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In an isothermal process, the whole heat is converted to work but not in a cycle so it doesn't break the 2nd law of thermodynamics. $$dQ=dW$$ But let's imagine a scenario with it. If we connect an isothermal system to a cold heat reservoir and get work from it and give the work to a reverse Carnot engine which transfers heat from that cold reservoir to a hot reservoir. Then we have a system that takes no external energy and transfers heat from a cold to a hot body. But this breaks the second law of thermodynamics.

Simply, if the Isothermal process can convert the whole heat from a cold body into work then that work can be given to a hot body as heat to make it hotter. According to Clausius's statement of 2nd law "Heat can never pass from a colder to a warmer body without some other change". So how can this be possible? Although the problem is not with the isothermal process because it's not a complete cycle, then where is the problem with this idea? How can heat be transferred from a cold to a hot body? enter image description here

Ziaul Hasan Hamim
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  • That is what the statements are trying to say. It cannot happen, so something is wrong. – naturallyInconsistent May 24 '23 at 06:50
  • "If we connect an isothermal system to a cold heat reservoir and get work from it" Please clarify your question to describe what's happening here. A cold reservoir can only be heated. It doesn't accommodate work collection. If you're envisioning some special source of energy (e.g., a high-pressure reservoir that pushes a piston), then it's not accurately described as a cold reservoir, and you're not describing a heat engine, and the Second Law must be more generally formulated (e.g., entropy can't be destroyed). – Chemomechanics May 24 '23 at 07:34
  • If we look at it simply the isothermal process can convert whole heat to work. A Carnot engine can transfer heat from a cool to a hot body using this work. So if you look at both as a single system. The isothermal process takes heat from a cold reservoir and converts it to work and using this work Carnot engine transfer heat from a cold reservoir to a hot reservoir. So there is no external work but still heat is transferred from the cooler body to the hotter body. – Ziaul Hasan Hamim May 24 '23 at 08:31
  • Entropy cannot be destroyed and cannot be converted to anything else, The reversible motion of entropy between temperature differentials is doing thermal work, and thus heat, especially the one in isothermal heat transfer, is not converted to work. The free energy of the system is converted to work in an isothermal process. – hyportnex May 24 '23 at 08:35
  • Yeah, that's what I want to know. Where is the mistake? Because according to laws it's not possible. – Ziaul Hasan Hamim May 24 '23 at 09:15
  • Is the cylinder in the isothermal process returned to its initial state (without being affected) at the end of the process? – Chet Miller May 24 '23 at 11:47
  • There's an "isothermal process" which is not cyclic that took heat from a cold reservoir and outputted work. But you didn't really even attempt to say how this would happen. Perhaps a cylinder at an extremely high pressure that you just expand for a really long time. Yeah. sure. You can extract work from a high pressure cylinder. And when you run out of pressure, you stop extracting work. Show that you can make the system return to its start point and you'll have an issue with thermodynamics. Until then you're just saying "I can extract work from a source of free energy"... kind of obvious – AXensen May 24 '23 at 13:20

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Note, not all isothermal processes keep the internal energy constant, so isothermal does not necessarily imply $\Delta Q = \Delta W$. However, for an ideal gas you do get $\Delta U = 0$ in an isothermal process, so in an isothermal expansion of an ideal gas, for example, one has that the heat entering, $\Delta Q$, is equal to the work done, $\Delta W$. This process could happen in the box marked 'isothermal process' in your diagram. The diagram shows a perfectly possible scenario. It does not violate Clausius statement of second law because the isothermal process is not cyclic.

In the scenario you have described, overall some heat is transferred from the colder to the hotter reservoir and the gas has expanded. So the entropy of the reservoirs has fallen but the entropy of the gas has gone up. Overall the entropy of the universe does not change.

What Clausius statement forbids is that this heat movement could take place with no net change in any other system apart from the reservoirs.

If instead of the isothermal process you propose any sort of cyclic process where the system there has no net change, then the whole scenario would then be impossible and it amounts to a neat proof that the Clausius statement implies the Kelvin statement.

Andrew Steane
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  • So heat can flow from a cold body to a hot body but not in a cycle? and there has to be some other positive entropy change which will at least balance out the negative entropy change due to heat going from cold to hot – Ziaul Hasan Hamim May 24 '23 at 16:30
  • @ZiaulHasanHamim yes the positive entropy change here is in the gas undergoing isothermal expansion. Its entropy goes up by $Q_1 / T_{\rm cold}$. – Andrew Steane May 24 '23 at 17:07
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The isothermal process solely converting heat to work is in contradiction with the second law. In fact, this is Kelvin's version of the second law. Your construction is a convoluted proof to show that Clausius' version implies Kelvin's version. It turns out that both versions are equivalent.

In fact, you don't need a Carnot cycle, you just need an isothermal cycle that converts work to heat at the hot reservoir (like Joule's experiment). The converse though requires a Carnot cycle. By using a cycle that brings heat to the hot reservoir, you use a Carnot cycle to bring back all the lost heat to the cold reservoir. The total cycle is therefore and isothermal cycle converting heat from the hot reservoir to work.

All of this is explained in more detail for example in Fermi's Thermodynamics.

Hope this helps.

