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Lecture 12. Entropy and The Second Law.

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OK are you guys doing the exam go up at the Comedy people
thought the exam was too it is many people thought the exam
was too hard move
so we're going to post scorer's later on today that it is 1 oppose storms Stephen G. that Mark Mac graded all these exams by themselves some very grateful to them for doing that and I don't
honestly have any idea how you did right now all find out later today just like you but don't worry will deal with the outcome of the matter what it is a little bit about OK ,comma unfortunately I did make 1 major mistake which was you know
I'm supposed to put 1 of those exam forms on the front of the exam and I always forget to do that for for some reason on the 1st exam of the quarter quarter after quarter so we're going to have to return these exams to you in discussion this week and so in your discussion section the teams cannot return yards OK and you can look at Opel
was the key later on today as well so you can look at the key compare that with higher exam was created talk to me OK but we do have a quiz Friday quiz for it's going to be about Chapter 15 which is all about the thermodynamic definition of entropy will be talking about
that today and Wednesday OK I'll tell you more about what's going to be on the quiz on Wednesday and now where
are we were right here right with it the on at this for 4 weeks now we had your midterm on Friday were coming to the end of this topic of statistical mechanics and thermodynamics starting Chapter 15 today and we're going to just sort of
zoomed for rose 16 and 17 wouldn't it had like 20 thousand feet we're gonna look at these chapters why we don't have time to devote all the time that we ought to to these 2 chapters because we have to talk about kinetics
reaction dynamics and so we're going to sort of skim these 2 Chapter 16 and 17 right so I believe it or not we are near the end of where we're going to be talking about this topic right here and I'm personal
personally happy about that because I find this topic difficult myself and you think it's difficult I think it's difficult to OK so we're in
a transition the chemical kinetics probably next week OK probably maybe someplace in the middle of next
week and we are pretty much
on track which is unusual for us right by now we usually trashed the schedule and were behind in and were starting to .period about but rather were pretty much hanging in there and pretty happy with where we are OK so just so you have an idea for a year and a need to read 16 and 17 you're going to have to have some knowledge of what's in those chapters but probably not everything
and I'll tell you what's important OK located Chapter
14 Chapter 14 was all about the 1st lot conservation of energy and reduce conserve an isolated system change in the internal energy is equal to 0 for any process OK the central concept is internal energy were constantly talking about that and Chapter 14 and in in connection with the 1st loss White and what we learned from the 1st
lies is to understand whether a transformation is allowed the transformation that doesn't conserve energy is not allowed In other words all right so we can tell immediately whether particular process is even possible or not but that doesn't tell us whether will
actually happen it only tells us it's an allowed process In the 2nd loss the 2nd lies the entropy and isolated system increases in the cost of spontaneous change we find out whether the process that allowed will also
happen all right not all allowed processes will happen all right we need to be able to sort allowed the processes that will not happen once at will
by the central concept here is entropy non-energy and the answers the question is the transformation spontaneous will it happen if it's allowed will it pretty important part question Babel answering chemistry I think you'll agree remember this we talk about entropy weeks ago but we talk about statistical entropy we put Nichols in Kinney Shoe Box 111 or we shut the box and we looked inside made sure we had 1 layer of Nichols OK and we saw this evolution in the distribution of heads and tails as we shook the Berkshire the Berkshire the box and we concluded that what was changing here is not the energy because the energy of this state is just a virtually identical to the energy of the state where the energy doesn't really depend on whether the coins are face upper face down those are energetically
degenerate states for the calling and yet we see this
system evolved consistently in this direction and what we concluded is that this is the direction of increasing W but the system is
optimizing on increasing W. wants a configuration that maximizes
so we concluded Friday isolated assembly we can always predict the direction of spontaneous changes that which W. increases and remember W's not not energy it's something so Boltzmann postulated that this parameters the entropy he defined it according to this equation here this is really a postulate right and this is the statistical definition of entropy is the statistical definition were talking a moment about the thermodynamic definition now we can already take the statistical definition and apply it to the expansion of the gas Lin try and tidy gather statistical entropy information entropy What is the probability of finding gas molecule in the entire volume of a close vessel I've got 1 molecule In a vessels right the probability that it's in the vessel's somewhere this vessel is airtight but the molecule can get out the probabilities but this or is 100 per cent chance that the molecules in the entire Basel I put it in there but it can't be anywhere else now what's the probability that that it's in half of the vessel what we know intuitively the answers 1 half-hearted we get that answer well if a time have the vessel the over 2 no but the 2 molecules the probably the balls or in the prime is a product of these 2 1 molecule probability now the words we can take the probability for molecule 1 multiplied by the probability for molecule to that's the same as the probability for molecule 1 just multiply these together that's the same squaring act OK and sulfur and molecules this is going to be squared anymore it's going to be this ratio to the nth power beyond OK so we're just using the statistical definition of entropy to arrive at this conclusion this is going to tell us something about volumes but if we can apply right here's an experiment about gas in this half of this vessel and guest in this half of this vessels India located the 2 halves of the container now I take away the barrier between the 2 hats what's going to happen we now with 100 per cent probability these 2 gasses are going to mix you don't have to wonder whether that's going to happen or not and we also understand that that's a manifestation of the
