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 II

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

- Intro 0:00
- Example I 0:09
- Example II 5:18
- Example III 8:22
- Example IV 12:32
- Example V 17:14
- Example VI 20:34
- Example VI: Part A
- Example VI: Part B
- Example VI: Part C

### Physical Chemistry Online Course

### Transcription: Free Energy Example Problems II

*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 dive right on in.*0007

*The first problem says, demonstrate that for a real gas the constant pressure heat capacity ×*0010

*the Joule-Thompson coefficient = RT²/ P DZ DT constant P,*0018

*where the entity is the Joule-Thompson coefficient and Z = PV/ RT, the compressibility factor for the gas.*0026

*Z which is pressure × volume/ RT is the standard variable that you run across in thermodynamics.*0033

*It is called the compressibility factor so we want to demonstrate that this is the case given*0042

*the mathematics and all of the equations that we have at our disposal.*0046

*Let us see what we can do.*0051

*We know from back when we did on energy and we first talked about the Joule-Thompson coefficient that we have the following relation.*0054

*Let us go ahead and actually do this in blue.*0063

*We have the following.*0065

*We have that CP C sub P × the Joule-Thompson coefficient = - DH DT at constant P.*0067

*This was the definition that we have for this particular relationship earlier on.*0083

*Let me go ahead and put Z = PV/ RT over here.*0088

*We have by the thermodynamic equation of state which we just a few lessons ago, we have the following.*0099

*The DH DT sub P = that molar V with a line over any variable that means per mole.*0107

*- T × DV DT sub P so we have that Z sub P μ of JT = negative of this which is the negative of this which is the negative of this.*0124

*I’m just substituting this expression in for that.*0144

*I'm going to flip these to change the negative so I end up with T DV DT sub P – V.*0147

*This is by the thermodynamic equation of state.*0158

*Let us go ahead and we see this thing here.*0162

*Let us go and find what DZ DT is when we hold P constant.*0167

*Here is my expression.*0172

*I’m going to differentiate this with respect to T holding the P constant.*0174

*When I do that, I get the following.*0179

*We do with over here separate, I get DZ DT holding P constant.*0183

*When I run the full derivative of this and I will go through it entirely step by step, I will just write up what the final result is.*0191

*It will be T × DV DT sub P - V/ T².*0199

*We get this thing for the derivative.*0211

*Now what I'm going to do is, I’m going to take this side and I’m going to move some things around.*0214

*I’m going to multiply by R, I’m going to multiply by T².*0220

*I’m going to divide by P so I’m going to end up with the following.*0224

*I end up with RT²/ P × DZ DT sub P = T × DV DT sub P – V.*0228

*This right here is that right there.*0250

*If this is equal to that and that is equal to this, that means this equal this.*0255

*I'm done, I’m just going to substitute this expression in for here and I end up with what it is that I wanted.*0262

*CP μ Joule-Thompson coefficient = RT²/ P DZ DT sub P which is absolutely what I wanted.*0269

*I use the original definition of the heat capacity × the Joule-Thompson coefficient*0285

*which was the rate of change of enthalpy with respect to temperature at constant pressure.*0290

*I use the thermodynamic equation of state.*0295

*I set these two equal to each other.*0298

*I have differentiated Z, manipulated that and then substituted this expression back here to get what it is that I wanted.*0300

*Nice and straightforward.*0308

*Relatively speaking, there are a lot of symbols floating around here.*0312

*Let us see what is next, calculate δ A, the change in the Helmholtz energy for the isothermal expansion.*0319

*We have something that is going to happen isothermally of 1 mol of an ideal gas.*0328

*This was nice and easy.*0332

*25° C from a volume of 20 L to a volume of 80 L.*0334

*We are going to expand this gas, this ideal gas.*0340

*We are going to keep the temperature constant and we want to know what the change in Helmholtz energy is.*0343

*We go back to our basic fundamental thermodynamic equations.*0349

*For DA, it is the following.*0356

*It is DA = - S DT – P DV.*0359

*It is an isothermal process so DT = 0.*0366

*This is 0 because it is isothermal that means that DA is just equal to - P DV.*0369

