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Lecture 13. The Carnot Cycle.

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before we get started can I ask a question How many
people took the chemistry during June work raises
his hand nobody took the chemistry Jerry How many people are going to graduate school you
guys didn't take chemistry during just the general
during just unjust OK
most of you know by now the midterms are posted in the key is posted you can pick up your examined in your discussion section Stephen NG
marketing Havel's next time I'll put the cover sheet on the exam properly and you'll get it back electronically it takes longer actually it's not the greatest system right now but that's the way we do it but if you have
questions about the grading of your exams don't be afraid to talk to Stephen NG in March the peoples of the 2 guys who did the grading and so fast them 1st if you have a question
or a conflict about way that was the way your exam was greater than come and see me or work it out
Chris forced Friday and in a moment I'm going to show you that these quizzes are very important to
us injury that just said
but let me just tell you I looked very carefully about what's going to be on the Clintons and what's going to be on it is the stuff at the beginning of the chapter in Chapter 15 start of the 1st two-thirds of the chapter which is all the way through page 496 and if I look at the exercises the stuff that you need to be able to do is sort of in this range of exercises right here 15 . 1 he
is already stuff for next week
OK so these 1st 19 problems in the exercises those any 1 of those would be a great quiz
OK now you don't a this is what the
history and look like for mid-term 1 I'm not happy with the new
1 so if you did better than a
49 you actually did pretty well I mean in this class right the means should be a B 4 of course that's not the case when you have
a straight scale these are arranged these are these and these are lower grades over here
but OK so that I don't know why
didn't go better than this 1 1 concern I have is that you get in the groove of
taking the quizzes with the multiple gas in there pretty straightforward problems sort of plug-in challenge problems on the quizzes and let's face it because there pretty short these quizzes and then when you get to the exam and it's harder problems of the workout out it is not used to that maybe I don't let me just say if you think I think this material is easy you're wrong and I'm not even a physical chemist by training on an analytical chemist and even know why got signed a teacher's class and me I had enormous difficulty with this material if you think of all the symbols look the same to you but the same to me to be a U S 2 W I mean I get wrap around my neck constantly right when I took that trip last week to Kansas 0 sittin on the airplane with my calculator working out problems when I got the airplane at about 10 pages of problems and try to try trying remember how to do this stuff all right so you guys realistically no that tender ages and I'm teaching the class I've had this material before mind you I mean it's just been a while all right you guys should have a hundred pages of notes 100 pages of problems worked out in your notebooks at a minimum wouldn't you agree so I'm
just saying that I don't know that you don't have a hundred pages but if you don't by
mid-term to is 3 weeks from Friday it's going to be just like mid-term 1 in terms of length and difficulty I mean I don't think the term 1 should have been too difficult on maybe mid-term to be a little bit less difficult but it's going to be pretty close to mid-term Winans on just tellin ya In these 3 weeks you
need to really buckle down if you haven't done this 100 pages of problems I'm really suggesting that you book
focus on doing the hard problems working them out when I think you realize that open book doesn't really behind you very much on this type of exam I mean if you don't know exactly for not really familiar with the book Museum in Beijing back and through at work and problems with on the book because you can do all these problems than having the book there's a pretty helpful right but if you're opening the book and like Koch and trying to find a problem like this 1 that's been worked out here in the book you just don't have time for that on this contest meantime is really an issue OK so I
know it's materials heart it's really hard for me to and the only
way around that and we do need to really learn this stuff the only
way to learn it is to do a lot of problems that's the only way to learn this stuff I try to make his good lectures but these
lectures are not going to teach you this material you gotta work the problems "quotation mark now fortunately the quizzes or are you saving due to a large extent this is the This is how you doing I calculated this right yesterday morning no quizzes were dropped in the generation of this score I
didn't drop any quizzes right so some of you missed quizzes some of you did poorly on 1 or more quizzes everybody virtually everybody would
move to the right on this diagram if we drop to quizzes or 1 OK so actually the sister histogram a worst-case scenario it would be further to the right if we had dropped quizzes OK so what
this means to us is that the quizzes or are in fact important right they are moving us from the mean
in this range over here to Amin in the lobby range and hopefully by the end of the quarter this meeting is going to move further to the right after we dropped quizzes and we do better on midterm too but that's the
