahead and do this red.0566

Things to notice, the coefficients 1 / T and P/ T are positive.0570

There are two independent ways of changing the entropy of the system.0587

The first way is changing the energy, that is one, and the other way is change the volume.0618

If you make changes to either or both, you change the entropy of the system.0630

Let us go ahead and continue here, add a little bit of room.0640

Since 1/ T and P / T are positive, if I hold V constant, and if I change the energy, if I raise the energy in other words if the change in energy is positive,0645

if this differential is positive, positive × positive is positive, the change in entropy is also positive.0668

If I raise the energy of the system, the entropy of the system is raised.0676

If I hold the energy constant and if I raise the volume of the system, this is positive and this is positive that means the entropy goes up.0681

If I raise the energy of the system, I raise the entropy of the system.0691

If I increase the volume of the system, I increase the entropy of the system and vice versa.0696

If I decrease the energy, I decrease the entropy.0700

If I decrease the volume, I decrease the entropy, that makes sense.0703

If I change the energy of the system, I'm agitating the system.0707

I'm putting more energy into it, it is becoming more agitated, it is becoming more disordered.0710

Again, that qualitative way of looking at entropy.0715

If I change the volume of the system and take it from here to here, the particles of the system0718

have much more room in order to move so now there is more chaos available to it.0723

This is orderly, this is disorderly, if I raise the volume of the system I actually raise the disorder of the system.0729

If I take the volume and make it smaller, I actually create more control.0738

I have lowered the disorder of the system so that is what is going on here.0742

Since 1/T and P / T are positive , if I fix the volume, fixing V and raising U energy raises S the entropy.0747

If I fix U, if I fix the energy and raise the volume then S rises.0776

Let me go back to black for this one.0796

We do not to control volume in experiments, we do not normally control energy and we certainly do not hold it fixed.0802

We do not control U in the laboratory.0816

Our next phase is let us see, if we cannot transform this equation, this fundamental equation of thermodynamics to one0825

that involves temperature and volume or temperature and pressure.0849

We want to be able to not worry about controlling the energy because we do not normally control them in a laboratory.0857

In the laboratory, we control temperature, pressure, and volume.0862

Can we take this equation, the fundamental equation of thermodynamics and mathematically manipulate it to give us something0866

that gives us the same expression but in terms of temperature and volume, and temperature and pressure.0873

The answer is yes, we can.0877

The first thing we are going to do is we are going to look at entropy as a function of temperature and volume.0880

We do this in blue, that is okay I will stick with black.0888

Entropy, which of course is the title of this particular session.0892

Entropy as a function of temperature and volume.0897

We will let S= S(TV) this just says that entropy is a function of temperature and volume.0909

We can go ahead and write the total differential of this expression then DS is going to equal DS DT, a partial of S0916

with respect to the first variable holding the V constant × DT + DS DV holding T constant DV.0930

This is just the total differential expression, we know this already.0942

Our task is that these differential coefficients right here, can we identify these differential coefficients with things0945

that we can easily measure or things that we can look up in a table, that is what we want.0954

We are doing the same exact thing that we did with energy.0960

We have energy as a function of temperature and volume.0963

Energy as a function of temperature and pressure those involved differential coefficients.0968

We related, we identify these differential coefficients with things that are easily measurable.0974

We are trying to do the same thing here, that is all we are doing or we are finding a way to express0980

this elusive quality entropy in terms of things that we can measure and we begin this way.0985

Let us see what we can do.0991

Let us go ahead and write the fundamental equation again.0993

The fundamental equation we have DS =1/ T DU + P /T DV.1002

Notice they are almost the same DS DS, DT here is DU DV and DV.1016

Here is the same, we need to start manipulating these set of equations.1021

Let us go ahead and recall that we had DU = DU DT V DT + DU DV T DV.1030

That was equal to the constant volume DT + DU DV T DV.1050

This is just this, we know this already from our work in energy DU = this.1059

I’m going to go ahead and take this expression for DU and actually substitute into here into the fundamental equation of thermodynamics.1066

