Lec 2 | MIT 5.60 Thermodynamics & Kinetics, Spring 2008

[@stoenchu122]

Thermodynamics is one of these subjects where the first person which introudce you the subject is very emportant. If he`s bad you`re gonna hate it. This is what happened with me. My profesor was trash and I hated thermodyanimcs because of him. 1 year later I started loving Thermodynamics because off all the videos with good profesors explaining it easily and understandable

[@moonmoonlight5343]

I can't agree more. My first thermodynamics teacher was TERRIBLE and I hated it because of her. Now I'm trying to learn to love it by watching these lecture and I love how this teacher makes me realize things i never thought about ( like where PV = nRT came from for instance )

[@Mr0Omar]

Did you finish all the videos?

[@crazyfunyt7111]

It's a 13 year old lecture , I don't know whether the professors would be alive or not but what he had done would live forever. It's so fresh like he is lecturing me at this instant.

[@jonahansen]

You must be pretty young.

[@paulcabuguason2546]

He just won the Nobel prize in Physics

[@fizzfizz6144]

Thermodynamics is not an easy class to teach for teachers and not easy to understand for students, but this professor made it easy to understand. He grabbed this class greatly so he can teach it easily and make students easy to understand.

[@enisten]

They just launched a new version of this course (taught by a different prof) on Edx. Check out MITx: 5.601x Chemical Thermodynamics I: Thermodynamics and Statistical Mechanics (course 1 of 2)

[@rodolfogomes6834]

Great lecture, very good teaching approach. The smooth and powerful way he went from showing that work is path dependant to a thermodynamic cycle was incredible.

[@varons0x4]

Many Thanks to MIT initiative for making these videos available for free also to the clear communication of it`s teachers. Kind Regards, from Brazil.

[@Pro-dq9ey]

Thanks to KZbin & MIT, it's become possible to enjoy such knowledge which can't be reached in China mainland!!!!!!!!!

[@amreshverma5747]

I just envy those guys, sitting there and listening to this remarkable man.

[@amitkala1000]

Amresh Verma

[@ToastingInEpicBread]

... like we are doing right now? Except that we can rewind and rewatch any part of lecture as much as we need to

[@salvo7842]

Amresh Verma Me too...

[@salvo7842]

Me too

[@VKHSD]

you should see my physics teacher he is pretty cool

[@parkerflop]

This guy is really good at explaining

[@Ash99Ob]

بالزبط :)

[@Ash99Ob]

من تونس

[@muhammadsaad4317]

No he's basically shit..

[@venkatadurvasula6379]

A little respect would go a long way

[@rasmusturkka480]

@@venkatadurvasula6379 he doesn't wanna go a long way, short will do

[@michalbotor]

sir your graceful explanation of the concept of the absolute temperature scale, ideal gas law, and the van der waals equation is beyond excellence.

[@80gsalgado]

MIT teachers make everything seem so easy. It is perfectly taught.

[@pancakes465]

These are such great lectures. I wish my professor taught like this.

[@kmosheng]

Wonderful lectures. Thank you very much for uploading these for the public to use! I love this professor's articulation and pace.

[@Abdulrahman-lt9iu]

I'm watching those videos again even after finishing thermo courses! Incredible professor

[@brycepatties]

Let me see if I can help. Remember what the definition of slope is, change in y over change in x. In this case, y is f(T)=pV and x is T. If you didn't multiply the slope by T, you would simply have a horizontal line. And, if you didn't multiply by T, the units wouldn't match up. Take a closer look at the derivation and follow what the units would be. That should help.

[@ashash23]

Now this legend is nobel laurette for 2023 ❤🎉🎉🎉🎉🎉 Kudos prof 👌🙏🙏🙏🙏

[@antoniosmith343]

Completing these engineering courses from MIT, Yale, and Cal state univ. system has taught me that I can compete with the greatest minds in the world. Very impactful to anyone who graces these lectures.

[@enisten]

They just launched a new version of this course (taught by a different prof) on Edx. Check out MITx: 5.601x Chemical Thermodynamics I: Thermodynamics and Statistical Mechanics (course 1 of 2)

[@clairer9551]

Fabulous resource allowing me to prepare for university during lock-down, Thank you for making these available to all.

[@herojha]

He is an awesome professor. I have been trying to understand the origin of Van der waals' gas equation for 2 years now and it finally makes sense to me!

[@tysmful]

WONDERFUL CLASSES! IF I HAD STUDIED AT MIT AND IN FACT THE MAJORITY OF ITS TEACHERS ARE SO GOOD AS THIS ONE, SURE, I SHOULD BECOME A GOOD PHYSISCIST TOO.