goal OK so that is actually
a little and we're hoping that this moves over to the middle of the range over here somewhere and I think that's a realistic objective for us right but it's
not going to happen on its own I mean turned to could be just as bad as mid-term 1 I hope not right Help me out here OK so
the way I calculated as in case you're wondering as I just ate added 3 quizzes together quizzes are out of 5 points and multiplied by a because each quizzes reality worth 40 points so that would be 120 plus 100 is 220 right at 4 I got this that so this is scale from 0 to 1 writes that the 100 per cent of the cost points if you got 5 and all the quizzes and a 100 on the mid-term the mid-term was actually worth
110 with the extra credit why you'd be over here somewhere
OK Dean Cage 1 and I calculated so you if you want to you can drop your own queers recalculate your whole life doing store and see where you compare on this
diagram right eye understand most folks are going to move to the right of little bit cocaine so we have
this guy Sadie ,comma he's interested in steam engines and at the time that he he's doing his research steam engines are about 5 per cent efficiency the Connell
limit for steam engine turns out you can run a steam engine or the steam that about 300 degrees C. and vote cold reservoir it's not room-temperature exactly the boiling point of water it's 100 degrees C OK so principle you could operate this thing that sort of 65 per cent efficiency that's the no limit for a steam engine or the schemes at 300 degrees C in the cold reservoir is this the boiling point of water right so that's really is quite a
bit lower than Nakano limit so he was asking 2 questions could do better and it would changing From wanted something else could that helps we
haven't really addressed that question it
OK so the question is 1 limits does nature place on the amount of work that we can derive from this temperature gradient right here
right nature doesn't allow us to get an arbitrary amount work under this temperature gradient there are well-defined limits that's what we're talking about that's why this is sort of an important subject so can't
derived this thing called the Cardinals cycle turns
out he didn't really talk about entropy in his writings you never use the word entropy the editor took it and that a 2nd guide to sort through everything that he had done to read formulated in a way that's useful to us in terms
of what we now call the cycle of thermodynamics all right but the wide this thing called the carnal cycle which is the most efficient existing cycle cable to bring in a given month energy in work he did there's another way to
look at this number if you inject a certain model work into the process of transferring heat From a cold reservoir hotter hotter reservoir the temperature difference that you're able to generate using a certain model work that's also it's the same thing
doing so in his cycle we start here on this I suffered and we do an ISO thermal expansion and an Indian batik expansion and isothermal compression and 80 better compression and the reason we care about this this looks like a completely artificial thing To be
focusing so much attention on right the weather we actually
in real life I want go from here to here or from here to here or from here to here but why are we going to care about this cycle will the answer turns out to be that any arbitrary cycle like this thing shown in red here any
arbitrary cycle that you care about can be decomposed into a large number of cycles right
this is a diagram from your book a no cycle right here here's another 1 right next to a right here but I think if you could you look at these 2 Connell cycles you can see that this 1 is going this direction this one's going this direction and so the 2 the
constitute to Colonel cycle that share that hedge
but what's going on in that it is just going to cancel right and this adds cancels between these 2 cardinal cycles in this said canceled between these 2 Connell cycles because what that had represents is the 2
cycles going along the same path in opposite directions right which you without
you cancel all of these path here is you're left with the pat on the outside of this thing that doesn't right this is nothing there's no O'Connell cycle next to it to cancel it out so I think you can see that here there's maybe 30 Connell cycles right and if you made this 3 million you could map
out this red circle pretty accurately with little tiny condo cycle OK so the point is is if we
understand how 1 of these cycles works we can generalize what we learned from 1 cycle to this
infinite number of cycles or a large number of cycles OK so
any cycle process like this is constrained by the no
considerations these Connell efficiencies that we're talking
about this applies to that's why we spend so much time studying it right we
constantly bump into this car no limit In chemistry OK so this is what
we said a lot of black politicians the heat engine efficiency is the work that we get out of it divided by the you he gets absorbed from the hot reservoir or that's what the hell
the efficiency is defined
OK and so in this case and 5 over 20 we said this on Monday and we can derive an expression for this deficiency which is 1 the temperature of the cold weather more divided by the temperature the hot reservoir murdered we get
that equation where can we
derive so this is just going back to the stuff that we did on Monday we can prove that this is right and the way that we do that as we ride out the work for each of the cycle cycle the should not I don't mean cycle I mean past stepped step 1 this I still thermal
