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### Free Energy Example Problems I

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 0:11
• Example I: Deriving a Function for Entropy (S)
• Example I: Deriving a Function for V
• Example I: Deriving a Function for H
• Example I: Deriving a Function for U
• Example II 15:18
• Example III 21:52
• Example IV 26:12
• Example IV: Part A
• Example IV: Part B
• Example IV: Part C
• Example V 33:45
• Example VI 40:46
• Example VII 43:43
• Example VII: Part A
• Example VII: Part B
• Example VII: Part C

### Transcription: Free Energy Example Problems I

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

Today, we are going to start our example problem sets for free energy.0004

let us just jump right on in.0009

Our first example it says given that G of TP = G super 0 T + nRT + LN T for an ideal gas,0013

use the fundamental equations and the definitions of the composite functions to derive functions for S, V, H, and U.0024

This is a really important problem.0034

It is a more important problem theoretically than is practically but I think it just depends on the particular situation you run across.0039

Remember, we said when we are talking about free energy in the last lesson that if you know the functional form of the free energy expression,0046

If you know what the function looks like, in this case we have this,0054

that we can derive all the other thermodynamic functions from this using the basic equations that we had at our disposal.0061

From this expression for free energy, we can derive the entropy, the volume, the enthalpy, and the actual energy of the system.0067

Again, this is really great way to play with the mathematics to get a sense of what is going on and just become comfortable with it.0077

A lot of the problems in the problem sets for free energy are going to involve a lot of mathematical manipulation.0087

The manipulations themselves are not difficult but they are mathematical manipulations.0095

Again, this is just the way of getting you comfortable with handling the equations0101

not just seeing them on a page but actually using them to derive other relations.0106

None of them are going to be very long, it is just you have to be very careful because of all mathematics there is going to be a lot of symbolism on the page.0110

It is just a question of busyness that is going to be the biggest problem here.0119

In any case, let us go ahead and jump right on in.0123

Let us go ahead and rewrite this.0128

We have G of TP = Z0 of T + nRT nat log of P.0130

This G with a superscript 0, this is just a standard free energy.0143

My free energy at a given temperature and pressure is going to equal the standard free energy + nRT ×0147

log of the new pressure whatever the pressure happens to be that is raised to 2 from the 1 atm.0155

Using the fundamental equations of the definitions of a composite function,0162

our fundamental equations for G is this.0166

We had DG = - S DT + V DP so we know that DG DT,0170

That made us slow down a little bit just to make sure we have all of our variables correctly, is equal to –S.0186

We know this already from that or we can just read it out from here, this is the total differential expression.0192

This is going to be the DG DT holding this variable constant.0198

From the equation above, from this equation, if I actually take DG DT holding P constant,0205

If I take the derivative of this directly I’m going to end up with DG DT.0223

This is just a function of temperature so it is a regular derivative not a partial derivative +0235

the derivative of this with respect to T holding P constant is going to be nR LN P.0241

From the fundamental equation I have this, from taking a derivative of this equation which is the equation they gave us I have this.0253

I’m going to set these equal to each other.0260

I have - S = D 0 DT + nR LN P but this, that, this thing right here this DG DT is just - S for the standard conditions.0262

DG DT under constant P = - S so DG standard DT is just - S standard, that is all it is.0295

We get - S = - S standard + nR LN P.0306

I can go ahead and I can leave it like that or I just can go ahead and rearrange things, multiply by -1.0316

S = S standard - nR LN P, there we go.0321

From this equation which is the equation for the free energy of the system,0330

I was able to do derive an equation for the entropy of the system.0338

It is equal to the standard entropy - nR LN P, that is what we are doing here.0344

That takes care of the S.0352

Let us go ahead and let us see what is next.0357

Once again, let me go ahead and write the equation DG = - S DT + V DP.0361

I know that DG DP under constant T = volume.0372

Let us see from the equation of the equation that they gave us, the functional form of the equation which is this one,0384

+ nRT LN P, if I take the derivative directly, let me see if I do the DG DP under constant T.0393

If I take the derivative of this holding temperature constant and take it with respect to pressure,0412

I get nRT × the derivative of LN P which is 1/ P.0419

I get nRT/ P.0427

I have this equal to nRT/ P from the fundamental equation I have the same thing is equal to V.0430

