Lecture 10. Jim Joule.
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MethionineHamActive siteWirbelstromverfahrenElectronic cigaretteLecture/Conference
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Ale MunicipalityAzo couplingLecture/Conference
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Surface finishingEntropyCalculus (medicine)CancerScreening (medicine)CheminformaticsWine tasting descriptorsLecture/Conference
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Process (computing)Gene expressionCigarDrainage divideBoyle-Mariotte-GesetzProcess (computing)Bottling lineSystemic therapyVolumetric flow rateWalkingWeinfehlerPotenz <Homöopathie>WursthülleGene expressionGrowth mediumBattery (electricity)Boyle-Mariotte-GesetzLeft-wing politicsLecture/ConferenceComputer animation
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ProlineDiamantähnlicher KohlenstoffTarWalkingProcess (computing)Sorption isothermDiamantähnlicher KohlenstoffErdrutschFunctional groupWine tasting descriptorsGene expressionPressureGrowth mediumLecture/Conference
08:32
IceNotch signaling pathwayDiamantähnlicher KohlenstoffTitanium carbideProlineTrihalomethaneSorption isothermProcess (computing)Diamantähnlicher KohlenstoffPressureGrowth mediumFormaldehydeLecture/Conference
10:07
Process (computing)PressureLecture/ConferenceComputer animation
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River deltaProcess (computing)Volumetric flow rateSystemic therapyLecture/ConferenceComputer animation
11:46
Metabolic pathwaySetzen <Verfahrenstechnik>Flammability limitSystemic therapySetzen <Verfahrenstechnik>KatalaseProcess (computing)NanoparticleDeath by burningLecture/Conference
12:47
GasOctane ratingMetabolic pathwayLecture/Conference
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GemstoneWine tasting descriptorsBeerBreweryLecture/Conference
14:17
BrauwissenschaftBeerSodium carbonateConcentrateWaterMachinabilityCoca-ColaCarbon (fiber)KohlenhydratchemieMixtureDeathPharmacySyrupCoke (fuel)Barley wineLecture/Conference
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BrauwissenschaftBeerProcess (computing)WaterBottling lineController (control theory)Sea levelTiermodellLecture/Conference
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BrauwissenschaftProcess (computing)ChemistrySea levelChemical engineeringBottling lineBeerAlkoholische GärungLecture/Conference
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BrauwissenschaftENUAlkoholische GärungProcess (computing)Bottling lineBreweryStainless steelBeerLecture/Conference
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BrauwissenschaftAgeingStainless steelAlkoholische GärungBeerBrauwissenschaftChemistryLecture/Conference
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BreweryChemistryGesundheitsstörungLecture/Conference
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BrewerySea levelBrauwissenschaftBeerProcess (computing)Lecture/Conference
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WaterDihydroergotamineBlood vesselDrop (liquid)Body weightLecture/Conference
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Sea levelWaterSea levelVacuoleReaction mechanismMan pageLecture/Conference
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UltraschallschweißenSea levelMustLibrary (computing)UrocortinSea levelBreweryLecture/Conference
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Sea levelMustSea levelStructural steelLecture/Conference
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IsothiocyanateTennesseeBody weightStickstoffatomWaterGlassesLecture/ConferenceComputer animation
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HamBrown adipose tissueMoleculeWaterWaterfallPhysical chemistryWaterLecture/Conference
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MoleculeNanoparticleMoleculeReales GasAction potentialLecture/Conference
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InflammationÖlBond lengthVan-der-Waals-KraftIonenbindungLecture/Conference
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Boyle-Mariotte-GesetzAction potentialMixturePressurePressureAtomic numberAmount of substanceReales GasAction potentialSystemic therapyBinding energyLecture/Conference
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PressureAltbierPressureGasReales GasMolar volumeBoyle-Mariotte-GesetzAtomic numberStress (mechanics)Lecture/Conference
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SunscreenLot <Werkstoff>NanoparticleMoleculePressureBlood vesselLecture/Conference
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Boyle-Mariotte-GesetzMan pageRiver sourceMoleculeAtomic numberLecture/Conference
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MoleculeCollisionAtomPressureContainment buildingFireLecture/Conference
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MoleculePressureMoleculeContainment buildingSystemic therapyLecture/Conference
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MoleculeGasAction potentialPressureGemstoneLecture/Conference
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WaterBoyle-Mariotte-GesetzBoyle-Mariotte-GesetzGasLecture/Conference
