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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

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