I will just go ahead and set them equal to each other so I get V = nRT/ P which we knew already.0437

We actually ended deriving it from the functional form of the free energy as a function of temperature and pressure V= nRT/ P.0446

This is just PV = nRT, this is the ideal gas law, this confirms it.0454

I have derived a thermodynamic property of volume from the free energy using these other equations that I have.0460

Using this equation and this equation.0468

Let us see what we can do, we have taken care of the S, we have taken care of the V, let us see if we can do something for the enthalpy.0474

The composite functions let us just go ahead and recap our definition for composite functions.0484

The enthalpy = the energy + PV.0490

Our Helmholtz energy which does not concern us in this particular problem but I will go ahead and write it down anyway as the definition.0495

G = DU + PV - TS which is equal to G= H – TS.0501

These are the definitions of the composite functions, enthalpy A this.0513

Let us go ahead and deal with this one because again they gave us G in the problem and now we want to find H.0517

Let us go ahead and take this, bring it over here.0526

We will rearrange this and we will write H = G + TS.0529

H = G is just G is this thing so we are going to write G of T + nRT LN P + T × S.0536

H = G standard of T + nRT LN P + T × we already know what S is, we just found it.0561

In the previous page that is just is equal to S standard of T – nR LN P.0574

I get H = G standard of T + nRT LN P + T S standard T – nRT LN P.0584

That cancels with that and I am left with the enthalpy of my system = the standard free energy of the system + the temperature × the standard entropy.0602

This is interesting.0625

I have G standard + T S standard.0627

You remember the definition of H was just G + TS.0632

G standard + T S standard is nothing more than H standard.0641

This is my equation, this is what I was looking for.0647

I was able to derive the functional value for the enthalpy of the system based strictly on the fact that0654

I was given the functional form of the free energy of the system and ends up being the standard free energy +0663

the temperature I happen to be at that moment × the standard entropy.0671

This just happens to equal the standard enthalpy.0679

In other words, the enthalpy of the system under standard conditions.0689

Again, because H = G + TS that is the definition of a composite function, we just rearranged.0694

H standard = G standard + S standard.0709

This is the functional form but it just happens to equal that.0716

Let us go ahead and see if we can derive something for U.0727

They also want U.0730

We have H = U + pressure × volume so U = enthalpy - pressure × volume.0732

U = enthalpy – we are dealing with an ideal gas so it is going to be H – nRT.0745

That is kind of interesting.0756

Actually, we want it, here we go.0760

U = H – nRT, we already found H.0762

H is this nRT so you already found that.0766

Here which happens to equal the energy of the system under standard conditions because U = H – nRT.0774

Therefore, U standard = H standard – nRT.0795

Therefore, let us see, I got U = H – nRT.0806

Under standard conditions, the H happens to equal H standard so this is just H.0812

Basically, I have got from here, I can stop here because I already got everything else.0822

If I want to continue, U = H – nRT, H = H standard nRT.0827

Therefore, U = H standard - nRT but that is equal to U standard.0835

Everything is falling into place nicely here.0843

U = H which is G standard T + TS standard T – nRT.0847

I can use the short version of it or I can use this version of it which is expressed in terms of the actual standard free energy.0866

Given the functional form of the standard free energy I was able to derive expressions for the entropy.0876

I was able to find the entropy, the change in entropy.0881

I was able to find the volume of the system.0883

I was able to find the change in enthalpy of the system.0885

I was able to find the change in energy of the system strictly from having the functional form of the free energy.0888

This procedure is the same, this is the general procedure, it will always work.0900

If you are given the functional form for free energy, what we did here will always allow you to derive the entropy change,0904

the enthalpy change, the volume and the energy change.0910

Using the thermodynamic equation of state and the Van Der Waals equation find DU DV sub T for the Van Der Waals gas.0921

DU DV sub T for the Van Der Waals gas, not the ideal gas.0931

Let us go ahead and let us write the thermodynamic equation of state.0936

In this particular case, we have got P = T × DP DT constant volume - DU DV T.0944

I’m going to rearrange this because now I'm looking for that.0961

I have got DU DV sub T = T × DP DT under constant volume – P.0966

The Van Der Waals equation P = nRT/ V - nV - AN²/ V².0983

Therefore, I have got this and I have got this so I'm going to find this DP DT.0998

I have P so now I’m going to take DP DT and I’m going to hold volume constant.1005

