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Lecture Comments (3)

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Post by Professor Hovasapian on November 4, 2015

Hello Jupil.

I hope you're king well.

If the mixture is of a fixed composition that does not change ( for example, 30% Methane Gas and 70% Carbon Dioxide gas), then the equations are the same: you do not have to consider anything else.

But if the composition is changing during a change of state, then a new set of equations is necessary: these equations are the same, except they have an extra term which involves something called "chemical potential", which we symbolize with a lower case Greek mu.

I have not done any lessons on mixtures, because, in general, they are not discussed in most P-Chem courses. However, I may consider adding several lessons on this topic in the future.

I hope I have answered your question satisfactorily. If I have misunderstood anything, please let me know.

Best wishes.


1 answer

Last reply by: Professor Hovasapian
Wed Nov 4, 2015 8:37 PM

Post by Jupil Youn on November 1, 2015

when you talking about U and H as functions of two variables, you mean U and H about a pure substance?  In case of mixture, what extra factors should I consider?

The Joule Thompson Experiment

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
  • General Equations 0:13
    • Recall
    • How Does Enthalpy of a System Change Upon a Unit Change in Pressure?
    • For Liquids & Solids
    • For Ideal Gases
    • For Real Gases
  • The Joule Thompson Experiment 18:37
    • The Joule Thompson Experiment Setup
    • The Flow in 2 Stages
    • Work Equation for the Joule Thompson Experiment
    • Insulated Pipe
    • Joule-Thompson Coefficient
    • Changing Temperature & Pressure in Such a Way that Enthalpy Remains Constant
  • Joule Thompson Inversion Temperature 36:26
    • Positive & Negative Joule-Thompson Coefficient
    • Joule Thompson Inversion Temperature
    • Inversion Temperature of Hydrogen Gas

Transcription: The Joule Thompson Experiment

Hello and welcome back to and welcome back to Physical Chemistry.0000

We have introduced this notion of enthalpy and today we would talk about something called the Joule-Thompson experiment.0005

Let us just dive right on in.0011

Let us recall the following facts, enthalpy H is a state function and that implies a state function is an exact differential.0014

Mathematically speaking it is an exact differential.0034

If we let, which we did, H = some function of T and P, we are saying that enthalpy is a function of temperature and pressure we have the following.0041

We have already done this.0053

We had that DH = a partial of H with respect to T at constant P × DT + a partial of H with respect to P at constant T DP.0056

We have is that exact differential state function and we can write it in this form depending on the variables that we choose for our function.0073

We already identified the DH DT at constant pressure with the heat capacity, 0085

the constant pressure heat capacity DQDT which was defined as the heat capacity at constant pressure.0107

This is the definition of the heat withdrawn from the system or the heat withdrawn from the surroundings.0117

We are always looking for this one, the heat withdrawn from the surroundings divided by the change in temperature of the system.0126

The heat withdrawn from the surroundings is the heat that goes into the system.0131

It is just a question of point of view.0134

That is the heat capacity, we identified with its partial derivative.0137

It is a change in enthalpy with respect to a change in temperature.0141

And this is an easily measure quantity.0145

The question is what about that?0163

What about the change in enthalpy with respect to pressure at constant temperature?0168

Let us see what we can do with this.0176

In other words, how does the enthalpy of the system change because the total change in the enthalpy is the sum of the two.0179

If we happen to hold one constant it does not matter but if we are not going to hold the temperature of the pressure constant, 0189

the total change in enthalpy is going to be the sum of the two changes.0195

How does enthalpy of the system change upon a unit change in pressure?0200

That is all it is, simple straight derivative, rate of change.0222

Let us begin with the definition of enthalpy, we have the H=U + PV.0227

The enthalpy of the system, the energy of the system + the pressure × the volume of the system, let us go ahead and differentiate that.0235

This is a huge differential here.0243

This is a product so we can use the product rules, this × the derivative of that + that × the derivative of this.0245

We have a differential form DH = DU + this × the differential of that P DV + this × the differential of that + V DP.0250

