Irreversible Heat and Work

The work associated with a reversible process is always less than that for an irreversible process between the same initial and final states. The opposite is true for heat.

[@adityagoyal3491]

fanatstic was this , am now even more knowledgeable than earlier , thanks old man . also i would also one day write backwards and still not make any mistake just like you did while explaining this

[@GulzarAhmad-sw1kh]

Excellent! this implies that a machine will have maximum efficiency if operated reversibly because during expansion it will deliver maximum work while during compression it will require least work.

[@RealMonkeyDKirby]

Seriously thank you for these videos....thank you so much. We get zero explanation in class just reads slides that repeat the book and just looking at derivations and proofs doesn't teach lol. Love the explanations

[@PhysicalChemistry]

Thanks, I'm glad you find them useful

[@gelomelo3426]

What a beautiful explanation, what a great professor 👏. Amazing 🌸🌸🌸🌸

[@fridmamedov270]

I just did not understant differential form. The differential paths are not the same, but the overall path is the same. Two differential points are not the same, so how we can say the same differential internal energy change for reversible and irreversible paths?

[@jalsamuel]

Thank you, sir. This video is really helpful

[@PhysicalChemistry]

You're welcome! Glad to help

[@mortezakhoshbin]

So the conclusion" patience is power!" If you want to get more work and do less of that be patient!

[@vicelabae]

If we consider an isovolumetric process, đw = 0 without any regard of the reversibility of the process i.e. đw_rev = đw_irr = 0. Then by first law, đq_irr = đq_rev. Does this constitute a contradiction? Thank you very much Sir

[@jupironnie1]

Need to explain why Carnot shows reversible work is maximum work i.e higher than irreversible work. Video based on graph indicates reversible work is higher than irrev work (as the irrev curve has lower area) and work done is a scalar energy value. Video statement at 1.23 is confusing even in a short while later at 2.15 using absolute values work rev is higher than work irrev. This algebraic manipulation is a source of confusion as it is against other thermodynamic textbooks. BUT it an innovative explanation which removes sign conventions discussion is other textbooks, unless there is reason for this innovative approach?.

[@PhysicalChemistry]

There are some videos on the Carnot cycle (kzbin.info/www/bejne/qavcd4Srhc6Db9U) and the Carnot Theorem (kzbin.info/www/bejne/o4qxdKibi8dsoKc). Notice that I'm using a different sign convention for PV work than you are: chemists and physicists usually prefer to define work as negative when a gas expands. So reversible work is *less than* irreversible work. (kzbin.info/www/bejne/jmGoZZqGq52pirM). This sign convention comes from the definition of PV work here: (kzbin.info/www/bejne/hHnKiIyOic-keas). Engineers often use the opposite sign convention (work for a gas expanding is positive). Both approaches are correct, you just have to adjust your expectation of whether PV work means work done "by" or work done "on" the system. This may indeed be contrary to other thermodynamic textbooks if you are mostly following engineering sources. From my point of view, of course, those engineering texts use a sign convention that is contrary to that of most of the books on my shelf.

[@bikashsahu4809]

Sir graphically it is understood that irr work done in case of expansion is more but wht is the actual reason behind it why system has to do more work for irr expansion than rev expansion plz sir explain it.

[@PhysicalChemistry]

When the expansion is reversible, the external pressure is the same as the internal pressure. So the pressure that the system is pushing against is the same as the internal pressure. This is the largest pressure that the gas can successfully push against, so it requires the most work to push against. For an irreversible expansion, the external pressure is *lower* than the internal pressure. So the gas is pushing against a lower pressure than in the reversible case, and does less work.

[@bikashsahu4809]

@@PhysicalChemistry sir then in both case rev work is more either compression or expansion it doesn't matter?

[@PhysicalChemistry]

@@bikashsahu4809 No, for compression the reversible path requires the *least* work. An irreversible path will require more work (because the external pressure must be larger than the system pressure)

[@bikashsahu4809]

@@PhysicalChemistry plz sir explain through a vdo not able to get it ,it is so confusing

[@PhysicalChemistry]

@@bikashsahu4809 The good news is that I have made a video trying to explain this as well as I can. The bad news is that it is the video you are already commenting on!