When I do that, when I take the derivative of this expression holding volume constant,1013

I'm going to end up with derivative with respect to T, it is just going to be this.1019

I'm going to get nR/ V –NB.1029

Again, constant and constant, NV is a constant so the derivative ends up going to 0.1033

I take this value and I plug it into there so I get DU DV constant T = T × nR/ V - NB – P.1038

P is just this, the whole thing, just put back in here - nRT / V - NB – N² A/ V².1058

That is going to equal nRT/ V - NB - nRT/ V - NB + N² A/ V².1072

That cancels and what I’m left with is DU DV constant T for the Van Der Waals gas = N² × A/ V², this is what we wanted.1088

Notice, this does not equal to 0 like the ideal gas.1106

Recall, when we had DU = CV DT + DU DV under constant T DV, this is the expression for the differential change in energy of the system.1111

It is going to be a function of temperature and the function of volume.1133

The DU DV, the Joules law for an ideal gas, this DU DV sub T that was equal to 0.1135

Because it was equal to 0, this term went away and the energy just became a function of temperature based on the constant volume heat capacity.1146

That was for an ideal gas, in this particular case this is a Van Der Waals gas.1153

For the Van Der Waals gas, this is no longer equal to 0.1158

When we use this equation for a Van Der Waals gas we actually have to include it.1161

When we integrate this, we have to integrate this term and we have to integrate this term, in order to find our δ U, our change in energy.1168

δ U is no longer just DU = the integral of CV DT from T1 to T2.1182

For a Van Der Waals gas, now you have δ U = the integral of this one which is CV DT from T1 to T2 +1198

the integral from volume 1 to volume 2 of this Du DV which in this case is N² A² / V² DV.1210

If we have an isothermal process for a Van Der Waals gas, isothermal means that this term can go to 0.1224

But now, the change in energy is not equal to 0.1232

For an ideal gas and isothermal process involves that there is no change in energy and it implies that δ U = 0 but only for an ideal gas.1236

Do not just automatically assume that just because we are talking about an isothermal process, the δ U = 0, it does not.1245

For an ideal gas and isothermal process this is 0 and this is 0 so δ U = 0.1256

Not for a Van Der Waals gas, for Van Der Waals you have to evaluate this integral right here.1262

An isothermal process implies that δ U = 0 only for an ideal gas.1271

That thing is, you get so many problems and questions that involve ideal gases.1284

A lot of students automatically equate isothermal with δ U = 0, that is not the case.1287

It is never the case.1294

This is the equation you want to know.1296

Let the equation tell you what is what.1299

Are you dealing with an ideal gas, solid, liquid, Van Der Waals gas, are you dealing with a different kind of gas, whatever it is.1301

Let us go ahead and take a look at example 3.1314

Calculate δ U for the isothermal expansion of 1 mol of nitrogen as a Van Der Waals gas from 10 dm³/ mol to 100 dm³/ mol.1317

It should be just 1 dm³ to 10 dm³.1330

Let us see what we have got, 10 to 100.1334

A, in this particular case is 0.141 so let us see what we have.1337

We always start with our basic equation, those are the ones that we want to use.1342

Those are the one that tell us what we are doing.1347

DU = CV DT + DU DV at constant T DV.1350

This is an isothermal process so isothermal means the DT is 0 so we can take that to 0.1362

We have got DU = this thing, DU now equal this thing, that is what we are going to integrate.1366

We got δ U = the integral from v1 to v2 of N² A/ V² DV = N² A × the integral DV/ V from v1 to v2.1381

Let us put our v², there we go.1403

δ U = 1 mol, this is 1 × A is 0.141 × the integral, 10 × 10⁻³ to 100 × 10⁻³.1406

The reason it is 10 dm³ to 100 dm³, A is expressed in Pascal’s.1431

When you are working with Pascal’s you have to work in m³.1441

Recall, volume must be in m³ when working in Pascal.1445

Again, it is pesky and it is annoying but you have to keep track of your units.1464

Take a look at what units you are actually given in the case of the Van Der Waals gas,1469

this particular unit is 0.141 using the Pascal and meters so it has to be in cubic meters.1473

