What would be the physical interpretation of the equation dG = Vdp − SdT? I was trying to understand the derivation of the formula $G=G^\circ+RT\ln (p/p^\circ)$ starting with the following relationships:
$$
\begin{align}
\mathrm dU&amp;=T\,\mathrm dS-p\,\mathrm dV\tag{1}\\
\mathrm dH&amp;=\mathrm dU+p\,\mathrm dV+V\,\mathrm dp=T\,\mathrm dS+V\,\mathrm dp\tag{2}\\
\mathrm dG&amp;=\mathrm dH-T\,\mathrm dS-S\,\mathrm dT=V\,\mathrm dp-S\,\mathrm dT\tag{3}
\end{align}
$$
Under isothermal condition $\mathrm dG=V\,\mathrm dp$ $(\mathrm dT=0)$. Therefore, the change in $G$ from the standard state is given by:
$$
\begin{align}
\int\mathrm dG&amp;=\int V\,\mathrm dp=\int\frac{nRT}p\,\mathrm dp\tag{4}\\
G-G^\circ&amp;=\ln p-\ln p^\circ\tag{5}\\
G&amp;=G^\circ+\ln\frac p{p^\circ}\tag{6}
\end{align}
$$
But I'm confused about the equation $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$. What does this physically mean?
I'm assuming $V\,\mathrm dp$ is some form of work done by the system, but that suggests $G$ of the system increases while it loses energy doing the work. I also have the same problem for $\mathrm dH=T\,\mathrm dS+V\,\mathrm dp.$ Can someone please explain what they really mean?
For a continuous flow system operating at steady state in contact with a constant temperature reservoir at the same temperature as that of the entering fluid, and with negligible change in kinetic energy and gravitational potential energy, the reversible shaft work per mole that the system does on the surrounding is equal to the integral of -vdP, where v is the fluid molar volume.  So $\Delta G$ per mole is equal to minus the shaft work per mole.