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OrlistatNational Rifle AssociationChemistryWine tasting descriptorsBallistic traumaHorse meatAnomalie <Medizin>Lecture/Conference
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WaterGemstoneChewing gumRiver sourceWaterStorage tankThermalquellePedosphäreLecture/ConferenceComputer animation
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CycloalkaneWaterPressureThermalquelleCombustion chamberProcess (computing)Lecture/Conference
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Process (computing)PressureContainment buildingMotion (physics)Lecture/Conference
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VerdampfungswärmeProcess (computing)GemstoneWaterProcess (computing)PressureIsoliergasStorage tankVerdampfungswärmeGolgi apparatusLeft-wing politicsVolumetric flow rateBattery (electricity)Lecture/Conference
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Columbia RecordsNational Rifle AssociationVerdampfungswärmeProcess (computing)Boyle-Mariotte-GesetzGemstonePressureDerivative (chemistry)SubstitutionsreaktionSpawn (biology)PainLecture/Conference
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GasDiamantähnlicher KohlenstoffPressurePressureGemstoneClaus processWursthüllePhysical chemistryGasDiamantähnlicher KohlenstoffLecture/Conference
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PressureDiamantähnlicher KohlenstoffENUSunscreenWine tasting descriptorsPressureCalculus (medicine)GemstoneWursthülleNuclear fissionRiver deltaSpawn (biology)Lecture/Conference
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Diamantähnlicher KohlenstoffPressureLexus LFAColumbia RecordsBattery (electricity)PressureSeparation processLecture/Conference
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Columbia RecordsGasPressurePressureFunctional groupSpawn (biology)GemstoneTrace elementLecture/Conference
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GasPressureIon transporterRefrigeratorOrganische ChemieSystemic therapyPressureAction potentialPhysical chemistryElectronegativityLecture/Conference
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Ion transporterRefrigeratorLecture/ConferenceComputer animation
Transcript: English(auto-generated)
00:05
Okay, so Quiz 3 was by far the hardest quiz I think I don't have to tell you that. It was much more difficult than one or two and I think you know two was harder than
00:21
one three was harder than two but I think we've reached the point where we've titrated now to the maybe past where we wanted to go. Okay so this is what the histogram looked like for Quiz 3. Now there's a mistake on the key that I posted this morning but on problem five I work out
00:41
the answer and it's actually B not D which D is indicated on the key but G and Mark found that error before I did. Okay so the quizzes should have been graded correctly. The correct answer to five is B not D. Okay?
01:02
So I'll try and make a change to the key before the end of the day. So here's where we are. So far we've had three quizzes and this is what the histogram looks like in other
01:21
words all these folks here are getting A's these folks here are getting B's and so on. Okay so going into Midterm 1 which is Friday you guys are doing really well which is good
01:40
that's the way we want to see it. So here's what's going to happen Midterm 1 is Friday it will cover all of Chapters 13 and 14. It turns out that this is just perfect all right accidentally all right this lecture that I'm going to give today will take us right to the end of Chapter 14 all right
02:02
then what's going to happen is you're going to review Chapters 13 and 14 and Lectures 1 through 10 on Wednesday all right we'll have a review for you all right Steven's going to give the lecture that I've written okay and then we'll take the midterm on
02:22
Friday all right there's a copy of last year's midterm exam and what you'll see what Steven will do on Wednesdays he'll review what's going to be on the midterm in some detail he will tell you how many questions there's going to be he'll tell you something about what those questions are going to be okay so Wednesday's lecture
02:43
is a rather important one to attend okay but let me just tell you that it's going to look a lot like last year's midterm and I already posted that a couple of weeks ago right last year's midterm is already on the announcements page so you can see what that is. Turns out that last year's midterm would have been a perfect midterm to give this
03:01
year right because there's nothing about Chapter 15 on the midterm there's nothing about entropy calculations on that midterm okay it's about really about Chapters 13 and 14 okay so what we're going to do today is finish Chapter 14 and then on Wednesday
03:23
we're going to review and then on Friday we'll take the midterm exam now the midterm exam is going to be open book open notes I'm not going to give you you know that shotgun page of equations that I showed you on the screen earlier where there was random equations
03:41
all over no alright open book open notes not open iPads or computers okay but you can use any calculator that you want.
