# Heat Capacity of Distinguishable Particles

## 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.

## 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!

## Answers and Replies

TSny
Homework Helper
Gold Member
Can you show how you are taking the limit of your expression as T goes to zero?

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

Your comment has made me realise 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 :)