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I'm trying to figure out how to show that the orange curves connecting the two green curves are infact adiabatic:

enter image description here

The set up is that I have an ideal gas, and I am searching for the loop when it's operating between two temperature such that it's work maximized (More details). I can argue why two segments must be isothermal , but I can't seem to find any way to justify the two adiabatic segments.

I know that adiabatic segment have the maximum work done relative to all other standard path in the $P-V$ plane, but I am not sure how to go from that to the required.

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The segments are adiabatic (no heat transfer) because the system needs to be brought from temperature of one reservoir to temperature of the other. This can't happen while in contact with any of the two reservoirs (the reservoir would act against it), so it happens without contact with any of them, and thus this process is adiabatic.

Such adiabatic process can be done in many ways, including irreversible ones where the process can't be represented as a curve in PV diagram, for example, sudden expansion to decrease temperature, or sudden compression to increase it. But that would not be a reversible process. Maximum possible work for given accepted heat from the hotter reservoir is achieved in any reversible process involving that accepted heat, and here it can be done by making both adiabatic processes quasi-static and also without friction. Quasi-static process can be represented by a curve in the PV diagram, and the condition of adiabaticity fixed its shape ($PV^\kappa = const.$).

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  • $\begingroup$ Thanks Jan. I think your answer is the one I was looking for. However, one point is not clear in it to me. Could you explain, the point "the reservoir would act against it" a bit more? $\endgroup$
    – Brian
    Commented Apr 29, 2022 at 19:05
  • $\begingroup$ If the system is in contact with reservoir at $T_1$, then any attempt to change temperature of the system to $T_2$ will be hampered by the reservoir, due to heat conductive contact. The reservoir will "try" to maintain temperature of the system at $T_1$ by accepting/giving off heat, and it won't be possible to get all parts of the working medium to temperature $T_2$ which is necessary for the next isothermal process at $T_2$. $\endgroup$
    – Ján Lalinský
    Commented Apr 29, 2022 at 19:08
  • $\begingroup$ The issue I have is , in one of the edge we are actually dropping the temperature, so wouldn't it be efficient to keep it near the cool one to cool it off faster ? And reversely near the hot one wheen we are trying to heat it up. I actually thought of the same argument you said, but I was stuck at the point I have metioned now. $\endgroup$
    – Brian
    Commented Apr 29, 2022 at 19:12
  • $\begingroup$ > "wouldn't it be efficient to keep it near the cool one to cool it off faster ?" -- What is "cool one"? The colder reservoir? $\endgroup$
    – Ján Lalinský
    Commented Apr 29, 2022 at 19:18
  • $\begingroup$ Yes the cold reservoir. The system being kept near the cold reservoir so it's cooled faster, and near hot for getting heated faster. $\endgroup$
    – Brian
    Commented Apr 29, 2022 at 19:19

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