expansion in step 2 is
the immediate batik expansion step 3 is the isothermal contraction step for the Vatican track OK and we have equations for each 1 of these terms and it should be obvious to us that these 2 terms right here are gonna cancel because my goodness they're identical except for the ordering of these limits TCU's here TCU's here right so reversals limits and is gonna
put a minus sign in front of 1 of these integral they're going to cancel but
some only luck with these 2 guys right here these volume ratios here are related to 1 another and
that's not obvious but the way that you can tell that they
are is this equation right here which is on Page 471 yearbook this describes the relationship between the volume at 1 point immediate bad V 1 and T 1 volume and temperature and the volume and this
was before as was before so I'm tired
of this .period 1 in this point 4 so that's the for allegedly chief for V 1 and T 1 medical bills of 2 . applied to OK so these because these 2 points are both on the same 88 bat this ratio has to be
satisfied were Damir is CPO receive OK in the same thing is true here tonight and so what sort of this as well this is to actually where officers following practices T 2 that's this temperature as the right here this is the 1 that's the size of the right here and these are the 2 volumes right here so this
actresses of certain Sosa OK
so if if I write these 2 equations and rearrange them I get this equality right here and this equality right here 1
from each 1 of those equations
and then I like divide this guy by this guy the key ones go away the teachers go away
right and I'm left with these volume
ratios and so I can simplify this into their right and once they've done that it's obvious that these 2 logarithmic terms are going to cancel 1 another the same
OK so what while rather than canceling I can add them and so I can express them both in terms of the 2 and the could express them both in terms of the three-year before I wanted to right so this is the expression that I get for for the work fair integrated over the whole
cycle now OK this is that
hot reservoir temperature the cold reservoir temperature in these 2 volumes OK that's the work for the whole cycle that keep its transferred in the hot part of the cycle is given by this equation right here this is just the expression for the Eisele thermal expansion that's occurring
in the 1st step of the cycle
OK and so that efficiency is just that whole thing divided by the whole thing according to this equation right here I just take the total work divide by then he gets transferred the hot part of the cycle biden should get the efficiency and so when I do that lot and is going to cancel when I look at this racial right lobby to over the 1 that's going to cancel rights on and be left with th minus 30 the over th In that simplifies right away that so we can prove this we
know enough to prove this remembering these equations is the hard part but you don't have to remember them
ready to look a lot OK now with the entropy change for each of the 4 steps of reversible Colonel cycle we needed from an anemic expression for excellent saves this will come back to this
later no other words change in the
entropy is the change in the he provided the process is reversible divided by the TV that the processes occurring right now this reverse ability issued turns out to be important that's the reason this is an equal sign Turns out
will come to that later OK so if I
want to know the Delta asked for this whole cycle just added the dollar as the steps the reason I can do that is because Delta
has is a state function case and so I wanted know what is the
whole cycle I can just add up adult as for these individual stats and I will get dealt as for the whole cycle right steps 2 and 3 steps to
inform rather sorry RED back right
step 3 years
reversible isothermal compression rights abuse 0 for
those 2 steps Frank that's why that's 0 but the entropy is 0 for those adiabatic steps because he was 0 all right for the eyes of isothermal steps which are the 1st 1 in the 3rd 1 part these are the centerpiece IQ is not 0 otherwise the BAT batik but in so what can we say about steps 1 and step 3 while we can look up to you all right 1st step 1 right it's going to be equal to according to
equations that we talk about on Monday and last Wednesday a week ago right this is the equation that
applies to me and I still thermal change in volume hopefully that which Miller that was eliminated from the mid-term are and what what's not totally obvious to me is that she was a goal of mine is W Art Howe we know that this is true well we know that DU equals DQ DW right we know you doesn't change during any Eisele thermal process right you doesn't change the words to change EU is equal to the constant volume he capacity that that the constant volume capacity is
defined as the change in you divide by changing as solid the capacity is
defined in so the change in you is just the heat capacity times dealt T right in Italy the formal process nobody is
0 right so we know use
not changing any excuse .period equal W rather DQ is getting people minus the W right refuse not changing so that's why there has to be true
don't take anything for granted when you're looking at this stuff and we try to prove it yourself that stuff
like this as simple as that seems is is true and
constantly having to do that myself OK so what are
we now now we we don't the 1st step 1 the 2 is bigger than the 1 it's an expansion right
we're going down there the bigger
