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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 Jun 29 '16 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 Jun 29 '16 at 18:09
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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 Jun 29 '16 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. $\endgroup$ – Chet Miller Jun 29 '16 at 19:14

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