# What meaning does Gibbs energy acquire when applied to solutions?

I have been picturing functions like enthalpy and Gibbs free energy as mathematical tools, which were defined in such a way as to be useful for some specific cases. I understood enthalpy as a useful tool for calculating heat when the only work being done was P-V and the system was closed and at constant pressure. Enthalpy being a state function can be represented as a function of intensive properties, this makes it useful to calculate heat, heat being a path function it is difficult to tabulate experimental values to calculate heat in other cases (different temperature ranges, different substances, etc), but for enthalpy it is not, it is easy to tabulate experimental values that do not depend on mass or path.

For Gibbs energy I did the same, i thought it was a useful function to determine if a process would be spontaneous or not, or in an equilibrium with quantities easier to measure and to control in experimental settings, those being temperature and pressure.

But now that I've been studying solutions all the meaning I had given to Gibbs energy has been thrown out the window. My text book just grabs Gibbs energy to study solutions without a particular justification to do so

If $$nG=f(T,P,n_1,..,n_j)$$ the total differential is

$$d(nG) = nVdp - nSdT + \sum_i \mu_i dn_i$$

Where a new a new property is introduced, the chemical potential. I can follow the math and everything, but what does Gibbs free energy mean here? Why it gets picked up to study solutions? Why not enthalpy or enthropy? Physically now what does $$G$$ mean? It's not work or heat, nor a criteria to determine spontaneity, I don't know what it is and neither why does it work. Not resolving these doubts makes it difficult to follow what happens in the rest of the chapter. Everything is build upon this function, but what is this function? A result of $$G=2 \frac{J}{mol}$$ in this context what would it mean? How does $$G$$ manage to keep everything together?

Another thing, this is an easier question to answer, why do thermodynamics books always use $$\equiv$$ to define such functions? Why not just $$=$$?

• Note that G,H,U,V,S are generally extensive properties. Then there are various intensive molar versions with explicit indexes. Commented Sep 9, 2023 at 5:50
• This Enthalpy being a state function can only be represented as a function of intensive properties is wrong. Commented Sep 9, 2023 at 6:00
• you're right, what I wrote wasn't quite right, I'll fix it Commented Sep 9, 2023 at 6:54
• And about $G,H,U,V,S$, I know they can be extensive properties or intensive properties, they have to be devided by mass in order to be intensive, but in general they are computed via intensive properties (at least on my experience) making them intensive for a "brief time" before multiplying by the systems mass Commented Sep 9, 2023 at 7:00
• Well, if you measure reaction enthalpy, you measure extensive enthalpy, dividing it by amount of substance to get the molar formation or reaction enthalpy. The point is the state variable symbols U,H,S,G,A with implicit indexes are implicitly considered as extensive variables with energy units. // Ask rather, why is equality operator often incorrectly used as the definition or assignment operator. Commented Sep 9, 2023 at 8:23

$$- \frac{\mathrm{d}G}{T} = \mathrm{d}S_\mathrm{sys} + \mathrm{d}S_\mathrm{surr} = \mathrm{d}S_\mathrm{tot}$$