The volume was given cubic decimeters so you have to convert to m³.1482

10 dm³ is 10 × -3 m³ when working with Pascal.1487

It is going to be the DV/ V² so we end up with δ U = 1² × 0.141 × 1/ V.1498

It is minus because this is going to be when we integrate this, the negative sign is going to be 1/ V -1/ V.1516

I'm just going to bring the negative over here 10 × 10⁻³ to 100 × 10⁻³.1522

When I evaluate this, I get δ U = 12.69 J.1530

An isothermal expansion of 1 mol of nitrogen as a Van Der Waals gas from 10 dm³ to 100 dm³ isothermally you end up with 12.69 J as your δ U.1544

Van Der Waals gas δ U is not equal to 0, it is equal to 12.69.1563

Let us see what we have got.1571

Derive an expression for the change in entropy with respect to volume under a constant temperature for a Van Der Waals gas.1577

Find an expression for δ S for the isothermal expansion of the Van Der Waals gas from v1 to v2.1586

Comparing the expression in part B with the analogous expression for the ideal gas, for the same increase in volume,1592

will the change in entropy be greater for the Van Der Waals gas or for the ideal gas?1600

Let us see what we can do.1608

Part A, I’m going to do this in red.1613

Part A, the equation for the Van Der Waals gas is nRT/ V – NB.1618

Let us make V a little bit better, - N² A/ V².1629

From one of Maxwell's relations said that DS DV at T = DP DT under constant V.1638

We are looking for DS DV and Maxwell's relations said the DS DV = the change in pressure with respect to temperature holding the volume constant.1661

I have the equation, the Van Der Waals equation.1676

If I hold the volume constant and if I take the derivative of P with respect to T this is just equal to nR/ V – NB.1678

There you go, that is my first part.1695

The rate of change of entropy as I change the volume for a Van Der Waals gas is nR/V – NB.1698

That is all, nothing strange is happening here.1704

It is just mathematics, simple derivatives.1706

Now, they want us to find an expression for the δ S.1712

Let us start with our basic equation again for DS = CV/ T DT + DS DV under constant T DV.1715

They say that this process is isothermal.1732

Isothermal so that goes to 0, DT is 0.1734

All I have is DS = this thing.1738

For the Van Der Waals gas, I just derived this thing, it is nR/ V – NB.1743

It was equal to nR/ V - NB × DV.1747

I just integrate this and I get δ S = nR × integral from v1 to v2 of DV V – NB.1755

I get nR × log of V -NB from V1 to V2.1773

I get my expression, I get nR × the LN of V2 – NB.1790

They just want the expression over v1 – NB.1798

For Van Der Waals gas, my δ S is this based on the fact that the DS DV is this.1803

That comes from this, the basic equation.1811

They want us to compare this with that for the ideal gas.1820

Let us go ahead and do that.1825

We can go ahead and just recall what the expression is for the ideal gas or we can go ahead and derive it.1828

Let us go ahead and derive it because I think it is always good to recall the things.1835

We cannot remember everything that is going on.1839

Maxwell's relations says the DS DV under constant T = DP DT under constant V.1843

For an ideal gas P = nRT/ V.1853

Therefore, this DP DT under constant V this thing is just equal to nR/ V.1858

DS = CV/ T DT this is our basic equation + DS DV under constant T DV.1872

We are dealing with an isothermal process so that goes to 0.1886

DS DV = DP DT, DP DT = nR/ V.1889

Therefore, DS = nR/ V DV.1895

When we integrate this, we end up with δ S.1903

For the ideal gas = n × R × the log of V2/ V1.1914

We just calculated that the δ S of the Van Der Waals gas = nR × the nat log of V2 - NB/ V1 – NB.1923

We can compare these two, between these two, the δ S of the Van Der Waals gas is going to be bigger than the δ S of the ideal gas.1939

The change in entropy of a Van Der Waals gas is going to be bigger than the change in entropy of the ideal gas.1956

When you subtract the same value, V2/ V1 V2 – NB V1 – NB.1964

You are subtracting the same value from numerator and denominator.1975

When you subtract the same value from numerator and denominator of a ratio, the value of the quotient goes up.1978

The value of the quotient actually rises.1996

The log of the bigger number is bigger.2000

Therefore, the δ S of the Van Der Waals, this expression is greater than this expression.2002

The value of the quotient rises as just a basic mathematical fact.2011

Let us see if we can make our way through this ocean of mathematics that we happen to be swimming through.2020

