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

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

### Free Energy Example Problems III

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
- Example I 0:10
- Example II 15:03
- Example III 21:47
- Example IV 28:37
- Example IV: Part A
- Example IV: Part B
- Example IV: Part C

### Physical Chemistry Online Course

### Transcription: Free Energy Example Problems III

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

*Today, we are going to continue our example problems on free energy.*0004

*Let us jump right on in.*0009

*I have to warn you, in this particular example problem set, this is going to be heavily mathematical.*0011

*Not heavily mathematical in terms of difficulty, a lot of it is just using equations at our disposal in order to come up with new relations,*0023

*that is really what it is all about.*0031

*Between energy, entropy, and the free energy, we have collected a fair number of equations.*0033

*We sort of close the circle on this discussion of thermodynamics.*0037

*But now we want to do is we want to take all of these things and put them together.*0042

*This is the part where it is going to be just strictly mathematical.*0047

*Most of it is manipulation, what is going to make it seem intimidating is just the symbolism on a page.*0056

*It is going to be a lot of symbols on the page and there is one of the lot equations that we are going to have to recall.*0062

*This is actually an excellent exercise in putting all of this together.*0067

*In any case, let us just jump right on in and see what we can do.*0072

*The first one, using the appropriate Maxwell relation and the cyclic relation among volume, temperature,*0076

*and entropy, demonstrate that DS DP sub V = Κ C sub V/ A T.*0083

*The rate of change of entropy with respect to pressure holding volume constant is equal to the,*0092

*This is the coefficient of thermal expansion, this is the coefficient of compressibility,*0101

*this is the constant volume heat capacity, and this is the temperature.*0104

*We want to find ways of expressing these derivatives, these relations in terms of easily measurable things.*0108

*Like kappa and α, I can look those up.*0115

*Constant volume heat capacity, I can either look it up or measure it directly.*0118

*The temperature, I can just measure it, that is what we want.*0121

*Let us see what we can do.*0126

*When I look at the collection of Maxwell relations that I have, the only relation that involves TV and S and P V and S is the following.*0129

*Let me actually go ahead and do this and blue.*0145

*It is this one, DT DV sub S = - DP DS sub V.*0147

*Well, what we want is DS DP sub V.*0162

*Basically, we want the reciprocal of this.*0174

*It is not a problem, let us take the reciprocal of that.*0177

*This is what we want, we have this relationship.*0181

*We recognize the PS and V, PDS and V, we just want to switch the P and S.*0186

*We want the P on top.*0190

*We want the rate of change of pressure with respect to change in entropy at constant volume.*0192

*We go ahead and we reciprocate this expression.*0199

*Therefore, DS DP at constant V = - DV DT of S.*0202

*Let us use the cyclic relation.*0220

*The cyclic relation between V T and S is the following.*0222

*DV DT S, VTS let me actually show you how I come up with this, in fact.*0229

*I have V T S, let me write those down V T and S.*0238

*Underneath them, I'm going to just choose and I’m going to go ahead and put T here.*0246

*I'm going to put S here, and I’m going to have V here.*0251

*I’m going to do that.*0256

*These VT, therefore this is an S, this is a TS, this is a V SV, this is a T.*0261

*That is how you derive the cyclical relation.*0267

*A cyclic relation among three variables in this case, volume, temperature, and entropy, I can do it that way.*0269

*List any three of them then underneath it just at random, list the other three, and then what you are left out with,*0275

*let us just go ahead and put the particular property that is held constant on the outside.*0283

*The TS TSV SVT and = -1.*0288

*We have this relationship and of course these are partial derivatives.*0292

*You can go ahead and put the partial derivative signs in.*0295

*That is the relationship that exists.*0299

*With this particular relations, we have this and let me go ahead and rewrite it so I can see a little bit better.*0301

*DV DT constant S, DT DS constant V, and DS DV constant S V T = -1.*0310

*This right here is just the reciprocal of DS DT at constant V.*0334

*The rate of change of entropy with respect to temperature at constant volume.*0361

*We know that is equal to the constant volume/ T from back in the entropy.*0367

*This one right here, this DS DV sub T, let us go ahead and bring this down here.*0375

*This is just DP DT sub V from another of the Maxwell relations.*0387

*We are just looking for what is equal to what.*0404

*It is like all of these things, all of these equations at our disposal we are going to pick*0408

*and choose to see how we can arrange this puzzle to get what it is that we want.*0410

