From Wikipedia: "In thermodynamic systems where the temperature and volume are held constant, the measure of useful work attainable is the Helmholtz free energy function" https://en.wikipedia.org/wiki/Work_(thermodynamics)#Free_energy_and_exergy

By definition, constant volume means that no PV work (expansion work) is done by the system. Because of this, is it not confusing to define Helmholtz free energy as a measure of the maximum useful/non-expansion work attainable at constant volume and temperature rather than just the maximum work attainable at constant volume and temperature, since all work done at constant volume is inherently useful/non-PV work? Adding the "useful" makes it seem like the useful work is separate from work done in general for a constant volume process. Am I missing something?

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    $\begingroup$ I don't understand. Isn't that the same thing? $\endgroup$ – Zhe Dec 29 '16 at 14:21
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    $\begingroup$ No. You are not missing anything. There are other kinds of work besides PV work, such as electrical or work in driving a shaft within a constant volume system. $\endgroup$ – Chet Miller Dec 29 '16 at 14:50
  • $\begingroup$ What I mean is that adding the "useful" is confusing because all work done at constant volume is inherently useful because constant volume means no PV work. $\endgroup$ – user28295 Dec 29 '16 at 15:02
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    $\begingroup$ I see your issue. But you can't just change the definition of "work" which is not inherently useful in general. Even though it is in this case, we might add the qualifier "useful" to disambiguate and be as clear as possible. But I agree, it's redundant given the context. $\endgroup$ – Zhe Dec 29 '16 at 15:13

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