Gibbs free energy and enthalpy relationship

by iScience
Tags: energy, enthalpy, free, gibbs, relationship
 P: 221 1.) Technically, you use $P dV + V dP$ for both of them, but at constant pressure, $V dP =0$, so that $dH = dU + P dV$. In the Gibbs free energy case, you have both constant temperature and pressure, so that both $V dP =0$, and $S dT =0$ giving us that $dG = dU + P dV -T dS$. 2.) $V dP$ and $\mu dN$ are not generally expressions of the same thing. The chemical potential is not uniquely defined by the pressure, since it is a function of temperature as well. Where the enthalpy is equal to the internal energy plus the work needed to displace the atmosphere around the system at constant pressure ($H = U + PV$), the Gibbs free energy is equal to this minus the heat taken (at constant temperature) from the surroundings to bring the system in thermal equilibrium with the surrounding atmosphere ($G = H-TS$). In this sense the Gibbs free energy is like the internal energy plus the work you actually need to put in given that you get this heat for free.