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I’ve seen many books considering reversible isochoric processes performed by an ideal gas. But is that possible?

Suppose I’ve a general isochoric process going from A to B, suppose I’m having a decrease of internal energy so that $$\Delta U = Q < 0$$, so I’m letting $Q$ get out of my system, now let’s take this process in the opposite direction, I’m having that heat spontaneously flow from the sorroundings to my system , against the second law . This works also if the process is quasi-static. Can someone show me an example of an isochoric process that can be reversed?

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To cool a gas reversibly, what you do is put the gas into contact with not just one, but with a sequence of constant temperature reservoirs, each one at a slightly lower (i.e., differentially lower) temperature than the one before. This allows both the gas and the sequence of reservoirs to experience a reversible change.

To reverse the process, you just reverse the sequence. That would consist of exchanging heat from the reservoirs to the gas, in sequence. The only difference here would be for the very first and very last reservoirs. But, in the limit of infinitesimal changes, this difference would become negligible, (and both the gas and the reservoirs will have been returned to their original states, without affecting the state of anything else).

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  • $\begingroup$ Perfect, so I just need to start from the hotter reservoir in the first path and from the colder one in the second right? So if I have an isochoric process there are several ways to do the same path on the PV plane and not every way is reversible? $\endgroup$ – Jacopo.R Jan 31 '18 at 13:57
  • $\begingroup$ In my judgment, I has nothing to do with the PV plane, since the volume is constant. It has everything to do with how the heat transfer is carried out. If it is carried out, for example, in a way where the gas is put in contact with a single ideal constant temperature reservoir, the process will be irreversible, with entropy being generated within the gas a result of the rapid heat transfer associated with large temperature gradients within the gas. $\endgroup$ – Chet Miller Jan 31 '18 at 14:25
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For a process to be reversible the intensive macroscopic properties i.e. pressure and temperature, of the system and its surroundings must differ by, at most, an infinitesimal amount. For an isochoric process, neither the pressure nor the temperature of the system is constant, so the pressure and temperature of the surroundings must also change accordingly. This is unsurprising for the pressure (an idea rigid box will do this 'automatically'), but for the temperature it generally requires more effort. In particular the (immediate) surroundings for the system cannot be our beloved reservoir at a constant temperature. never the less there are, at least in theory, setups that achieve this, for example in the ideal Stirling cycle.

If the system and the surroundings have the same temperature at each point in the process then we have \begin{align} \mathrm{d}S_\mathrm{sys} &= \frac{\mathrm{d}Q_{\mathrm{rev}\,\mathrm{sys}}}{T} \\ &= - \frac{\mathrm{d}Q_{\mathrm{rev}\,\mathrm{sur}}}{T} \\ &= - \mathrm{d}S_\mathrm{sur} \end{align} so there is no net change in entropy.

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  • $\begingroup$ So in order for this process to be reversible I need a reservoir always in thermal equilibrium with my system? But in order to get this I need to extract heat from somewhere else right? Thinking about stirling engine it has a lower efficiency of the Carnot cycle, so it cannot be reversible $\endgroup$ – Jacopo.R Jan 31 '18 at 12:07
  • $\begingroup$ The idea Stirling cycle with regeneration is reversible (and so must have the Carnot efficiency) . While it is true that you must do something to the surroundings to the surroundings, that something does not have to be heating $\endgroup$ – By Symmetry Jan 31 '18 at 12:17

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