We are going to start with that.0265

We have the following.0271

We have the DH = DH DT sub P DT + DH DP at constant temperature × DP.0273

What else we have the following.0290

We also have the differential for U as a state function is exact differential.0291

Let me go ahead and write that you already identified this.0298

Let me write this as the constant pressure heat capacity.0300

This thing is that thing.0305

I’m going to write this as C sub P and DU we had a constant volume heat capacity × DT + the other thing that we talked about.0306

Remember when we talk about the Joules experiment, DUDV T DV.0317

We are going to go ahead and substitute this expression here, this expression here, but rearrange some things and play with this mathematically.0325

We have the following.0333

We have CP DT + DH DP sub T × DP = the constant volume heat capacity × DT +0337

Let me make sure my T's are correct here and let me have a lot of symbols floating around.0354

Let me make sure that they are correct and legible.0358

+ DU DV sub T × DV + P DV + V DP.0364

We have this very long equation but let us see what we can do with this.0375

We see that DV and DV is common here so let us go ahead and put these terms together.0382

I will work in blue, I hope you do not mind.0388

We have got CP DT + DH DP, this is just mathematical manipulation, that is all what is going on here.0392

DP =CV DT + DU DV T + P.0405

I’m going to put these two together × DV + V DP.0417

Holding T constant, it implies that DT = 0, the change in temperature is 0 if we hold the temperature constant.0426

The things at T they go to 0 so that becomes 0 and that becomes 0.0444

What we are left with is the following.0451

We are left with DH DP sub T × DP = this.0453

Let us see here we have DU DV, I should call it a little bit slower just to make sure that I'm getting all the symbols done correctly + P DV.0465

Mistakes are very easily to make when we have this many symbols floating around + V DP.0478

Let us go ahead and divide by DP.0484

Divide everything by DP and what you end up with is DH DP sub T = this stays,0491

DV ÷ DP that is just DV DP, I can write that as a partial derivative.0512

We are holding temperature constant so this is that and V DP divided by DV DP is cancel.0517

We are just left with V.0523

We were able to extract an expression for this DH DP, this is a general equation.0526

This DU DV is exactly what it means.0535

I do not think it make a lot of sense physically but at the very least you want to assign some physical meaning to the partial derivatives.0538

This is just the rate of change in energy with respect to volume.0545

This is the pressure, this is the rate of change of volume with respect to pressure, that is all it is.0550

Do not get too intimidated by these symbols.0557

This white side is a little strange, granted you know it is a little difficult to wrap your mind physically but 0559

as long as you can assign some meaning for the partial derivatives and know what is going on, at least it makes it a little bit more tractable.0567

This is a general equation that we were looking for.0575

We cannot do much with this but it is nice to see where it came from.0582

This is a general equation and it applies to all systems, gas, liquid, solid, it does not matter.0585

This is the general equation for all systems.0597

For liquids and solids, the first term, in other words this term, is usually very much smaller than this term.0604

For all practical purposes, we can ignore that term altogether.0627

For liquid and solid, the first term is usually much smaller than V.0632

Then I will just say the second term which is the actual volume of the system.0645

Therefore, for liquid and solids, the following approximation is valid DH DP of T = the actual volume of the system.0660

If I know the enthalpy of the system, the rate of change in enthalpy per unit change in pressure but 0674

I’m holding the temperature constant for a liquid and solid, it is often just = to the particular volume of liquid or solid.0684

That is all this is saying.0696

It gives you some numerical value.0698

Remember we are trying to figure out what this says because we have that differential for the energy.0702

The enthalpy of the system has to do with the constant pressure heat capacity + any change in pressure.0710

If that changes the enthalpy a little bit, here you go for a liquid or solid, you can just take that 0717

number as partial derivative actually = the volume of the system, the liquid or the solid.0722

