HW_4_2010_2_SJedit

# HW_4_2010_2_SJedit - APh161 Physical Biology of the Cell...

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Unformatted text preview: APh161: Physical Biology of the Cell Homework 4 Due Date: 5 pm, Friday February 12, 2010 Comments from RP to Class: This fourth homework explores several problems I have done in class in much greater detail. 1. Free energy scale of entropy of chemical interaction in k B T units. The goal of this problem is to get you to examine entropy, free energy and chemical potentials from a few different angles. In addition, the results of part (b) will permit you to do the problem on the critical micelle concentra- tion later in the homework. NOTE: in this problem I am a little imprecise with my free energy definitions, using the Gibbs free energy G in the first part and the Helmholtz free energy F in the second part. (a) Entropy per molecule due to configurations. Imagine the chemical reaction A + B AB . The goal in this part of the problem is to estimate the free energy scale due to entropy when a single reaction occurs changing the number of molecules from N A , N B and N AB to N A- 1 , N B- 1 and N AB + 1. Estimate the free energy difference by using Δ G =- T ( S f- S i ) in conjunction with the Boltzmann expression for the entropy S = k B ln W . As a result, we have Δ G =- k B T ln W f W i , (1) where W f and W i are the multiplicities for the system in the final and ini- tial states, respectively. Hence, to obtain our estimate, all you need to do is construct the ratio W f /W i . Use the lattice models favored in class in order to write this result. Make sure to explain how you convert from the lattice model which involves the number of ligands and the number of lattice sites to a representation in terms of concentrations. Once you have a formula for this free energy, make a numerical estimate of its value by using some “typical” concentrations for reactants and products. (b) Entropy per molecule including the distribution of kinetic energy as part of the entropy. In this second part of the problem, we do something a 1 little more sophisticated for the same reaction by considering the contribu- tion to the entropy coming from the ways of distributing the kinetic energy. That is, all of the molecules in our solution (or gas) share a certain amount of total kinetic energy. But there are many different ways that they can do this sharing and this provides a contribution to the entropy. This result will prove useful to us in the next problem when we work out the critical micelle concentration. In this case, we will use the expression for the relation between the Helmholtz free energy and the partition function, namely, F =- k B T ln Z . To compute this, we use the fact that the total partition function is given by Z tot = Z N 1 N ! , (2) where Z 1 is the partition function of a single-molecule. This partition func- tion can be computed in turn as Z 1 = 1 h 3 Z d 3 q Z d 3 p e- βp 2 / 2 m , (3) where h is Planck’s constant, q is the generalized coordinate and p is the momentum for the molecule of interest. The first integral is just an integral over all the configurations of the system and simply yields...
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HW_4_2010_2_SJedit - APh161 Physical Biology of the Cell...

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