I saw the lecture of thermodynamics on the MIT opencourse ware . In that lecture , the professor said that change in internal energy in an adiabtic process is caused by only reversible work , and not by irreversible work . He said that the irreversible work can be zero and non zero as well , but it don't accounts for change in internal eneegy . Please explain how .
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$\begingroup$ The professor talked about this at 36:15 in the following video youtu.be/RrVq7Yduz2g $\endgroup$– AbbasCommented Dec 19, 2021 at 3:40
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$\begingroup$ 36:15 is the time in the video at which he said . $\endgroup$– AbbasCommented Dec 19, 2021 at 3:41
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The statement you mention is plainly wrong. It was noticed in one of the comments (by Andréas Bertoni).
If it was right, the professor would be contradicting himself, since, when he introduced the first principle there was no mention of reversible or irreversible processes. Moreover, it would be impossible to give a thermodynamic account of the Joule paddlewheel experiment he briefly described before, if the irreversible process should be excluded.
Actually, the important content of the first principle is that the sum of exchanged head and exchanged work is always a function of the state whatever process is used.
answered Dec 19, 2021 at 5:13
GiorgioP-DoomsdayClockIsAt-90GiorgioP-DoomsdayClockIsAt-90
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$\begingroup$ I had mention the duration of the video where the professor said so in the comment section of my video .Please check it . $\endgroup$– AbbasCommented Jan 16, 2022 at 7:40
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$\begingroup$ @Abbas When I replied, I had seen the video and I identified when he made the statement. It is a wrong statement. Mistakes can appear even in MIT lectures. We are human beings. $\endgroup$ Commented Jan 17, 2022 at 2:21
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The statement (as well as some others) is clearly wrong.
Perhaps what he had in mind is that the change in internal energy is not the same for reversible versus irreversible work, and that’s because an irreversible and reversible adiabatic process cannot connect the same two equilibrium states.
For example, for the same volume change irreversible expansion work is less than reversible expansion work and therefore the decrease in internal energy is less for the irreversible expansion.
Hope this helps.
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$\begingroup$ I had mention the duration of the video where the professor said so in the comment section of my video .Please check it . $\endgroup$– AbbasCommented Jan 16, 2022 at 7:40