Heat Capacity of Distinguishable Particles

[BOAS]

Homework Statement

A system of N distinguishable particles, each with two energy levels. The lower energy level has energy equal to zero, and the higher energy level has energy $\epsilon$. The higher energy level is four fold degenerate. Calculate the heat capacity.

Homework Equations

The Attempt at a Solution

I have an expression for the heat capacity, but I think I have a problem that it does not go to zero as temperature goes to zero, and was hoping to get some help on this matter.

The partition function for such a system is given by

$Z = \Sigma_s g_s e^{-\beta E(s)} = 1 + 4e^{- \beta \epsilon}$, where $\beta = \frac{1}{k_B T}$

The average energy of the system $\bar{E} = - \frac{1}{Z} \frac{\partial Z}{\partial \beta} = \frac{4 \epsilon e^{- \beta \epsilon}}{1 + 4 e^{- \beta \epsilon}} = \frac{4 \epsilon}{e^{\beta \epsilon} + 4}$ and the heat capacity is given by $C_v = \frac{\partial \bar{E}}{\partial T}$, so substituting back in the expression for $\beta$

$C_v = \frac{\partial}{\partial T} \frac{4 \epsilon}{e^{\frac{\epsilon}{k_B T}} + 4} = \frac{4 \epsilon^2 e^{\frac{\epsilon}{k_B T}}}{k_B T^2 (e^{\frac{\epsilon}{k_B T}} + 4)^2}$

It looks to me like my expression blows up at T=0.

Any help is appreciated,

thanks!

 

[TSny]

Can you show how you are taking the limit of your expression as T goes to zero?

 

[BOAS]

Can you show how you are taking the limit of your expression as T goes to zero?

Your comment has made me realize I was being stupid, and just plugging zero into my equation.

I have now used l'hopital's rule and seen that the limit is indeed zero. Thanks for the nudge :)