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I'm aware that through deriving Gibbs free energy to infinitesimal changes, we could get the formula: $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$, giving that Gibbs free energy is pressure-dependent.

However, while dealing with the definition of Gibbs free energy: $ΔG = ΔH - TΔS$ (and $ΔH$ must be held at constant temperature, am I wrong?) most textbooks stated that it is held under constant pressure.

I still couldn't understand the relationship between Gibbs free energy and pressure and why: $\mathrm dG = V\,\mathrm dp$.

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  • $\begingroup$ $G$ is a function of both pressure and temperature. Only at constant temperature is $dG=Vdp$. At constant pressure $dG=SdT$. $\endgroup$
    – porphyrin
    Jul 2, 2019 at 8:52

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$G$ is defined as $H - TS$, and $H$ is $U + PV$.

Since in closed system you have: $\mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V$, it follows that:

  • $\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P$
  • $\mathrm{d}G = -S \mathrm{d}T + V \mathrm{d}P$
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