[@mourerj]

KrzysztofLorek-- He's right -- the triple point is defined as the point where (l), (s) and (g) coexist in dynamic equilibrium. It's invariant, only occuring at a definite T and P (273.16 K and 6.11 mbar). It's beyond our control, thus it's a good definition for T. Imagine a phase diagram; the triple point marks the lowest pressure allowing a liquid phase -- if the slope of the (s) to (l) boundary is > 0 it marks the lowest temp that allows liquid to exist, T(critical) will be the highest.

[@david020506]

a) the maximum pressure in the flask can be estimated by means of an state equation using T=100 celcius degrees and 300 ml using the right system of units. b) n = PV/RT using pressure calculated in a. c)The final pressure must be 1.00 atm

[@ahndhcn24ijokdnh6c]

Lecture is really helpful for the engineer too. I've forgot some important basic concept. This lecture remind such things.

[@engmohamedabohassan1736]

From Egypt... I salute MIT........ Good work 😍😍

[@enisten]

They just launched a new version of this course (taught by a different prof) on Edx. Check out MITx: 5.601x Chemical Thermodynamics I: Thermodynamics and Statistical Mechanics (course 1 of 2)

[@wadesammohamed]

@@enisten so

[@niladrichakraborty9915]

Absolutely fantastic & crystal clear lecture by the Professor 👌🙏

[@dorothydorothy111]

i love u!!! you are the best teacher in chem i have ever seen! u make it interesting and talk about the history of how things came to be. Thanks a LOT LOT LOT

[@hugoalvarez2721]

What a great professor. Incredible class! His students are privileged enough

[@juanrada66]

Gracias al MIT por estas clases saludos desde Venezuela! thank Prof.

[@david020506]

well the confusion with signs was solved by the IUPAC: they recommend to consider positive heat and work going into the system. So the first law for a closed system is basically Q+W= energy change of the system.

[@FWBarnes]

One of the best teachers I've ever seen!

[@AydaBouaziz-lj8wp]

Summary Gas ideal law n. R.T = p.V equation of state a) Z compressibility factor Vbar real /V bar ideal .. Z.R.T=p.vbar b) add virial expansion by Taylor series B(T) & C(T) so : p.vbar/R.T = Z=1+ B(T)/vbar + C(T)/vbar +... c) van der waals : (p+a/vbar).(vbar-b)= R.T b volume/mole a attraction First law many types of Work take expansion W = -F.l and F = pext.A so W = -pext.◇V (volume changes).. đw=-pext.dV the bar that we care about the path.. W=integral đw 0( the changes of w by cooling and heating) Wtotal =w2-w1

[@saii221]

29:04 (just a time stamp for myself)

[@alexandercampbell7917]

Why is a linear interpolation and not an exponential one used for temperature?

[@hassaanbaiq3316]

The cause to linear interpolation is to generate a reference point which doesn't care about the environment

[@luisdmarinborgos9497]

Is more intuitive. Although exponential relationship between energy and temperature is frequent in physical chemistry (activation energy, free energy, etc)

[@shutter.snitch]

I simply fell in love with this professor !!

[@ameerzaaba7264]

this lecturer is good, love those explanation about work n pressure..if i was taught this way probably i might have scored my thermo exam with higher marks..in b4, thank alot MIT!

[@BigBen866]

This guy is a Beast!!! I dreamed of being an Engineer with this kind of knowledge and intelligence when I was a kid🤔

[@emyyunos2501]

you are very good. im going to cry. thank you. pray for my next week, ive exam on thermodynamics

[@TheBhawnayadav]

it was no doubts very good or say excellent and the way he explain things are really nice ( by diagrams )...... I hope I'll learn more as i go more into it. Thanks MIT

[@chip5893]

Thanks MIT.

[@DaveJohnsonsuvam]

better to stay at home watch these lectures ,with a cup of coffee maybe , relax ,than go to college !!!

[@TheBhawnayadav]

It was really good especially their way of explanation by diagrams and graphs........ I liked that and hope I'll get many more things to learn as I go in...... Thanks MIT

[@benYaakov]

At 26:11 , if we have to see probability of another molecule present nearby in the volume, then why V^2 is used? Why only 'V' can't be used?

[@MrParaguay12]

This is so awesome, Chemical Kinetics is rad!