Here is what I get when I do that I get DS =1/ T × CV DT + DU DV T DV + P/ T DV multiply this out,1076

this is equal to CV/ T DT + 1/ T × DU DV sub T DV + P/ T DV.1106

The combined terms, this is a DV and this is a DV, I have T in the denominators so I can go ahead and combine that.1125

I’m going to write this as CV / T DT + 1/ T × DU/ DV T + P DV.1132

Again, there is nothing strange happening here, this is a basic algebra is was what I'm doing, it is not in calculus.1155

Now we have this equation, I'm going to call this equation 2.1162

This is from the fundamental equation of thermodynamics and the expression for energy that I substituted into it.1169

Notice I have DT and DV.1178

I have this thing and I have this thing.1182

Equation 1, let us write the total differential expression which was DS DT DT + DS DV T DV, I will call this equation 1.1184

Since both equation 1 and the equation 2 express DS in terms of DT and DV, DS DS DT DT DV DV that means that equation 1 and 2 are identical.1210

Once again I have 1 equation that derive one way and I have another equation which is from1242

the total differential expression DS DT DV DS DT DV they are equivalent.1247

Therefore, this thing is this thing and this thing is this thing, that is what I'm going to do next.1253

This is that DS DT holding V constant = CV/ T.1268

I have achieved what I wanted, I wanted to find a way to express this in terms of something that I can measure.1280

I have identified it with something that is measurable but I can measure the constant volume heat capacity divided by the temperature.1285

It is fantastic, that is one of them.1291

The other one, this is equivalent to that.1294

It looks a little bit more complicated and is more complicated but again we are going to simplify that one too.1297

We have DS DV T =1/ T × DU DV T + P.1303

The change in entropy per change in temperature, in other words, when I change the temperature how does the entropy change?1324

That is how it changes.1333

Here if I change the volume how does the entropy change under constant temperature?1335

That is how it changes.1340

I was able to identify this and this with things that are now easily measurable, this was our task.1342

Let us deal with these one at a time.1351

At constant volume the rate of change of entropy with respect to temperature = CV/ T.1354

Since CV/ T is always positive because the constant volume heat capacity is always positive, the temperature is always positive.1377

The reason is because CV is always positive and at constant volume,1394

if I hold the volume of the system constant an increase in temperature implies an increase in entropy.1409

Increase in entropy that is what this says.1433

If I raise the temperature, I raise the entropy of the system.1437

The temperature dependents of entropy, in other words the extent to which entropy depends on temperature is very simple,1443

it is just the constant volume heat capacity divided by the temperature.1451

The temperature dependence of entropy is simple, it is just the constant volume heat capacity CV ÷ T.1459

Let us see what we have got, so we have DS = DS DT V DT.1495

I’m just writing the total differential expression + DS DV T DV at constant volume.1506

Constant volume that means DV is 0.1517

This is 0 because these constants, if we happen do that what we end up with is the entropy = this thing which we know is this thing = CV / T DT.1519

If I integrate this, I end up with the change in entropy = the integral from temperature 1 to temperature 2 of CV / T DT.1537

In a particular system, if I hold a volume of the system constant and just raise the temperature this gives me a way of finding the entropy.1552

I just integrate the constant volume heat capacity divided by temperature / the temperature range that I do.1561

Let us go ahead and see an example.1569

Example 1, I think I have here we go.1578

Here is my example and we will go ahead and do this in black.1582

1 mol of nitrogen gas is heated at constant volume so we have a constant volume process from 350°K to 600°K what is the δ S for this process?1584

The constant volume heat capacity is 3 Rn/ 2 and is the number of moles.1596

Nice and simple, we just calculated it.1603

When you do these problems you want to start with the most general situation and work your way down.1609

You do not want to just memorize equations for a particular set of circumstances, that is not going to work with thermodynamics.1615

There are far too many equations to actually memorize and there are far too many variations of the circumstances.1623

You are just going to make yourself crazy, you are going to cause yourself a lot of stress and you will make it worse.1629

You want to start with the most general situation and then flow from what the problem says,1640

go ahead and derive what you need which is what we did with the energy.1644

The one of the reasons that we did all the problems that we did with energy and we are going to be doing the same.1648