04:00
Any questions about that? Any questions about midterm 1? Okay so G and Mark started to talk to you about adiabatic processes in an adiabatic
04:20
process the heat flow is turned off right the system's in a thermos bottle there's no flows of heat in or out for any infinitesimal step of the process there can be work right
04:43
so we know normally the internal energy is given by heat plus work alright and an infinitesimal change in the internal energy is then an infinitesimal change in heat or an infinitesimal change in work but for an adiabatic process it's all about the work right the system is in a thermos bottle another expression for DU can be obtained from
05:09
the definition for the constant volume heat capacity which we talked about some weeks ago right the constant volume heat capacity is just the partial derivative of the internal energy with respect to temperature evaluated at constant volume right that's why it's
05:24
called the constant volume heat capacity okay and so once I've got this heat capacity I can get I can express it as a derivative according to this equation here and so for an ideal gas DW is simply given by alright we know the work is minus PDV and if I
05:49
just substitute for P from the ideal gas equation I get this equation right here alright because P is just in our T over V and since internal energy and work are equal to one another
06:05
for an adiabatic process that means that these two things have to be equal to one another alright that equals that for an adiabatic process and so I can just set these two things equal to one another like this and then I can divide through by T and
06:22
I'll eliminate it then from this side and it will pop up over here on the left hand side and then if I integrate these two expressions now I integrate this one from T1 to T2 this one from V1 to V2 here's what that integral looks like I'm just integrating from T1 to T2 V1 to V2 the integral is the same in other words here I'm integrating 1 over V
06:45
here I'm integrating 1 over T right so I get log T2 over T1 log V2 over V1 and then I can just divide through by this heat capacity now I'm just doing a little algebra to clean
07:01
this up and when I do that and I take the exponential of both sides I get this expression right here alright which is an equation that we're going to use to describe adiabatic processes alright it's your equation 14.37 okay so this NR over CV that's going to pop up as an exponent when we take the exponential of both sides now notice that I flip this
07:27
over alright here it's T2 over T1 V2 over V1 but I flip this over so now it's V1 over V2 so I can get rid of the minus sign that's all I'm doing there alright
07:42
I'm not skipping any steps okay so this is an adiabatic reversible process involving an ideal gas so we've got this expression for an adiabatic process that we just derived earlier we derived this expression for an isothermal process we can use whichever one
08:02
we need G and Mark walked you through this slide on Friday alright we've got two isotherms here alright here's an isotherm that applies at T1 here's an isotherm that applies at
08:21
T2 alright what we're plotting here is the pressure is a function of volume as we do an expansion these are isothermal expansions here on these two isotherms we use the word isotherm because they're constant temperature okay if we want to look at an adiabatic
08:41
process that starts here for example alright it's going to cross over it's not a constant temperature process what's constant is heat okay and when we do an expansion that means that the temperature has to fall so change in P with an expansion is larger for an adiabatic
09:03
process than for an isothermal one because temperature decreases for the adiabatic process the temperature is going down during this adiabatic process alright obviously that's not happening with the isothermal process alright so in other words if I look at T1
09:22
I go from this initial volume to this final volume alright here's the change in pressure that happens at T1 everyone see that here's T1 right here's V1 alright we're going from here to here right and the pressure difference would only be this shown by this
09:42
yellow bar here or if we wanted to look at the pressure difference at T2 right from the isothermal process here's T2 so here's its pressure difference given by this yellow bar here even smaller alright and if we look at what the adiabatic process does
10:02
it's going from V1 to V2 over this much larger range of pressures much larger range of pressures here because temperature is changing too in this process. Okay now it's
10:26
not obvious from these two equations but there's a more profound implication of this adiabatic business alright that we've sort of we've said it but we weren't very explicit
10:44
about it alright DU equals DW because DQ is zero right and that has important implications obviously if I integrate DU from some initial internal energy to some final internal energy I'm just going to get the final minus the initial because U is a state function alright
11:08
the difference between any integral alright it's just going to be the final minus the initial alright and we're going to just call that delta U alright in an adiabatic process that
11:20
has to be equal to W right if this process if this change in the internal energy is happening in a thermos bottle where there can't be any flows of heat only work can occur alright then this change in the internal energy has got to be given by just the amount of work that
11:42
was done for as long as it was adiabatic work it was work that was done on the system when it was in a thermos bottle in other words the adiabatic work is different from all the other types of other types of works because it is a state function just like