*This thing is equal to that from another one of Maxwell's relations.*0417

*Furthermore, DP DT sub V happens to equal Α/ K that came from the coefficient of thermal expansion and compressibility.*0430

*You recall the coefficient of thermal expansion A =1/ V DV DT that and Κ =-1/ V DV DT that.*0485

*When we arranged these, we ended up with that Α/ Κ.*0503

*You know what, I will go ahead and actually do it real quickly here just so you actually see it.*0510

*This is from the previous lesson but that is not a problem, we can go ahead and do it here.*0517

*I have pressure, volume, and temperature.*0521

*The cyclic relation would look something like this DP DT constant V × DT DV constant P × DV DP constant T = -1.*0525

*I get DP DT constant V × based on this DV DT = VΑ.*0547

*DT DV is 1/ VΑ, same thing here.*0558

*DV DP that is just - VΚ = -1.*0564

*When I move this over to your side, these cancel and I get Κ/ Α.*0572

*When we move it over to the other side I end up with the following.*0579

*I end up with DP DT sub V = Α/ K.*0582

*That is where the Α/ Κ came from.*0589

*Let us go back and finish what we started.*0591

*We have DV DT sub S, DT DS sub V, and DS DV sub T = -1.*0599

*This becomes DV DT sub S × T/ CV × Α/ Κ = -1.*0617

*We get DV, therefore, DV DT sub S = - Κ CV/ T Α.*0635

*This DS DP sub V = -DV DT sub S = - Κ CV/ T Α.*0655

*Therefore, I have DS/ DP sub V = Κ CV/ T Α which is exactly what I wanted.*0673

*I have the rate of change of entropy with respect to pressure holding volume constant = Κ × CV/ T Α.*0693

*What do we just do and why is it important?*0706

*We just threw a bunch of mathematics, why is this significant?*0709

*Recall, that we have the following expressions.*0713

*Recall the following, we had DS = CV/ T DT + Α/ Κ DV, where entropy was a function of temperature and volume.*0720

*We also had DS = CP/ T DT – V A DP.*0743

*In this particular case, this is entropy expressed as a function of temperature and pressure.*0758

*What we have done here is deriving an expression for how entropy changes with respect to pressure under constant volume.*0765

*In other words, we have expressed entropy in terms of pressure and volume.*0775

*What we have done is derive an expression for how entropy changes with pressure under constant V.*0786

*In other words, we have expressed entropy as a function of pressure and volume.*0816

*Later in the lesson, we are going to derive an expression for so what we have done is we found DS DP sub V.*0828

*Later in the lesson, we are going to do DS DV holding that constant.*0836

*That is what we have done here.*0843

*We have expressed entropy as a function of temperature and volume.*0845

*We have expressed it as a function of temperature and pressure.*0849

*We are expressing it as a function of pressure and volume.*0853

*Now, we have all three variables accounted for all constraints.*0857

*In this case, we are holding the pressure constant.*0861

*In this case, we are holding the temperature, pressure, the volume constant.*0865

*In this particular case, we end up holding the temperature constant, that is what we have done.*0868

*We have found a way to express the change in entropy with respect to pressure and volume.*0873

*Before, we had temperature and volume, temperature and pressure, now we have pressure and volume.*0880

*That is what we have done based on things we already knew in addition to the equations that we derived.*0886

*That is why this is significant.*0891

*And this is the theme that is going to run through this particular problems set.*0893

*Let us go ahead and continue on here.*0900

*In a lesson on the thermodynamic equations of state we were able to finally break down the following.*0905

*DU = CV DT + this thing DV.*0910

*Do you remember this was DU DV constant T.*0917

*The Joules law for an ideal gas that is equal 0 but it is not an ideal gas it does not equal 0.*0925

*We have to account for it.*0930

*Given this demonstrate the following DU = this thing DT + DP.*0932

*Why is it significant?*0938

*We express energy in terms of temperature and volume.*0940

*When we are talking about energy for temperature and pressure it was DH,*0944

*it was enthalpy, temperature and pressure or the variables for enthalpy.*0949

*Based on this, we can go ahead and now express a change in energy in terms of temperature and pressure also.*0954

*We do not have to be in enthalpy, we can just deal straight with the energy itself, that is why this is significant.*0961

*Let us go ahead and run through the process here.*0969

*The way we are going to do this, this is DT and this is DV.*0973

*We want DT and DP so I need to expand DV in terms of temperature and pressure.*0980

