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

Attached Files

Internal energy question.doc (94.0 KB, 3 views)

Studiot

Does this help?


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

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.