12:01
U it would have to be it's equal to U alright so that means for an adiabatic process you know what we saw earlier is if we if we take a normal process and we do PV work if we break the work down into smaller chunks we can do less of it to get from an
12:24
initial state to a final state remember that we took the brick and we ground it up to make it into tiny particles and we can add tiny particles and then we can do the minimal amount of work alright but if the work is adiabatic there's only one way to do it right it's a state function you can't there's only one way to do the adiabatic
12:45
work there's only one adiabatic work answer it's a state function just like you does not depend on the path because all adiabatic paths between UI and UF have to be identical right so adiabatic work is special right it's a state function just like the internal energy
13:11
okay now James Jewell we're gonna come back to this adiabatic work business but we're
13:23
gonna take a detour this is James Jewell he was born in 1918 in a little town outside of Manchester England Manchester England is right here this is England okay and he
13:43
was born in Salford which is right over here okay Manchester is a big city but in those days not so big Salford is actually a pretty good sized town itself but not when he not in 1918 it's tiny little town every town in England had its own brewery in fact
14:01
that's true for most of Europe at this point in time right every little town in Germany has its own brewery usually just one all right and so every town has its own identity in terms of the beer that you can drink there all right and it's still true to this day to a large extent
14:20
and let me just tell you that making beer is not like making soda right to make Coke right Coca-Cola you there's some chemists who go to a lab they come up with this mixture of flavorings and you know they make a flavor packet right and what you if you go to Joe's
14:45
on the green and you buy a coke right there's a a thing of concentrated syrup that gets diluted with carbonated water all right and so to make coke you just need the flavoring you need carbonated water and you need sugar and it's dead easy to mix them together right every time
15:05
you get coke out of a machine it's just mixing these things together for you right the syrup actually has the water it has actually has the sugar in it in beer you start with barley and hops all right there's no flavor packet that you dilute with water to get beer that has a
15:25
particular flavor all right every beer that you buy everyone a world has to go through all of these different processes and so if you think about it it's totally amazing that when you buy Heineken it always tastes exactly like Heineken you know what I'm talking about it's got the
15:41
distinctive Heineken taste right they're making billions of bottles of Heineken and every they have to do all of this to every single bottle all right so what that means is that there is exquisite control I mean at a level that you might not even believe I'm sorry yes so there's
16:11
a lot of process chemical engineering that goes into making this absolutely reproducible for every bottle I mean it is really a chemical engineering feat that this can be done and this
16:23
this was not something that we've learned in the 20th century this is learned in the 18th century maybe even the 17th century how to reproducibly make beer that tastes the same every time one of the keys is here all right fermentation all right the temperature here has
16:51
to be controlled all right and this is true to a lesser extent in these earlier processes too but in this fermentation process all right the temperature has to be controlled with better
17:02
than one degree C precision okay and that's not enough to get the beer that tastes the same every time I mean a lot of other things if you ever want it I don't know if you're fascinated by such things but if you've ever if you've never been to a big commercial brewery all of them will give you a tour or you go to a Budweiser brewery and you can get a tour of the
17:23
Budweiser brewery you walk through that place it is spotlessly clean you can't believe it all right there is it's like a hospital in there all right you look down at this five acres worth of stainless steel kettles and bottles and there's like two guys running a factory of five acres
17:43
making billions of bottles of Budweiser it's all computer automated all right and process control is taking care of this whole everything that happens here is happening in an automated way all right and you know Budweiser's beech beech wood age you know you look down there's a guy
18:04
shoveling beech wood into a stainless steel vat that's about the size of this room right there's actually beech wood in this stupid uh fermentation right beech wood so what what does any of this have to do with thermodynamics
18:24
Jim Jewell his dad was a brewer in Salford his dad made the beer in Salford and like I said there was only one guy who did that and he understood chemistry from being a brewer's kid
18:41
right he learned how to run the brewery in fact he did run the brewery after his dad had some medical problems he went to school in Manchester for two years with his brother Benjamin studied with a guy named James Dalton anybody know that name what did he do atomic theory
19:07
all right studied with James Dalton for two years and then there was some sort of health problems back home and he and his brother went back to run the brewery he never had any more college education than that
19:21