*We are talking about an ideal gas so an ideal gas is PV = nRT which means that P = nRT/ V.*0380

*I can go ahead and put this expression in for that and I get DA = -nRT/ V DV.*0391

*Then I go ahead and I integrate my expression and I end up with δ A = -nRT.*0404

*I will go ahead and do this here V1 to V2 of DV/ V.*0416

*We have seen this 1000 times already, nRT LN.*0423

*This is not pressure, we are not doing δ G here.*0429

*This is V2/ V1.*0432

*We just put our numbers in δ A = - 1 mol and we have 8.314 J/ mol °K.*0436

*I’m going to go ahead and leave the units off, I hope you will forgive me.*0449

*25° C so this is going to be 298 °K and the log of 80 L.*0452

*As long as the units match we are okay.*0462

*20 L we do not have to make any conversions.*0464

*And if I have done my arithmetic correctly, which I hope you will confirm -3435 J/ mol for this ideal gas.*0467

*That is the change in Helmholtz energy δ A = this.*0481

*This much energy is available to do a certain specified amount of work under the conditions of isothermal change in volume.*0487

*Calculate δ G for the isothermal expansion of 1 mol of an ideal gas at 400°K from a pressure of 7000 kPa to a pressure of 100 kPa.*0504

*It was going from a really high pressure situation to a low pressure, it is expanding.*0520

*Again, the process is going to be isothermal and we are going to do this at 400°K.*0525

*It is the same exact thing as before, except now, instead of Helmholtz energy we are calculating the change in free energy.*0531

*This is going to be a question of temperature and pressure.*0537

*You remember Helmholtz energy is under conditions of constant temperature and volume.*0541

*The Gibbs free energy is under conditions of constant temperature and pressure.*0548

*Since, most of the things that we do in chemistry happen under constant temperature and pressure,*0552

*when we really do not have to walk anything, we just run the experiment.*0556

*Temperature is constant, pressure is constant, this is why the Gibbs energy is important in chemistry.*0559

*We go back to our fundamental equation of thermodynamics in the form of DG = - S DT + V DP.*0568

*Again, this is 0 because we are talking about isothermal process.*0584

*Therefore, we have the DG = V DP.*0591

*We will take our ideal gas law PV = nRT and we solve for V.*0596

*This is nRT/ P and we substitute this into here and we integrate.*0601

*We get DG = nRT/ P DP.*0608

*When you integrate this expression you end up with the following.*0618

*The same thing as before, you get DG = nRT the log this time it is going to be the final pressure - the initial pressure.*0620

*P2/ P1, I'm sorry final pressure/ initial pressure.*0630

*We get 1 × 8.314 × 400°K × log.*0638

*The final pressure now was 100 kPa and this is 7000 kPa.*0648

*The units are the same so you do not have to make any conversion, it is not a problem.*0654

*If I did my arithmetic correctly, I should get - 14,129 J and this is because it is 1 mol we can go ahead and put the per mole.*0658

*I actually put the 1 mol here so technically this is -14,129 J.*0672

*Since, it is 1 mol let us go ahead and put the per mol.*0678

*We can do that, not a problem.*0682

*There you go, that is δ G.*0685

*For this particular process, there are 14,129 J of free energy available to do work.*0687

*That is the maximum amount of work that this particular process, this expansion can accomplish.*0698

*Will it accomplish this much work?*0704

*No, the maximum has never reached.*0706

*It is an ideal, it is the maximum that you can get.*0709

*If everything were perfect and there is nothing lost as heat, then yes you can recover this much work.*0712

*But again, this is the maximum amount of work that is derivable from this process.*0718

*That is what δ G is, it tells you how much work you can extract from a process that is why we call it free energy.*0727

*It is free energy, it is energy that you can use freely to do work because the other energy of the system*0735

*that is not available for work is tied up in the entropy of the system.*0742

*Things seem to be moving along pretty nicely here.*0753

*Use the following form of the Van Der Waals equation to derive an expression for the δ G of 1 mol*0756

