Partition Function for Thermodynamic System

[Hart]

Homework Statement

I. Finding the partition function Z.

II. If the middle level (only) is degenerate, i.e. there are two states with the same energy, show that the partition function is:

Z = (1+exp(\frac{-\epsilon}{k_{B}T}))^{2}

III. State the Helmholtz free energy F of the assembly in part II.

IV. Show the entropy of the assembly in part II is:

S = 2Nk_{B} ln(1 + exp^({\frac{-\epsilon}{k_{B}T}})) + \frac{2Nk_{B}(\frac{-\epsilon}{k_{B}T}) exp(\frac{-\epsilon}{k_{B}T})}{1+exp(\frac{-\epsilon}{k_{B}T})}

Homework Equations

Partition function for a system that can exist in energy levels \epsilon_{1},\epsilon_{2},..,\epsilon_{i},.. etc. is defined as:

Z=\sum_{i}exp(\frac{-\epsilon_{i}}{k_{B}T})

The Attempt at a Solution

Part I:

Part II: Don't know how to do this! :|

Part III:

Free energy: F=-Nk_{B}Tln(Z)

Each atom replaced by 3 oscillators..

.. therefore: F=-3Nk_{B}Tln(Z)

Define: Z = \frac{exp(-\frac{\theta}{2T})}{1-exp(-\frac{\theta}{T})}

.. hence after substitution:

F=-3Nk_{B}Tln(\frac{exp(-\frac{\theta}{2T})}{1-exp(-\frac{\theta}{T})})}

Which rearranges to:

Z = \frac{exp(-\frac{\theta}{2T})}{1-exp(-\frac{\theta}{T})}

Part IV: Don't know how to do this! :|

 

[Hart]

Ok, after looking at this question details again earlier today..

The partition function for a system that can exist in energy levels \epsilon_{1}, \epsilon_{2}, ... , \epsilon_{i}, ... is defined as:

Z = \sum_{i}exp\left(\frac{-\epsilon_{i}}{k_{B}T}\right)

Suppose that the energy levels of each particle in an assembly of N localised particles in a solid are: 0 , \epsilon, 2\epsilon

Therefore attempts at the solutions:

I.

Input into the expression: exp\left(\frac{-\epsilon_{i}}{k_{B}T}\right)

the values of 0, \epsilon, 2\epsilon which results in three similar equations.

Then using the fact that the equation for the partition function Z contains the sum of these values, simply add the three calculated values together. Therefore deduce:

Z = \sum_{i}exp\left(\frac{-\epsilon_{i}}{k_{B}T}\right) = 1 + exp\left(\frac{-\epsilon}{k_{B}T}\right) + exp\left(\frac{-2\epsilon}{k_{B}T}\right)

Which is the value of the partition function.

II.

Z = 1 + exp\left(\frac{-\epsilon}{k_{B}T}\right) + exp\left(\frac{-2\epsilon}{k_{B}T}\right)

If the middle level is degenerate then the middle term in the expression of Z derived (i.e. therefore the second term) appears twice. Therefore simply multiply this term by 2, and calculating the resulting new value of Z (proving the stated value of Z to be shown):

Z = \left(1 + exp\left(\frac{-2\epsilon}{k_{B}T}\right)\right)^{2}

III.

Helmholtz Free Energy of the value derived in part II:

F = - Nk_{B}T\ln{(Z)} = Nk_{B}T\ln{\left(1 + \left(\frac{-2\epsilon}{k_{B}T}\right)\right)^{2}}

.. which I don't think I can do anymore with other than state as that.

IV.

S = - \left(\frac{\partial F}{\partial T}\right)_{V} = -Nk_{B} \left\frac{\partial}{\partial T}\left(T\ln{\left(1 + \left(\frac{-2\epsilon}{k_{B}T}\right)\right)^{2}} \right)

.. and then just need to calculate:

\left(\frac{\partial F}{\partial T}\right)

.. which should give the expression to be shown.


Is this all correct?