I’m starting to lose my own way here.2025

Example number 5, to a first approximation the compressibility factor Z of the Van Der Waals gas is Z = PV/ RT.2030

The compressibility factor of any gas is just the pressure × the volume/ R × T.2037

It is equal this thing, from this and the thermodynamic equation of state to demonstrate that the change in molar enthalpy2045

with respect to pressure under constant temperature for a Van Der Waals gas = B – 2A / RT.2054

Again, this is nothing more than mathematical manipulation.2060

Let us see what we can do.2064

Let us go ahead and rewrite this again.2067

We go ahead and go back to blue.2072

I have got PV and again when this line over it just means molar, it just means V is nothing more than this V line, this V bar is nothing more than V/ N.2074

That said so, if I just want to multiply everything by N, I end up getting V.2086

It is not a problem, this is just the molar volume instead of the volume.2088

PV/ RT = 1 + B - A/ RT × P/ RT.2095

Let us see here, they want this thing.2109

The thermodynamic equation of state is the following.2115

Thermodynamic equation of state says that DH/ DP = V - T × DV DT.2119

I will put these over, it does not really matter, DT under constant P.2136

This is equal to this so I need to find DV DT from this equation and plug it into here and see what I end up with.2144

I’m going to rearrange this equation and solve for V.2153

Let me go ahead and do that over here.2157

I get PV =, I’m going to multiply everything by RT.2160

I’m going to get RT + B - A/ RT × P.2165

I’m going to go ahead and divide by P so I'm going to get V = RT/ P + B - A/ RT.2173

Once again, rewrite the DH DP = V - T × DV DT that is the thermodynamic equation of state.2188

Now, I have this so let me go ahead and find DV DT while holding P constant.2207

I’m going to find this from this equation, holding P constant I end up with R/ P - -A R/ RT².2217

When I take the derivative of this with respect to T, this derivative is 0 and I'm going to get - and – that is going to be +.2245

We are good.2254

This - and this -, when I take the derivative of this with respect to T, I get this -.2256

It is going to equal R/ P + AR/ R² T².2262

I’m just going to go ahead and cancel the R so I will just do it now.2273

A/ RT², 1 R cancels 1 R leaving just 1 R on the bottom.2275

My DH DP T = V - T × DV DT which is this R/ P + A/ RT².2283

This is equal to V, V is equal to this whole thing so I’m going to put that back in.2310

It is going to be RT/ P + B - A/ RT –,2315

I’m going to go ahead and distribute, this is RT/ P - A/ RT².2326

RT/ P cancels RT/ P and I get the DH/ DP under constant T = this - that - that = B -2A/ RT².2339

What happened here?2373

I think I have lost my way in this mathematics.2373

Let us see what is going on.2376

Let us see if I copy everything correctly here.2377

There is our T – T, I'm sorry this is here and I distribute the T.2379

I forgot to distribute the T over this one.2389

This is T there, sorry about that.2391

There we go, let me go ahead and erase this part and let me do this part again.2397

This RT/ P cancels this RT/ P.2400

We have B - A/ RT and this T cancels that - A/ RT.2403

There we go, -2A/ RT.2410

There you go, that takes care of the first one.2417

The change in molar enthalpy with respect to a change in pressure at constant temperature for a Van Der Waals gas = B – 2A/ RT.2424

Example 6, using the same expression from example 5 for the compressibility Z of2448

the Van Der Waals gas which was this thing, show that DS DP sub T = this thing.2457

By one of Maxwell's relations, we have the following – DS DP sub T which is actually what we want.2473

That is actually equal to DV DT sub P.2486

This is the same thing that we did before, we are going to take this equation.2503

We want to take the derivative of V with respect to T so we are going to rearrange this equation2516

just like we did before in the previous example.2523

We ended up with the following, we ended up with V = RT/ P + B - A/ RT.2526

When we take the derivative of this with respect to T holding P constant so that DV DT holding P constant we end up with R/ P + AR/ RT².2538

We go ahead and put this in for here and we are going to end up with DS DP under constant T =...2571

Let me just rewrite this again, DT under constant P which is actually equal to – R/ P + AR/ RT².2593

This one was not that bad, this was just using Maxwell's relation and taking the derivative of the Van Der Waals equation.2611

