Infinitely long molecular zipper

[subzero0137]

Homework Statement

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A "molecular zipper" has two rows of molecules, and each row has a large number of monomers. A monomer from one row is weakly linked to a monomer in the other row. The zipper can unzip from one end by breaking the bond between pairs of monomers. A bond can be broken only if the bond to its left is already broken. It requires energy ε to break each bond, i.e. for every broken bond, the zipper absorbs energy ε. Thus, for every extra bond that is broken, the zipper can be regarded as moving up on the energy ladder by ε.

Treating the zipper as infinitely long, find expressions for the energy, the Helmholtz free energy, heat capacity and entropy of such a zipper at temperature T. [Hint: A zipper can have n broken bonds at T. Each value of n corresponds to one microstate. You need to find the partition function Z at temperature T. You may assume $$a + ar + ar^2 + ... = a \sum_{0}^\infty r^n = \frac {a}{1 - r}$$

Homework Equations

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$$Z = \sum e^{- \frac {ε_{i}}{kT}}$$

The Attempt at a Solution

I'm trying to find the partition function Z. I think ##ε_i = iε## here, so

$$Z = e^{ \frac {-0ε}{kT}} + e^{ \frac {-1ε}{kT}} + e^{ \frac {-2ε}{kT}}... = \frac {1}{1 - e^{ \frac {-ε}{kT}}}$$

but I'm not sure if it's right because if it can only have n broken bonds at T, should it not be a finite series?

 

[mjc123]

Read the question again. "Each value of n corresponds to one microstate". Each molecule has its own value of n, and there is a distribution of n values in the sample. (n corresponds to the index i in your equation for Z. It is not the limit of i. The limit of i (or n) is infinity.)

 

[subzero0137]

Read the question again. "Each value of n corresponds to one microstate". Each molecule has its own value of n, and there is a distribution of n values in the sample. (n corresponds to the index i in your equation for Z. It is not the limit of i. The limit of i (or n) is infinity.)

Thanks for the reply. So I'm on the right track with my partition function?

 

[mjc123]

Looks like it.