*of a gas as it is compressed isothermally from 1 atm to Pa atm.*0762

*We are going from 1 atm to some random Pa atm.*0767

*We do not want actual number, we want an expression.*0771

*In the previous problems, we work with the ideal gas, the PV = nRT.*0775

*We are going to do something with the Van Der Waals gas.*0779

*We are going to see what happens when we deviate slightly from ideal behavior.*0782

*Z = PV/ RT this is the compressibility factor and this is the version of the Van Der Waals equation expressed in this form.*0789

*Let us write out, we have got this expression.*0799

*I’m going to go ahead and express volume.*0807

*I’m going to go ahead and move the RT over here and I'm going to divide by P.*0814

*I’m going to rewrite this expression as V = RT/ P + B - A/ RT.*0818

*This is just simple algebra, moving things around so I can leave V alone on that side.*0830

*We are looking for δ G so again we use the same relation that is universally available.*0835

*It is DG the fundamental equation of thermodynamics for free energy,*0841

*the relationship is DG in other words the total differential for this is - SDT + V DP.*0848

*This is isothermal so we can go ahead and dismiss this term because DT is 0.*0860

*We have DG = V DP.*0865

*V is right here so now I take this expression and I put in here like I did for the ideal gas.*0871

*The ideal gas is just nRT/ P.*0878

*Here it is a little bit more complicated because we are dealing with a Van Der Waals gas.*0880

*What should I do here?*0898

*That is fine, I’m going to rewrite it.*0900

*We have DG = the integral from P1 to P2 of V DP = the integral from P1 to P2 of this expression RT/ P + B - A/ RT.*0902

*Do not let the symbolism intimidate you, it is just stuff.*0924

*And notice, you are integrating with respect to P.*0927

*There is no P in this one, the P only shows up here.*0932

*This is a simple integration, this is just a constant.*0934

*Do not let the symbols intimidate you.*0937

*When we do this, we are just going to integrate this expression because the integral of the sum is*0940

*the sum of the integrals we end up with the following.*0947

*We end up with RT LN P/ 1 because we are 1 atm + B - A/ RT × δ P -1.*0950

*This is a very simple integration.*0969

*It is the integral of this with respect to P and the integral of this with respect to B.*0970

*This is just a constant so you end up with just the integral of DP which end up being the DP which is P-1 final – initial.*0974

*You have your expression.*0983

*For an ideal gas, for a Van Der Waals gas this is the expression.*0985

*For the ideal gas, we have this part P2/ P1.*0992

*The Van Der Waals gas, there is a little bit of an adjustment to it.*0999

*Notice, you are actually adding some so the amount of free energy that you have for an ideal gas*1002

*is going to be a certain amount like our previous problem.*1007

*The amount of energy that you have for a Van Der Waals gas is actually going to be a little bit less.*1010

*You have a negative number and you are going to add some to it.*1015

*A real gas behaves in a non ideal fashion, that is all this is saying.*1019

*For a Van Der Waals gas this is the expression for δ G.*1027

*The Van Der Waals constants for oxygen are as follows.*1037

*A= this, B = this, find δ G for the isothermal expansion O2 as a Van Der Waals gas at 400 °K from 7000 kPa to 100 kPa.*1040

*What it actually do, example problem 3 except now we are going to treat it as a Van Der Waals gas*1051

*instead of an ideal gas to see what the difference is.*1058

*We have our expression for the Van Der Waals gas DG = RT LN P2/ P1 + B - A/ RT × P2 - P1.*1063

*When I put my numbers in, δ G is going to equal 8.314 and we are doing this at 400°K.*1085

*We are taking the logarithm, we are doing 100 kPa/ 7000 kPa + then we have the B which is 3.18 ×*1096

*10⁻⁵ - 0.138 which is A/ R which is 8.314 and T which is 400.*1111

*This is just numerical stuff × final is 100 -7000.*1129

*I’m going to multiply it by 10³ because we have to express it in Pascal.*1140

*These are in kPa and I need the actual value to be in Pascal, that is why I multiply by 10⁻³.*1146