LPZ
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  • thank you for saying this "The isothermal process solely converting heat to work is in contradiction with the second law" although this is not "In fact, this is Kelvin's version of the second law." because Kelvin's version includes an adiabatic stage to close the cycle in which isothermal is followed by an adiabatic. Your statement is still true because heat never converts to work, instead only entropy as it is moving from a higher to a lower temperature is the one producing work, and entropy does not convert instead it is conserved in that motion and only its temperature changes. – hyportnex May 24 '23 at 13:05
  • The exact quote from Fermi's book is: "A transformation whose only final result is to transform into work heat extracted from a source which is at the same temperature throughout is impossible." I haven't seen a version with the necessary adiabatic stage, do you have a source to recommend? I don't quite understand your last sentence though. What I mean by isothermal process is one with only one heat source which is what the OP meant I think. In French it is called "monotherme" but I couldn't find an analogue in English. There is no higher/lower temperature. – LPZ May 24 '23 at 14:30
  • Fermi was careful to include this clause "...whose only final result is to transform..", which means that the engine must return to its original state, this is the cycle, and since the "heat" exchange in monothermal meaning thermal energy exchange with only one fixed temperature reservoir all other interaction the system may have within the cycle is by definition adiabatic. The impossibility of that kind of a cycle is Kelvin's postulate. – hyportnex May 24 '23 at 17:19
  • Since $dF=-SdT+\delta w$, $\delta w = \sum_k Y_kdX_k$, in an isothermal process $dT=0$, hence, $dF_{T=const} = \delta w$ meaning that in an isothermal process the work done on the system is equal to its variation of free energy. In words, this just means that "isothermal work potential makes work" and not "isothermal heat makes work". Any other interpretation of the $dF_{T=const} = \delta w$ demands that one must introduce an internal and not reversible entropy generation, so called compensation, to compensate for the destruction of the entropy coming with the "heat" but converted to work. – hyportnex May 24 '23 at 17:24
  • Why the down votes, especially without any technical comment? Either LPZ is right or he is wrong; if he is wrong then it should be pointed out where and how just as a matter of minimum common collegial courtesy. – hyportnex May 24 '23 at 17:41
  • Oh ok, from the diagram I thought that the OP was considering a cyclic monothermal process. So you do use monothermal in English as well (there wasn't the corresponding wiki page)? Yes, I agree with you that for a non cyclic isothermal process, it isn't Kelvin's statement. Thanks for the support! – LPZ May 24 '23 at 17:55
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    It is a rare usage, but for example, in his book "Thermodynamics of Systems in Nonequilibrium States" 2ed, 2002, and also in his journal articles of which there are many, Tykodi uses both monothermal and polythermal to describe interactions with a single or with several thermal reservoirs of different temperatures. You may find this also relevant https://physics.stackexchange.com/questions/765097/is-tds-dq-true-for-processes-involving-mass-transfer/765104#765104 – hyportnex May 24 '23 at 18:09
  • Ok thanks, that was an interesting take – LPZ May 24 '23 at 19:14
  • I have written several, equally unpopular answers on this matter with varying details with the same point of view, see, e.g., https://physics.stackexchange.com/questions/707625/what-stops-entropy-from-forming-if-you-extract-100-of-heat-flow-and-use-it-for/707654#707654, https://physics.stackexchange.com/questions/755110/is-there-a-4th-law-of-energetics-how-is-energy-conserved-in-a-stationary-flow, https://physics.stackexchange.com/questions/755053/how-the-temperature-used-in-second-law-of-thermodynamics-and-ideal-gas-law-equiv/755060#755060, – hyportnex May 24 '23 at 19:36
  • also: https://physics.stackexchange.com/questions/762677/is-it-possible-that-a-system-undergoes-an-irreversible-change-while-the-surround/762696#762696, https://physics.stackexchange.com/questions/742232/why-isnt-heat-conduction-considered-a-form-of-work/742245#742245, https://physics.stackexchange.com/questions/763179/on-definitions-of-interactions-in-thermodynamics-changing-the-energy-of-a-closed/763191#763191, https://physics.stackexchange.com/questions/759539/do-chemical-potentials-stay-constant-in-chemical-equilibrium/759566#759566 – hyportnex May 24 '23 at 19:38
  • you can find more original here: http://actachemscand.org/volume.php?select1=1&vol=3 under Bronsted memorial edition, pp1208-1238 – hyportnex May 24 '23 at 19:51
  • @hyportnex I downvoted because it started "solely converting heat to work." But the question says the isothermal process is "not in a cycle." It doesn't say solely converting heat to work. Could imply something like a high pressure cylinder expanding, but using thermal contact to retain its temperature. It's using a source of free energy to get work. Comments with downvotes aren't required here. I just did so because this person clearly knows what they're talking about, so I didn't see any point starting a comment-section-debate, but they didn't interpret the question the same way as me. – AXensen May 25 '23 at 22:00
  • @AXensen I know that "Comments with downvotes aren't required here.", and I do not like it, this is why I mentioned it in my comment. Anyhow, I still believe that any reasonable and reasonably thoughtful argument deserves that if in error then the error be pointed out, just as a matter of collegial courtesy, nothing else. – hyportnex May 26 '23 at 01:22
  • @AXensen By the way, I am delighted to read your "using a source of free energy to get work", most books and most people, here, too. would say that the "heat absorbed is completely converted to work", or some such. I completely disagree with that view, see my comment above. – hyportnex May 26 '23 at 01:28