That was nice and simple.2618

Example 7, this one is a little interesting.2624

Using the results from the previous two problems for DH DP and DS DP, calculate δ H and δ S2630

for the isothermal pressure increase of CO₂ as a Van Der Waals gas from 1 mPa to 15 mPa at 298°K.2639

These are the values for the Van Der Waals constants for CO₂ and they ask us to run the same calculation of 598.2649

They say compare the values you calculated for 298 to those obtained for an ideal gas.2658

We just found expressions for a Van Der Waals gas for the change in enthalpy with respect to pressure,2666

the change in entropy with respect to pressure, we could integrate these expressions to get expressions for the actual change in enthalpy and2671

the change in entropy and when you compare that to the ideal gas.2679

Let us see what we have got.2688

We know what δ H is and we want to know what δ S is.2690

Let us go ahead and write down what we got.2697

Our CO₂ gas, we are taking it from 1 mPa, we are taking it to 15 mPa and we have our value of.2709

A = 0.366 m 6 Pa/ mol² and B = 42.9 × 10⁻⁶ m³/ mol.2720

From the previous example, we found that DH DP at constant T = B - 2A/ RT.2742

Therefore, I have got DH = V -2A / RT DP.2755

When I go ahead and integrate this expression, I’m going to get δ H = B - 2 A/ RT × the integral from P1 to P2 of DP and it is going to = B -2A/ RT × δ P.2768

Our δ H is going to equal, again this is molar.2801

We have got D – 2A.2808

Let us just plug in all of our values so we have B which is going to be 42.9 × 10⁻⁶ -2 × 0.366/ 8.314 × 298 × δ P.2810

It is going to be 15 -1, I’m going to do × 10⁶ because this is mPa.2837

We are working in Pascal’s in order to make sure that the units match.2847

15 -1 is 14 mPa, it is 14 × 10⁶ Pa.2855

When I do this, I get δ H of -3536 J/ mol that is all this line means molar.2861

There we go, that is the δ H based on what we calculated for DH DP and integrated it.2874

For the DS, we work out that DS DP under constant T = - R/ P + RA/ RT².2886

Therefore, DS = when I move P over I integrate the expression, I'm going to get from P1 to P2 of this thing.2906

This thing R/ P + R A/ RT² DP.2919

DS = - R × the integral from P1 to P2 of DP/ P.2932

This - comes out so this becomes a - and this becomes a - that distributes over both.2943

- A / RT² and I cancel one of the R × the integral P1 to P2 DP.2949

Therefore, our DS = - R LN P2/ P1 - A/ RT² × δ P.2967

We put the values in, δ S = -8.314 × the nat log of 15/ 1.2984

The mPa cancels mPa, the units cancel, I can leave the numbers alone.2996

-0.366/ 8.314 × 298² × 15 - 1.3003

And again, this one × 10⁶ and there is no cancellation here.3017

I’m going to do these separately, - 22.5 - 6.9.3024

Therefore, the entropy change is -29.4 J/°K mol.3033

That takes care of the δ S.3047

Let us see what is next, they ask us to run the same calculations part B under a new temperature.3052

It is the same exact process same, same exact thing we just did the only difference is instead of 298 for the T, we are going to put 598 for the T.3061

What we end up with are δ H, I will go ahead and put 598.3068

It is going to equal- 1460 J/ mol and δ S = - δ S at 598 = -24.22 J/mol °K.3076

Let us go ahead and compare the two values.3118

Let us see, C wants us to compare the values for the ideal gas and the Van Der Waals gas.3125

The δ S for an ideal gas = - R LN P2/ P1 that was the first term.3130

We will do it in just a second.3144

It is - 22.5 J/°K mol3147

The δ H for the ideal gas = 0.3158

It is 0 because DH/ DP T = 0 for an ideal gas, remember Joule Thompson coefficient for an ideal gas.3168

I will put Joule Thompson.3185

In case you are wondering where this came from, right here for the ideal gas, just look back on lesson 21.3193

Lesson 21 which is the entropy changes for an ideal gas.3207

Notice, this expression right here this was the first half of the expression for the Van Der Waals equation that we just did in part A.3218

The Van Der Waals expression, remember I did into -22.5 - another value.3228

That -22.5, that takes care of the ideal gas portion of the Van Der Waals gas.3235

Because it is a Van Der Waals gas, there is actually an additional term.3240

It is no different than the Van Der Waals equation.3244