In these lessons, I’m doing one or two example problems.1653

I’m saving all the example problems for the end of this particular unit where we are going to do a huge bulk of problems over and over again.1656

Do not worry about just having 1 example, you are going to be doing a lot of problems but let us see.1665

The best way to handle this is the way we just handled it.1671

We know that the change in entropy in general = DS DT V DT + DS DV T DV this is the most general equation.1675

This is the equation that you want to now.1691

You also know this, we also identified this with CV/ T DT + DS DV DV.1694

The general equations we want to start off with is, this is what you want to memorize, this one right here.1704

The problem tells us that the volume is constant.1711

The volume is constant these just drops off to 0 so what we are left with is DS = CV/ T DT.1716

We want δ S so let us just integrate this.1724

Therefore, we have our equation δ S = the integral from temperature 1 to temperature 2 of CV/ T DT.1728

We have all the information that we need.1738

We have CV so let us go ahead and do it.1740

Let us go ahead and move over here.1745

Therefore, δ S = the integral of temperature is from 350 to 600°K.1747

Our CV is 3 Rn/ 2 so it is 3 Rn/ 2.1758

We do it this way, we have 3 Rn/ 2/ T DT.1769

3 Rn/ 2 is a constant so what it comes out 3 Rn/ 2.1778

The integral from 350 to 600 of DT/ T we can do this, this is very simple.1786

This is 3 × R which is 8.314 × n which is the number of moles we have 1 mol here.1796

The integral of DT/ T is just Ln.1808

It is going to be Ln of 600/ 350 and when I solve this I get 6.72 J/°K.1812

There you go, that is my δ S.1826

If I have 1 mol of nitrogen gas and I take it from 350°K to 600°K, the entropy change for that is 6.72 J/°K.1831

It is all based on this, the general equation that applies to every single system solid, liquid, gas, it does not matter.1842

This is the equation that you want to memorize, the same thing that we did for energy.1850

Once you have this, the problem will allow you to knock things out.1853

From the general and then strained we end up with a really simple integral.1859

This way we do not have to memorize this equation for this particular situation, that is not going to work with thermodynamics.1864

I promise you do not go down that path it will cause you a lot of heartache and it will affect your grades in a bad way.1870

There are just too much information for you to memorize.1877

The nice thing about thermodynamics is there is only a handful very few general equations that you have to memorize.1882

You remember that from what we did with energy, there was only the definition of work, the definition of energy, and a definition of enthalpy, and the two basic mathematical relations.1889

That is only 5 to 6 equations that you have to memorize, that is better than any other class.1901

Definitely do it this way.1906

We have dealt with the temperature dependence now we have to deal with the other one.1910

The other one that was not quite so simple.1913

The volume dependence of entropy, in other words the rate of change of entropy might change the volume.1916

That is not so simple.1922

Let us stick with black here.1924

The volume dependence of DS at constant T is not so simple.1927

We are going to have to fiddle with that mathematically.1949

It was DS DV at constant T =1/ T × P + DU/ DV sub T.1956

I think I actually reversed them, it does not really matter. I think I have the other one first.1975

DU/ DV sub T + P does not matter what order you do it in.1984

Let us go ahead and call this equation 3.1989

Let us see what we have got.1995

Here is what we are going to do, let us go ahead and recall this equation, the one that we just had.2005

DS DT the temperature dependents under constant volume that was equal to CV/ T.2014

I’m going to differentiate this equation with respect to volume holding the temperature constant.2023

We differentiate with respect to volume holding T constant.2037

I’m going to differentiate this and it becomes the following.2048

Differentiate the whole thing so it becomes, I'm going to take the derivative with respect to volume of this,2051

it is going to equal, the T is constant 1/ T D CV DV.2062

Write the derivative of this with respect the volume =1/ T DV of DU DT and all it done here2071

is I have use the definition of constant volume heat capacity D DV of CV is the D DV of this.2088

You remember the constant volume heat capacity is actually equal to the change in energy / the change in temperature that is the definition of heat capacity.2096