we don't W's minus the DVU that's true always right so we know that's going to be negative because Peter is bigger than the the victims bigger than the 1 and is a minus sign in front of us think so Delta these
positive but there's a minus sign
but then excuse that positive if work is negative Tuesday the positive activity the opposite of work I remember we're talking about Hugh this is the time
to high-temperature because retired
step 1 along the top that I suppose OK and if you look at this guy this this this guy is exactly like this guy except the having out TCU's and it's a compression not an expansion and so on OK so that
we get this equation right here get this equation right here were trying to conclude something about this doping tests for the whole cycle the sum of all 4 steps what we have to remember is that we showed backlight 10 flights that the 2 over the 1 equals the 3 review for and if you make that substitution these 2 terms equal and opposite because check this out to sea it is in the numerator it's going to cancel the cheesy in the denominator same thing with th that cancels with that so these terms if that equals that these 2 terms were to be equal 1 another right while there was no obvious
to me that that had to be truthful and that's why I went through in the slides so that means that this unified
approves since this really is a
square box In other words the entropy is not changing here or here right in the temperature obviously is not changing here in here and so if you make a temperature into the diagram about
colonel cycle it's a square box right now Everything that
we said applies to only reversible prophecies where we can talk
about reversible process so far right in real life processes and virtually never reversible but because to be reversible Teddy infinitely slow so what 1 more
the stuff about surreal reversal of efficiency remembers was not knowing if the process is irreversible it's going to be less efficient than a reversible process or at best equal efficiency quite limited the efficiency of the irreversible process is bounded by if you will the efficiency of the reversible process OK so what does that mean well it means that if I write this efficiency in terms of feets here's the efficiency of the irreversible process here appears the efficiency of the reversible process here if it's reversible nominal rights Red on it to make
sure that we understand and remember that it's reversible OK so these
ones are going to cancel so we've that this guy right here all right and now what you need to remember is that these skew CEC here those guys negative by QC and QC on reversible oddball negative but we showed that back and
slide 36 I have no idea what's like this 45 honey you know
that all outstanding OK so if
I write these cues as negative I've got a flip that equality sign around here is greater than so years less than a year degraded right greater than or equal
to I make these negative about a
foot that equality sign around right and these reversible feats can be equated with the temperature of the cold and hot reservoirs OK and so if I certainly there the goal of this I've got used is greater than or equal to 2 to sign than in other words if I just crossed multiply here I've got the heat transfer from the cold resident was aboard abided by its temperature and heat transfer from the hot reservoir its temperature and is a minus sign in front of this guy because it's negative but at so for
everything that so this applies to an irreversible process that I will both of these guys over the same side I can put a 0 over here and say in general this has to be true that would divide value ladies heat from the cold weather on hot weather work which can be less than or equal to 0 for
at any irreversible process now any process but this applies to reversible or irreversible processes because there's an equal
sign here this is called the classiest inequalities
whose closet but In other words if
I'm in a way Q of achieved for a series of you'll recall this is security that applies for step 1 is security that applies for steps variety that we've
integrated for the whole cycle
right so if I integrate overhaul cycle security that's gotta be less honorable 2 0 this is really the most general statement
of the classiest inequality
but he is the 1st German that we talked about right and what he did as he took on ideas and translated them into what we now call the cycle
of thermodynamics he made entropy and he tried to give it the symbol thank good 1
but not sorry political no
workers were to get best out into Santa that I don't think
so now you can memorialize qualities with the teacher of your own right and you can order this T-shirt online at a Web site that I will show you but lot look at this right here right now I'm sorry that was due on the energy versus constant
sorry that was in classes the so
this is totally wrong and I would ask that with me you are joining in silent protest by not purchasing
24 dollars T-shirts inaccurate
and expenses OK so this is the Claudius equality but the secular will cycle of namics unless are right now is an obvious to you that this comes from well that was not obvious
to me right so but see if we can figure
that out now very here's how we can show that this is right Wright State 1 state too where to go buy some irreversible path between state 1 state to inevitable back on reversible path right irreversible reversible very
simple now according to closets
inequality these 2 steps I can write to terms like this for the these 2 steps on legal forms state 1 state too the DQ over achiever that to back from state to state 1 except it's going to be but the causes the inequalities says hate if there is any reverse area if there are any