*When I do this, I get δ G = -14,129 which is the same value that we got for the ideal gas.*1159

*This part is the ideal gas portion of it and then I have + 66.9, this is the adjustment that we make.*1169

*Our final δ G for oxygen as a Van Der Waals gas is -14,062 J/ mol.*1182

*We have less energy available in this particular expansion once it expands as a Van Der Waals gas.*1194

*You can imagine a real gas is going to have even less free energy available.*1200

*The maximum amount available is the -14,129 that comes from ideal behavior.*1206

*Treating it as a Van Der Waals gas, we have to make a little bit of adjustment.*1213

*If we use other equations of state, whatever they happen to be, we are going to end up with less and less and less.*1216

*The real gas is going to be the least of all, that is all that is happening.*1223

*Let us see what we have got.*1231

*1 mol of the following substances at 298°K are subjected to isothermal pressure increases from 1 atm to 100 atm,*1236

*calculate δ G for each substance.*1246

*An ideal gas and liquid water whose molar volume is 18 cm³ / mol and iron metal with density of 7.9 g/ cm³.*1248

*Let us just jump right in and do these.*1260

*Let me go back to blue here, no worries.*1263

*DG = -S DT, you always want to start with your fundamental equations.*1269

*Those are the ones that you want to know, to memorize.*1278

*From those, based on what the problem is asking whether it is isothermal or isobaric, whatever, you knock off terms, you adjust terms.*1280

*You do not have to learn 50,000 equations.*1291

*There is a handful of equations that you should know, that is your starting point.*1293

*Not to mention the fact that you are always consistent when you do your problems.*1297

*You are always starting at the same point and the problem will go in the direction that*1300

*it needs to go based on the constraints that you are putting off of the problem is putting on it.*1304

*The same equation.*1309

*It is a nice way of learning the equations because you are only starting with them.*1311

*DG = - S DT + V DP this is isothermal so that goes away.*1315

*We already know what this is, this is for an ideal gas.*1321

*Let us just go ahead and integrate this so we can just write δ G = nRT LN P2/ P1.*1324

*Nice and simple.*1336

*When I put these particular values in here, I get δ G = 1 mol 8.314 temperature at 298°K, nat log of 100 atm / 1 atm.*1337

*100 atm is actually a lot of pressure.*1360

*When we do this, we end up with δ G = 11,409 J/ mol.*1366

*The δ G is positive, this is not a spontaneous process.*1381

*Clearly, it is not a spontaneous process.*1385

*I have to do work on the system to compress it from 1 atm to 100 atm.*1388

*If the other way around, 100 atm and I just let it expand that would be a spontaneous process.*1396

*It is going to go from high pressure to low pressure.*1402

*I do not have to do anything, it will expand by itself.*1404

*I have to put the pressure on to actually compress it.*1406

*That takes care of the ideal gas.*1413

*Let us go ahead and see what we have for liquid water.*1415

*I have got some space available that is not a problem.*1434

*Part B well, I know already that δ G = the integral from P1 to P2 of V DP, that just come from the equation from part A.*1438

*For a liquid, I have to put a lot of pressure on a liquid to change its volume.*1453

*When we are talking about liquid and solids, for all practical purposes,*1461

*unless you are talking about a huge pressure difference and even then, the volume changes very little.*1465

*For all practical purposes, the volume of liquid or solid is constant.*1471

*It is not like that for a gas, you know that from experience.*1475

*For a liquid and a solid, we can treat the volumes constant so we can actually pull this out of the integral sign.*1478

*δ G is just equal to whatever the volume happens to be × the change in pressure P2 - P1, this is our equation.*1483

* We just need to find the volume of the 1 mol of water and I multiply by the pressure difference.*1492

*This is a really easy problem.*1498

*We just have to make conversions because the molar volume given to us was in cm³/ mol.*1501