The molar volume of liquid and solids is usually very small.0731

This DH DP molar at constant T can be ignored.0760

In other words, that heat capacity, the change in enthalpy that comes from the heat capacity term, 0774

the CP DT that is going to dominate the change in enthalpy of the system.0781

The molar volume of liquid and solid is so small but for all practical purposes, that second term of the differential, remember we had this DH = CP DT + DH DP TDP.0788

This term is going to dominate.0811

For liquid and solids, the molar volume is so small that for all practical purposes, this can be ignored.0812

It really does not matter for liquids and solid, this really not that big deal.0818

It is only under conditions of really high pressure that this value, this thing actually becomes significant.0825

In general, that is not going to happen.0835

When I say high pressure I mean extremely high pressure.0837

It is very specialized, you will probably never run across it in any of the work that you do unless 0839

you are specifically working under high pressure conditions for liquid and solids.0844

Let us go ahead and talk about ideal gases.0850

For ideal gases, it is easy.0854

It is the same thing that we have for the Joules experiment.0858

For ideal gases, this value DH DP sub T it just = 0.0861

In other words, if you change the pressure, the enthalpy of the system is not going to change.0869

If you want to go ahead and see a quick derivation of that, other than that is also important but that is fine, you can do it very quickly, it is pretty short.0877

We start with a definition U + PV.0888

For ideal gas PV = nRT.0892

We have H = U + nRT.0897

It is very important to be able to actually make these substitutions when you are dealing with the problem.0901

If we go ahead and divide by n and we put molar quantities, we have H = U + RT.0907

Any variable that has a line over it just means that you divided by the number of moles so it becomes a molar value.0916

This is a molar enthalpy J/ mol, molar energy is J/ mol.0923

That is all that is going on.0929

Let us divide by M.0931

We already know that from ideal gas, the energy is just a function of temperature.0934

If this is just a function of temperature and we have R × T, it turns out that from ideal gas the enthalpy itself is only a function of temperature, 1 variable.0938

It is a function only of temperature not pressure and volume.0949

Therefore, if we go ahead and differentiate with respect to pressure, you end up with DH DP T.0952

The differential this with respect to P there is no P involved in U so this is 0.0967

This is 0, that is it.0972

That is the derivation, this is what is important.0974

From ideal gas, that second term, this term right here is 0, it drops off.0977

For an ideal gas, the change in enthalpy of the system is gone strictly by this and this.0982

You do not have to worry about this term.0985

That is what we are talking about here.0990

The enthalpy depends on this term and this term.0991

For liquids and solids, this is really small.0996

For all practical purposes, we can ignore it unless otherwise stated.0998

For ideal gases we can ignore it completely because that is 0.1001

It is only that it matters.1005

If I want to measure the enthalpy change, I go ahead and take the temperature change and multiply by the constant pressure heat capacity, 1007

that gives me the change in enthalpy.1013

This is the differential form.1015

Let us talk about real gases.1021

For real gases, this DH DP sub T is not 0, not definitely.1025

It is small but it is measurable.1046

When we did the Joules experiment, you remember we were trying to figure out this thing.1051

We are trying to figure out DU DV.1056

If I change the volume of the system how does the energy of the system change?1060

Joules experiment it has 2 balls connected by something.1065

The gas expanded under free expansion, we measure the change in temperature of water.1070

As it turns out it was sensitive enough.1074

The heat capacity of the water was too much and the heat capacity of the gas was too small.1078

It was far below any measurable level.1082

It is not actually 0 but really it is so small that essentially is 0.1085

For Joules experiment that ended up being 0.1090

For the Joules constant experiment, they were able to actually do this experiment, a variation of it, and actually measure a change.1093

Let us see what it is that they did.1103

We said that this was 0 for the Joules experiment so we want to find out what is the change in enthalpy upon a change in pressure?1107

