3.012 PS Issued: Fall 2003 Graded problems due:

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1 3.012 PS Issued: Fall 2003 Graded problems due: Graded problems: 1. Planes and directions. Consider a 2-dimensional lattice defined by translations T 1 and T 2. a. Is the direction [hk] perpendicular to the plane (hk)? Always? Sometimes? Never? b. Let s now extend our consideration to three-dimensional lattices. Is the direction [hkl] perpendicular to the plane (hkl)? 2. Assignment of Miller-Bravais indices. The following process leads to unambiguous determination of the Miller-Bravais indices (hkl) for a rational lattice plane in terms of its intercepts (in units of a, b, and c) on the edges of the unit cell. The method is based on the fact that the intercepts of the first plane from the origin are 1/h 1/k 1/l and those of the n th plane from the origin are n/h n/k n/l. Therefore, for any rational plane in the stack one can: (a) find n/h n/k n/l; (b) take the reciprocals h/n k/n l/n; (c) clear the common factor 1/n to obtain hkl. The indices hkl, therefore, should contain no common factor. However, one often sees diffraction patterns with peaks indexed in terms of the indices of the d hkl that gave rise to them. Some of these are assigned indices such as 222, 400, or 550 (to name a few) that obviously contain a common factor. What s going on here? Don t these people know how to assign Miller-Bravais indices properly? Please provide an explanation of this situation using Bragg s law. 3. 1D diffraction. We direct a beam of X-radiation with wavelength l = 1 Å at an angle of incidence m = 52 o onto a real one-dimensional crystal consisting of a row of identical atoms spaced at a lattice constant a = 3.5 Å. A 10x10 cm square piece of film is placed in an orientation normal to the lattice row (see sketch of the arrangement below). a. At what distance from the crystal must the film be placed if a record of the complete diffraction cone is to be recorded on the film? b. Sketch qualitatively (but realistically) the distribution of intensity that you would expect to see on the film, remembering that the electrons scattering the radiation have a spatial distribution about the atom that is comparable to 1 Å (with a lattice constant of 3.5 Å, the atomic radius will be 1.75 Å). c. Will a record of any other diffraction cone appear on the film? PS 4 1 of 14 10/15/03

2 4. Identifying symmetries. Two patterns follow that are translationally-periodic and which possess symmetry elements of various sorts. a. Draw on the patterns a set of dots that represent lattice points (bold, plump little dots, please, so that they may be readily discerned by a grader!). Connect four of them to define a cell. b. Draw in, using standard symbols (bold lines for mirror lines, n-gons for rotation axes), the loci of all symmetry elements that you can find. (Note: As everything is repeated by translation you need do this only at the periphery and within the area encompassed by one unit cell). c. Identify with shading the asymmetric unit- that minimal area of the pattern which, upon specification of the symmetry of the pattern, would permit generation of the entire pattern PS 4 2 of 14 10/15/03

3 3.012 PS 4 3 of 14 10/15/03

4 3.012 PS 4 4 of 14 10/15/03

5 5. Interpretation of the second law. Mortimer problem 4.46, parts a-f. (a) false the entropy of the system + surroundings must increase in a spontaneous (irreversible) process occurs, but the entropy of the system alone may increase or decrease. (b) False same as part (a). (c) True. (d) False- the entropy of the system + surroundings remains constant in a reversible process. (e) False- same as part (d). (f) True. 6. The free energy of a rubber band[dji1]. If a rubber band is stretched adiabatically, its temperature increases (try it! Stretch a rubber band and press it immediately against your lip). If we assume the rubber band maintains a constant volume (though its shape changes), we can write it s fundamental equation as: du = TdS + (kl)dl where k is an elastic constant for stretching the band (equivalent to the elastic constant of a spring), and l is the length of the rubber band. Measurements show the elastic constant k is always positive and increases with increasing temperature. (a) Derive the Helmholtz free energy for the rubber band at constant volume. df = du - TdS - SdT df =-SdT + (kl)dl - PdV at constant volume, dv = 0 df =-SdT + (kl)dl (b) Derive the Maxwell relation for this material from the Helmholtz free energy and use it to predict whether the entropy increases, decreases, or stays the same when the rubber band is stretched isothermally. The Maxwell relation is obtained by differentiating first with respect to temperature, and second with respect to length: Ê F ˆ Á =-S Ë T l Ê 2 F ˆ Ê (-S)ˆ Á = Á Ë l T Ë l T PS 4 5 of 14 10/15/03