Equals 1/ T I’m going to put this expression together as D² U DV DT.2108

Our final expression, I’m going to put this together this is D² S DV DT = 1/ T D² U DV DT this is just mathematical manipulation.2125

What I did is I took this one, a differentiated with respect to volume holding T constant2159

and I’m going to take this equation and I'm going to differentiate with respect to T holding V constant.2167

I’m going to take this equation which I will rewrite so I have got DS DV T = 1/ T × let us go ahead and write it the way that I have here DU DV T.2176

Let us go ahead and differentiate this with respect to T and this time we will hold V constant so we end up with the following.2201

This is going to be a little long so I got DDT of DS DV T and I'm going to put that together as D² S DT DV2217

that is going to equal when I differentiate this I'm going to end up with the following.2239

I’m differentiating with respect to T so it is going to be this × the derivative of that + that × the derivative of this.2243

It is the product rule so was going to be 1/ T × differentiating this thing with respect to T holding V constant +2253

differentiate this with respect to U DV T – 1/ T² × P + DU/ DV sub T.2270

I know, do not worry about it.2291

S = S (TV) and DS is an exact differential.2300

You remember from our previous math lesson that the mixed partials are equal which means that D² S DV DT = D² S DT DV mixed partials.2312

I just found the square D ⁺S DV DT and I found D² S DT DV.2331

I'm going to set those two equal to each other so I get the following.2336

I get 1/ T D² U DV DT = 1/ T DP DT V + 1/ T D² U DT DV – 1/ T²2341

I’m just taking this from the previous page, I’m just setting these two things equal to each other -1/ T² × P + DU/ DV sub T.2369

1/ T² DV DT, 1/ DT² DT DV.2394

The V's cancel because again mixed partials U is a state property, U is a function of temperature and volume.2398

Therefore, the mixed partials are equal because this time for U, mixed partials are equal.2410

We have 1/ T DP DT sub V =1/ T² I just move this thing over there 1/ T² × P + DU DV T.2429

We have this thing, 1/T DP DT sub V = 1/ T².2462

We also had this which was one of our equations.2469

We had DS DV sub T = 1/ T.2473

I’m going to make a little change before I do this one.2485

I'm going to go ahead and cancel one of the 1/ T and I’m going to end up with DP DT sub V = 1/ T P + DU / DV T.2489

Do not worry you do not have to know this derivation, it is only the final result that we need but2508

it is important to see the derivation especially at this level of your study of science.2514

You need to know where this is actually coming from.2519

You need to be comfortable with recognizing where the mathematics is going.2523

It is very important.2527

We have this, now we also had from one of the previous equations this thing we had DS DV sub T =1/ T,2530

this is 1 of our basic equations + DU DV sub T.2548

well this and this are the same which means these two are equal.2556

What we end up with is, we go back to black.2566

Our differential coefficient DS / DV T is actually equal to DP/ DT V.2573

This is what we wanted.2587

We related one of the differential coefficients to constant volume capacity / T.2589

With some mathematical manipulation, the other differential coefficient the one which is volume dependent,2594

the change in entropy or the change in volume is actually equal to or identifying it to the change in pressure over2599

the change in temperature when we hold the volume constant.2607

This is easy to measure, this is not easy to measure, this is easy to measure.2611

This is telling me that whatever this numerical value is it happens to be the rate of change of entropy with respect to volume.2615

Let us go ahead and rewrite that so we have a DS DV at constant temperature = DP DT at constant volume.2626

This is our relation that we wanted.2641

Now we have a simple expression or a simpler expression at least than what it was for the volume dependence of entropy at constant temperature.2646

This DP/ DT constant V on the right side is very easily measurable and that is what we want.2683

We want something that is not easily measurable.2691

We do not measure entropy directly but this is my easily measurable.2696

We are identifying this with that, that is what we want.2700

It is very easily measurable.2706

Recall from example 2, in the previous math lesson, we said when we are discussing the cyclic rule we ended deriving this relationship.2716

We end up deriving the DP/ DT under constant volume is actually equal to Α / K,2743

the coefficient of thermal expansion divided by the coefficient of compressibility.2751

