# Is there an easy way to perform changes of representation given a thermodynamic state function?

I am struggling with this because there doesn't seem to exist a regular pattern to follow, and each case is dependent on the specifics of the exercise. What I mean is the following:

For a closed, monocomponent, closed system, we know $$S=S(U,V,N)=A(UVN)^{1/3}$$ to be a valid state function, with $$A>0$$. Find the fundamental equation in the representation of the Gibbs potential.

I can kind of state the steps to follow:

1. Find the natural variables of said potential. In this case, the Gibbs potential is given by $$G = U - TS + PV$$, therefore $$dG = d(\color{red}{U} -\color{blue}{TS} + \color{green}{PV}) = \color{red}{TdS - PdV + \mu dN} \color{blue}{- TdS - SdT} \color{green}{+PdV +VdP}$$ $$dG = -SdT +VdP +\mu dN \Rightarrow G=G(T,P,N)$$
2. Because we know $$G = U - TS + PV$$, we need to find $$U(T,P,N)$$, $$S(T,P,N)$$ and $$V(T,P,N)$$ (or any other function in each case that allows us to perform the substraction, like $$V=V(T,P)$$.

How could I start? I know I can use the differential form of each potential in certain cases to find certain derivatives that would help (like $$(\frac{\partial U}{\partial V})_{S,N} = -P$$), but my textbooks uses all sorts of procedures which imply taking a hard-to-foresee route. This would not be a problem if I had more time, but I cannot spend thirty minutes trying to find the correct procedure to go from $$S(U,V,N)$$ to $$S(T,P,N)$$ in my exam. Is there a faster, more direct way? Could jacobians be somehow used in this scenario? Or is it just a matter of practice?

Thanks!

You can start with it and notice that $$\frac{\partial G}{\partial T} = -S$$. I believe this is a good starting point and expressing the other variables in terms of $$N,T,p$$ should follow easily from here.
• Wait, I think I could proceed from there, but in this case I can't think of a way to keep solving it. I know that $\frac{\partial G}{\partial T} = -S =-A(UVN)^{1/3}$, but how does that help me? Sorry, I'm very tired today and can't seem to find an answer. Does it have to do with taking second derivatives with respect to the variables we already have and using Schwarz's theorem to find helpful Maxwell relations? – Manuel Jun 4 at 18:27