# Can the work in a isochoric process be non-zero?

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 $Q=\Delta U$?

• In any isochoric process, work done on the system due to boundary movement is zero not net work. $$W_{\textrm{net}}=W_{\textrm{bounary}}+W_{\textrm{other}}$$ – lucas Jun 29 '16 at 17:58
• These figures from “THERMODYNAMICS An Engineering Approach, Fifth Edition, by YUNUS A. CENGEL and MICHAEL A. BOLES” can help you! – lucas Jun 29 '16 at 18:09