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By the first law, we know $\text{d}U=\delta Q+\delta W$ and, on adiabats, we know $\text{d} U=\delta W$. But what is $\delta W$ for irreversible adiabatic processes? Take a thermally isolated container. If I stir it, I increase it's energy but clearly $-p\text{d}V$ is $0$ because the volume didn't increase. This leads me to believe that $-p\text{d}V$ only applies for reversible adiabats. Fair enough. This means the definition of work depends on the path even within just the adiabatic curves but, for a general thermodynamic system, what is this definition? Or must be it be defined for every new system?

Edit: To clarify my question, consider processes that are only quasistatic which are well-defined continuous paths on the state space. For reversible processes, we can go straight from looking at the path in the state space to the work done if we know it is adiabatic. For irreversible adiabats, is this possible for a general thermodynamic system or must we always appeal to the physical realization of such process to determine it?

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You need to distinguish between $dW=pdV$, or boundary work between an adiabatic system and its surroundings and $dW=$ “stirrer” work between an adiabatic system and its surroundings which will not result in volume change (unless the increase in temperature is sufficient to expand the system contents). Stirrer work was the basis of the famous Joule experiment where he demonstrated the equivalence between heat transfer to a system and mechanical work transfer to a system (both resulting in temperature increase).

So you don’t need to define a new system. You only need to be clear on the type of work transfer. In addition to stirrer work you can do electrical work (place a heater in the system and supply electrical energy by wiring to the system) to increase the temperature of an adiabatic system.

Hope this helps.

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  • $\begingroup$ Assume the process is quasistatic which is a well-defined continuous path on the state space. By looking at this path alone, can one know the work function? This seems to be only possible for reversible. I'll add an edit to my original question $\endgroup$ – Aakash Lakshmanan Dec 18 '18 at 21:48
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You are right to conclude that $\delta W =pdV$ doesn't in general apply for an irreversible adiabatic process, such as work done by a rotating paddle, or a gas expanding into a vacuum. But I don't see why this makes you dissatisfied with the definition of work. The work done on a system is calculated using, essentially,

Work = external force x distance moved by force.

This applies both to the slowly moving piston (reversible) and the paddle (irreversible). In both cases the work, if done adiabatically, can be equated to the increase in the system's internal energy, that is$$(\delta W)_\text{adiabatic}=dU.$$ The expansion of a gas into a vacuum is an interesting case. Clearly $pdV$ is not zero, yet the gas does no work. And, with adiabatic walls, there is no change in internal energy.

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  • $\begingroup$ Say we disregard nonquasistatic processes. My dissatisfaction with this is that all quasistatic processes are well-defined in the state space and for reversible processes, the work was nicely defined for paths in this state space. Now it seems however we must look beyond the state space to think about what the work would be which is a problem for general thermodynamic systems. This leads me to believe that the work function is something that must be given apart from the state space and can't be defined within it unless it is reversible. $\endgroup$ – Aakash Lakshmanan Dec 18 '18 at 21:46
  • $\begingroup$ "This leads me to believe that the work function is something that must be given apart from the state space " I think that this is what I was trying to say. There's not much we can do about it! $\endgroup$ – Philip Wood Dec 18 '18 at 21:54
  • $\begingroup$ So there is no way to translate from path in a state space to the work without appealing to its physical realization? $\endgroup$ – Aakash Lakshmanan Dec 18 '18 at 22:28
  • $\begingroup$ I think this is so.. $\endgroup$ – Philip Wood Dec 18 '18 at 22:52
  • $\begingroup$ @PhilipWood I'm not sure it's correct to say that "$dW=pdV$ doesn't apply for an irreversible adiabatic process". How about a constant (external) pressure adiabatic exapansion process? For such a process, the work done is $p\Delta V$. $\endgroup$ – Bob D Dec 19 '18 at 2:47

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