- #1
skrat
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Homework Statement
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.
Homework Equations
##<E>=<E_p>=\frac{\mathrm{d} (\beta F)}{\mathrm{d} \beta }##
##e^{-\beta F}=\sum_{n}^{N}c_ne^{-\beta E_n}##
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 length 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!
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