[@richardfush8881]

the guy should get Nobel prize for the most confusing ever presentation of Boyle's law :)

[@mohammedwissam2685]

He just did :)

[@Stejarulpufos70]

The first 15 min of this video are probaly the most confusing 15 min of anything on internet right now.

[@eubiw3666]

haha speak for yourself 😅

[@xue0chao]

he defined the temperature and the pressure that will keep water in equilibrium between the three phases as the triple point. that's his definition.

[@madass888]

I like this way of teaching. More and more teachers at my university are using powerpoint, wich sucks, because it feels more like a presentation than teaching.

[@Pdpfire]

add/ remove heat from system at state 1(p1,v1), according to values to reach at state 2(p2,v2).... physically means put a hot or cold body near system to add/remove heat from system! hope u get d thing! :D

[@PaulMagnusSrensenClark]

At 46:00 mind blown. I saw so many of these squares without realising this.

[@vasanth844]

These lectures must be seen by the Indian liberal arts college professors to remind them how teaching is actually done , not by reciting from a book like a story teller. And expect the same in the exams.

[@jashmodi]

Very true. And it’s not like this even in the best of the IITs

[@pragalbhawasthi1618]

I totally agree.

[@Xplosiv3s]

جزاك اللهُ خير

[@ayushsharma9008]

Question ❓🙋🙋 _______________________ At 32:50 When the system is in equilibrium and has constant volume... Then we pull the piston... We work on the system..then we get Expansion of the gas.. Then is the work positive or negative?

[@RobinRobinRobinOrro]

Work would still be positive ! If you pull on the piston, the pressure would get "negative". So we get : W =-pdV =-(-p)dV =pdV

[@ayushsharma9008]

@@prachiargulewar2409 1 ... Means

[@ayushsharma9008]

@@RobinRobinRobinOrro thanks

[@oscarcardenas9272]

Hi, pressure can't be negative, so, the work is W=-p*dV, if you pull piston the work isn't done by gas, if pressure of internal gas is greater than Pext, the piston moves up and work is an expansión work with negative sign.

[@JosefinaMoraesArraut]

I don't understand The following: if The linear interpolation between The two reference points used to define The thermometer scale is an arbitrary choice, how can it lead to something fundamental, The concept of absolute zero?

[@sangkookyun578]

Thank you for MIT lecture, butI would like to point out the sign of work for compression and expansion.If we get work from system to surroundings like turbine expansion work of power plant, the sign of work should be (+).For compressionto system, the value of work (-).But, he explained the reverse.

[@Ottmar555]

That is the older convention. Modern one preferred by the IUPAC is the "Customs convention" If enery enters the system, it's positive, if energy comes out of the system, it's negative.

[@kshitijsingh3323]

The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.

[@AhmedMohamed-ei6tc]

Thank you for this very wonderful course content ❤.

[@mattheoswho1010]

I didn't get what was said at 19:45. "The corrections are going to be more important the bigger the volume. The lower the pressure, the further we are from the ideal gas regime." While earlier it was mentioned that this is the condition for an ideal gas.

[@omarashraf3194]

VDW equation of state isn't the ideal gas law its the deviation from it , that's why we are adding correction factors corresponding to the volume (b) and to the interactions between molecules (a). and also the Z has the same target there exist other equations of state such as peng robinson , nrtl .... all these equations to get more accurate results ( to those experimental ) to avoid the catastrophe he mentioned of using the ideal gas law in designing a chemical plant or an engine . best.

[@OpsOfCod]

Im using 'introduction to chemical engineering thermodynamics (7th edition)' by j.m smith and boy it is a hard read. Cant follow or make out some of their mathematical expressions. This really helped me break the ice on the virial equation of state matter. Thank for the video, i hope you guys also post a video on cubic equation of state.

[@keropoklekor007]

@qian225 yes, you are right. my prof also thought that.. work done on the system is negative sign. and work done by the system is positive.. but in this situation its relate with volume which is negative.. thus work done on the system= - FL = -PA(-V) thus turns out = work is positive.

[@hingidkorats3556]

Excellent camera person and lecturer

[@ThisOneGoes211]

It goes to show that MIT is really a prestigious school!

[@AA-ef2nw]

14:09 How does the fact that we're taking the limit of p going to zero make ideal gas it true for any gas? What's the relation between the two concepts?

[@oscarcardenas9272]

It's a result from experiments, each gas have a constant for each temperature that is equal to PV when P tenders to 0.

[@burtub]

At 45:45 he says total work is negative. Doesn't it mean that the work is done ON the system, and not by the system?

[@timuroner387]

No ,he defined positive work as work done on system so negative work would mean that the system is doing work on surrondings.