Here is the setup.1118

Let us go ahead and draw this out here.1122

Let me draw a little bit lower.1125

Based on what they had, they had an insulated pipe.1142

There was a pipe and the pipe there is insulation around it.1145

No heat can actually cross the barrier.1149

The DQ actually ends up being 0.1153

It is insulated so heat can escape and no heat can come in.1156

This thing right here, this is a porous membrane.1160

In the actual experiment themselves, what they use was just a silk scarf.1169

Imagine taking a pipe and just sticking a silk scarf in there.1172

Basically, what they are trying to do is this.1176

They are going to push some gas through this silk scarf.1178

They have a pressure gauge attached here and they also have a temperature gauge attached here.1183

You know there is a certain pressure on the side of the silk scarf.1194

There is a certain temperature inside that pipe.1196

There is a gas that is moving in this direction.1199

They also have a pressure gauge here and a temperature gauge here.1202

You know from your experience that air is going to pass through the silk scarf but on the other side it is going to be a lot less pressure.1208

And that is exactly what happens is there is going to be a certain pressure so there was going to be 1217

a certain amount of gas here and it is going to be under P1 V1 T1.1221

It is going to be a certain temperature, a certain pressure, and a certain volume of gas.1227

If we pick a certain volume of gas, let us say 1mol, it is going to have a certain temperature and pressure which are going to be read right off the meter.1232

On the other side, that same molar gas, the same thing, I mean it is going to now have a new pressure, a new volume, and the new temperature.1239

You know this already.1253

You stick something to a pipe, you force something through it, the pressure on this side is going to be higher than the pressure on this side.1254

If we measure a change in pressure, measure the change in temperature, we can come up with some numbers and that is what we are doing.1261

Let us go ahead and do that.1268

We have to deal with this mathematically.1272

The pipe is insulated so no heat can actually transfer during this process.1275

The pipe is insulated so no heat flows.1283

What we are going to do was we are going to basically take 1 mol of gas and we are going to push it through this membrane.1296

There is 1 mol of gas at a certain pressure and temperature.1302

We are going to push it through the membrane and we are going to measure its pressure and temperature on the other side, the same mol of gas.1305

Again, because this is obstruction, there is going to be a pressure drop on the other side.1315

You know this already just from your intuition tells you.1319

The thermometer, we will measure the temperature difference.1326

1 mol of gas has to pass through this obstruction.1330

We are going to imagine it in two ways.1334

I'm going to imagine that I'm taking this 1 mol of gas and squeezing it all into this one member, that is going to be the pushed.1337

From this membrane, that same mol of gas has to go this way, it has to expand.1345

All I'm doing is, I'm going to be describing mathematically the flow in two stages.1350

I'm going to be imagining it as completely crushing it from volume 1 down to volume 0 and from volume 0 to a new volume.1356

I just mathematically, I need to break and break it down to the two stages because this 1 mol of gas has to flow through the membrane.1365

I can actually do anyway that I want to.1372

We will imagine the flow in two stages.1379

This I how I think about it, you do not have to think about it this way.1387

Basically, you can think about it as a certain volume of gas that is being pushed this way, there is a certain pressure that is being applied.1390

The pressure is measured by this gauge.1396

That same gas has to push against the gas that comes after it.1398

Therefore, it is going to be pushing against this pressure which is this is the flow of gas but the pressure inside is pushing against the flow.1402

What we are going to be doing is we are going to be measuring the work done in pushing it this way and work done in expanding that way.1411

That is all we are doing here.1416

One, is going to be a compression from volume 1 to 0.1420

And two, it is going to be the expansion from 0 to volume 2.1433

I’m going to compress it and then it is going to expand.1444

I’m just taking it a fixed mass of gas and I chose 1 mol.1448

The work done in taking it from here to here and compressing this gas into the porous disk is going to be the integral V1 to 0 of P1 DV.1455

Basically, because this pressure, the pressure is what is pushing it into, like this pressure here is the pressure here.1471

The work being done, the external pressure that was pushing that gas and compressing it, it is P1 DV.1481

Work 2, it is going to go from a 0 volume to V2 and the work that is done here, this gas going from 0, 1491

imagine all the gas is right there in the porous disk, it needs to expand against an external pressure.1506

