Internal Energy as a function of U(S,V,A,Ni)

[mcoth420]

A general thermo question...
for the function describing internal energy U(S,V,A,N)

U=TS-PV+γA+μN

please explain how the total differential is

dU=TdS-PdV+γdA+μdN (for a one component system)

Basically why is dT=dP=dγ=dμ=0? Is it because they are intensive or potentials?

Thank you,

M

 

[Studiot]

Does this help?


\begin{array}{l}<br /> U = U(S,V,.N,A...) \\ <br /> dU = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N,A...}}dS + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_{S,N,A}}dV... \\ <br /> T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N,A...}} \\ <br /> P = {\left( {\frac{{\partial U}}{{\partial V}}} \right)_{S,N,A}} \\ <br /> {\rm{etc}} \\ <br /> \end{array}

I will leave you to fill in the bits for moles and area or other quantities.

 

[mcoth420]

Thank you for your reply...what is this technique called? I am working with Legendre transforms and this is similar...

Another thing...how can you derive T=partialU/partialS or others without knowledge of the internal energy equation?

 

[Studiot]

This is the Gibbs Formulation.

Carington : Basic Thermodynamics P 187ff : Oxford University Press

Also in many Physical Chemistry texts.