[@michalbotor]

regarding the currently accepted sign convension of the work. for me it all makes sense, because in thermodynamics we're looking at everything from the perspective of the studied system. if we, the surrounding, do work on the system we are transfering energy into the system consequently increasing its internal energy, on the other hand if it's the system, that does the work on us, the surrounding, it has to do so at the cost of its own internal energy, it has to lose it, consequently decreasing its internal energy. capiche? :)

[@mralanduriez]

I'm doing Thermodynamics on M.I.T from RJ, Brazil. Congratiolations and thank u all

[@pvr1963u2]

Now I can see what I missed in my college! Great gesture - sharing such videos!

[@jayjaytheclever7124]

It's not a lecture. It is a concert, a thermodynamics concert by Keith Nelson. I wish I had a professor like him.

[@petarbul]

When choosing two points as reference and saying one is zero Kelvin and the other is triple point, how is the absolute zero defined and why do we choose the triple point over any other when we still have to determine the pressure and temperature for it?

[@srpenguinbr]

18:40 wait. if you substitute the definition of z: (I'll call v bar real just V and v bar ideal just v) pV=(V/v)RT p=RT/v pv=RT It goes back to the ideal gas law, so...?

[@vatan4504]

Z is a ratio, in ratios If 6:3 = 2:1, that doesn't mean that 6=2 and 3=1, it is property of ratios, you just can't do that 😅

[@abhinandanmehra7765]

Anyone kindly explain me how at 4:07 sir told that,this expression is a property and it changes with temperature.

[@roger72715]

The expression is a property becoz the terms in it Pressure and Molar Volume are properties of a pure substance and so their product must be a property. Boyle found out through experiments that this property is purely a function of temperature 🌡. Its his finding.

[@dragoonxd]

I believe that since the system is returning to equilibrium, the pressure of the outside must equal to the pressure inside

[@musedpony8939]

since PV = U [N*m = J], and as P-->0 we can practically imagine the minimal pressure being that produced by a single gas 'particle' (actually a mole of gas in this case?), is it valid to consider limp->0(PV/n) as the definition of generic energy density, or U/n? And therefore we'd expect this energy density to be a function of temperature, with higher temperature corresponding to gas of higher kinetic energy, though of course it wouldn't be a function of the particular gas because the definition is for a generic particle and based on very basic assumptions from some laws (Avogadro, Boyle, Charles).

[@enisten]

They just launched a new version of this course (taught by a different prof) on Edx. Check out MITx: 5.601x Chemical Thermodynamics I: Thermodynamics and Statistical Mechanics (course 1 of 2)

[@oscarcardenas9272]

PV is a work to substance fluids, it's a one type of energy of many, and U is the internal energy of the study substance.

[@yasinthabimal5339]

Sir your lecture is gold.. (y)

[@keropoklekor007]

@2kotok work done on a system is negative, bcoz in thermodynamics. System is needed to do the work. thus if system do the work its should be Positive. if work is do on the system then its should be negative.

[@smartmanjanu]

this lecture clears your every confusion .. like it!!!

[@zerobeginning]

awesome lecture, interesting, in dept, and it does explain quite a bit.

[@pisnicsergey5861]

Thank you for this great course!

[@JTGaffley]

Thank you so much MiT!!!

[@riturajmishra8948]

I have a question. How is it possible to choose quadratic interpolation while PV varies linearly with T?

[@shayannaveed8775]

I guess the linear variation came because they chose linear interpolation. Not the opposite. So if we choose quadratic interpolation, the relationship would be quadratic

[@shaunfedrick4602]

31:35 said every person ever. He double checked and everything and still had to change signs lol.

[@shazirahmani]

Is modified rankine cycle lectures available by this guy?

[@tyjhit]

39:19 min confused me about the definition of reversible processes. Anyone can explain?

[@StashGameplays]

you can return to your original state without applying an external energy (no energy in the system gained or loss = equilibrium)

[@sherry8444]

Awesome attempt at chad stride in the thumbnail 👍

[@enisten]

They just launched a new version of this course (taught by a different prof) on Edx. Check out MITx: 5.601x Chemical Thermodynamics I: Thermodynamics and Statistical Mechanics (course 1 of 2)

[@looktonature]

Unbelievable clarity! I pay heaps for an education where to the quality of the profs is shit. This is a great act by MIT and an amazing prof.

