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We know that heat is an energy of higher entropy (Diffused/Dispersed Energy). Work is the energy of lower entropy (Concentrated form of Energy). We know that entropy cannot reduce on its own, but we see that in an isothermal process, the heat seems to convert into work with 100% efficiency. How is that possible? Doesn't it violate the Law of Entropy?

Is an Isothermal process really possible which can convert heat into work with 100% efficiency?

Some say that it is possible in theory but not practically possible. My Question is, when such a process like an isothermal process violates the idea of Entropy so how could it be even theoretically possible?

Kindly help.

Qmechanic
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Devansh Mittal
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2 Answers2

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Heat from a single reservoir can, in principle, be converted to work entirely - during isothermal expansion done reversibly. 2nd law does not prohibit this. It says only that it is not possible to have a cyclic process where the system returns to the original state, and have net heat conversion to work with efficiency higher than the Carnot efficiency.

In isothermal expansion of gas, the gas does work and its volume increases. At this stage, there is no limit to efficiency, it can be 100%. However, this isothermal expansion can't be maintained indefinitely in practice, at some point it has to change, and the system has to be brought back to the original volume or close to it.

During the return, the system has to give off some heat to lower temperature reservoir, so that the system can return to the original state (so that expansion and working can continue after that). This limits efficiency of the conversion of heat into work.

Ján Lalinský
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  • Thanks for your response. My question is following. We know that heat is an energy of higher entropy (Diffused/Dispersed Energy). Work is the energy of lower entropy (Concentrated form of Energy). We know that entropy cannot reduce on its own, but we see that in an isothermal process, the heat seems to convert into work with 100% efficiency. How is that possible? Doesn't it violate the Law of Entropy? – Devansh Mittal Aug 16 '23 at 06:42
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    @DevanshMittal Because the process is not cyclic, the final state of the system is not the same as the initial state. In particular the final entropy of the system can increase to accommodate the entropy transfered in by the heat – By Symmetry Aug 16 '23 at 09:11
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    @DevanshMittal entropy transfer is not associated with all energy transfers, especially not with work. Entropy is not attached to all kinds of "energy in motion". Entropy comes into the system with the incoming energy due to heat transfer, but it stays in the system when energy transfer out of the system via work happens. Flux of entropy is not associated with flux of total energy, only to flux of energy via heat transfer. – Ján Lalinský Aug 16 '23 at 11:02
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    @DevanshMittal 2nd law and its implications for entropy does not mean energy can't be stripped of entropy, it means entropy of an isolated system can't decrease. "Energy in motion", incoming via heat and outgoing via work, is not an isolated system, 2nd law does not prevent entropy disassociation with energy, remaining in the system, and energy transforming into other form with zero entropy via work. – Ján Lalinský Aug 16 '23 at 11:09
  • Thanks a lot for your response. I must confess that my doubt is still not completely clear to me. Let me share my understanding with you and you can respond if I have understood correctly or not. 1. In a reversible process, the entropy should not change. This is why it is reversible. 2. In an isothermal expansion, heat of higher entropy seems to be converting into work of lower entropy, but at the same time, entropy of the gas will be increasing due to expansion, so net entropy will not change and hence, it is possible. Have I understood it correctly now? – Devansh Mittal Aug 16 '23 at 13:19
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    Yes, you've got it. In more detail, 1. In a reversible process, total entropy of the whole isolated system does not change. Entropy of a part of the system (e.g. of the working medium) can change. 2. In an isothermal expansion, a part of energy due to heat transfer converts to work (in case the medium is ideal gas, this part is the whole 100% of the heat, but <= 100% in general). Reversible work does not transfer "low entropy", there is no entropy transfer associated with reversible work. Only if the work is done irreversibly (e.g. friction), it creates additional entropy. – Ján Lalinský Aug 16 '23 at 14:12
  • Wonderful. Thanks a lot. – Devansh Mittal Aug 16 '23 at 16:24
  • Can every process be done reversibly in principle? Or there are some processes which cannot be done reversibly, even theoretically? – Devansh Mittal Aug 16 '23 at 16:34
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    It depends on how you define "process" and what bodies are accounted for in it. If reversibility status is part of the definition, then one process cannot be done both reversibly and irreversibly. Either it is reversible, or not. On the other hand, if process is defined via path of medium in space of thermodynamic states, then the same path can be traversed reversibly, and irreversibly, depending on the other bodies taking part in the process. – Ján Lalinský Aug 16 '23 at 16:52
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    E.g. isothermal expansion can be reversible, if there is no friction between the piston and walls of the container. If there is friction which does not heat the medium, only the environment, then the expansion can still be isothermal, but it is irreversible. – Ján Lalinský Aug 16 '23 at 16:54
  • Ján Lalinský : Your responses are truly amazing. I read all your responses again now, after a few days and with a lot of other studies on the subject and they make a lot more sense to me now. Thanks a lot for them. – Devansh Mittal Sep 01 '23 at 12:51
  • I have accepted your answer. I wish I could interact with you in some way apart from this platform (over some chat or something). You understand my questions in the spirit I ask them and respond accordingly. Thank you. – Devansh Mittal Sep 01 '23 at 12:53
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Is an Isothermal Process really possible? Heat cannot convert into work with 100% efficiency!