Therefore, DP/ DT sub V that is right there so we have DS/ DV T this is the V not U = Α / K, there you go.2756

A and K I just love those, they are very simple just using your handbook or I measure it.2773

Either this equation or this equation, this expresses the volume dependence of entropy.2779

Let us say a few more words here.2790

Let us do our summary and recap, in case we have lost our way in all this crazy mathematics.2795

Let us start on the next page here, I’m going to go ahead and do this one in blue.2802

In case we have lost our way, we have a fundamental equation of thermodynamics.2815

I’m just going to call it the fundamental equation of thermal F E T.2821

It is said that DS =1/ T DU + P/ T DV.2826

We also said that for entropy as a function of temperature and volume our differential expression gives us DS = DS DT V DT + DS DV T DV.2840

We want to identify these differential coefficients with things that are easily measurable or things that we can look up in a handbook.2866

We wanted to identify the differential coefficients these things with quantities easily measured or already tabulated and we did.2880

We had DS DT, the temperature dependence of the entropy at constant volume is just equal to the constant volume heat capacity divided by that.2926

The volume dependence of entropy under constant temperature happens to equal DP DT2939

under constant volume which also happens to equal Α / K.2949

This is the temperature dependence of the entropy.2955

This is the volume dependence of the entropy.2957

K is always positive, Α is mostly positive, when we say mostly, with the exception of water and a couple of other things.2965

For most situations and by most I mean like 99.999% of situations, Α / K is positive.2982

Positive what does this mean?2995

It means that if I increase the volume, I increase the entropy.2998

I knew that already but this just confirms it mathematically.3009

It confirms it in another way, that is what is going on here.3012

The total differential expression which is what we start off with DS DT V DT + DS DV T DV can now be written DS = CV/ T DT + Α / K DV.3023

If I'm dealing with the system and, by the way this applies to every systems gas, liquid, solid,3064

under any set of circumstances this is the most general equation.3069

If I want to know what the change in entropy of the system is, under conditions of change in temperature, change in volume, this is the equation that I use.3073

If volume is held constant in a particular problem this term goes to 0.3082

If temperature is held constant in a particular problem this term goes to 0.3087

I just do it with this term.3091

Α and K, I can easily look up or the problem will give it to me.3092

C, V, or T I can either measure or the problem will give it to me, that is what makes this beautiful.3096

This is the equation that you want to memorize this and this.3102

This is the general version that you want to start off with your problems and this is the version3105

that actually involves what it is that we have identified this with.3112

This is identified with this, this is identified with this.3119

This is what we wanted, we want to express entropy in terms of variables that we do know temperature, pressure, volume, things like that.3122

We wanted to identify the differential coefficients once we had the expression for how entropy behaves3131

in terms of things that we can actually measure, that is all that we have done here.3136

The rest is just mathematical manipulation.3140

Please, do not think that is mathematical manipulation or something that was just obvious.3144

It took a lot of months, years, hours, for us to figure out the stuff out, for people to figure the stuff out in the 19th century.3148

It took a lot of time.3157

What we are getting is the result of all that.3159

It looks like let us just do this, let us differentiate here, let us identify this with this.3161

A lot of paper, a lot of time has been spent trying to put this together.3167

Now one final word, for liquids and solids the volume dependence, the second part of the equation3173

the volume dependence of entropy is usually so small that it can be ignored.3195

In a particular problem, unless they tell you to specifically include this for anytime you are dealing3215

with a system which is a liquid or solid because automatically take that to 0.3223

You cannot do it with a gas so but you can with a liquid a solid for the most part, unless the problem tells you otherwise.3227

Let us go ahead and revisit Α = 1/ V DV DT under constant pressure and K happens to equal -1/ V DV DP under constant temperature.3236

This is the coefficient of thermal expansion.3257

This is the coefficient of compressibility.3260

This is one of the equations and this is pretty much what you want.3265

You want to know this equation, you want to start your problems with that equation.3270

Thank you so much for joining us here at www.educator.com.3274

We will see you next time, bye.3276