uh but he was interested you know he had an amateur's interest in science and that really was driven by understanding how to make the brewery run more efficiently right you want to understand what are the limits and efficiency that we can achieve in this brewery right so we can maximize our profits he was thinking on a very practical level
19:43
so one of the things you want to understand is what is the relationship between work and heat right now that's a pretty profound thing to want to understand when you're brewing beer all right but he was smart enough to understand that that was an important thing for him to
20:01
appreciate in terms of the brewing process all right and so the most famous experiment that he did involved taking a weight right using that to drive an agitator inside a vessel of water all right and imagine how hard this experiment would be you put a weight here
20:23
you drop this weight this thing spins like crazy in a bucket of water and that's a thermometer you're going to measure the temperature change are you kidding me all right the temperature is going to change all right not by much all right he could measure the temperature to one
20:41
two hundredth of a degree fahrenheit he was the only guy in the world who could make that measurement all right and he measured it he saw a reproducible temperature change all right and he could correlate the temperature change with the distance that this mass fell and with the size of his mass and the volume of the water he figured all of that out the quantity
21:05
of work that must be expended at sea level in the latitude of Greenwich in order to raise the temperature one pound of water weighed in vacuole by one degree fahrenheit from 60 to 61 degrees fahrenheit is equivalent to the mechanical force associated with raising 772.55 pounds
21:22
through one foot he measured it to five sig figs he measured it to one part in ten thousand you know it is inconceivable how difficult it was to do that and when he went around and gave talks in England and elsewhere in Europe nobody believed him all right because he would
21:45
show data he would say you know here's my data all right I'm measuring 17 hundredths of a degree change and they would go there's no way you can do that reproducibly all right nobody else could make temperature measurements with this level of precision but where did he learn how
22:00
to do that in the brewery all right he'd been making precision temperature measurements for years all right you got to do that to run the stupid brewery all right he took it to a new level mind you so that number is right there on his gravestone
22:23
all right he showed the equivalence between work and heat that's a pretty important thing to appreciate and I know what you're thinking yes you too this number is still available to you if you should want it nobody not everybody's going to honk at you if you have this on your
22:46
license plate you know you've got to have it imagine what kind of nerd it has to be to recognize no not too likely okay so that's not the experiment we care about he tried to do a
23:03
harder experiment he's less famous all right this he's really famous for the experiment where the weight falls you get one degree change 700 okay he tried to do another experiment right he tried to do this experiment right here he pressurized one bulb this is these
23:21
are these great these uh tan things here are glass bulbs he pressurized one right with some nitrogen the other one evacuated all right and then he put a thermometer in this water bath and he opened up this valve right and foom all right the gas goes from the bulb
23:41
to this bulb and he's looking for a temperature change here and nothing happened okay so the question is first of all what was he thinking all right he was convinced that if he did this he would see a temperature change
24:01
all right why right this guy knew a lot of physical chemistry all right and he knew that there should be a temperature change if he did this carefully enough if the volume of water here was small enough it wasn't right in his experiment he should see a temperature change so let's see if we can understand what he was thinking what could he
24:27
have been what could he have been thinking all right how many people have seen this before this is a leonard jones 612 potential good where in our earth did you see it steven don't put
24:42
your hand up who put their hand up where did you guys see this g cam okay what is it all right it's it's the energy between two molecules right that are not going to form a covalent bond it's it's
25:03
a non-covalent bonding interaction that we're talking about here okay and so there's an equilibrium bond distance let's say we've got two neon atoms all right if they're far apart this is the distance between them here on this axis here this is the energy right now for an ideal gas there's no energy all right an ideal gas this is
25:26
the energy that you get as a function of distance all right in other words in fact in an ideal gas we assume that the gas is a point particle we don't even assume it has any volume okay so there's no interaction energy as we bring a gas molecules close together nothing happens
25:43
okay but in a real gas there's a long range attraction right that's what this is and then there's a short range repulsion that's what this is and this repulsion is approximately r to the 12th and this attraction is approximately r to the 6th so if i write that this
26:01
red line here is the sum of the repulsive potential plus the attractive potential in other words if i add this dash line to this dash line right here i get this red line and that's what is given by this equation and that's the leonard jones 6-12 potential okay so there's what is