6 Reversing the order of differentiation, we have: Finally, equating the two derivatives: Ê F ˆ Á = kl Ë l T Ê 2 F ˆ Ê (kl)ˆ Ê k ˆ Á = Á = lá Ë T l Ë T l Ë T l Ê 2 F ˆ Ê 2 F ˆ Á = Á Ë l T Ë T l Ê S ˆ Ê k ˆ -Á = lá Ë l T Ë T l The Maxwell relation says that the entropy of the rubber band decreases if the rubber band is stretched isothermally- because the derivative on the right hand side is positive, given that k increases with increasing temperature. (c) If the rubber band is stretched adiabatically, does the internal energy increase, decrease, or stay the same? (Show why). In an adiabatic process, the entropy change ds = 0 (dq rev = 0). This leaves: du = (kl)dl Since dl, k, and l are all positive in a stretching process, the internal energy must increase PS 4 6 of 14 10/15/03

7 Optional problems: 7. A crystal that is cubic, with lattice constant a = 10 Å is irradiated with Cu Ka radiation (l Cu Ka = Å). A side-view of the experiment and the sphere of reflection that is employed in the Ewald Construction to interpret the experiment is shown below. The vectors that are used to describe the reciprocal lattice are a 1 * = l/d 100 = l/a 1, a 2 * = l/a 2, and a 3 * = l/a 3. The reciprocal lattice vectors are parallel to the directions of the corresponding a 1, a 2, and a 3. The reciprocal lattice vector s hkl that must be brought onto the surface of the sphere of reflection to generate diffraction by d hkl is given by s hkl = ha 1 * + ka 2 * + la 3 *. The crystal is rotated about an axis normal to the incident beam to satisfy the diffraction condition. a 3 * is oriented along r the rotation axis and a 1 * is initially parallel to S o, the direction of the incident beam. a. What is the maximum value of l for which a diffraction peak can be generated? b. What is the maximum value of h for which a peak will be generated? c. What is the total number of different diffraction peaks of the class hk0 that will be generated as the crystal is rotated through 360? ( ±h ±k and ±l are to be considered different indices). d. Through what angle in a counter-clockwise direction must the crystal be rotated in order to generate the 320 diffraction peak? PS 4 7 of 14 10/15/03

8 8. More practice with symmetry recognition. Repeat problem 4 for the two (more difficult) periodic patterns provided below PS 4 8 of 14 10/15/03

9 3.012 PS 4 9 of 14 10/15/03

10 9. The second law applied to heating processes[dji2]. a. One kg of water at 273K is brought into contact with a heat reservoir at 373K. When the water has reached 373K, what is the entropy change of the water, of the heat reservoir, and of the universe? b. If the water had been heated from 273K to 373K by first bringing it into contact with a reservoir at 323K and then with a reservoir at 373K, what would have been the entropy change of the universe? c. Explain how the water might be heated from 273K to 373K with almost no change of entropy of the universe. The heat capacity of water can be taken as: C p L = J/mole K (a) First, the entropy change in the system is obtained using the relationship between heat capacity and entropy: 373 L nc P L DS system = Ú dt = nc P ln = 4184 ln =1305.8J /K T n is the number of moles of water (needed to convert from the heat capacity per mole to the total amount of entropy in the system). To calculate the entropy change of the reservoir, we need to know how much heat was passed from the reservoir to the system during this process: = dq P dt C L 373 \q L system = Ú nc P dt = 418,400J 273 Then the entropy change of the reservoir is given by: dq ds = rev T -q \D S system -418, 400 reservoir = = = J /K 373K T reservoir Finally, the entropy change of the universe is simply the sum of the entropy changes in the system and the reservoir (nothing else in the universe changes in this process): DS universe =DS system +DS reservoir =184.1J /K PS 4 10 of 14 10/15/03

