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I'm reading Blundell's Thermal concepts in physics and on page 169 he explains that the different thermodynamic potentials are actually the same quantity, the change in which will take on different forms depending on which variables of the system are fixed, namely the "Availibility" that can be shown to always tend to decrease:

$$ A = U + p_0V - T_0S$$

$$ dA = dU + p_0dV +-T_0dS $$

With $p_0$ and $T_0$ the constant pressure and temperature of the environment.

He explains that for a system of constant entropy and volume, $dA = dU$, since $dV = dS = 0$. And that, in seeking equilibrium (or minimizing its availibility), the system minimizes the internal energy $U$.

$$dU \leq 0$$

But if $dU = -pdV + TdS$ then shouldn't it be impossible for the internal energy to change at all by the logic of $dV = dS = 0$?

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  • $\begingroup$ Your textbook may be older. What it calls "availability" is known in modern usage as Gibbs Free Energy, symbol $G$. The symbol $A$ is sometimes used for a different quantity called Helmholtz Free Energy, which are related by $G = A +pV$. You can avoid a huge amount of confusion by adopting modern terms and notation. $\endgroup$
    – RC_23
    Commented Oct 4, 2022 at 23:22
  • $\begingroup$ Also, I'm not sure I personally agree with the statement "different thermodynamic potentials are actually the same quantity." If by "same quantity" he means "are equal to the work required to change the state of the system under various constraints," then I agree. But the way it's said seems careless $\endgroup$
    – RC_23
    Commented Oct 4, 2022 at 23:28
  • $\begingroup$ @RC_23 you are confusing Gibbs free energy that is indeed a state function with availability, also called exergy, that is not a state function for it also depends on the environment in which the system is embedded. In the example above $T_0$ and $p_0$ are environment parameters and are independent of the system's internal parameters.. $\endgroup$
    – hyportnex
    Commented Oct 6, 2022 at 1:16
  • $\begingroup$ If you find Blundell obscure then read Pippard "ELEMENTS OF CLASSICAL THERMODYNAMICS" pp100-102, very clear description. $\endgroup$
    – hyportnex
    Commented Oct 6, 2022 at 1:20
  • $\begingroup$ I see, maybe I was glossing over a nuance. I thought exergy $E$ was $E=G–G_{Env}$. Is that only a special case? $\endgroup$
    – RC_23
    Commented Oct 6, 2022 at 3:36

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