# How to explain Gibbs free energy is a pressure-dependent state function?

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: $$\mathrm dG = V\,\mathrm dp$$.

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

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