# Why is pV work “useless” in Helmholtz free energy?

According to standard proofs found in any analytic thermodynamics book, at constant T:

$$dF > dw$$

ie, the change in the Helmholtz free energy gives the maximum amount of energy which is free for a system (at constant T) to convert to work.

However, this is frequently followed by the statement comparing it to Gibb's free energy and stating that "dF also includes the unextractable (and hence useless) pV work, in contrary to dG at constant T and P".

How comes work by volume change is considered useless? Isn't that exactly what we normally use to drive pistons and engines in most standard thermodynamics problems?

• The statement could be clearer. Some work is unextractable. Not all of the $pV$ work. – garyp Apr 1 '18 at 13:20
• Doesn't the standard proof also assume a constant-volume system? – Chet Miller Apr 1 '18 at 15:49

## 2 Answers

When environment does work to a system through pV, the system internal energy changes according to the energy conservation law $dU=dQ+pdV$. If it is adiabatic, then $dU=pdV$.

By definition, $F=U-TS$, or $$dF = dU - TdS$$

From this, because some of the energy should be thrown away in the form of entropy, not all dU can be extracted . The maximum available work is dF, which is less than $dU$. So some of input $pdV$ is not extractable.

By the way, there is one error in your post. It should be $dF \le dW$, which really repeats the statement that not all input work is extractable.

• Isn't dF=dU-TdS-SdT=-SdT-PdV? – Chet Miller Apr 1 '18 at 16:00
• for constant temperature $dF=dU-TdS$ – user115350 Apr 1 '18 at 16:01

If the only heat transferred to the system is from a heat reservoir at a constant temperature $T_R$, and the initial and final temperatures of the system are equal, and equal to the temperature $T=T_R$ of the reservoir, then, from the Clausius inequality: $Q\leq T_R\Delta S$. Therefore, from the first law for a closed system, $$\Delta U\leq T_R\Delta S-W$$and, from the definition of F, $$\Delta F\leq -W$$ or $$W\leq -\Delta F$$The work done by the system on its surroundings is less than or equal to the decrease in the Helmholtz free energy. For a rigid container in which no P-V work can be done, the criterion for equilibrium is that F has reached a minimum.