irreversible steps
that's not the equal sign goes away I see how that's less than a equals to be equals only applies if everything's reversible to this injustice so this is just going to be less than 0 now because of this guy OK results since the sex there's room to work Our so since this is reversible we article the DS only because it's
reversible right and so on I
can slip these limits here if I put these limits I think you'll agree I gotta make this plus sign a minus sign right and now I can move this guy over the other side I only done that I just try this equation right here right so this is the 2nd laughter millenium acts but it said that what markets so was it's what it makes it says his 3 kinds of processing 1st of all if Diaz is greater than DQ over tea integrated over the whole
cycle right that cycle
can be spontaneous and it's
greater than DQ over tea it's also right if these 2 things are equal on another it's reversible N if U.S. is less than the purity it doesn't happen spontaneously you can
make it happen but doesn't happen spontaneously this is the
elegant ,comma OK if we're talking about an isolated system
and isolated system is 1 where there's no heat flow in in a rout no mass slowly OK and
so if I won understand what these inequalities translate into in an isolated system I just substitute 0 4 kiddo because there's no heat that can go on another system of its isolated location so all of these inequalities just turn into I just substitute 0 for this the I get that OK so the
secular government in Annex allows us to classify a process as 1 occurs spontaneously or 1 that doesn't very important as to us as chemists look what are some
examples here some simple examples and this is the kind of problem that's right at the beginning of the exercises from exercise 1 exercise 19 I think it'd be looking at the reversible phase transition there vaporization sublimation melting all current particular temperature with a defined and felt that by Delta H in this case the phase transition occurs if the phase transition requires reversible 80 Delta as is people to Delta age over tea so for example here we have a plot of land to be on axis and temperature on this axis we are feeding frozen solid
entropy increases until we get to a phase transition where melting occurs and there's a stepwise increase In the entropy at the melting point but what is that Delta asked its that right if we know the Delta age we just take this temperature but we can calculate what that stepwise increase in interviews and now repeating the liquid yet to be continues to go up now we get to the boiling point and by golly this would be a different Delta H that applies the adult age vaporization part we divide by that temperature we get that change in the interview right there OK so the stepwise increases in entropy happen
as we leaving a solid
through this phase transition by melting and boiling but also be reversible heating or cooling of aghast right Diaz DQ reversible over that means that he capacity can be used to calculate the DQ multiply that by by the tear in the great divided by 2 you were going to get DS art we use the constant pressure he capacity that the constant pressure process or constant volume the capacity that the constant volume process that parts pretty easy but at OK so far I evaluate this integral but it's an integral won over to somebody like the final Liberty initial that's for the Delta S is going to be equal to s goes up with the logarithm Of this ratio 2 years of that looks like Delta plotted against that racial right there TF over but this is a logarithmic increases are just different capacity OK it should be goes up great that's why when you look at this diagram right here you were talking about again last right in principle that's what this is right here see
entropy going up for this gasses we heated it right it's happening in accordance With
this equation right here right which would predict a logarithmic increasing the entropy with temperature right so we can
calculate that very easily we've got these good that this equation
right here we can also talk about the expansion compression of the gas by DG is my esteemed dailies PTV a but so the change in the entropy it's a security again the Quran t rather and in this case is the was completely but in so if the gas is ideal why we can recast that in this form right here within just talk about the final initial volume Wikinews that equation to calculate the change of volume so a constant TV is inversely proportional to be writes I can express this in terms of volume on terms of pressure either 1 right so I can calculate the Delta as for that process to that's another example from the beginning of chapter 15 those exercises OK so because as like uses different you can add bit a multiple universes yes so any
random process that you're thinking about you can break up into steps yourself you can decompose any process you wanted steps calculate the Delta as for those steps you will always get the right final Delta asked because it's a state function it doesn't matter what happened take right you will get the
same right Delta as you calculate the Delta a Beechcraft correctly and add them couple the end of change in argon gas at 25 deg C in 180 in a container volume 100 cubic centimeters compressed 5 cubic centimeters scientists and media 45 so we compress the ghastly compress are gone and we needed