entropy of system increasing right that's a direct manifestation of the 2nd law here
they notice I've switched from blue to red to purple and beautiful show this profitable increase entropy we know it does can we show that it does here's the Boltzmann laws a single scalar W the change in the entropy is the final entropy the system minus the initial entropy the system because entropy is a state function
so that the had to be associated with the final volume that Santarpia she with the initial volume and we can just use Bolton's La and plug-in Bolton side of this equation right years we've got the final state final Member States in the initial member states so this is a minor this is a subtraction I can just write this blog as WF over WI but that's the changing it be right there and we have an equation that relates that these number of states with this volume we just showed that W sub and for the prime is given by this equation right here so we can just substitute this guy guy for that In term just like him for that V F and V I play the role of the prime OK
so I'm going to make the primacy
of In the numerator and make the prime VI over the in the denominator part and these 2 these are just going to cancel for us and so we're just going to end up with the above revealed by that I can move that in and out from the logarithm the case of a change in the entropy is going to be Kerry and the number of molecules log the emperor VI the and I've got a term like that for a Delta S for gas a and dealt as against beat because there's 2 gas is involved here right and there's a
VF and of the eye both gasses right there if I talk only gas as a right and I allowed it to you expand into a larger volume I think you'd agree it would do that spontaneously OK so if there was no gas be we could arrive at the same conclusion there
just would be no 2nd term here this would be 1 and a lot too right all the
balls were to be spontaneous processes a expanding into a vacuum if we take the
barrier weight and a and B spontaneously mixing if we take the barrier when this is a positive number we just need to know and the calculated but we can calculate this Delta but it's positive that means this is a spontaneous process yeah going all
the way back to lecture former Art was instead of a change in entropy we wish to calculate the absolute and people model Tommy gas we actually talked about this briefly In lecture for we
derive an equation for this purpose we started out with
this equation right here at Excuse me and we did a bunch of extra steps to market and talk to you about you can see selector for the restless derivation and we derive this thing cul-de-sac attach Rhode equation all right and the thing is this is not a Delta Air snow it's an absolute
entropy usually for calculating a change in the entropy but here for a model atomic gas we've got an equation that tells us the absolute entropy salsa prettyimportant equation that can anchors
or entropy calculation up and so this is just the residual end of this derivation and so we can use this architecture equation to calculate the standard launcher be of something like neon gas this is actually right a lecture well that's the thermal wavelength what is this business of intermediaries yes yes yeah yes show that that was true then we can plug everything in that had that mass has to be in units of kg never forget right the mass of the neon molecule has to be in units of kg and so that Neon has a massive 20 . 1 8 grams from all but that's 20 . 1 8 tends to minus 3 kilograms per mole but OK other goggles number blob loss left the mass of a single neon atoms in units of kg yes yes OK and so on we can calculate the formal way blankets always very very short we expect to get absurdly small number like this will never be surprised by that when you calculate in the formal wavelength and then we can plug everything into the equation in S is equal to 130 jewels for Calvin from all those of our units of Mullah entropy jewels for Calvin promote OK so this is just straight lecture for
Emily putting it in here to remind you that we can calculate the at absolute entropy of a model atomic ideal model atomic gas because that's something you might need to do hypothetically on aqueous OK yeah this is
Sadie Cornell he is the 1st French guy that we've talked about this
quarter then I could say something bad about the French but
I won't right he was interested in steam engines he was a mechanical engineer and he wanted to understand how to make them
more efficient right now as we go through and we look at these people who contributed to the thermodynamics and statistical mechanics they all have practical motivation for what
they were doing right they were just
interested in pure science and they were interested in pure science but they needed to get an answer to make something happen better make better be here if you're jewel make it more efficiently make a better
steam engine if your car now you want to know 2 things was the energy available from a heat engine steam engine is a type of heat engine is an unbounded there some fundamental limits involved from harder-line known my steam engine is operating as efficiently as I could possibly expect it to but
always there no limit to how efficient it could be he was doing
this stuff actually in In the 1822 misses before Jewell had figured out the 1st of all writes a conservation of energy wasn't the of defined concept at this point time he was thinking about entropy before conservation of energy was even worked out all right and the 2nd thing is can a steam engine be made more efficient by changing the working fluid using something besides a wise as the working fluid in the steam engine when was thinking about that
