For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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### 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
- Adiabatic Changes of State
- Adiabatic Changes for an Ideal Gas
- Adiabatic Changes for an Ideal Gas
- Equation for a Fixed Change in Volume
- Maximum & Minimum Values of Temperature
- Adiabatic Path
- 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

- 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

### Physical Chemistry Online Course

### Transcription: Adiabatic Changes of State

*Hello and welcome back to www.educator.com 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 www.educator.com.*2149

*We will see you next time, bye.*2151

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 1:47 AM

Post by Van Anh Do on February 15, 2016

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.