# Derive Internal Energy from Thermodynamic Identity

Tags:
1. Jan 10, 2015

### SalfordPhysics

1. The problem statement, all variables and given/known data
For a single molecule, derive the internal energy U = 3/2kBT
In terms of the partition function Z, F = -kBTlnZ
Where Z = V(aT)3/2

2. Relevant equations
Thermodynamic identity: δF = -SδT - pδV
p = kBT/V
S = kB[ln(Z) + 3/2]

3. 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.

2. Jan 10, 2015

### gabbagabbahey

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

3. Jan 11, 2015

### 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

Last edited: Jan 11, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted