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The system shown in gure 1 below contains four types of molecules X, Y, Z, and O




Problem set anser Key;Please can you answer to all the questions in the problem set 5;Thank you;Attachment Preview;PS5.pdf Download Attachment;Chemistry C130/MCB C100A Problem Set 5;due March 7, 2014 (15 pts);Please turn in your homework at the beginning of class on Friday.;Reading: Chapter 7;1. (3 pts);(a) The system shown in gure 1 below contains four types of molecules X, Y, Z, and;O that can be arranged in any way on the lattice, subject to the constraint that;each site can be occupied by at most molecule. Calculate the entropy of this system.;You may nd it helpful to study Box 7.1 in the textbook.;Figure 1: Model of a gas mixture for problem 1(a).;(b) How much work is done in compressing one mole of an ideal gas, reversibly, from a;starting volume of 1 L to a nal volume of 250 mL at a constant temperature of 293;K? What is the change in entropy?;(c) If the increase in entropy indicates the direction of spontaneous change, how can a;system ever undergo a process that results in a decrease in the entropy of the system?;2. (3 pts);Imagine you have a container, like the one pictured in gure 2 below, whose individual;boxes are each occupied by a particle of type A or type B. There are NA type-A particles;and NB type-B particles initially segregated by a partition. Every box is occupied, so;that the total number of boxes in the container is N = NA + NB. When the partition is;removed, the A and B particles can exchange positions, leading to mixing of the particles.;(a) Calculate the entropy of this system before and after removal of the partition. What;is the change in entropy of the system?;(b) Write S from part (a) in terms of the mole fractions xA = NA /N and xb = NB /N.;Is the entropy change you calculated intensive or extensive? How do you know?;1;occupied by A;occupied by B;A and B can mix;partition;no partition;Figure 2: NA particles of A and NB particles of B, initially separated by a partition, on a;grid with N = NA + NB boxes.;(c) What does S tell you about whether the two species will spontaneously mix?;3. (3 pts);One of the simplest models of biopolymers like proteins or nucleic acids is a random chain;(or, equivalently, random walk) on a cubic lattice. In this model, each unit (or bead) of;the biopolymer occupies a site on the lattice. Each site can be occupied by at most one;bead, mimicking the steric constraints faced by real biopolymers. Consecutive, bonded;units of the chain must occupy neighboring sites on the lattice (diagonals dont count as;neighbors).;For simplicity, we will consider a two-dimensional chain of six beads on a square lattice.;When two non-consecutive beads occupy neighboring lattice sites, there is an attractive;energy. We can think of this as corresponding to a favorable van der Waals contact or;hydrogen bond. Some example congurations for this chain are shown in gure 3 below.;(a) For this relatively short chain, the allowed congurations can be divided into three;groups with dierent energy: those with no bead contacts (E = 0), those with one;(E =), and those with two (E = 2). Find all congurations with two bead;contacts.;(b) The density of states (E) gives the number of congurations with energy E. You;have already enumerated congurations with E = 2, so you know (2). Given;that () = 22 and (0) = 41, calculate the partition function of the chain.;(c) Congurations with two bead contacts tend to be more compact than those with one;or zero. We can think of these two-contact congurations as dierent conformations;of the folded state of a protein. At what temperature is the protein folded at least;99% of the time?;4. (4 pts);An enzyme has many microscopic congurations, or microstates, available to it. We can;divide these microstates into three groups: active, inactive, and unfolded. For simplicity;2;E=;E=0;E=;2;Figure 3: Example congurations for problem 3. Beads colored red are involved in favorable;contacts.;we will assume that microstates in a group have the same energy. Thus, our model enzyme;has three energy levels, and each energy level has a multiplicity (or density of states) equal;to the number of microstates in that energy level.;(a) Based on your knowledge of protein structure, which energy level would you expect;to have the largest multiplicity? Why?;(b) Suppose the active and inactive states each have a multiplicity of one. In other words;there is only one microstate that is folded and active, and only one microstate that;is folded and inactive, all other microstates are unfolded. You measure the enzyme;activity to be 10% at 300 K. Assuming the unfolded state is so infrequently populated;at this temperature that you can neglect it, what are the energies EA and EI of the;active and inactive states, respectively, relative to the lowest-energy state? What is;the value of the partition function Q?;(c) The temperature is now raised to the melting temperature of the protein, Tm = 325;K. Can the unfolded state still be neglected when calculating the partition function?;What is the value of Q now?;(d) As a rough estimate, you guess that the energy of the unfolded state is 10 times the;energy of the active state, EA, you found in part (b). What is the multiplicity of the;unfolded state? Is this reasonable?;(e) What was the change in entropy of the system when the temperature was increased;from 300 K to 325 K?;5. (2 pts);Recall that the partition function of a system is given by;Q=;i;exp (Ei);3;where = 1/kB T is called the inverse temperature, and is introduced here for convenience.;The probability of being in microstate i is then;pi =;1 Ei;e.;Q;(a) The average energy, or expected value of the energy, is given by the following formula;from probability theory;E=;pi E i.;i;Show that;E =;ln Q.;(b) The variance of the energy is dened as;var(E) = (E)2;where E = E E is the deviation from the mean value. In words, the variance is;the average value of the squared deviation from the mean, or mean-square deviation.;It is straightforward to show that;(E)2 = E 2 E;2.;Given this, show that;(E)2 =;2;ln Q.;2;This problem shows that the partition function contains useful information that can be;extracted by taking the appropriate derivative. We will use the derivatives above in this;weeks extra credit problem.;6. (3 pts extra credit);In last weeks extra credit problem, you derived an expression for the partition function of;an ideal gas particle in one dimension. The generalization to three dimensions is relatively;straightforward and leads to the expression;V;q= 3;h;2m;3/2.;As before, = 1/kB T is the inverse temperature, which is more convenient to work with;in this problem than the absolute temperature T.;If we have an ideal gas of N particles, the full partition function is simply the product of N;one-particle partition functions, or q N. This expression is valid for a gas of distinguishable;particles, for which an interchange of two identical particles leads to a distinct microstate.;In quantum mechanics, however, identical particles are not distinguishable, so that an;interchange of two particles does not lead to a dierent microstate. To correct for this;4;we divide by N !, which is the number of permutations of N particles. Thus, the correct;partition function for an N -particle ideal gas in three dimensions is;Q=;1N;q.;N!;In this problem, be sure to show all work in detail. You may nd the derivative formulas;in problem 3 to be helpful.;(a) Calculate the average energy of an ideal gas.;(b) Calculate the mean-square deviation in energy of an ideal gas.;(c) The root-mean-square deviation, or RMSD, is simply the square root of the meansquare deviation;E =;(E)2.;Note that the square root is taken after computing the variance. The RMSD of the;energy, E, characterizes the typical size of energy uctuations in the system. More;meaningful is the ratio of this quantity to the average energy, E/ E, which is like;a signal-to-noise ratio. How does this ratio scale with the number of particles N in;the gas? What does this tell you about uctuations in thermodynamic systems, for;which N is on the order of Avogadros constant?;(d) In contrast to thermodynamic systems, which are macroscopic in size, the cellular;environment is very small. In some cells, a particular protein might number in the;dozens or hundreds. Given what you found in part (c), what can you say about the;importance of uctuations in the cellular environment?;5


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