*Let us expand DV in terms of DT and DP and collect terms.*0988

*We are expressing volume as a function of temperature and pressure so the total differential expression is the following DV = DV,*1007

*Temperature and pressure, I’m sorry.*1024

*We have got DV DTP DT + DV DPT DP.*1028

*DV DT sub P that is just Α V, the coefficient of thermal expansion DT.*1046

*DV DP sub T that is just the coefficient of compressibility.*1057

*It ends up being - Κ × V DP.*1062

*I'm going to go ahead and put this thing DV, this expression up here into this.*1070

*I'm going to collect terms.*1081

*Here is what happens so I get DU = CV DT + Α T - Κ P/ Κ × Α V DT - Κ V DP.*1083

*DU = CV DT + I’m going to separate this out.*1115

*This is going to be Α T/ Κ - P × Α V DT – Κ V DP.*1121

*DU = CV DT + this times this, this times that.*1136

*This is going to be Α² TV/ Κ DT and this times that, - Α TV.*1146

*The Κ and the Κ cancel that is going to be DP – P A V DT + P Κ V DP.*1166

*Let us collect terms, here is the DT, here is the DT, here is a DT.*1183

*I’m going to do this one in red, so we see it.*1192

*I have a DT term, I have a DT term, I have a DT term, this was going to be CV + Α² TV/ Κ - P A V × DT.*1205

*I have a DP term and a DP term so it is going to be + P K V - Α TV × DP which I think it is exactly what we wanted if I’m not mistaken.*1222

*It is actually, all we have to do is now factor out the V so we have DU = CV + Α² TV/ Κ – P Α V DT + V × P Κ - Α T DP.*1245

*This is what we wanted.*1270

*Notice how everything is expressed in terms of things that are measurable.*1276

*Heat capacity Α, temperature pressure Κ, pressure Κ Α, temperature volume, everything is there.*1280

*Everything is easy, I found a way.*1287

*I have derived a way to express energy in terms of temperature and pressure.*1289

*It used to be enthalpy but if I want to deal strictly with energy I have this option.*1296

*Let us move on to the next problem here.*1307

*Using the result from the previous problem, demonstrate that near 1 atm of pressure*1309

*at a change in energy with respect to pressure holding the temperature constant is approximately equal to – VT Α,*1315

*Then using the following data for water at 20°C, calculate the actual value of DU DP sub T.*1322

*Here we have an Α for water, we have the Κ for water, we have a smaller mass, and we have its particular density.*1331

*Let us see what we can do.*1337

*Let me go back to blue here.*1341

*We have DU = CV + TV Α ²/ Κ – P V A.*1343

*This is DT + V × P Κ - T Α DP.*1360

*At fixed temperature means that DT equal 0.*1372

*A fixed T implies that DT equal 0, this is isothermal so that goes to 0.*1377

*All I’m left with is that term.*1384

*We have DU= volume × pressure × Κ - temperature × Α × DP.*1387

*Let us go ahead and first of all, let us go ahead and calculate.*1399

*Let us do this in red.*1402

*Let us go ahead and find PK.*1403

*That is fine, I can do it here.*1408

*Let us find this term first so we can compare the two.*1409

*PK = the pressure is 1 atm and the Κ is 45.3 × 10⁻⁶/ atm.*1413

*I end up with 45.3 × 10⁻⁶.*1431

*Let us go ahead and find TA.*1436

*TA the temperature is 293 °K because it is at 20° C and Α is 2.07 × 10⁻⁴.*1440

*I end up with a pure number of 0.0607.*1453

*Notice, 45.3 × 10⁻⁶ is 6.07 × 10⁻².*1458

*In this particular case, the P × K is a lot less than the T Α.*1470

*Because that is the case, we can pretty much ignore that term.*1478

*Therefore, we have what we wanted, the DU = VT A – VT Α approximately equal to,*1481

*I’m using the approximate because we ignore this, not exact but this is what we wanted.*1495

*It was very simple, this goes away then compare these two.*1501

*You can ignore this because it is so much smaller than this number.*1505

*This is what you get, the change in energy at around 1 atmosphere = the volume × the temperature × Α under isothermal process.*1508

*This is really simple nice and easy.*1519

*Let us see what we have got here, this is what we want to prove.*1526

*Let me double check and I think we want to express it this way.*1531

*DU DP under constant T = - VT Α.*1541

*There you go, sorry that should be a DP there.*1556

*This and this are the same thing.*1560

*I can just move the P over here, I'm holding T constants so I can just go ahead and expressive in partial differential notation.*1563