but so I told you Cornell
he's French but Maxwell
Scotland in yeah
Maxwell got jewel you guys know that when England did American yes were in there were in the same or about him Boltzmann Austrian yes the this would be a good quiz
questions OK so there's there's a
statistical definition of entropy that's the Boltzmann equation and there's a thermodynamic that definition which is this maybe will the derived this equation on Wednesday I don't know takes much 6 flights to do that
all right the change in the entropy is the change in the heat for reversible process divided by the temperature were going to talk about what happens when the process is not reversible on Wednesday right but today we're always going to be talking about a reversible process write the change in the entropy is the change in the heat that flows edifying temperature that's what the entropy years that the thermodynamic definition OK so wanted ,comma now do he worked out how
much work you can extract from a temperature gradient right a statement and that is is the temperature gradient you've got steamed that maybe 150 degrees C U Got the ambient temperature which is roughly 20 degrees C you got it huge temperature gradient and you're extracting work from that your you're pulling the train with the temperature gradient
for all practical purposes OK and so he devised single Licano cycle we talk about the colonel cycle because it's extremely important it places an upper limit on the amount of work that we can extract from a temperature gradient alternatively it tells us how much temperature how much he weakened pump with a given amount of work right that is at a very practical problem that we need to be able to solve right Connell cycles the motivation existing cycle cable converting a given amount of thermal energy into work or conversely it's the largest temperature difference we can establish using a particular model work legs so if you've heard of a heat pumps all right
a heat pump is it's if its efficiency is bounded by the efficiency of Okano cycle we have to be able understand now I heeded in
extracts work from a temperature gradient here's a hot temperature this would be like the steam in the steam engine here's a cold temperature this would be like the ambient temperature outside the steam engine all right this delta T is the temperature gradient that we're talking about th minus the OK and some of that can be extracted his work right similar temperature gradient can be extracted his work with the Cardinals cycle tells us is how much how much work is it possible to extract from this temperature gradient so here it is all right here's a pressure volume traced this is an Eisele thurmond T-1 and here's an affirmative to Biden pressure volume space all right we're starting here at this point in cycle OK so this is step 1 step too steps 3 and step I haven't labeled them that way all right but we're starting here so this is step 1 right step 1 is an eyesore thermal compression step 2 is an 80 better compression steps it's an thermal expansion step forwards and Adia batik expansion I still thermal obviously from the eyes of the AT batik obviously it's not on the ice right so that helps you remember it's 80 batik that's a nice of them also OK so that's an isothermal upstart that I say expansion L expansion expansion compression compression I so thoroughly diabetic isothermal lady about OK so the 1st thing we have to know is what this not there's an
infinite number of ah would be able write some conclusions about the Mets General and the same for all of them but I think you can in
principle I could stop this year the duty that expansion here then moved here the duty better compression here and I could concoct any number of these cartels cycles what we know
for sure step 1 is greater than 0 but because I'm doing an expansion of an isothermal expansion work is less than 0 that means he's going to be greater than 0 then is a nice formal compression but workers would be greater than 0 for that that means cues can be less than 0 and I'll show you why that's true in just a 2nd OK queue of course is 0 for these 2 steps to the radio back these are the things that we know for sure OK so how efficient is the heat engine any the danger not just a Okano cycle but the efficiency is defined as the work that performed divided by the heat that's absorbed from the hot reservoirs right and he
engenders a hot pot and cold parts the in your car the hot pot part is inside the cylinder the called part is outside the engine right that's the delta T that matters internal combustion
engine right in the steam engine there's this temperature the stain and there's the temperature outside the steam engine but what we care about is the heat that's transferred from the hot reservoir the inside of the cylinder a stain but this guy OK so there's some numbers here and the units don't matter could be killer jewels OK but we wanted to calculate the efficiency of this particular heat engine right here it's all right I'm transferring 5 units From the Heart residue reservoir or I'm sorry I'm extracting 5 units of work from the hot reservoirs and dividing by 2018 untold number of units that were transferred out of the hot reservoir right so that's the work right there that 20 years the number coming out of the hot the efficiencies 25 per cent right you with
me OK now Okano cycle
well it turns out that Oconnell cycle has efficiency of can by this equation right here where that's the cold temperature the cold reservoir that's the temperature this kind and that's dividend hot reservoir temperature
that reservoir all right the
efficiency depends on what these 2 temperatures are and it can be higher than that so let's see
if we can prove that this is true
