Question

As you know, U(S.V.N) is a first-order homogeneous function.

a) If we switch out for T using the appropriate Legendre transform, show that the resulting thermodynamic potential is also a first-order homogeneous function. In general, the Legendre transform preserves homogeneity.

b) Consider the potential F(T.V.N). Starting with F suppose we want to switch out T with its conjugate variable. What is the variable conjugate to T?

c) Use the Legendre transform on F to construct the appropriate function that switches out 7' for its conjugate. Show that the resulting function is simply U(S.V.N). From this we see that since F itself is the Legendre transform of U, applying the transform successively twice gives us back the original function. Symbolically, if we represent the transform operation with S, then S(S(U)) = U

Answer 1

The student's question involves how the Legendre transform, applied to thermodynamic potentials, preserves homogeneity and how applying it twice on internal energy U(S,V,N) returns to the original function.

Step-by-step explanation:

The question is about thermodynamic potentials and how they transform under a Legendre transform, specifically concerning internal energy, U, as a function of entropy (S), volume (V), and particle number (N), and the free energy, F, as a function of temperature (T), volume (V), and particle number (N).

In (a), when replacing entropy (S) with temperature (T) using the Legendre transform (which replaces a variable with its thermodynamic conjugate), we obtain the Helmholtz free energy (F). This transformation preserves the order of homogeneity; hence F(T,V,N) is also a first-order homogeneous function. In (b), the variable conjugate to temperature (T) is the entropy (S). In (c), applying the Legendre transform to F, replacing T with S, we return to the internal energy function U(S,V,N), demonstrating that successive Legendre transforms revert to the original function, analogous to applying the transform operation, S, twice: S(S(U)) = U.