Gibbs free energy and enthalpy relationship

1. iScience

379
***i just realized that i posted this in the wrong section; could a moderator perhaps move this thread to the classical physics section please? Sorry for the trouble***

enthalpy is the energy available if a system with some defined volume were to be annihilated and have the atmosphere collapse inward where the atmospheric pressure remains constant. hence..

dH= dU + PdV (i like sticking to differentials)

Gibbs: this one takes place at constant temperature and pressure. I don't understand why we do the following for the gibbs:

G=U+PV-TS ----> dG=dU+d(PV)-d(TS) ----> dG=dU+PdV+VdP-TdS-SdT

Question 1: the cases for both the enthalpy and gibbs free energy involve the environment being at constant pressure hence the PV term right? (if i'm wrong please correct me) so why do i use PdV for one of them (enthalpy case) and PdV+VdP for the other (the gibbs case)?

Question 2: VdP looks like the case where we are shoving/compressing air into a fixed volume (perhaps a container or something...), so i was wondering, could this term could be somehow combined with the μdN term? it doesn't seem necessary to keep both terms around since they seem like different expressions for the same thing.

Last edited: Dec 28, 2013
2. jfizzix

324
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.

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