Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Physical Chemistry
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books

Bookmark and Share
Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 1:47 AM

Post by Van Anh Do on February 15 at 09:47:27 PM

Professor Hovasapian, could you please explain why we have dU/dV = 0 for ideal gas? Thank you so much!

1 answer

Last reply by: Professor Hovasapian
Sun Sep 13, 2015 9:40 PM

Post by Shukree AbdulRashed on September 11, 2015

What is a fixed change in volume? How does this imply that it's a constant? Thank you, sir.

Adiabatic Changes of State

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
  • Adiabatic Changes of State 0:10
    • Adiabatic Changes of State
    • Work & Energy in an Adiabatic Process
    • Pressure-Volume Work
  • Adiabatic Changes for an Ideal Gas 9:23
    • Adiabatic Changes for an Ideal Gas
    • Equation for a Fixed Change in Volume
    • Maximum & Minimum Values of Temperature
  • Adiabatic Path 18:08
    • Adiabatic Path Diagram
    • Reversible Adiabatic Expansion
    • Reversible Adiabatic Compression
    • Fundamental Relationship Equation for an Ideal Gas Under Adiabatic Expansion
    • More on the Equation
  • Important Equations 32:16
    • Important Adiabatic Equation
    • Reversible Adiabatic Change of State Equation

Transcription: Adiabatic Changes of State

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

Today, we are going to start talking about adiabatic changes of state.0004

Let us just dive right on in.0009

Let me go ahead and change this into blue, it is my favorite color here.0013

If no heat flows during a change of state, during the change of state then DQ = 0.0021

No heat is flowing so there is no change so DQ = 0.0042

When this happens, we call this change adiabatic.0048

We had isothermal which was when the temperature stays constant and DT = 0.0064

At the constant pressure process we call a constant pressure it is isobaric, that is when there is no change in pressure so DP = 0.0075

Isochoric or isometric process which we actually just called constant volume process, that is when DV =0.0089

We have a circumstance where there is no heat that is lost so DQ = 0, that is called adiabatic.0097

Those are our most important things.0107

We could call this isothermal but we tend instead of saying isochoric or isobaric, we tend to just say constant pressure and constant volume.0115

In this particular case, we know heat is flowing DQ = 0.0123

The statement of the first law was this, that DU = DQ - DW that is a statement of the first law, the conservation of energy.0129

DQ = 0 in which you have is DU = - DW.0143

For a finite change, you have δ U = - W.0154

Again remember, this δ W does not make sense.0160

Work is not something that the system possesses.0163

It is a quantity, it is a path function.0165

In an adiabatic process as it turns out when DQ = 0, it says that work turns into a work behaves as a state function.0168

Do not think about it that way.0180

There are books that actually say that.0181

They say work is a state function under an adiabatic process.0183

Work is not a state function, never a state function.0187

Mathematically, under an adiabatic condition would be Q = 0.0190

The quantity of work happens to = the change in energy, that is all this is saying.0194

Do not confuse the two, energy is a state function.0199

Work and heat are not state functions.0201

Let us see, we can also write it like this that work = - DW DU or W = - DU.0207

That might actually be a little bit more sense to you, if you like.0220

What this says is the following.0226

Again, anytime we are faced with any kind of mathematics we want a sense of physical meaning to it.0229

What this says, if you do not stop and assign physical meaning to it, the mathematics of thermodynamics is going 0233

to get away from you very quickly and it is just going to end up being just a bunch of symbols on a page.0239

That means absolutely nothing.0244

This is sort of a historical problem with learning thermodynamics.0246

It is just mathematically, symbolically gets away from you very quickly.0252

Just stop and assign these things meaning.0257

This is what this is saying when we see this and when we see this.0259

It says that the work produced in the surroundings in an adiabatic process = decrease in energy of the system.0265

That sort of makes sense already.0302

If I’m not going to allow any kind of heat to flow, the work that the gas was expanding this way is doing work on the surroundings.0306

Work is being produced in the surroundings.0315

Upon expansion, the work is being produced in the surroundings that means the systems could lose energy.0317

It is losing it as work instead of losing it as heat.0324

That is all that this is the same.0327

Assign some meaning, stop and say this means this and this means that.0329

Decrease in energy of the system means that how do we measure energy and temperature?0340

It means a decrease in temperature of the system.0356

The decrease in energy of the system means a decrease in temperature of the system.0375

When a gas expands under adiabatic conditions no heat is allowed to flow.0383

If no heat is allowed to flow, the work of the gas is doing on the surroundings then the energy of the system is going to decrease.0387

