For anyone having problems understanding the two equations: f(dw+dq) = 0 and dU=dw+dq, the clue is in 5:37. If you heat the cylinder you are adding heat dq (in this case positive quantity due heat added to the system) and then the cylinder does work on its environment dw (in this case negative quantity due to work done by the system on the environment). This path determines the path dependent values of dw and dq. Sum of the integrals (f(dw+dq)=0) is therefore going to be 0 as negative positive balance as everything should be in the universe. But in real world you are seeing that for a certain amout of heat input you are getting less work output. That heat is stored as internal energy U and the temperature inside the cylinder rises but not as much as you expected. The internal energy changes from U1 to U2. This internal energy depends upon the balance of dw and dq (input/output relation of heat/work). So, change in U is dU=dw+dq. Thus U is path independent and it depends upon the balance of dw and dq. These both equations are same thing: one is for modelling the law and the other is for fitting the same equation with accounting for internal change.
I am 65 and retired and really enjoy these lectures at my age this is clearly not the first time I have seen thermodynamics and I think some familiar concepts ( dU, q and w) are really well done bearing in mind that the reality ( increasing the efficiency of steam engines) preceded the theory [ unlike atomic physics for example] The link between work and heat - are the cannon boring experiments - by Benjamin Thompson, Count Rumford interesting guy - born in England and died in America - Count of the Holy Roman Empire.- his personal paper turned up in the midwest a few years ago. Joule - anyone coming up to Manchester - Joule House Salford, near the Lowry Gallery his private House is rather well hidden on Bury New Road
adilson cantamissa Junior bro..I'm also a chemical engineering student from RGIPT,India.i am little bit confused about my branch.can you help me in sorting my problem by just answering few questions?
Thank you thank you . RU-vid has become quite a powerful tool in ones pursuit for specific knowledge. I appreciate M.I.T ( and a few others ) for not using a pursuit of pertinent information as a business transaction . This is truly a foundation of an intellectual revolution.
This lecturer interpreted concept well, however, if some example question were provided, it will help us have better understandings. Specially, heat transfer is applied widely in both mathematics, physics and chemistry area, which is extremely useful for engineers. If more examples in this lecture, not only students will have better understanding on this concept, but also engineers will take benefits from this to real life application.
+Xiao ENGL314 I agree with you and feel the same way. Unfortunately as in all things there are limited resources. The one I am referring to is time. This lecture series covers concepts. If they had too many problems built into the videos it would be hundreds of hours longer to get though the content. If your serious about really learning the Materials I recommend looking on MIT website for problem sets and possible investing in a Good text book on the subject. Follow the text and use these lectures when your stuck or just as a supplementary information.
I think he is wrong at 36:35 In Joule's free expansion experiment, the gas goes through an irreversible process where it expands in an adiabatic system with no external pressure. That means the heat is zero as well as the work done by the system to the surroundings. The internal energy only changes when heat is transferred or work is done (dU=dq+dw) and it is only a function of temperature for ideal gases. That makes sense with the fact that the temperature of the gas doesn't change in the free expansion (internal pressure equal to zero, considering ideal gases). The variation of the internal energy (dU) of the ideal gas in a free expansion IS zero. (If the gas is not ideal (real), for an isothermal process dU is equal to the integral of the internal pressure for an interval of volumes). It is true that dU is a function of state (independent of the path) and an exact differential. In an adiabatic process the heat is zero and so dU=dw, being both exact differentials and therefore both functions of state (an exact differential cannot be equal to only one inexact differential). In a reversible or irreversible adiabatic expansion, the change of the internal energy is equal to the reversible work, except for free expansions (when dU is zero).
'In a reversible or irreversible adiabatic expansion, the change of the internal energy is equal to the reversible work, except for free expansions (when dU is zero).' 1) So, say if I do an irreversible adiabatic expansion, to find delta U I'd need to envisage a rev path then find w for that path? I thought First Law always works regardless rev or not? 2) If two points can be connected by an irrev path, they must be able to be connected by a rev path? Me myself think that dU = dq, and just find w (of the actual path, not necessarily rev) for delta U. The rev one is for dG and dA, not necessary for dU and dH.
@@thelearner3962 Im with you. I think that change in U can be also for irreversible work. We dont have to find a reversible path for calculating change in U, thats entropy
At 18:50 the statement that there is no pressure is only true if the weight of the piston is considered to be zero (otherwise, there is pressure = weight of the piston / area of the piston).
Based on the principle of conservation of energy. The object at a certain height has the potential to do work due to gravity. SO in free-fall the potential energy decreases as height decreases and kinetic energy increases by that amount(assuming a vaccum where no energy is lost to air friction). At the bottom the kinetic energy will be max and the potential energy will be zero. So KE at the bottom= PE at the top. This is the principle of conservation of energy.
Even if there is no external pressure,some work has to be done because piston moves upward due to the force from gases. So some energy is wasted to move the piston,so some energy is converted into work.
36:35 He may have been totally confused at the moment. When calculating the entropy dS=dq(rev)/T, which will appear after a few lectures, this reversibility is important.
Better explanation, why work is equal to 0 when gas expands to vacuum is the force to the piston from the gas is equal to 0 because there is no resistance from surroundings.
The string is only so long and when it reaches its peak point it has a certain amount of KE which is equal to the amount of PE height at the point where the string length has reached its max. The friction of the fluid is not the one stopping it. The counter force that the string exerts on the weight is what stops it.Although the weight still has PE at that point KE is at its max just before the string counter force takes over and at that point the only accumulated KE is PE h it fell.
@31:35 - he says that max work out required max heat in => that doesn't make sense. here is the wikipedia article: en.wikipedia.org/wiki/Reversible_process_(thermodynamics) The second paragraph in this link states that "In an ideal thermodynamically reversible process, the energy from work performed by or on the system would be maximized, and that from heat would be minimized; heat cannot fully be converted to work and will always be lost to some degree (to the surroundings). The phenomenon of maximized work and minimized heat can be visualized on a pressure-volume curve, as the area beneath the equilibrium curve, representing work done. In order to maximize work, one must follow the equilibrium curve closely." So does this mean that the professor stated something incorrect? Or the wikipedia article is incorrect?
when your lips make a smaller hole the air tends to flow faster and thus cool down, but if you blow air at the same velocity as when your lips are open wide the emp should be the same
@markomiz You are correct. I concede. I misinterpreted the diagram. I assumed the wheel was actually a pressurized turbine, the difference of pressure between which and the external would have contributed to the resistant forces.
i could just pick one doesn't matter because the divergence u2-u1 is the same, irrespective of the integration constant, anyway my Professor picked this: u=CvT for Cv=const, u0=0 J/kg for T0=0K
@markomiz As the weight approaches the lower limit of its movement, it's kinetic energy approaches 0, and its potential energy approaches its maximum. Therefore, in the ideal-model-world of physics, kinetic energy is negligible.
@DerSchreiFalke the floor point is irrelevant. But if it will make you happy replace "floor" with "centre of gravity of the body pulling the weight down" For the kinetic energy to reach 0 the resistance to motion by the fluid on the paddle must be greater than the gravitational force on the weight - impossible, think of a simpler case; pushing a box on a floor with set friction - however you push the box on the floor, the frictional force can never exceed the the pushing force. think about it
Shouldn't the relation at 4:29 have a work term as well? And why isn't work included in the equation at 6:42? There is clearly expansion work happening
Why in free expansion which is an adiabatic irreversible process. dU is not 0? Work done is zero and dq is 0 so dU should be 0 why the work done has to be calculated for reverible adiabetic process?
For those who watch these videos regularly. Can you please answer me? Are you a uni student? if yes, isn't your prof&the textbook enough to understand the concepts? or just these lectures are great. Also, what is the university that you study in? if you are not a student. please write why you are watching the lectures pleeeease. Thanks.
yeah, Im a studen. I watch these videos because the lectures are so damn fine!! and i wanna brush up my knowledges. I didnt like the lectures in my university.
I'm not a student, but i see the lecture because are great, in fact i don't speak English very well,but I can't find lectures in my language than be half of this jewel
@peritech21 Well, he quite clearly showed a while before, that dUv is is dqv, for constant volume, so if you just remembered that, it should be all clear, I guess.
At 36:30 he explained an adiabatic expansion which expanded irreversibly. He said that - you remove the stops and the piston slams against the other stops and did no work. But he argued that change in internal energy is not zero. But these are the exact conditions of joule's experiment. In joule's experiment there was no change in temperature of the gas after the expansion(considering nearly ideal gas) so the change in internal energy is zero. Also why only reversible work can give change in internal energy and not irreversible work. Can anyone comment on this.
I've just stopped at 38:00 to scroll through the comment section looking for an answer to exact same concern. Woul prefer the lecturer to spend much more time for talking about what is the internal energy and what are reversible vs nonreversible processes with technical examples. It was tought the same way on my univeristy but the devil is in the details.
Hello, suppose That a gas expanded adiabatically and irreversibly from volume V1 to volume V2. Now to do that, that gas can have an infinite number of non - reversible processes. Hence, the work done by irreversible process is path dependent. Since, ∆U is equal to work done, hence ∆U should also be equal to work done as Q is zero. Hence, if you consider a non reversible process, then change in internal energy, which, if is equal to work done by non reversible process, is also path dependent, which is a contradiction because ∆U is not path dependent. Hence, it must be reversible work because for adiabatic reversible change, the work done does not depend on path.
The sign convention in the video is strange for me. As a physics student, I learned that the work done by the system is positive. So we should have dU = Q - dW. I hope it does not confuses me.
@markomiz I'm... sorry? When was a floor mentioned? The diagram and lecturer only specify a lower bound of motion. However, I'll explain. Because the weight's velocity depends not only on the acceleration due to gravity but the internal pressure on that fluid, it will likely, at some point, decelerate as it moves toward its lower bound. Kinetic energy is 0, when it stops moving--at the lower bound of its motion.
For the adiabatic case, he claims that internal energy in the system changes. Why does internal energy change? Also, can the same isothermal example be used? It's not adiabatic; there's a heat bath.
@markomiz Yeah definitely..Then again i don't think he actually ever says that the heat converted is mgh, he just says that the work done in bringing the mass from h to 0 is mgh. He does say work is converted to heat but he doesn't explicitly say mgh is converted, does he? But i agree that the work converted to heat should be mgh-KE of mass at 0
Here lecture has some mistake. Again work out from system is positive value.When dQ goes in the system, certain work obtained, and remain energy in system is dU. dQ-dW=dU. if dUsystem=(+), dU universe =(-), Not dU universe = -q-w, dusystem=q+w. Hurge mistake, I think.
thank you for reply, dU=0, so the max work out means the max q in? if so, I think I will understand it. and I think, "dU system =q+w" is OK, w is the work from system to surrounding, and so does q. dw= (-)pdV, but dU= dw+dq
There is the sign of Q, W through boundary. First, about Q. Q is going in system, sign is (+), come out then -Q. Work is in. then -W, come out +W.I can say if anyone(system) is so angry by heat-up(+Q) and hit someone(+W), vise versa.Let's go the1st law, Q(+100) goes in the system, W(+70) produced, then 30? that is internal energy. Express the equation, dQ-dW=dU system. 100-70 =30Not dU = q + w. no 30=100+70 There are two kind of external works, expansion or compression work.Always sign of expansion from system to surrounging should be (+), reversely compression to system is (-). " du=q+w" is not O.K. du=q-w.
Work done is area under graph . Now from graph of reversible process work , it is clear that we can draw more graphs whose area is bigger than reversible one .so how maximum work is obtained by reversible work m
in a process from state A to state B, the maximum work is the reversible work, the another possibilities have less area because are rectangles with height equal to final pressure process or final pressure of step.
11:00 when he says the work done by the paddle wheel is mgh... wrong, mgh=work done by paddle wheel + kinetic energy of the weight at the bottom of movement.