[@GustavoMerchan79]

Greate lecture! two questions: 1) how would it physically be a a path from (p1, V1) to (p2, V2) with no work (w=0)? 2) is pressure measured as absolute presure, or is it relative to the pressure of the environment? Thanks,

[@markhurley0405]

Nobody replied to you in 7yr buttttt. Transitions through p1,v1 have to be done with no work else energy is leaving or entering the system. If this is true then p1v1 =/= p2v2 as there is a change in energy and itll come as a p or v change Pressure is measured as both! There are diff pressure measurements, you have to check what is your reference points.

[@blindsk8r490]

It's all about that: e/c^2 sqrt(-1) PV/nR

[@Dr.Sortospino]

27:05 How Italians teach Thermo

[@joshkooo]

Can someone please explain me why that work total is negative i think if the volume decrease work must be positive because this is the work done by the surroundings

[@pancakes465]

Coşku Moler By convention, negative work is defined as work done by the system. That means that the system loses energy to the external environment. Similarly, positive work is defined as the work done on the system. That means that the system gains energy. To summarize: Work>0: compression of system (the gas) Work

[@bombasteininfinity]

Thanks a lot MIT team, God bless you.

[@UsmanAli-mn1mn]

why molecule of real gas attracted by other molecules but not repelled by other molecules when it's going to strike the wall

[@TheTspahiu]

These are very helpful lectures

[@yankumar5280]

thanks for sharing MIT OpenCourseWare

[@altbeb]

I do not know if anyone is motoring this, it is 13 years old, but the ideal gas law, does it still hold for temperatures close to abs zero, at these temperature, a gas would no longer act as a gas, l understand it is a limit - but

[@kokfahchong1867]

Absolute temperature "function" is a wrong concept since ideal gas or real gas where its volume will never attain to zero at very low temperature because it will freeze to a lump of solid eventually when the temperature is below 50 Kelvin. Likewise, ideal gas or real gas at very low temperature such as below 50 Kelvin where its pressure will never attain to zero because once it will freeze to a lump of solid where the area the solid is sitting on will exert pressure on it; therefore the pressure of ideal gas or real gas will never attain to zero pressure or volume even at very low temperature. Since absolute temperature "function" is not a linear function at all then all "work" concepts that based on it will be wrong. A machine can do positive work because there is a constant supply of fuel that cause dissipation of dynamic photons through combustion to heat up the gas within a cylinder to expand its volume to drive a piston. A particular design of a machine not just any design is important so that it can harness the work done by combusted fuel gas within the cylinder to do positive work. Clearly what it shows is that we should focus our attention on those photons because they are the ones that force to do the work by the system. Work done by the system is dependent on its paths because its movement is driven by dissipated heat through the combusted fuel. Thermodynamic should focus on photons rather than relying on graphs and some equations to explain how work is done by the system to do useful positive work. If the graph is right, when the process goes in a full cycle or circle back to the same point where it started it should have done zero work by the system, but this is not true since the dissipated heat or photons have been released during the process by combusted fuel therefore the work done by the system is no longer zero. mx+c=y is the only known function. Therefore it is pointless to explain research findings in thermodynamic by using equations. It is paramount to pay close attention to see how dissipated dynamic photons from burning fuel to do work by the system to produce useful work for us. Thermodynamic should focus the attention on photons. If you are interested in real discoveries, I would recommend you to read my book, the Unification Theory - Volume One and you will be amazed with lots of new, interesting discoveries. In God I trust.

[@smashadams9811]

Agreed. This was a helpful lecture, thank you!

[@praveensanap]

@qpham26 yah! That's the problem with a lot of people they should make subtitles for that

[@daveweisz248]

Also, I have a new theory. Black holes are negative Kelvin. On the other side is another universe, everything bounces back and forth between this until everything in the universe reaches a singularity in one black hole and can't exist which is the big boom. Over and over. I'm not physicist but can someone prove me wrong?

[@thomasmckenzie11]

He references the limit as pressure goes to zero of pressure times molar volume as a constant. How is it possible for the limit of a product of two numbers to not go to zero when one of the multiplicands goes to zero?

[@murtadhamaythamalbahadily7973]

excuse me can some one help me , when he derive this formula w = p delta V to 👉 dw = p dv , why he consider p as a constant and didn't derive it like this dw = p dv + v dp ????

[@oscarcardenas9272]

Hi, work by definition is all can be done by energy, and it's an area of pressure times distance (int [ F*dl ]), the volume work is similar (int [P*dV]). you can't derive it how you say. If P is constant you cant integrate as constant and get the first result in the video.

[@xkidofemericax]

Thank God for MIT...

[@sphtut]

MIT forever!