Finding the partition function

[S_Flaherty]

Homework Statement

Consider a solid of N localized, non-interacting molecules, each of which has three quantum states with energies 0, ε, ε, where ε > 0 is a function of volume.


Question: Find the internal energy, Helmholtz free energy, and entropy.

Homework Equations

Z = Ʃe^-E(s)/kT
U = -N(dlnZ/dβ)
S = U/T + NklnZ
F = U - TS = -NkTlnZ

The Attempt at a Solution

For the internal energy would I just multiply N by the average energy?
So U = N((0 + ε + ε)/3) = 2Nε/3?

I also know there are equations for the three of these values that require the partition function, Z.

I know Z = Ʃe^-E(s)/kT so would this just be
Z = 1 + e^-ε/kT + e^-ε/kT = 1 + 2e^-ε/kT
U = -N(dlnZ/dβ)
lnZ = ln(1) + ln(2e^-ε/kT) = ln2 - εβ, so U = Nε...

Which one of those solutions for U is the correct one? Or am I wrong in both cases?

 

[qbert]

Second is right technique, although you've mathed wrongly.

for three states: 0, 1, 2
U = avg energy = E(0) P(0) + E(1) P(1) + E(2) P(2)

For E(0)=0, E(1)=E(2) =ε,
and P(energy) = exp(-energy/kT)/Z
P(0) = 1/Z, P(1)=P(2)=exp(-ε/kT)/Z

 

[S_Flaherty]

Second is right technique, although you've mathed wrongly.

for three states: 0, 1, 2
U = avg energy = E(0) P(0) + E(1) P(1) + E(2) P(2)

For E(0)=0, E(1)=E(2) =ε,
and P(energy) = exp(-energy/kT)/Z
P(0) = 1/Z, P(1)=P(2)=exp(-ε/kT)/Z

So, I get U = 2εe^-ε/kT/Z

I just figured out something I did wrong with Z. dlnZ/dβ = (-2εe^-εβ)/(1 + 2εe^-εβ) so U = (2Nεe^-εβ)/(1 + 2εe^-εβ) ? Or did I mess something up?