This lecture should have explained more clearly about the equations... Only writing so many equations does not help understanding. After listening this lecture, I am confused and need to look up the text book : (
So, if the work is irreversible, with the conservation of energy of the system, does work by the system on its surrounding not occur, or cannot be calculated. Even with a vacuum as the external pressure, the work done by the system on its surrounds is till measurable I think.the force of the pressure and the distance the piston travels still amounts to work. Does the work of the system on its surrounds equate to an I crease in the entropy of the surrounds, and hence to the universe?
Hi, the P_ex and dx equation is equal to work done by the system, if work is irreversible, the work done by the system will be the fix P_ex times distance, but if work is reversible (slow down of P_ex) will be the ln of pressures times another terms (maximum work) and lost energy (second law) will be the difference between W_r and W_irr (in this case).
Yes, if entropy increases with every process, then reversibility could be obtained on a small scale, but not on the universe (or other closed system) as a whole.
What he means by that is that they aren't testing to see if they know the integrals. They could use very complicated integrals in the problems, but that would be counterproductive to learning thermodynamics. How can you learn the thermodynamics when you are working hard on the calculus?
When you blow on the hand with your mouth opened the air will flow hot, While when you decrease the hole of your lips the air will flow cold, why does this happen?
I am confused at 12:09. It seems like an Irreversible adiabatic expansion. So shouldn't (T2 - T1) be (W/Cv) and not (W/Cp). As (U2 - U1) will be Cv(T2-T1) and also (U2 - U1) = W.
Super late reply, so apologies for that, but the reason that Cp is used instead of Cv also caused some initial confusion for me. However, the way I see it is that a process has to be either constant volume, or constant pressure. If we are dealing with a completely sealed container, then it will be a constant volume process, therefore Cv. If, however, that isn't the case, then we assume it is a constant pressure process. The idea of it being adiabatic does make one think that it should be completely sealed, but since that isn't explicitly stated, then we go with a constant pressure process.
Hi, of course, the work quantity depends of path (way to pass from state A to state B) and u is a state function (independent of path), if work is equal to u is because you get the maximum work (reverisble work: without energy losses), this is a unique and better path and correspond to internal energy change.
Kindly explain me anyone at 37:59 what does proffesor trying to elaborate about measurement in calculations and entropy. I'll be thankful to that person
an ideal gas du=(đu/đT)"at const v" dT + (đu/đv)"at const,T" dv .... we have u2-u1=Cv(T2-T1) + integral (from v1 to v2) of (đu/đv) "at const T"dv then if i want u for Cv=const i get this u=Cv(T-T0) + integral (from v to v0) of (đu/đv) "at const T"dv + u0 then for a ideal gas, at temperature T, and volume v, and Cv=const u=Cv(T-T0)+u0 where u0 is the integration constant,
In the first lecture, it has been said that we need two coordinates to define a system, and this time we took these two variables. There's no upper hand of these two, we could have taken P and T as well but that won't give us what we want.
Kindly explain anyone that at 44:00 why proffesor had written CvdT in place of dq because only spefically in case of constant volume it will imply we could'nt generalize this result
he didn't generalize--what he did was show that taking dU at constant V gives rise to a term that includes the heat capacity at constant V, so we can then rewrite the original formula in those terms. I admit it is a bit disjointed since he talked about other stuff in between this section and his initial discussion of heat capacity
@DerSchreiFalke the weight speeds up as it goes towards the floor (acceleration), how then, you absolute genius, can the kinetic energy be approaching 0?! Try balancing the energies of the water and weight before and after.
maths is a language of science you can't explain or poove anything without working on maths............ working hard on calculus will make it easy to understand and explain the complications of thermodynamics or whatsoever
I do not envy the man that has to teach thermal physics. Prof. Levin does a great job with the intro. lectures though. This presentation seems to focus too much on the operational aspect of the physics, not on the insight; again, this stuff is non-intuitive, so it is not a criticism of the speaker.
Every thermodynamical properties can be find (in these examples) if you know two state functions, in this case he selects these functions; the state principle says that the number of independent variables is equal to a energy independent ways in which a system can change (heat and compressible work for this class).