The external pressure against which it is expanding is pressure 2.1512

It is the pressure that is on the other side of this obstruction.1519

It has to push against that in order to actually expand.1522

So that is equal to pressure 2 × the change in volume.1526

The total work that is done in this process is work 1 + work 2 = this V1 to 0 of P1 DV + the integral from 0 to V2 of P2 DV.1535

This is equal to the pressure is constant, pressure is constant, so it comes out so what you end up with is P1 × - V1 + P2 × V2.1560

This is – so let me flip this around so = P2V2 - P1V1.1578

Let us go ahead.1589

The insulated pipe, we said that there is no heat flow so that implies that DQ = 0.1595

The first law expression for this, the change in energy of the system = Q – W.1607

If DQ = 0, Q = 0, that means that δ U = -W.1615

Δ U is U2 - U1 = - the work which is P2V2 - P1V1.1627

We have U2 - U1 = P1V1 – PQ V2.1643

Let us go ahead and rearrange this.1654

We have to bring this over there and bring this over there.1657

I have U2 + P2 V2 = U1 + P1 V1.1662

What is U + PV?1672

That is the enthalpy.1674

Therefore, the molar enthalpy in the final stage = the molar enthalpy of the first stage.1676

This shows that the Joules Thomson experiment or expansion is an isoenthalpic process.1688

The enthalpy of the system does not change when the gas passes through that obstruction.1695

The Joule Thompson expansion because it is an expansion, the gas is more pressure here, is less pressure here, the gas is going to expand.1706

As the pressure goes down the volume goes up.1723

Volume 2 is bigger than volume 1.1726

Expansion is isoenthalpic.1730

In other words, the enthalpy is constant through the process.1740

The enthalpy is constant through this change of state.1744

We see a change in pressure from one side of the pipe to the other, from one side of the obstruction to the other.1763

There is a difference in temperature.1767

We also see a change in temperature.1771

We see a δ T and let us take the ratio of the two.1775

We take the ratio of δ T / δ P. 1784

The P down in the denominator and we defined something called a Joules Thompson coefficient which is μ sub JT = the limit has this δ P goes to 0 of this ratio.1789

This ratio as we take δ P smaller and smaller, we are going to arrive at some number.1807

δ T/ δ P that is DT DP constant enthalpy, this is called the Joule constant coefficient.1815

And it is a measure of the extent to which the temperature changes upon a change in pressure.1836

It is a rate of change at which the temperature of a gas changes upon a change in pressure of that gas that is what this measures, 1843

that is what the Joule constant coefficient does.1855

Again, we were after this, we were after DH DP, a change in enthalpy under a change in pressure.1858

Let us go ahead and see if we can find that now given this thing.1865

We calculated as Joule constant coefficient which is δ T δ P or DT DP.1869

Let us go back to our differential statement, CP DT + DH DP sub T DP, this is just a statement of the change in enthalpy 1883

with respect to a change in temperature and a change in pressure.1901

We can always change the temperature and pressure in such a way that the enthalpy stays constant.1906

Experimentally, we can do that.1912

It is always possible to vary the temperature and the pressure which are the two variables in such a way that the enthalpy remains constant.1920

The Joule constant effect is an isoenthalpic process so we do that.1946

We will discuss what enthalpy means.1950

The constant enthalpy that means that the change in enthalpy = 0 that is great.1952

Or the differential form the DH = 0.1959

Therefore, I’m going to put this over here I get 0 = CP DT + DH DP T DT.1963

I’m going to divide by the DP.1982

When I divide by the DP I get 0 = CP DT DP constant enthalpy + DH DP T.1984

I’m going to go ahead and move this over the other side.2004

When I do that, I end up with the following.2006

I end up with DH DP sub T = - CP × DT DP H.2009

Well this is the constant pressure heat capacity.2024

This DT DP we just found as the Joule Thompson coefficient.2028

It is the change in temperature upon a unit change in pressure of the system at constant enthalpy.2032

