# Thermodynamics - (polymer) caluclate the average potential energy

1. Dec 21, 2013

### skrat

1. The problem statement, all variables and given/known data
A polymer of $10^{20}$ molecules each 2nm long is hanged from the ceiling. The other end of the polymer is attached to a $m=4\cdot 10^{-10} g$ load. Calculate the average potential energy! Temperature is 300 K.

2. Relevant equations
$<E>=<E_p>=\frac{\mathrm{d} (\beta F)}{\mathrm{d} \beta }$

$e^{-\beta F}=\sum_{n}^{N}c_ne^{-\beta E_n}$

3. The attempt at a solution

if $e^{-\beta F}=\sum_{n}^{N}c_ne^{-\beta E_n}$ than

$e^{-\beta F}=\sum_{n=0}^{N=10^{20}}c_ne^{-\beta E_n}$

Now, I'm not sure, but I think that $E_n$ which is potential energy of one particular molecule should be $E_n=mgln$ if l is lenght of one molecule.

therefore $e^{-\beta F}=\sum_{n=0}^{N=10^{20}}c_ne^{-\beta mgln}$

This is now $e^{-\beta F}=\frac{e^{-\beta mgl(N+1)}-1}{e^{-\beta mgl}-1}$

so $\beta F=ln(e^{-\beta mgl}-1)-ln(e^{-\beta mgl(N+1)}-1)$

and

$<E>=<E_p>=\frac{\mathrm{d} (\beta F)}{\mathrm{d} \beta }=mgl(\frac{(N+1)e^{-\beta mgl(N+1)}}{e^{-\beta mgl(N+1)}-1}-\frac{1}{e^{-\beta mgl}-1})$

BUT this gives me $E_p=9.23\cdot 10^{-21}J$ which is only 2.35 nanometers... I siriously doubt that is the case :/ Does anybody know what's wrong here?

Thanks for all the help!

Last edited: Dec 21, 2013
2. Dec 21, 2013

### skrat

okej everything above is completely wrong.

I think it goes something like this: $e^{-\beta F}=\sum_{n=0}^{N}e^{-\beta E_n}$

where $E_n=-mgh(n)$ but the key here is to write h as function of n... Now that's all i have..