Oh my god. I cannot even begin to explain how elegant this video is. It’s a hidden goldmine of explanations that actually raise the understanding of thermodynamics on a fundamental level. Bravo.

the best thermodynamic explanation ever made thumb up

I have always struggled with this topic, thank you for this very clear exposition. One thing that used to confuse me, was I thought that in the Carnot engine, gas had to flow from the hot to the cold reservoir. But from your video (and from considering a toy Stirling engine) it is clear that only heat needs to flow from the hot reservoir into the piston and then from the piston into the cold reservoir (with some of the heat energy converted into work). In many heat engines gas does flow from the hot to the cold reservoir but it is not necessary to consider for this analysis. The same amount of gas can remain in the piston throughout the cycle.

This is so great I love it

Love this series, thanks a lot for your affort, time & energy putting in this masterpiece!

Great lecture series. I subscribed this channel. Thank you very much Sir. From Indonesia🇮🇩🙏

Amazing video thank you for sharing this information

I think at 18:41, you shoud write the extra ∆P∆V which would then approximate to 0 when taking ∆P and ∆V as dP and dV.

Hello sir. This serie is the best explanation of thermodynamics I have ever watched. I also watched your serie about optics which was very helpfull. I wonder if you made a lecture note on thermodynamics as you did for optics. Because the lecture note on optics was amazing. Thank you sir

@SanderKonijnenberg

Жыл бұрын

Thanks for the kind comment. I'm happy to hear these videos have been helpful. I do hope to write more lecture notes at some point, but unfortunately I'm currently not working on that. Too many interesting things to do, too little time ;)

@nosms6581

Жыл бұрын

@@SanderKonijnenberg thank you sir again anyway and keep doing these videos they are very helful

@27:28 the simplified form of clausis inequality may have an error. You have used Th and Tc and so that is the same as the original equation where you got a positive result. However, you get a negative result the second time and say that you are not considering the Temperature of the reservoirs as before. But the equation says otherwise. You say you are considering the temperature of the surrounding. And indeed that is what you say in the integral equation. I believe what you mean is Qin/Ts -+ Qout/Ts 0. But Qout/Ts is 0 because Th > Tc and Qin and Qout are opposite. And by convention Qin is positive and Qout is negative. It's the denominator Tc > Th that makes the whole thing positive. Excellent video, though. I don't think I've ever seen such a comprehensive treatment. You almost sound more like a chemist or mathematician and not a physicist. For whatever reason, I find physicists tend to be more sloppy with the book keeping than are chemists and mathematicians.

@SanderKonijnenberg

6 ай бұрын

Thank you very much for this comment. I hope to learn from it, but I'm afraid I don't fully understand it yet at the present moment. - '[You] say that you are not considering the Temperature of the reservoirs as before. But the equation says otherwise.' Is this referring to what I say at 27:03? My point there was that in the equation, we don't consider the *entropy* change of the reservoirs, because the signs of Q_in and Q_out are defined with respect to the engine. I.e. a positive Q_in would imply a *negative* change in entropy in the reservoir, but the term in the equation would be *positive* ,so the equation does not express the entropy change in the reservoir. And since the equation does not express the change in total entropy (i.e. of both the engine and the reservoirs), it does not contradict the Second Law of Thermodynamics. - 'It's the denominator Tc > Th that makes the whole thing positive.' I'm not sure I follow. Should it not be TcTc, the negative term Q_out/Tc dominates the positive term Q_in/Th, making their sum negative. Could you perhaps indicate in which of these three claims you find something objectionable?

9:30 Why is this process reversible, when at 7:53 it is said that moving a piston connected to a hot reservoir is an irreversible process. It probably has to do with the fact that the temperature is kept constant when expanding the gas, but I do not see why this leads to a reversible process? Also, keeping the temperature constant while compressing the gas when it is connected to a hot reservoir should not be possible at all, or what am I missing here?

@SanderKonijnenberg

Жыл бұрын

The assumption that is made here, is that the process is 'quasi-static'. Indeed, it sounds somewhat self-contradictory: how can there be a process when the system is static? The idea is that the process occurs so slowly, that while you're changing the system, it's always in equilibrium. E.g. if you compress the gas by the tiniest amount, then the temperature is raised by the tiniest amount, but this amount is so tiny that the excess heat is immediately dumped into the reservoir, so that the gas has the same temperature as the reservoir again. Therefore, if you compress the gas sufficiently slowly, it is able to remain in thermal equilibrium with the reservoir during the compression (i.e. the gas always has the same temperature as the reservoir). Because the system is always in equilibrium, no extra entropy is generated (assuming there's no friction).

