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I came up with a doubt regarding isochoric irreversible processes.

Question: Is it always true that, for any isochoric process, reversible or not, the work exchanged by the system is zero and the heat exchanged is $Q=\Delta U$?

I'm asking this because, in a exercise on thermodynamics trasformations of a gas, there was to be considered an "isochoric irreversible transformation in which the tank containing the gas is thermically isolated and work is done on the gas with a fan of negligible thermal capacity, the gas goes from $T_a$ to $T_b$".

Now if the tank is isolated $Q$ should be $0$ but that cannot be, since the gas changes its temperature and the process is isochoric. Furthermore it is said that work is done on the system, but the process is isochoric, how can that be?

Nothing else is specified on the trasformation so in my view it can be a case where it does not matter at all how the process is done, as long as $V_{final}=V_{initial}$ the process is isochoric and the total work done on the gas will be zero (maybe some positive and some negative), but still I don't see how the gas can exchange heat in this case.

So do I have to care about it or, in any isochoric trasformation I can be sure that $W=0$ and $Q=\Delta U$?

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    $\begingroup$ In any isochoric process, work done on the system due to boundary movement is zero not net work. $$W_{\textrm{net}}=W_{\textrm{bounary}}+W_{\textrm{other}}$$ $\endgroup$
    – lucas
    Commented Jun 29, 2016 at 17:58
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    $\begingroup$ These figures from “THERMODYNAMICS An Engineering Approach, Fifth Edition, by YUNUS A. CENGEL and MICHAEL A. BOLES” can help you! $\endgroup$
    – lucas
    Commented Jun 29, 2016 at 18:09
  • $\begingroup$ $\delta W = P\textrm{d}V$ is valid only for hydro-static systems. An hydro-static system is, by definition, always at equilibrium and has P, V and T defined at every moment. If you want to demonstrate that $\delta W = P\textrm{d}V$ you need the system to have pressure defined everywhere and in every moment, with the same value everywhere and equal to the outside pressure. The case you are describing doesn't satisfy these requirements. This is my understanding at least. $\endgroup$
    – HomoVafer
    Commented Jul 15, 2022 at 16:27

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P-V work is not the only kind of work that can be done on the contents of your system. In the case of your fan example, the fan is doing work on the gas within the container by exerting force on it through a displacement (of the fan blade). The kinetic energy imparted to the gas by the fan is then converted to internal energy by viscous dissipation (a damping effect). So the net effect is that the work done by the fan increases the internal energy of the gas.

In Joule's famous experiment, he ran a paddle wheel inside an insulated container of water, by attaching the paddle wheel to a rope and pulley arrangement, driven by a descending weight. The change in potential energy of the weight was equal to the work that the paddle wheel did on the water (and the change in internal energy of the water). He measured the rise in the water temperature as a result of the viscous dissipation of the mechanical energy. By doing this, he was able to establish the equivalence between the work done in N-m (Joules) and the change in internal energy of the water. And he was thereby able to determine the relationship between the change in internal energy and the temperature rise.

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  • $\begingroup$ Thanks for the answer! If I compare this case with the usual situation of a gas that expands moving a piston I have a doubt. In the case of the gas that moves a piston up, I can say that the gas does work on the piston, precisely against weight force (the work done by the gas is equal and opposite to the one of weight). But here, with the fan, can I say that the gas does work on the fan? My guess would be yes, since if a power is delivered to keep the fan moving it means that something is stopping it, doing work, and that work is done by the gas, and it's negative. May this be correct? $\endgroup$
    – Sørën
    Commented Jun 29, 2016 at 18:58
  • $\begingroup$ Yes. The gas is doing negative work on the fan (force opposite to the displacement) and the fan is doing positive work on the gas. $\endgroup$ Commented Jun 29, 2016 at 19:14
  • $\begingroup$ I'd be interested in seeing how explicit calculations of this process would be carried out. $\endgroup$
    – Babu
    Commented Sep 11, 2020 at 20:46
  • $\begingroup$ @Buraian This would involve solution of the complicated fluid dynamic and thermodynamic partial differential equations, probably using a rotating frame of reference (if the fan is vertical and symmetrically placed within a cylinder), and would require computational fluid dynamics to solve. Plus, additional approximations would be required if the flow were turbulent (which would almost certainly be the case). $\endgroup$ Commented Sep 11, 2020 at 21:04
  • $\begingroup$ Maybe a simplistic example for the curious would do well $\endgroup$
    – Babu
    Commented Sep 11, 2020 at 21:19
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Most text books implicitely assume no work can be done if you don't change the volume. Somehow, they mean the only possible macroscopic interraction with the system (a gas) is done by moving the piston.

Nothing else is specified on the trasformation so in my view it can be a case where it does not matter at all how the process is done, as long as Vfinal=Vinitial the process is isochoric and the total work done on the gas will be zero (maybe some positive and some negative), but still I don't see how the gas can exchange heat in this case.

Beware, this is false however. Some work can be done when the volume is changed in a non reversible way. Example :

  • perform a Joule free expansion by multiplying the volume by 2: no work is done at all
  • get back to the original volume with an adiabatic reversible compression: you do some positive work

This transformation produces work even though it starts and end at the same volume. A true isochoric process needs to keep the volume constant at all time.

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Disclaimer: I am just a student; take the following answer with a pinch of salt and please correct me if I am wrong! I think the confusion here is due to the definitions of each term not being clear and well,..definite.

Definition of an adiabatic process:

An adiabatic process is one in which the energy transfer(i.e. a change in internal energy U) to or from the system is done only via a restricted form of work that can be expressed as PdV, EdP (E is electric field, P is electric dipole moment, and EdP is the electric work ) etc. The form of the expression is

(a macroscopic intensive force-like state variable of the system) x differential of(a macroscopically measurable extensive displacement-like state variable of the system);

this product has the dimension of energy and it is possible to see that these terms are simply give sum of mechanical or electrical work done on each particle of the macroscopic system due to mechanical or electric forces.

OR more simply, an adiabatic process is any process in which there is no change in entropy S. Therefore $dQ = TdS = 0 $.

An adiabatic process is NOT defined as any process which is carried without with the system surrounded by an "adiabatic" wall (thermally isolated).


Any energy transfer process that cannot be expressed as,

(a macroscopic intensive force-like state variable of the system) x differential of(a macroscopically measurable extensive displacement-like state variable of the system)

will involve a change in entropy and can therefore be more correctly referred to as "heat transfer" ($dQ = TdS$) rather than "work done".

In the question, it is said that "work is done on the gas with a fan". This work cannot be expressed as (a macroscopic intensive force-like state variable of the system) x differential of(a macroscopically measurable extensive displacement-like state variable of the system). Using a fan to change the internal energy involves a change in entropy of the system is therefore more correctly referred to as a form of "heat transfer" than "work done". Though the system is surrounded by an "adiabatic" wall(thermally isolated), this is not an adiabatic process.

Reference:

Herbert Callen, Introduction to thermodynamics and thermostatistics. Chap 1.

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