# Homework Help: Partition function of simple system

1. Apr 29, 2014

### unscientific

1. The problem statement, all variables and given/known data

A molecule has 4 states of energy -1, 0,0 and 1. Find its partition function and limit of energy as T → ∞.

2. Relevant equations

3. The attempt at a solution

$$Z = \sum_r e^{-\beta E} = e^{-\beta} + 2 + e^{\beta}$$

$$U = -\frac{\partial ln(Z)}{\partial \beta} = \frac{e^{-\beta} - e^{\beta}}{e^{-\beta} + 2 + e^{\beta}}$$

As $T→\infty$, $exp(-\beta) \approx 1 - \beta$ and $exp(\beta) \approx 1 + \beta$.

Thus,
$$U \approx \frac{(1-\beta) - (1+\beta)}{2 + (1+\beta) + (1-\beta)} = -\frac{\beta}{2} = -\frac{1}{2kT}$$

The equipartition theorem should take over with Energy = 4 * (1/2)kT = 2kT = 2/β.
But instead i'm getting -β/2.

2. May 2, 2014

bumpp

3. May 2, 2014

### skrat

Ammm, okay...

Your calculation is right, at least I got the same result for $U$.

my comment about the equipartition theorem: My experiences are that you really have to master thermodynamics to completely understand this theorem. Lots of results can be "guessed" if you truly understand the concept. I was never that good at it therefore I always had to do the long calculations.
Ok, now to tell something that is actually useful:

from http://chemwiki.ucdavis.edu/Physical_Chemistry/Statistical_Mechanics/Equipartition_Theorem (Degrees of freedom):
"The law of equipartition of energy states that each quadratic term in the classical expression for the energy contributes ½kBT to the average energy."

Let's take a molecule of ideal gas for example: One molecule has in fact $6$ degrees of freedom. $3$ of them precisely describe it's position and are called coordinates (x,y,z), the other $3$ are of course components of momentum (note that momentum is quadratic in energy $E_k=\frac{p^2}{2m}$). Each component of momentum therefore contributes $\frac{1}{2}kT$, so the average energy of molecule of ideal gas is $\frac{3}{2}kT$.

I guess all I am trying to say is that you have no quadratic degrees of freedom and therefore your calculation using equipartition theorem is wrong.

ps: Keep in mind that I never mastered that theorem. I hope I didn't just make a fool out of myself.

4. May 2, 2014

### Fisica

try expanding the exponential in different form

5. May 2, 2014

### Fisica

Exp(-x) = 1/ exp(x) = 1/ (1+x)

try in this form and show what you get, i hope this work

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