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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

Transcription: Free Energy Example Problems II

Hello and welcome back to 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 gas1002

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

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

Take care, bye.1876