let's prove this because that's the central thing that Sadie Connell was able to all right we want calculate the work for each of our 4 cycles OK this is step 1 2 3 and 4 all right this is I still thermal expansion there is a view backed expansion this is societal formal compression this is 80 better compression where would recognize those equations yeah and that's also this is problem to one-year exam now this guy is the opposite of that God by because I look at the limits of the integration that's TCU th situations seems to seize about the negative of battles 2 terms are gonna cancel exactly what we're left with are these 2 guys OK then to further simplify them we have been noticed that these 2 data points here rely on an media back so that means that these temperatures and volumes are related to 1 another through this equation right here were damaged as the ratio of the constant pressure and constant volume he capacity these 2 guys right here these 2 data points are also related through an analogous equation because there are also located on an EU-backed right and so this equation holds true for any too temperature volume data points on any that OK so using these 2 equations icon but and this is just a statement of the 1 on the left this is a statement of the 1 on the right and I can make substitutions in fact I can divide this by this guy and I get this very simple the expression that I can use to substitute now into this equation right here but and substituting added to this to obtain that right the total workers is minus an temperature difference multiplied by lot to overview 1 the transfer fee the 1st compression and so that the the total work for the whole cycle OK in the transfer fee the 1st compression is that that was just the 1st term that we wrote and since we already agreed that the efficiency of the heat engine is just that divided by that wide as it turns out if you make a substitution any canceled terms we get this equation right here but try yourself so I he publishers to maintain the temperature of a billion 18 degrees C when the outside temperature it at minus-5 forfeiture public's work must be expanded to obtain a jewel of the it is a jewel of
a lot of fear a little heat pretty big unit of the critics
unit and here's the efficiency right how much work must be extended to obtain a jewel of the right here's are now expression for the cartel efficiency insult but at the end solving for work Art workers just equaled at UH times this kind right but have been so the fact that a jewel Odsal Q is a jewel of the what 1 one-liners these temperatures are of course converted to that Calvin scale from those 2 guys all right like calculators 0 . 0 7 9 jewels pretty small number of jewel ah but we were agreed
jewels a pretty good pretty big energy unit OK so the work
necessary to pop 1 jewel of heat from minus 5 C to 18 degrees C I think you'll agree that in the thermodynamic Lee uphill to
do that because that's
colder than that right that's going to be 79 Miller jewels 1 to 379 Miller jewels very simple OK what is
the entropy change for each of the 4 steps reversible Carnell cycle there's
another question concerning the McConnell cycle we need a thorough limits
person for S yeah right obviously if we can convert this differential into odd Delta but there we just compare attention to these Delta that the change in entropy during step 1 and step 2 instead of 3 in step 4 some of those is going to be equal to the change in the entropy for the whole cycle that's what we want now all right because as a state function but am so steps 2 and 3 are idiomatic secure a 0 for those and that means S 0 because as the security all right and
so on if Q 0 because that's
an batik step step Tuesday diabetic and stepped forward Xavier about those 2 change in interviews are also going to be 0 you can tell that right away but that's not true for steps 1 3 by steps 1 and 3 are gonna be Q 1 divided by th are we can make these told the Riverdance because this process is occurring entirely at th it's occurring in an isothermal way process story is going to be Q 2 0 over PC because that process occurring entirely on the cold BT that a cold I but and so if we we know what
that pressure volume Connell cycle looks like but if we
rewrite the cartel cycle in terms of temperature into the it's a box but there is no change in temperature but there's a change in enter before process 1 process to workers with no change in entropy but a change in temperature between the hot the cold pretty soon saying so
the carnal cycle represented as a temperature of entropy process is a box tends to be helpful
thing at least for me to remember
helps me remember that the entropy is not changing in process to process for right it's constant because no he flow OK
so steps to worried about execute assume that means there's so let me emphasize 1 thing about so that these 2 guys cancel that I proved that currently the slider navigate proves that part of that guy and that guy are equal and opposite what that New
yes but I don't think I adequately prove that that has to be the same distances I mean another it's the very sorry this has to be the same distance as there weakest only to prove that you I guess
on Wednesday now Everything that we've
said has so far pertaining to a reversible process but real processes are not reversible reversible process happen infinitely slow real
processes happen at some finite ring so we have to address what that means for
us in terms of thermodynamics a real process these verses reversible processes we're going to get to that on Wednesday were the talk 1st of all about this equation right here and then we'll talk about what happens when the process is not reversible OK and I think actually this is all I've got you know get 20
minutes left China somehow I managed to go through these 75 slides faster and I thought I would does anybody have questions yes so
steps to inform you that the staff should 3 stories OK so we'll see
you on Wednesday thank you
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Metadaten

Formale Metadaten

Titel Lecture 12. Entropy and The Second Law.
Serientitel Chemistry 131C: Thermodynamics and Chemical Dynamics
Teil 12
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 10.5446/18945
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 12. Thermodynamics and Chemical Dynamics -- Entropy and The Second Law -- 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 0:04:00 Energy Is Conserved for an Isolated System... 0:06:51 Entropy 0:15:53 Carnot Cycle 0:24:18 Efficiency 0:27:52 Data on an Adiabat 0:33:50 Temperature-Entropy Diagram 0:34:48 S is a State Function

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