OK we don't have any equation for heating and compressing but we have equations for each 1 by itself and so we can just feet and then compressed or compressed and then he at so the
total less for this process is just going to be the Delta as for compressing plus the adult as for heating this is process 1 will call isothermal compression will do that 1st will compress
1st and then wall heat without changing the volume but isothermal compression we expect dealt as to be negative for that based on the equations that idea showed you and Delta best forward as a volumetric heating that's going to be positive member that's a logarithmic increase right and to be goes up log rhythmically with temperature rights on this case we don't know how if this is going to be a spontaneous transformation or not we don't know Delta as is going
be positive or negative because we don't want term
that's positive wanted it's negative we don't know which ones bigger OK so it's useful to actually think that through and just say no if these were both positive about negative say alright dealt S is definitely going to be
an overall positive overall negative OK so process
was compression a constant temperature that this equation right here and all I've done is substituted for the number of malls and to substitute PV over RTE so don't know doesn't tell me how many moles of GOP I can calculate that from the information I've got here right I can just cancel these are both better not all right they don't cancel for forgot sake because I need this to be malls and that is Jules per Calvin per mole right it's different arts this was .period only 206 this 1 is 8 . 3 1 4 5 1 OK and if I do the dimensional analysis by golly that's that's the way I needed to be because I want you to be jewels for Celtic Trempealeau careful a case of applied the numbers in 180 m . 1 liters .period already 206 to 298 . 1 5 8 . 3 1 4 5 1 volumes boom boom boom -minus . 1 0 2 jewels Calvin good we expected it to be minus why do I have any intuition about the size of this number 0 I have
absolutely no clue if that makes sense I mean the fact it's not 10 of the 20 reassures me alone
rights I don't know now we can do step 2 all right that the equation that we need for stepped to only changing the temperature constant volume that's the heat capacity right there right for an ideal gas which we can expect Oregon to be pretty darned ideally I think you'd agree we know that he capacity be for constant volumes 3 are it's a constant volume
the capacity of any ideal gas that's the number of malls
right 3 are over to U.S. units will be yes me to multiply by right so I always think about the units but the units of this the capacity of the 3 are over to her jewels Calvin from also got a multiplied by malls to
get rid of that because I lot the units to be jewels for Calvin OK
so that's a number moles reuse before that's the cost .period the capacity we know British national filed temperatures are yes yes yes there's there's yes temperate Calvin temperatures and this is the dealt as we calculate its a lot smaller right positive we expected it to be positive OK and so then the toll billed as just some that's the what we do 1st he did and they were changed .period right now we can work with for a change from falling from their main source U.S. OK until that this sort of Of those 2 things of minus . 0 9 8 7 jewels for Calvin do I have any clue where that number makes sense now I really don't have any
intuition about these entropy changes all right but I can say
because that's a minus sign but this is not a spontaneous process right that's what the minus sign tells me it's not a spontaneous process and we don't care about the magnitude to but the most important thing is to understand that who can my
goodness it's exhausting his name OK so far study for the quiz on Friday we'll see that
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Metadaten

Formale Metadaten

Titel Lecture 13. The Carnot Cycle.
Serientitel Chemistry 131C: Thermodynamics and Chemical Dynamics
Teil 13
Anzahl der Teile 27
Autor Penner, Reginald
Lizenz CC-Namensnennung - Weitergabe unter gleichen Bedingungen 3.0 Unported:
Sie dürfen das Werk bzw. den Inhalt zu jedem legalen und nicht-kommerziellen Zweck nutzen, verändern und in unveränderter oder veränderter Form vervielfältigen, verbreiten und öffentlich zugänglich machen, sofern Sie den Namen des Autors/Rechteinhabers in der von ihm festgelegten Weise nennen und das Werk bzw. diesen Inhalt auch in veränderter Form nur unter den Bedingungen dieser Lizenz weitergeben.
DOI 10.5446/18946
Herausgeber University of California Irvine (UCI)
Erscheinungsjahr 2012
Sprache Englisch

Inhaltliche Metadaten

Fachgebiet Chemie
Abstract UCI Chem 131C Thermodynamics and Chemical Dynamics (Spring 2012) Lec 13. Thermodynamics and Chemical Dynamics -- The Carnot Cycle -- Instructor: Reginald Penner, Ph.D. Description: In Chemistry 131C, students will study how to calculate macroscopic chemical properties of systems. This course will build on the microscopic understanding (Chemical Physics) to reinforce and expand your understanding of the basic thermo-chemistry concepts from General Chemistry (Physical Chemistry.) We then go on to study how chemical reaction rates are measured and calculated from molecular properties. Topics covered include: Energy, entropy, and the thermodynamic potentials; Chemical equilibrium; and Chemical kinetics. Index of Topics: 0:00:06 Carnot Cycle 0:14:59 Efficiency 0:21:05 S is a State Function 0:26:43 Efficiency of Irreversible Processes 0:30:13 Clausius Inequality 0:33:42 The Second Law of Thermodynamics

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