Derive Internal Energy from Thermodynamic Identity

[SalfordPhysics]

Homework Statement

For a single molecule, derive the internal energy U = 3/2k_BT
In terms of the partition function Z, F = -k_BTlnZ
Where Z = V(aT)^3/2

Homework Equations

Thermodynamic identity: δF = -SδT - pδV
p = k_BT/V
S = k_B[ln(Z) + 3/2]

The Attempt at a Solution

U = F + TS
δU = δF + δT.S + δS.T
= -SδT - pδV + δT.S + δS.T
δU = -pδV + TδS

(δS/δU)_V = 1/T

However, don't have a variable U in S to differentiate with respect to.

 

[gabbagabbahey]

Hint: don't forget the basic rules of differentiation, namely $\left(\frac{\partial}{\partial U}\right)_{V}S(Z) = \left(\frac{\partial Z}{\partial U}\right)_{V}\left(\frac{\partial}{\partial Z}\right)_{V}S(Z)$

 

[SalfordPhysics]

I still don't see how I can solve it given that I don't have a term U to use.
The only thing I thought was solving p.dV and then substituting using U = -pdV but thing started getting messy.
Can't go for (dS/dU)_V because again, no U.

Solved it, and I was just massively overcomplicating things. Just needed to use S and F known in U = F + TS