I am going to rewrite it at the second line.2039

DHDP for real gas = - CP × the Joule Thompson coefficient.2045

DH DP which is what we wanted for this term, for our real gas.2060

It happens to equal from this.2066

The constant pressure heat capacity of the system is very easily measured.2069

You just take the heat that goes into the system divided by the change in temperature of the system.2073

The Joule Thompson coefficient is very easily measured.2077

I multiply the two, put a negative sign in front of it, and I end up with this term right here.2080

I just take this, put it right here, and if I want find out what the change in enthalpy of the system is,2085

upon a change in pressure and a change in temperature, the temperature component is this one.2091

The pressure component is this term.2097

I add them together and I get the change in enthalpy.2099

This is what we are after, right here, this second term.2103

We wanted to find a way to identify it with something that we can measure.2107

Sure enough, we can measure both of these very easily.2111

That is all what is going on here.2115

You, yourself do not have to derive this.2118

As long as you can follow the derivation, that is all that we ask.2122

At your level, that is all that is going to be asked of you, to be able to assign some meaning to this.2124

You will understand what these things mean.2130

It is like the change in enthalpy per unit change of pressure, constant pressure heat capacity, what the Joule Thompson coefficient is, 2132

which is actually a change in temperature per unit change in pressure.2141

Always assign physical meaning, you do not have to worry about doing the derivation.2145

Again like I said before, it is not like somebody sat and said let us do it this way.2149

No, they went this way and they have tried this mathematical path.2153

They tried this partial derivative, that partial derivative, until they actually saw some relationships.2157

What you are seeing is the final process that actually did give them something fruitful.2162

You do not have to do the derivation but as long as you follow the derivation, that is what is important.2168

Again, we can measure this and measure this.2175

We can calculate this very beautiful.2177

Let us go ahead and close this discussion out of the Joule-Thompson experiment, the Joule-Thompson effect.2181

The Joule-Thompson coefficient is positive at and below room temperature for all gases except hydrogen and helium.2189

Hydrogen and helium are the only two gases that have a negative Joule-Thompson coefficient.2200

This means that hydrogen and helium become hotter upon Joule-Thompson expansion.2207

All the other gases that pass through that pipe, upon Joule-Thompson expansion, the gas will cool.2212

Intuitively notice this just from your own experience with the expansion of gases and what you did in General Chemistry.2218

When a gas expands, it tends to cool.2224

The Joule-Thompson expansion which is keeping the enthalpy of the system constant, it is strange that hydrogen and helium 2228

actually end up getting hotter when the volume expands.2237

That is very odd.2240

Every gas, including hydrogen and helium, every gas has a temperature above which the Joule-Thompson coefficient becomes negative.2243

The Joule-Thompson inversion temperature.2254

At room temperature, carbon dioxide gas, if you do a Joule-Thompson expansion is going to cool but above a certain temperature, 2259

what that happens to be for carbon dioxide gas, I do not know.2267

Above that temperature, if you do a Joule-Thompson expansion it would actually get hotter.2271

It is called the Joule-Thompson inversion temperature.2276

For hydrogen gas, the inversion temperature is -80° C.2280

Below this temperature, hydrogen actually cools.2286

We said above which, in a particular case, under normal circumstances hydrogen has a negative Joule-Thompson coefficient, below this temperature.2290

80 is the inversion temperature: -80° C is the inversion temperature for hydrogen gas.2301

Below this, it actually cools.2307

If I'm doing a Joule-Thompson experiment for hydrogen gas during a Joule-Thompson expansion, 2311

if I run it below about negative -80…-90° C at that point upon expansion a hydrogen gas cools.2317

Again, at or below room temperature.2325

It is going to be the most of what we do.2331

All gases had positive Joule-Thompson coefficient.2333

That means if the pressure drops, the temperature drops.2338

If the pressure rises, the temperature rises.2341

It is positive, they correlate.2343

They have the same sign.2344

The δ T and δ P, they have the same sign.2346

Only hydrogen and helium are switched.2348

Thank you for joining us here at

We will see you next time, bye2354