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.
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
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 )
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.
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)
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.
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!
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.
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.
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)
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.
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.
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
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.
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
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)
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!
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.
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? :)
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?
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
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
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.
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
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?
Is more intuitive. Although exponential relationship between energy and temperature is frequent in physical chemistry (activation energy, free energy, etc)
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
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.
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.
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.
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?
@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.
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.
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
He used the glass of water as a deliberate wrong example and explained why it isn't in equilibrium. In his explanation he stated that the presseure of the water vapor is not equal of that in the water or the ice so there is no equilibrium.
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.
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.
Here my teacher is still teaching me that work done on the system is negative and work done by the system is positive. It's so confusing which one is right and which one is wrong!!😢
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
@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.
He is wrong about the convention to define the sign of work. If the work is done by the system, then it should be considered to be positive, (i.e expansion). On the other hand if the work is done on the system, then it is considered to be negative, (i.e compression).
Jyson AW you are right the prof. is making confusing during the example.....as we know work is positive when it is done by the system. and the converse.
+Freddy ACA no, he's right about it. I would recommend you watch it again, actually he explains it pretty well. Basically, the work will be negative if the work is DONE BY the system; the work will be positive if the work is done ON the system.
This makes me wonder if "convention" is real, because I've ready my textbook "Applied Thermodynamics" by T.D. Eastop and A. McConkey and it states the convention as being the opposite of what you said. That is to say, engines do negative work to themselves according to my textbook and this lecture.
Freddy ACA what you are saying about convention is totally wrong the convention professor wrote is absolutely correct. the expression of work done is "-P×deltaV"
From my intuition, on the right side of the equation we have Q + W. When energy leaves the system as heat, Q is negative. Likewise, when energy leaves the system as work, W should also be negative. Energy introduced into the system (heating, compression) is positive, energy leaving the system (heat loss, expansion of a gas within a piston/cylinder system, etc) should be negative.
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).
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)
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.
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)
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,
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.
Hi sir, @13:50 I was quite confused by the equation. I understand that the slope of the graph is = (dy/dx) = f(t) at triple point/ Temp at triple point but why do you multiply it again with the Temp?
I think he made a mistake. The hot pack would be a closed system, not an isolated one, because if it were isolated then we wouldn't be able to feel it get hot.