The temperature of the system is going to decrease.0395

It is expanding and it is going to cool.0397

You know this intuitively, we are just making it very rigorous.0401

If work is destroyed in the surroundings, if surroundings does work on the system that is what destroyed work means.0407

If work is destroyed in the surroundings, in other words if this is negative, then this negative becomes positive.0421

Then the energy, thus, temperature of the system increases.0435

Adiabatic means DQ = 0.0455

That is what adiabatic means, no heat is allowed to flow.0458

For pressure, volume, and work.0464

For pressure, volume, work, DU= - DW.0474

I have said that already, it is right here, adiabatic condition. 0481

DU = we know what DW is.0486

DW = - P external DV, pressure, volume, work, is just pressure × change in volume.0490

It is the same.0499

Let us go to the next page here.0501

This says upon expansion, when a system expands the DV is positive.0505

It implies that the DU is negative, increasing if the gas expands its volume increases, the energy decreases.0520

That is equation the same.0529

That is what that says.0535

For the compression, we are just watching the signs.0539

If a change in volume upon compression is negative, the final volume is less than the initial volume.0545

Therefore, final - initial we are going to get a negative number.0553

That is going to imply that the change in energy of the system is going to be positive.0556

That is what is happening here.0561

The adiabatic changes for an ideal gas, let us recall a few things.0565

The general expression for the energy of the system is = CV DT + DU DV T DV.0584

This is the general equation.0601

Well for an ideal gas, this thing is equal to 0.0603

The DU DV sub T = 0.0611

Therefore, what we had is that the change in energy = CV DT.0616

We have that relationship.0623

Well, under adiabatic conditions, we just said that an adiabatic condition implies that DU = - the work which is = - P external × DV.0625

Therefore, DU = this, DU = this.0646

Therefore, this = this so we have - P external DV = CV DT.0651

This is the fundamental equation for an adiabatic change.0666

The external pressure × the change in volume is = constant volume heat capacity × the change in temperature for an adiabatic process.0670

Let us see if we can fiddle around with this a little bit.0678

Let us go ahead and for a fixed volume change which is DV, the change in temperature T which is the DT becomes directly proportional to the external pressure.0681

This is what looks like.0726

I’m going to divide everything by CV here.0728

I get DT = - P external × DV ÷ CV.0731

If we fix the volume, for all practical purposes it is a constant.0740

This is the heat capacity so this is a constant, that means that the change in temperature is going to be proportional to the external pressure.0746

Constant and constant this is just going to be some constant K × the external pressure.0758

It as a direct proportionality.0765

Let us go ahead and write on the next page and start again.0773

I have DT = - P external DV / CV.0780

Notice that P external DV is just work, that is just the pressure volume work.0789

Work is a path function which means that I could follow any path that I want, an infinite number of paths.0812

Therefore, the change in temperature can take on an infinite number of values.0819

Work is a path function, this implies that DT can have an infinite number of values.0829

Work concerned with maximizing and minimizing the change in temperature.0851

We are concerned with maximum values and minimum values of this.0863

Again DT = this thing.0873

This thing is just work.0876

Work is a path function that means that a particular path I take determines what this number is.0877

It says there is an infinite number of paths, I have an infinite number of values of work.0882

We divide that work by the constant volume heat capacity, I have an infinite number of possible values for the change in temperature of my system.0886

I want to find that particular path that allows me to maximize and minimize this change in temperature.0893

It was the extreme values that are the most important, not really much in between.0901

Recall that max min values, remember when we were working with isothermal processes.0907

Recall that max min values for work, they happened along specific paths.0917

In the case of the isothermal process, it actually happens along that thing that we call the isotherm itself.0935

Specific path, that is why we called it reversible.0943

We call this path irreversible path when we said not only in isothermal expansion but irreversible isothermal expansion.0951

We are doing the process adiabatically.0962

We call this path reversible but we are not holding T constant here.0966

This is not an isothermal process, this is an adiabatic process but we are not doing things isothermally here.0976

Therefore, in order to maximize the temperature drop negative sign, upon expansion and minimize the temperature rise upon compression,1001

We must be moving along a specific path.1049

Let us see if we can give the path a name.1051

We must be moving along a specific path that maximizes and minimizes a specific path in both directions.1055

This path is called and adiabatic.1080

In isothermal process we talked about moving along the isotherm that was outlined.1089

We are going to be talking about moving along the adiabat and I will actually draw it out right now.1094

