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I came up with a hypothetical equation of state, where one of the consequences is that $U=G$, where $U$ is the internal energy and $G$ is the Gibbs free energy. I wanted to ask the chemists, is there a fundamental problem with this? Are there any real systems where this could happen? Thank you for your time

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$G=U+p \cdot V-T \cdot S$ by its definition and is also often called free enthalpy, as $G = H - T \cdot S$ and $H = U + p \cdot V$, where $H$ is enthalpy. So if $G=U$, then $S=p \cdot V/T$.

$pV/T$ is about constant at usual conditions for real gases, or even exactly constant for ideal gases ( $pV/T=nR$ ). It decreases with $T$ for condensed phases at constant pressure, or can dramatically grow with $T$ for condensed phases at constant volume.

Entropy $S$ increases with $T$ via $\mathrm{d}S=đQ/T=C(T) \cdot \mathrm{d}T/T$.

So $G=U$ could be true just accidentally for particular conditions, where both functions $S=f(T,p)$ and $pV/T = g(T,p)$ cross each other.

$G=U$ has no special meaning, but that the non free part of energy $T \cdot S$, blocked for work usage by the second law of thermodynamics, is accidentally equal to the $pV$ term.

$G=H$, or $A=U$ would be a different case, as it would mean a system must have zero entropy, what is not possible for $T \gt 0$.

( $A=U-TS$ is the Helmholtz (free) energy )

Poutnik
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