26:20
the what is the leonard jones 6-12 potential tell us well it tells us that there's an interaction energy epsilon and it tells us there's an equilibrium bond distance req right that's req is not in here right but the minimum of this curve is at req the equilibrium bond distance this is a van der waals bond that we're talking about here
26:44
these energies 100 wave numbers how big is 100 wave numbers well pretty how much how much thermal energy is there in wave numbers at room temperature 200 all right so is neon going to be a gas or a liquid at room temperature a gas
27:06
all right because it's only got neon atoms are only held together with 100 wave numbers of binding energy right if there's 200 wave numbers of thermal energy they're going to be out the door right they're going to be out here they're going to be a gas this is a liquid
27:26
you with me now yes for an intimate for an ideal gas the intermolecular potentials everywhere equal to zero that's this dash line right here at high pressures of course you're here in other words if you take the gas and you
27:43
pressurize it as hard as you can you eventually bang up against this repulsive wall here okay and the pressure that you measure for the system is higher than you would expect based on v n and t in other words if you if you measure v the number of moles of gas and the temperature and you calculate what the pressure should be in a real gas if you press
28:05
on it hard enough the pressure is higher than you could ever achieve for an ideal gas at those same volumes number of moles and temperature likewise if you're at sort of normal pressures here all right if you measure this pressure it's actually lower than you would expect
28:24
for this volume this number of moles of gas and this temperature right a real gas has a lower pressure here and a higher pressure here because it's not ideal all right this is its non-ideality expressing itself okay
28:43
now let me recall for you that there's something called the compressibility factor which is just the actual pressure times the molar volume divided by r t right that should be really big p not little p okay so in other words for an ideal gas this would just be one wouldn't it
29:01
all right but for a real gas it's not one it can be higher than one at high pressure and lower than one at moderate pressures okay so if i look at what this compressibility factor is here's that compressibility factor again come on don't fail me now all right at high pressures this is pressure on this axis here right this is the compressibility factor here
29:24
here's one this black line is one okay at high pressures the compressibility factor is greater than one why because these gas atoms are banging into one another all right you can't compress the gas anymore because it's got finite size and in a real in an ideal gas we assume the gas
29:43
particles have no size at all they're just point vertices in space okay at low pressure right the compressibility factor is less than one because gas molecules are exerting an attraction on one another at long distances okay and that
30:06
lowers the pressure right in the absence of those attractions there'd be more pressure on the molecules are attracting one another they are reducing the pressure that those gas molecules
30:23
are applying to the outside of the vessel okay now so these two regions this is in a region where repulsions dominate the gas behavior this is a region where attractions dominate the gas behavior okay got all that now we're going to do a thought experiment
30:49
right this is what I just told you is what Jim Jewell knew intuitively right and probably you knew it too so now he does this thought experiment in his head let's say that we've got
31:05
some gas molecules at a normal pressure now I think you can appreciate that in a gas atoms are moving around okay and there's every possible intermolecular distance because there's collisions occurring okay but imagine the average nearest neighbor distance let's say that let's say
31:35
you could calculate that how do how would you do that take a bunch of snapshots of where all the molecules are all right and then calculate what the nearest neighbor distance is for each
31:45
atom all right you need a supercomputer all right take the average of that the average nearest neighbor distance you with me let's say it's here all right at a particular pressure and now let's make the make the pressure lower all right the only way to do that would be to suck
32:07
some of the gas out of the container or to increase the volume one of the two right okay take the molecules and normal and expand them to a lower pressure that puts us here this would now be the new average nearest neighbor distance okay I think you can see if I make the
32:26
pressure lower on average gas molecules are going to be further apart right okay so if I start here at high pressure and I end here at low pressure this energy is the work required to
32:40
separate these molecules I've got to do that work for every molecule in the container you with me and the question is where does that energy come from because I'm I've got to put this energy into the system because I got to go from here to here right that energy's got to come from
33:04
somewhere where does it come from well to first order it comes from the gas itself right the gas can give up energy if the gas is at finite temperature
33:21
all right and that temperature is characterized by kinetic energy of the gas molecules if the kinetic energy of the gas molecules goes down the velocity of the gas molecules decreases the temperature will go down that was Jim Joule's insight right this is what he understood
33:42