Theoretically, heat can be completely converted to work in a reversible isothermal process. That does not violate the second law of thermodynamics. But it is not possible to completely convert heat into work in a cycle, per the Kelvin-Planck statement of the second law:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir.

What this statement essentially says is that, in a complete cycle, some heat must be rejected to another (lower) temperature reservoir. That means all the heat absorbed from the high temperature reservoir in a reversible isothermal process is not available to be completely converted to work in a complete cycle.

We know that entropy cannot reduce on its own, but we see that in an isothermal process, the heat seems to convert into work with 100% efficiency. How is that possible?

The entropy does not "reduce on its own" during the reversible isothermal expansion. You need to consider the energy transferred to the system by heat separately from the entropy content of that energy. All of the energy transferred is converted to work, but the entropy is not reduced.

Although the internal energy and temperature (for an ideal gas) of the system is unchanged, the entropy of the system (gas) is increased due to the increase in volume of the gas during expansion work.

Hope this helps.

Bob D
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  • Thanks for your response. My question is following. We know that heat is an energy of higher entropy (Diffused/Dispersed Energy). Work is the energy of lower entropy (Concentrated form of Energy). We know that entropy cannot reduce on its own, but we see that in an isothermal process, the heat seems to convert into work with 100% efficiency. How is that possible? Doesn't it violate the Law of Entropy? – Devansh Mittal Aug 16 '23 at 06:42
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    @DevanshMittal are you thinking that disorganized energy is converted to organized energy in isothermal expansion work (i.e. entropy is converted to work)? – Bob D Aug 16 '23 at 10:26
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    @DevanshMittal I think I now see your difficulty. See update to my answer. – Bob D Aug 16 '23 at 11:03
  • Thanks a lot Bob for understanding the core of my question. Thanks for updating your answer also. I must confess that my doubt is still not completely clear to me. Let me share my understanding with you and you can respond if I have understood correctly or not.
    1. In a reversible process, the entropy should not change. This is why it is reversible.
    2. In an isothermal expansion, heat of higher entropy seems to be converting into work of lower entropy, but at the same time, entropy of the gas will be increasing due to expansion, so net entropy will not change and hence, it is possible.
     – Devansh Mittal Aug 16 '23 at 13:16
  • Have I understood your response correctly? – Devansh Mittal Aug 16 '23 at 13:17
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    @DevanshMittal 1. In a reversible process the entropy of the system can change. It is the sum of the changes in entropy of the system plus surroundings that does not change (equals zero). For the reversible isothermal expansion the entropy increases by Q/T. But the entropy of the surroundings decreases by -Q/T, for a total entropy change of the system plus surroundings of zero. – Bob D Aug 16 '23 at 14:07
  • As I explained in the update to my answer the entropy transferred into the system by heat increases the entropy of the system, in this case by Q/T. So the entropy transferred into the system is retained by the system. The entropy is not “converted into work”. The energy is converted into work. I think your confusion is rooted in the idea that the entropy of the system does not change. It does.
    • – Bob D Aug 16 '23 at 14:08
  • Thanks for your response. I completely agree that entropy of universe is constant in a reversible process and not of the system. My core doubt is, how disorganized energy like heat is converted to organized energy like work with 100% efficiency in an isothermal expansion? – Devansh Mittal Aug 16 '23 at 15:00
  • I understand the entropy decrease of universe is equal to entropy increase of the gas, so total entropy is constant, but work which is energy of lower entropy is also produced, so doesn't that create an imbalance in entropy? – Devansh Mittal Aug 16 '23 at 15:01
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    @DevanshMittal The "disorganization" as you call it is transferred to the gas not to the work being done. But not as an increase in temperature, but as an increase in the number of microstates now available to the gas molecules due to the increase in volume. – Bob D Aug 16 '23 at 15:47
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    @DevanshMittal Do you or do you not agree that the increase in the entropy of the gas equals the entropy transferred to the system by heat? – Bob D Aug 16 '23 at 15:56
  • Yes I agree now. Thanks a lot. Great explanation. – Devansh Mittal Aug 16 '23 at 16:26
  • Can every process be done reversibly in principle? Or there are some processes which cannot be done reversibly, even theoretically? – Devansh Mittal Aug 16 '23 at 16:35
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    @DevanshMittal in reality there are no truly reversible processes. If a process cannot be carried out infinitely slowly (quasi-statically) and absolutely without friction then, by definition, it cannot be done reversibly. Even if it could be done quasi-statically, It is impossible to completely eliminate friction. – Bob D Aug 16 '23 at 16:54