*Let us go ahead and do the calculation.*1571

*Volume = 1 cm³ = 0.99821 g and we have 18 g/ mol and we have × 10⁻³ dm³/ cm³.*1573

*That cancels and that cancels, I'm left with a volume equal to 0.018O3 L/ mol because dm³ is a liter.*1601

*Therefore, our change in energy with respect to pressure under constant temperature conditions,*1618

*in this particular case let us be specific, 20°C for H2O = - VT Α = -0.01803 × 293°K × Α which is 2.07 × 10⁻⁴.*1627

*I end up with this being equal to 0.001094, this is going to be L atm.*1651

*I need to convert that to Joules, I do not need to but Joules is pretty standard.*1661

*20°C H2O = 0.111 J/ atm, the unit is J/ atm.*1668

*Joules on top energy, pressure on the bottom, this is not J/ mol but J/ atm.*1683

*At 20°C for water, if I change the pressure by 1 atm, I increase the energy by 0.111 J.*1689

*That is what this says, the rate of change of energy with respect to a unit change in pressure is equal to this under these conditions.*1703

*Nice and simple and straightforward.*1712

*Let us get down to some serious business.*1715

*Example 4, we know the DS = CP/ T DT – V Α DP.*1724

*We know this from our expression for entropy, for more chapters on entropy.*1730

*Given this we want to show that the rate of change of entropy with respect to pressure under constant volume is this.*1735

*The rate of change of entropy with respect to volume under constant pressure is this and then -1/ V ×*1742

*the rate of change in volume with respect to pressure under constant S = this.*1750

*Notice, SPV SPV SPV we are establishing relationships now between entropy, pressure, and volume.*1755

*Let us go ahead and do this given that.*1766

*Let us see what we can do.*1771

*Part A, we want to show that DS DP sub V = Κ CV/ T Α.*1773

*Notice, our variables are SP and V.*1799

*Now we have DS = CP/ T DT – V Α DP.*1802

*In other words, S is a function of temperature and pressure.*1816

*We want entropy to be a function of pressure and volume so we expand the same way we did before in the previous problem.*1825

*In this lesson, we expand DT in terms of DP and DV so we will let temperature be a function of pressure and volume.*1839

*For our total differential, we get DT =DT DP constant V DP + DT DV constant P DV.*1859

*We go ahead and put this expression into here and we collect terms and multiply it.*1880

*We get DS = CP/ T × DT DP sub V DP + DT DV sub P DV – V Α DP.*1889

*DS = CP/ T × DT/ DP sub V DP + CP/ T × DT DV sub P DV – V Α DP.*1918

*Let us see how I want to do this.*1949

*Let us go ahead and collect terms here.*1953

*I will do it on the next page.*1955

*DS = CP/ T DT DP.*1960

*I really hope I'm keeping my variables straight here because there are so many symbols on this page.*1971

*It is enough to make you absolutely crazy.*1978

*I understand if you are feeling crazy about this.*1981

*+ CP/ T DT DV sub P.*1984

*My God this is crazy.*1993

*This is just DS DP holding V constant, this thing the total differential.*1998

*DS DP sub V = CP/ T × Κ / Α - V Α.*2020

*We also have the following relation, we also have the other relation which I will do in red.*2038

*We also have from the lesson on the thermodynamic equations of state, we have CP = CV + TV Α²/ Κ.*2045

*Therefore, we are going to put this expression into here and we are going to get DS DP sub V = CV + TV Α² ÷ Κ/ T × Κ/ Α - VA = CV/ T +,*2071

*I’m going to separate this out + V Α²/ Κ × Κ/ Α – V Α = Κ CV/ T Α + VA – VA VΑ.*2104

*Therefore, I get my final expression DS DP sub V = Κ × CV/ T Α.*2134

*This is what we wanted.*2148

*There we go, long semi painful process all in order to get the change in entropy with respect to a change in pressure under constant volume = Κ CV ÷ T Α.*2150

*That is all we have done here.*2164

*Let us move on to part B, we want to show that DS,*2168

*We want to do a change in entropy with respect to volume holding pressure constant.*2179

*We want to show that is equal to CP/ TV Α.*2184

*For part A, we had the following.*2190

*We had DS = CP/ T × Κ/ Α - V Α DP + CP/ T × DT DV sub P and this is DV.*2192

*This part right here, this is just DS DV sub P, that is what we wanted.*2215

*DS DV sub P = CP/ T × DT DV sub P =C sub P/ T × 1/ V Α.*2230

*This is just the reciprocal of the rearrangement of the coefficient of thermal expansion Α.*2248