@StefanHoffmann84

Жыл бұрын

@@SanderKonijnenberg I see. I guess my misunderstanding came partly from the (wrong) assumption that the hot reservoir is always hotter than the gas, but if I got you right, it just maintains some "stationary"/constant temperature. But what still confuses me is what happens when the reservoir is intially hotter then the gas, then the heat will flow to the gas, which increases the pressure on the piston. Now, imagine I move the piston slowly/quasi-static until some endpoint where no heat flows between the gas and the reservoir. Then this process should not be reversible, because when the gas is compressed (even slowly), the reservoir cannot "spontaneously" take the excess heat that is generated, as both the gas and the reservoir started from the same temperature. But isn't this the situation in the intial part of the Carnot process? Initially connecting a hotter reservoir to the gas? So, why is this still considered to be reversible, when it seems unplausible that the process will go in the other direction?

@SanderKonijnenberg

Жыл бұрын

@@StefanHoffmann84 Indeed, the reservoir is assumed to stay at the same temperature, regardless of whether you put heat into it or extract heat out of it. Regarding the reversibility: it is the Carnot *cycle* that is reversible. The initialization should not be considered part of the cycle that follows. It is true that raising the temperature of a cold gas by extracting heat from a hot reservoir is irreversible, but during the Carnot *cycle* there is no point where the gas is connected to a reservoir with a temperature that is different than the gas'.

@StefanHoffmann84

Жыл бұрын

@@SanderKonijnenberg Ok. I guess I got it. That is the reason why the piston is moved slowly, so that the assumption is fullfilled that the gas and the reservoir have the same temperature during the process. Thank you!

@ 31:25 A "closed isolated" system??? An open system, but no flow of particles??? You always surprise me, Sander.

27:26 is little confusing to me. What i understand is first you defined entropy as [ heat given(+) or heat taken(-) to system] / T . Which always increases as you showed from 2nd law assumption. Then you said "in clausius inequality we are calculating entropy of engine not reservoir moreover we are dividing it by temperature of reservoir, hence it does not contradicts that entropy increases " In this statement their are lot of holes in my mind, the way you said it ... I feel like it means that in clausius inequality we are not measuring the entropy at all since dividing given heat to engine by temperature of reservoir is not how you defined entropy and hence it does not contradicts 2nd law. But another teacher actually defined entropy and derived 2nd law using clausius inequality infact what your 2nd law states is what actually he referred to as the clausius inequality. I know different teacher call things with different names. But the confusion from 27:26 remains in my mind and it's unclear what you actually meant. Thanks. Very insightful video.

@SanderKonijnenberg

6 ай бұрын

Perhaps the following quote helps (from hyperphysics.phy-astr.gsu.edu/hbase/thermo/clausius.html ): 'The Clausius Inequality applies to any real engine cycle and implies a negative change in entropy on the cycle. That is, the entropy given to the environment during the cycle is larger than the entropy transferred to the engine by heat from the hot reservoir.' In other words, Clausius' inequality does not state that the total entropy (i.e. of both engine and the two reservoirs combined) decreases, because that would contradict the Second Law of Thermodynamics. Evaluating the *total* change in entropy would take into account the change in entropies of both the engines and the two reservoirs. But what is calculated here is the negative of the entropy transferred from the engine to the reservoirs (so the fact that it's equal to or less than 0 indicates that the entropy of the two reservoirs combined remains constant or increases).

@intuitivelyrigorous

6 ай бұрын

@@SanderKonijnenberg Okay now i think it makes sense with what I was taught previously is similar to following integrals in worded form : (entropy given to engine) + (entropy taken out of engine) (the entropy given to engine)-(entropy given to reservoir) (the entropy given to engine) ( Considering reservoir as surrounding, engine as system and processes in surrounding to be reversible. Equality will hold true if Q/Tsurr is also reversible) We can write Q/Tsurr < ∆S Which is our second law.