11 For the first step, heating to 323K: (b) We solve this part in a similar manner, by accounting entropy changes in each step. 323 L nc P L DS system = Ú dt = nc P ln = 4184 ln = 703.7J /K T q L system = Ú nc P dt = 209,200J 273 DS reservoir = -q system = -209,200 =-647.7J /K 323K T reservoir and in the second step, heating from 323K to 373K: 373 L nc P L DS system = Ú dt = nc P ln = 4184 ln = 602.1J /K T q L system = Ú nc P dt = 209,200J 323 DS reservoir = -q system = -209,200 = J /K T reservoir 373K DS universe =DS system +DS reservoir = 97.3J /K (c) As you see from parts (a) and (b), heating the water step-wise by sequential equilibrations with many reservoirs having slightly increasing temperature reduces the total entropy change of the universe. In the limit of an infinite number of reservoirs, the total entropy change of the universe becomes zero (the process is reversible). 10. Gibbs free energy and Helmholtz free energy at equilibrium. Mortimer problem (a) True only for processes at constant pressure and constant temperature. The Gibbs free energy was designed to simplify constant T,P calculations. (b) True only for processes at constant volume and constant temperature. (c) False- the second law dictates only the behavior of the entropy at equilibrium. (d) True for closed systems (no external work applied on the system) PS 4 11 of 14 10/15/03

12 (e) True for closed systems (no external work applied on the system). (f) False- work can still be done on the surroundings. (g) True- the entropy of the surroundings is only affected by heat transfer: ds = dq/t. (h) True. 11. The equivalence of maximizing entropy and minimizing internal energy. A pair of blocks initially at different temperatures are placed in thermal contact. The blocks have constant total volume. Using an analysis similar to our entropy-based calculations in class, show that if the total entropy of the blocks A and B is held constant, then du total = 0 at equilibrium will require T A = T B at equilibrium. Our constraint in this analysis is: = S A + S B = constan t S total = ds A + ds B = 0 ds total ds A =-ds B We now calculate the change in internal energies occurring in this process using the fundamental equation for internal energy: du A =T A ds A du B = T B ds B du universe = du A + du B =T A ds A + T B ds B du universe = (T A -T B )ds A If equilibrium in this system is defined by the condition duuniverse = 0, then to satisfy this condition for any arbitrary small change in the entropy of block A, we must have: T A - T B = 0 or T A = T B which agrees with our earlier result obtained from the entropy point of view PS 4 12 of 14 10/15/03

13 12. Identifying stable phases. You are given below measured enthalpy and entropy data taken at 1 atm pressure as a function of temperature for two different phases (A and B) of a new material. Only one of these phases is the equilibrium phase at any given temperature, the other is metastable. a. At T = 350K, which phase is stable, and which is metastable? b. Show why the metastable phase will be driven to spontaneously transform to the stable phase at this temperature. Data: H A = 5.91(T 298) + 2.9x10-3(T ) cal/mole S A = log(T/298) + 2.9x10-3 (T 298) cal/mole K H B = (T 395) x10-3 (T ) cal/mole S B = log(T/395) x10-3 (T 395) cal/mole K We consider a situation of constant temperature (350K) and pressure (1 atm), thus the Gibbs free energy is the right tool to identify the stable phase. The stable form of the material will be the one with lower Gibbs free energy. We can readily determine the Gibbs free energy per mole using the given enthalpy and entropy data because: G H - TS Shown below is a plot prepared by simply calculating G from the given H(T) and S(T) data plugged into a spreadsheet. At T = 350K, the free energy of the A phase is 4408 J/mole, and the B phase is B 4313 J/mole. Since G A is lower than G, the A phase is stable at 350K G (A phase) G (B phase) G (J/mole) T (K) (b) The second law states that processes increasing the entropy of the universe will occur spontaneously. At constant temperature and pressure, this law translates as: a decrease in Gibbs free energy of the system will increase the entropy of the universe, and thus occur spontaneously PS 4 13 of 14 10/15/03

14 A Since G B > G, the process of transforming the B phase into A phase will be driven to occur spontaneously at T = 350K (it would lower the free energy of the system) PS 4 14 of 14 10/15/03

3.012 PS Issued: Fall 2003 Graded problems due:

3.012 PS Issued: Fall 2003 Graded problems due: 3.012 PS 4 3.012 Issued: 10.07.03 Fall 2003 Graded problems due: 10.15.03 Graded problems: 1. Planes and directions. Consider a 2-dimensional lattice defined by translations T 1 and T 2. a. Is the direction