*There you go, we have it.*2256

*We have DS DV sub P = CP/ TV Α.*2258

*Part A was a change in entropy with respect to pressure constant volume,*2269

*this is a change in entropy with respect to volume under constant pressure.*2275

*We are expressing entropy in terms of pressure and volume.*2280

*This is really beautiful stuff here.*2285

*Let us stop and consider what it is that we have done here, this is really important.*2290

*From the expression of entropy we have the following.*2300

*We have that entropy is a function of temperature and volume and we had the following total differential DS = CV/ T DT + A/ K DV.*2303

*We also had entropy as a function of temperature and pressure and we had DS = CP/ T DT – V Α DP.*2318

*We have S = a function of now temperature volume, temperature pressure, pressure and volume.*2336

*Same thing we did before for energy.*2351

*We have entropy in terms of pressure and volume.*2353

*For entropy equaling a function of temperature, pressure, and volume, we have the following expression.*2358

*DS = Κ CV/ T Α DP + C sub P/ T V Α DV.*2365

*We now have expressions for entropy under all three constraints.*2380

*This is temperature and volume, we are holding pressure constant.*2385

*In this case, this is temperature and pressure, we are holding volume constant.*2389

*We have an expression for entropy, in terms of pressure and volume, we are holding temperature constant.*2394

*This is what we have done here.*2400

*This is what all the mathematical machinery has brought us to.*2403

*It seems a little random all over the place but this is what it means, this is the big picture.*2407

*We have a way of finding entropy under any constraint with simple temperature, pressure, and volume.*2412

*This is absolutely fantastic and absolutely beautiful.*2418

*Let us go ahead and finish it off with part C and see what it is that we are looking for.*2423

*Let me see should I do it in this page or the next?*2429

*I will go ahead and do on the next page.*2431

*Part C, we want to show that -1/ V DV DP under constant S = Κ CV/ CP.*2434

*By the cyclic rule on volume pressure and entropy we have the following.*2454

*DV DP sub S × DP DS sub V × DS DV sub P = -1.*2468

*Therefore, DV DP sub S =, I'm just going to go ahead and multiply by the reciprocal of these.*2486

*I end up with the following - DS DP sub V × DV DS DP DV DS.*2494

*P DV DS that is correct, we have that.*2514

*We just found this, we found DS DP and we found DS DV in parts A and B.*2518

*Let us just put them in here.*2523

*-Κ CV/ T Α, this DV DS is just the reciprocal of the DS DV.*2525

*That is just going to be TV Α/ CP.*2535

*T cancels T, A cancels Α, we are left with DV DP sub S =- Κ CV × V/ CP.*2540

*I'm going to move the negative, I’m going to divide by V so I get -1/ V × DV DP.*2559

*This is a V not a U, S = Κ CV/ CP which is exactly what we wanted.*2570

*We found DS with respect to pressure.*2581

*We found DS with respect to volume.*2585

*We are expressing the final one, now we are putting the volume and the pressure.*2587

*The rate of change of volume with respect to pressure holding the entropy constant.*2592

*This is absolutely beautiful.*2597

*This thing is also written sometimes this way.*2599

*This is also written -1/ V DV DP sub S = Κ/ Γ.*2603

*Remember that Γ is equal to the ratio of the constant pressure heat capacity to the constant volume heat capacity.*2618

*That is what we have done here.*2629

*We have expressed entropy in terms of temperature and pressure.*2631

*You have expressed it in terms of temperature and volume.*2634

*Now, we have expressed it in terms of temperature and volume,*2636

*we found the rate of change of entropy in terms of pressure, constant volume, that was part A.*2640

*We did the rate of change of entropy with respect to volume under constant pressure, that was part B.*2647

*We went ahead and SPV SVP now we did at the rate of change in volume with respect to pressure under constant entropy,*2656

*that is what we have done here.*2665

*If we want the rate of change of pressure with respect to change in volume, all we have to do is reciprocate this and take the reciprocal here.*2666

*We have closed the circle on all of the equations.*2673

*We are now able to express things like entropy and energy and enthalpy strictly in terms of temperature, pressure, and volume, and this Α and this Κ.*2676

*This is absolutely extraordinary, it is more than a minor miracle that is not even possible but there it is.*2688

*In any case, thank you so much for joining us here at www.educator.com.*2696

*We will see you next time, bye.*2699

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