I will draw the isotherm so that you can see what it is it is actually going on with respect to the isotherms.1101

We are going to go ahead and do a pressure volume diagram.1106

We have pressure on the Y axis, we have volume along the X axis.1110

I’m going to draw 2 isotherms.1114

This is 1 isotherm and we call this temperature 1, that is the temperature 1 isotherm.1117

Let me draw another one.1125

This is going to be the temperature 2 isotherm.1125

I have a state 1.1131

My state 1 is going to be P1V1 and T1.1135

I’m going to go ahead and watch what happens.1140

Let me expand this a little further.1144

I have got my P1 here, I’m going to go ahead and put this here.1150

This is state 2 right here.1156

This is state 2, this is my P2V2 and T2.1160

We are not holding temperature constant.1167

We are not holding pressure constant.1170

We are not holding volume constant.1171

What we are doing is we are making a change in state adiabatically.1173

We are not allowing any heat to flow.1177

Therefore, the path that followed this one, this is what we call and adiabat.1181

This is pressure 2, this is volume 1, here is what is happening.1197

If I start a certain state pressure 1 volume 1 T1, if I expand and go to another pressure volume and temperature.1204

Under adiabatic conditions, I can take any path I want.1216

I can go this way, I can go this way, I can go any way I want.1218

The path that maximizes the temperature dropped, notice I'm going from one isotherm.1224

This is one temperature is going to go to another isotherm, a lower temperature.1232

If I follow this path which is called an adiabat, that is going to maximize my temperature drop.1237

If I'm doing a compression from a higher volume, volume 2 to a lower volume, if I go along this path, if I go this way, 1248

that is going to minimize the rise in temperature.1260

This is what we call a reversible adiabatic process.1265

When we talk about the isotherm, we called the moving along the isotherm reversible, isothermal and reversible.1268

This is going to be adiabatic and reversible.1274

If it is adiabatic and reversible, you are moving along the adiabat.1277

That is what you are doing.1284

State 1 state 2, initial state final state, temperature 1 temperature 2.1286

Adiabatic means that there is no heat allowed to flow.1291

That is what is happening here.1296

For a fixed change in volume, in other words going from volume 1 to volume 2, 1298

the reversible adiabatic expansion produces the maximum decrease in temperature.1318

The maximum temperature drop.1340

I will just say decrease in temperature.1344

The importance of being reversible adiabatic.1355

The reversible adiabatic compression expansion we are moving this way.1365

Compression we are moving that way.1377

The compression produces the minimum increase in temperature.1380

Remember, when you are working with isothermal process, we said that if you moved along the isotherm reversal process, 1402

upon expansion you are going to end up doing the maximum amount of work upon expanding the gas from this volume to that volume, 1414

if you move along the isotherm in order to keep the temperature constant.1422

It is going to require, if you are compressing it the minimum amount of work that you have to do 1426

in order to compress that gas from volume 2 to volume 1 is going to be minimized.1431

Those values happen to be the same.1438

In an adiabatic process, you are not traveling along the isotherm.1439

You are traveling along an adiabat.1442

What changes is now you are not going to get the maximum minimum work.1444

It is a reversible adiabatic expansion produces the maximum decrease in temperature and irreversible adiabatic compression produces 1450

the minimum increase in temperature going from state 1 to state 2 upon expansion.1459

Going from the state 2 to state 1 upon compression, that is what is going, it is a change in temperature at this point.1465

Adiabatic and isothermal, definitely differentiate between the two.1472

It is going to be very important.1476

When you are moving along an adiabat, you are moving between isotherms, that is what is happening.1478

Reversible, we already know what reversible means.1486

Reversible does not just mean moving along a certain path, that is true as far as the PV diagram is concern, 1489

Yes, that means moving along a specified path whether it is an isotherm or adiabat.1496

Mathematically, reversible means that any given moment the external pressure is actually = the pressure of the system which we call P.1501

P without a subscript is the pressure of the system itself.1516

Therefore, as we have - P DV = CV DT.1520

We have -P external DV = CV DT that is the expression for an adiabatic process.1531

The P external in reversible process means that the external pressure happens to be in equilibrium with the pressure of the system.1536

It is very small changes.1544

For an ideal gas, the pressure = nrt / V.1547

Let us go ahead and substitute that in.1559

We will go ahead and put this in to P portion.1560

We end up with - nrt/ V DV = CV DT.1566

We are going to go ahead and divide by T.1579

When we divide by T we get - nr DV / V = CVDT / T.1584

I’m going to go ahead divide by n so we deal with molar quantities.1598

We end up with is - R DV / V = CD DT with a line over it, it means molar.1601

Something CV divided by n is the DV with a line over it, molar quantity over T.1616