all right if I increase the pressure I should pull molecules apart along this potential here and they should get colder okay now the only question is am I good enough at measuring temperature and designing the experiment so that I can measure that delta T and it turned out he
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wasn't he got delta T equals zero which is the right answer for an ideal gas or for an ideal gas these interactions don't exist so we don't expect there to be any heat flows right you expand an ideal gas an ideal gas is here all right there's no difference between this point
34:27
and this point for an ideal gas in terms of its energy so there's no heat flows all right you with me on that okay so he had to he didn't have to but he met when he was
34:44
giving one of these talks where everybody sort of laughed him out of the room this guy William Thompson was at one of these talks he was a hot shot chemical physicist over in Scotland it's not too far from Manchester all right few days on horse this guy's name
35:08
he was knighted all right he became Lord Kelvin yes that Kelvin all right so together you know he goes and he he goes and talks to this guy after he gives one of these
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James Joules gives one of these talks where he's measuring like 17 parts per hundred temperature change and people are just shaking their heads and Thompson's in the audience and he comes and sees him and says look I think I can help you do this experiment in a way that we can measure this this temperature change I agree I think you're right it's happening
35:40
all right it's happening I think what we're going to do is we're going to make the gas its own thermal bath all right what he was doing before is counting on the transmission of this thermal energy in these two spheres to this water tank all right and water's got an enormous heat capacity okay and so you know you have to put a lot of heat into water
36:04
to get the temperature to change very much okay but if you're clever you can use the gas as its own thermal bath all right so here's what they did high sorry here's a defined
36:23
pressure on this side a defined pressure on this side this is high pressure this is low pressure this is an orifice a tiny needle pinhole between these two chambers I've drawn it big here but it's just a pinhole it's got to be small you'll see why in a second because what we do is we
36:44
transfer this gas at high pressure to this low pressure region and we make sure that these two pressures remain constant during the process okay so this is non-trivial experiment even today you would need feedback control of the pressure to do this properly I
37:00
don't know how they did it in those days all right they must have had some sort of pressure transducer I mean I don't know what that could have been all right this is an insulated container so there's no heat flows outside of the container all the heat has got to be transferred from here to here or vice versa all right that's key to making this experiment work
37:23
okay and so you go you blow this gas through this orifice right it's hissing as this happens right can you imagine that and you coordinate the motions of these two pistons so that these two pressures remain constant and you measure temperature in both of those
37:42
two compartments all right and James Jewell by golly knew how to measure temperature all right so you see how the experiment is different now we're going to what there's a tiny thermal mass here compared to what he was doing before he had a water tank you know good luck all right here this has got a tiny heat capacity by comparison so does this
38:05
all right now if there's a temperature change you're much more likely to be able to measure it so here's how the math works work on the left hand side p1 times delta v we know that minus p1 mind you on the right hand side same thing
38:22
for simplicity to say we push up you know so let's push the piston all the way in then this v is just equal to that difference it's that whole volume there and this v is just that whole volume right there right so that's v1 that's v2 here the final volume is bigger
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than the initial volume here the final volume is smaller than the initial volume all right so the total work is the sum of the work on the left side and the work on the right side all right that's p1 v1 this is positive because that's smaller than that
39:02
all right here it's negative because that's bigger than that okay and so we get minus delta pv for that difference okay since the apparatus is insulated this is all insulation here q equals zero and so u is equal to w and that's minus delta pv
39:23
and rearranging this equation we notice that this is just equal to that if i move that over to the left hand side and they're both equal to q all right so this is an isenthalpic process all right delta h or q either one is equal to zero okay and what they measured is the
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change in temperature at constant pressure for a process that involved no change in enthalpy no heat flows all right that's the joule thompson coefficient
40:02
all right the change in temperature with pressure and what joule measured in his first experiment was zero he didn't see any change in temperature with pressure all right and that's the right answer for an ideal gas okay so later on we're not going to derive this equation now but suffice it to say
40:23
all right later on we will derive it it's not an issue for midterm one and so if you just believe me for the time being that this equation is correct look we can evaluate this derivative right here all right what's the partial derivative of volume with respect to temperature for an ideal gas well it's just nr over p right right even i can do