*We need to either deal in cubic meters if we are dealing in Pascal’s or in this case we are dealing with atmospheres,*1507

*you want to work in liters which is the same as a dm³.*1513

*We just need to make some basic conversions.*1517

*Let us go ahead and do that first, let us find out what our volume is.*1519

*18 cm³/ mol × I’m just going to do the conversion so you see it.*1523

*1 dm is 10 cm, I have cubic centimeter which is cm cm cm.*1534

*You remember, when you are dealing with squares and cubes and power to the 4th, you have to cancel each unit.*1539

*I have 3 cm so I have to multiply by 3 times, it cancel cm 3 times.*1545

*I apologize for being so elementary but I think it is always nice to remember these things because it is easy to forget.*1553

*1 dm/ 10 cm, when I do this it is cm cm cm cancels that.*1561

*When I do that, I get 18 × 10⁻³ dm³.*1569

*Dm³ is a liter.*1576

*The volume = 18 × 10⁻³ L.*1578

*Well δ G = V × P2 - P1, volume is 18 × 10⁻³ L and my difference is 100 atm and 1 atm.*1584

*When I do this, I’m going to get δ G = 1.782, but this is liter atmosphere.*1604

*Liter atmosphere is not a Joule, I need to convert it to Joule.*1611

*The conversion factor is 8.314 J to 0.08206 L atm.*1614

*You just use the two values for the gas constant, J/ mol °K, L atm/ mol °K.*1624

*The mol °K cancels, you are left with a conversion factor, liter atmosphere cancels liter atmosphere, I get the δ G =180 J/ mol.*1630

*That is not a lot compared with the previous number which was in the thousands.*1642

*What is the number specifically?*1649

*It is 11,409, so 11,500, 180.*1651

*Clearly, pressure does not have the same effect on liquid and solids that it has on a gas.*1657

*This is a very small difference.*1663

*100 atm is huge.*1666

*We are putting a lot of pressure on this water, very little change.*1667

*The δ G that is pretty insignificant.*1671

*Let us go ahead and do part C here.*1676

*We are going to treat it the same way.*1680

*We have δ G = V × δ P which is the 99.*1683

*This time they gave us the density, we are going to have to do a couple of extra conversions.*1690

*They said that the density was 7.9 g/ cm³, we need to find the molar volume.*1697

*We are going to multiply this by iron is 55.85 g/ mol so now we have cm³/ mol.*1708

*We multiply by 10⁻³ in order to, that is dm³/ 1 cm³.*1719

*We can cancel that so we end up with 0.00707 dm³/ mol,*1728

*which is the same as 0.00707 L.*1744

*δ G is going to be 0.00707 L × 99 atm which is 100 - 1 and we end up with δ G of 0.700 L atm.*1748

*We do the multiplication, multiply by a 8.314 ÷ 0.08206.*1770

*We get a δ G is equal to 70.9 J/ mol, even less.*1776

*A solid response to pressure increases even less.*1785

*Free energy changes with volume.*1797

*When you change the volume of something, the free energy available is going to change.*1800

*A gas is very sensitive to volume changes, the pressure changes, volume changes which is why you have 11,500.*1805

*Liquids and solids are virtually non responsive to volume changes upon when you change the pressure.*1814

*You can squeeze them as much as you want, the volume is not the change that much.*1822

*Because the volume does not change that much, the free energy does not change that much.*1825

*It is 71 J/ mol is completely insignificant.*1830

*A 100 atm is a lot.*1833

*70.9 for a solid and 100 that we got for the liquid, they are actually kind of high relative to the fact that they are liquid and solids.*1835

*When you are dealing with liquid and solids, most of the time you are talking about changes of like 1, 2, 3 J/mol in J not in kJ, J/ mol.*1846

*Clearly, liquid and solids demonstrates that they just do not respond to pressure increases.*1857

*Their volume does not change all that much, but we knew that anyway from experience.*1865

*Thank you so much for joining us here at www.educator.com.*1870

*We will see you next time for a continuation of some example problems in free energy.*1873

*Take care, bye.*1876

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