We are going to go ahead and integrate both sides.1625

When we integrate, we end up with the following .1630

I will do it over here -R × the integral from volume 1 to volume 2 DV/ V = .1634

If this is constant that also comes out of the integral so we get T1 T2 DT / T and you end up with the following.1653

You get -R × LN V2 / V1 = constant volume heat capacity molar N T2 / T1.1667

There you go for an ideal gas under adiabatic expansion this is the fundamental relationship 1686

between R, the volume, the temperature, and the constant volume heat capacity.1694

Let us play with this equation a little bit more.1700

For an ideal gas that molar constant pressure heat capacity - the molar constant volume heat capacity is = R.1705

If I go ahead and divide by CV, I end up with CP / CV -1 = R / CV.1720

What we are seeing is CP / CV that is = the thing that we called γ.1738

γ -1 = R/ C sub V.1749

I go back to my equation, I divide by CV and I get -R / CV × the natural logarithm of volume 2 / volume 1 = nat log of temperature 2 / temperature 1.1759

This is = this so I’m going to derive this as - γ -1 × LM V2 / V1 is = LM of T2 / T1.1777

I end up with, I’m going to go ahead like this - LN of V2/ V1 γ -1 = LN of T2 for T1.1792

Let us see if we can play with as little bit more and see where it actually takes us.1814

This - allows me to flip these 2 so I get log of V1 / V2.1820

The γ -1 = the log of T2 / T1.1829

Therefore, I have log and log so therefore, I am left with V1 / V2 ⁺γ -1 = TQ / T1.1837

Or if I like I can go ahead and cross multiply in order to put the 1 and 2 together.1852

T1 V1 ⁺γ -1 = T2V2 ⁺γ -2.1857

Using PV = nRT, the equation I just wrote, I can fiddle around with that the expression between T1 V1 T2 V2.1872

I also can extract using PV = nRT.1882

We also get P1 V1 ⁺γ = P2 V2 ⁺γ.1887

This is an expression between pressure and volume.1898

We will also get T1 ⁺γ P1¹ - γ = T2 ⁺γ × P2¹ –γ.1903

This is just fiddling around with that one equation to get different relationships.1920

We ended up with a relationship we used on the previous page was temperature volume.1926

This is pressure volume, temperature pressure.1931

That is all that is going on here.1933

A little bit of a recap.1937

The important equations that we need or that we are interested in when we discuss things that are adiabatic.1941

Adiabatic means DQ = 0 or Q = 0.1957

Either one was fine depending on if you are talking about the differential version or the complete finite version.1963

What you end up with is DU = -W or δ U = - W.1969

This is DU -DW.1977

For a reversible adiabatic change of state, we can have an adiabatic process we can have a reversible adiabatic process.1984

Remember, we can have an isothermal process or we can have a reversible isothermal process.2011

That reversible puts a little bit more of a constraint on the situation.2015

For an irreversible adiabatic change, now for an ideal gas we had CV DT = - P DV as the basic equation.2021

You can go ahead and take care of the integration.2040

Once you integrate, you end up with the following.2042

You end up with CV × LM T2 / T1 this is the basic equation.2045

What we did just like this stuff right here, you took a little bit further.2052

We just rewrote this in a different form – R LN V2 / V1.2059

Personally, I like to stick with this equation.2069

This is just a figure that you see in books.2070

I would not worry about it too much.2073

This is the important equation right here.2074

Or if you prefer, because this molar thing is just = the heat capacity divided by n, 2078

you can go ahead and put this into here and get this particular version of the equation.2092

You can get CV × LM of T2 / T1 = - nr × LM of V2/ V1.2098

Like most kids actually they like to see the n, the mol. 2115

I suppose working with this thing, this little line over these more values, I like that too.2119

I actually prefer to see that mol there.2124

It is just a personal taste of mind.2127

So that is about it, that pretty much takes care of the adiabatic process any time you see anything was done under adiabatic conditions, 2129

you want to set DQ = 0 or Q = 0, this is going to be the equation you work with, 2136

under an adiabatic reversible condition or reversible adiabatic changes of state.2141

In the case of an ideal gas, this is what you are dealing with.2146

Thank you so much for joining us here at

We will see you next time, bye.2151