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

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

### 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
- Entropy As a Function of Temperature & Volume
- Fundamental Equation of Thermodynamics
- Things to Notice
- Entropy As a Function of Temperature & Volume
- Temperature-dependence of Entropy
- Example I
- Entropy As a Function of Temperature & Volume, Cont.
- Volume-dependence of Entropy at Constant Temperature
- Differentiate with Respect to Temperature, Holding Volume Constant
- Recall the Cyclic Rule
- Summary & Recap

- 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

### Physical Chemistry Online Course

### 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 derivatives*0029

*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 remember*0056

*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 able*0196

*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 equal*0232

*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

*Let us go ahead and start with our first law.*0275

*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 changes*0493

*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 system*0718

*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 one*0825

*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 something*0866

*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 S*0916

*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 things*0945

*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 express*0980

*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 from*1242

*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 will actually affect your understanding at a very bad way.*1635

*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

*We have the following, it was this.*1954

*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 here*2071

*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 constant*2159

*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 DV*2217

*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/ T²*2341

*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 but*2508

*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 over*2599

*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 DT*2939

*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 version*3105

*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 behaves*3131

*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 equation*3173

*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 dealing*3215

*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

1 answer

Last reply by: Professor Hovasapian

Tue Sep 15, 2015 2:54 AM

Post by Shukree AbdulRashed on September 14, 2015

Hello. 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 answer

Last reply by: Professor Hovasapian

Sat Jun 20, 2015 5:59 PM

Post by Joseph Carroll on June 11, 2015

Say 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?