40:44
that derivative right plug that into dv dt at constant p and that's zero all right if i plug if i substitute for v i just get this whole thing again all right so for an ideal gas the
41:03
joule thompson coefficient should be zero and that's what he measured in his first experiment shouldn't have measured zero but that's what he did measure all right so they were very happy with thompson's strategy for doing this experiment together they got joule thompson coefficients that were not zero right they could see exactly the the physics that they
41:24
hoped they would see right the gas got cooler when it jetted through this orifice all right and they could measure it for different gases this is what the data looks like today when we have all the modern conveniences to make these measurements with high precision
41:42
all right these joule thompson coefficients are positive okay we can look them up there are tables of joule thompson coefficients for different gases correlate in different temperatures the joule thompson coefficient of air for example
42:04
300 degrees kelvin and 25 degrees 8 25 atm is equal to 0.173 kelvin per atm all right it's really known to very high precision if a joule thompson expansion is carried out from a pressure of 50 atm to a pressure of 1 atm estimate the final temperature of the initial
42:23
temperature is 300 degrees kelvin okay so we're starting off at 300 degrees kelvin and 25 atm and we're going to expand into a pressure of 1 atm all right that's a pretty big pressure change factor of 50 all right we should be able to measure something
42:45
uh now the easiest way to do this calculation which is the way that we'll do it right now is just by assuming that this joule thompson coefficient is invariant over this range of temperatures all right that's not true joule thompson coefficient does depend
43:02
weakly on temperature all right but let's just assume for the time being that it does not in this case because we want to get a feeling for how big these temperature changes are all right 50 atm okay so the delta t that we're going to measure is just the joule thompson coefficient times delta
43:20
p because we're just going to linearize this partial derivative okay so that's really easy because then it's just the joule thompson coefficient times delta p by golly and uh that's going to give us directly our delta t and so there's the joule thompson coefficient change in pressure is 50 atm i got minus eight degrees kelvin so i'm going to go from 300 k to 292 k which is a cooling of eight degrees now that's not a lot of cooling is it it's really
43:53
not a lot of cooling and uh you know when you think about the experiment that joule and thompson had to do you know this device that's got the two pistons the reason it works so well is
44:06
because it doesn't have very much thermal mass the gas inside this these two pistons doesn't have a lot of thermal mass and so a small heat change changes the temperature quite a bit but the downside is that if you stick a thermometer in those pistons there isn't a lot
44:23
of there isn't great thermal contact between the gas and the thermometer right the thermometer has got a lot of heat capacity all right to change its temperature it's gonna so somehow they they figured all this out right somehow they had thermometers that had us they were
44:41
they must have been tiny thermometers otherwise there'd be no way they would even see anything right because this is pretty this would be a pretty heroic experiment 50 atm is is enormous pressure isn't it all right and so you know we're only measuring eight degrees kelvin here
45:09
these guys were really very very good experimentalists okay so what am i showing you here this is actually a plot from your chapter 14 and what it shows is the temperature as a function of pressure okay and
45:27
what it shows is the temperature goes up and then it goes back down all right and it's the joule thompson coefficient is the slope of these isenthalps right these are isoenthalpy
45:42
traces okay so the joule thompson coefficient is pa is rather negative here positive here and negative here right so at any particular pressure there are two inversion temperatures the sign of the joule thompson coefficient changes here
46:01
and then it's positive this is the region that we were just talking about this is the so-called normal region for the behavior of the gas and then as we go through this bottom part of this sideways parabola right it inverts again now do i have a is there a nice intuitive way to think about this no not for me and if you have one i would like to know what it is
46:30
because this has never been an intuitive concept to me why does the joule thompson coefficient change signs like this right how can it be positive at high pressure and sorry negative
46:43
at high pressure negative at low pressure but positive in between right the fact that it's positive here that part i can understand that's what that's the explanation that i gave earlier we got this potential we're going to move from here to here we've got to put energy into the system to make that change has to work this way that those physics apply here other physics
47:05
here and here okay so now what we've just learned is how to design a refrigerator it turns
47:22
out and i'm not going to walk you through this because we're out of time and you don't need to know this for the midterm exam but it's in your book actually all right your book walks you through how a lind refrigerator works and maybe we'll talk about it after the midterm exam
47:43
okay