h1. Question 1 *An iron rod of length* <math>l</math> *and made up of* <math>N</math> *iron atoms is held in vacuum* (<math>P=0</math>) *under uniaxial tension with a constant force* <math>F</math>*, at constant temperature* <math>T</math>*.* <p> </p> *(a) What is the characteristic potential for this set of thermodynamic boundary conditions?* <center> <br> <math>dE = TdS - pdV + Fdl + \mu dN</math> <br> <math>E = TS - pV + Fl + \mu N</math> <br> </center> Since there is a vacuum, <math>p=0</math>. <p> </p> The controlling variables are <math>T</math>, <math>F</math>, and <math>N</math>. <center> <br> <math>\phi = E -TS - Fl</math> <br> <math>d \phi = -S dT - l dF + \mu dN</math> <br> </center> Write in terms of a Legendre transform of the energy. <center> <br> <math>- \beta \phi = \frac{S}{k} - \beta E + \beta F l</math> <br> </center> *(b) Write down the appropriate partition function for this ensemble (call it* <math>\wedge</math>) *and express the characteristic potential in terms of this partition function.* <center> <br> <math>\wedge = \sum_j \exp [- \beta E_j + \beta Fl_j]</math> <br> <math> -\beta \phi = \ln \wedge</math> <br> <math>\phi = -kT \ln \wedge</math> <br> </center> *(c) Express the thermodynamic variables* <math>l</math>*,* <math>S</math>*,* <math>\mu</math>���*(chemical potential of* <math>Fe</math>*), and* <math>E</math> *(the internal energy) as a function of the partition function.* Start with the equations of state derived from an expression of <math>\phi</math> and differentiate to express thermodynamic variables as a function of the partition function. <center> <br> <math>l = - \left ( \frac{\partial \phi}{\partial F} \right )_{T, N}</math> <br> <math>l = kT \left ( \frac{\partial \wedge}{\partial F} \right )_{T, N}</math> <br> <math>S = - \left ( \frac{\partial \phi}{\partial T} \right )_{F, N}</math> <br> <math>S = k \ln \wedge + kT \ln \left ( \frac{\partial \wedge}{\partial T} \right )_{F, N}</math> <br> <math>\mu = - \left ( \frac{\partial \phi}{\partial N} \right )_{T, N}</math> <br> <math>\mu = - kT \left ( \frac{\partial \ln \wedge}{\partial N} \right )_{T, F}</math> <br> </center> Derive an expression of <math>E</math> with equations determined above. <center> <br> <math>E = \phi + TS + Fl</math> <br> <math>E = -kT \ln \wedge + kT \ln \wedge + kT^2 \left ( \frac{\partial \ln \wedge}{\partial T} \right )_{F, N} + kTF \left ( \frac{\partial \ln \wedge}{\partial F} \right )_{T, N}</math> <br> <math>E = kT^2 \left ( \frac{\partial \ln \wedge}{\partial T} \right )_{F, N} + F kT \left ( \frac{\partial \ln \wedge}{\partial F} \right )_{T, N}</math> <br> </center> h1. Question 2 *a) Derive an expression of the fluctuations of the volume of a system at constant* <math>T</math>*,* <math>P</math>*, and* <math>N</math>*.* <center> <br> <math>\frac{\overline{V^2} - \overline V^2}{\overline V^2}</math> <br> </center> Consider the isothermal-isobaric ensemble and a related partition function. <center> <br> <math>\Delta = \sum_j \exp \left [\frac{-E_j}{kT} \right] \exp \left [\frac{-pV_j}{kT} \right]</math> <br> </center> Follow a three step procedure. * Multiply both sides by the partition function * Differentiate with respect to conjugate of mechanical variable * Divide by partition function <center> <br> <math>\mbox{Step 1}</math> <br> <math>\Delta \overline{V} = \sum_j V_j \left [\frac{-E_j}{kT} \right] \exp \left [\frac{-pV_j}{kT} \right]</math> <br> <math>\mbox{Step 2}</math> <br> <math>\Delta \frac{\partial \overline{V}}{\partial p} + \overline V \frac{\partial \Delta}{\partial p} = \frac{\partial}{\partial p} \left ( \sum_j V_j \left [\frac{-E_j}{kT} \right] \exp \left [\frac{-pV_j}{kT} \right] \right )</math> <br> <math>\Delta \frac{\partial \overline V}{\partial p} + \overline V \left [\sum_j \left ( - \frac{V_j}{kT} \right ) \exp \left [ \frac{-E_j}{kT} \right] \exp \left [\frac{-pV_j}{kT} \right] \right ] = \sum_j \left ( - \frac{V_j^2}{kT} \right ) \exp \left [\frac{-E_j}{kT} \right] \exp \left [\frac{-pV_j}{kT} \right]</math> <br> <math>\mbox{Step 3}</math> <br> <math>\frac{\partial \overline V}{\partial p} + \overline V \left ( - \frac{\overline V}{kT} \right ) = \frac{ - \overline V^2}{kT}</math> <br> <math>\frac{\overline{V^2} - \overline V^2} = -kT \left ( \frac{\partial \overline V}{\partial p} \right )</math> <br> <math>\frac{\overline{V^2} - \overline V^2}{\overline V^2} = \frac{-kT}{\overline V^2} \left ( \frac{\partial \overline V}{\partial p} \right )</math> <br> <math>\frac{\overline{V^2} - \overline V^2}{\overline V^2} = \frac{kT}{\overline V \left ( \frac{\partial \overline V}{\partial p} \right )</math> <br> <math>\frac{\overline{V^2} - \overline V^2}{\overline V^2} = \frac{kT}{\overline V} \kappa </math> <br> <math>\kappa = - \frac{1}{V} \left ( \frac{ \partial \overline V }{ \partial p } \right )</math> <br> </center> *b) Evaluate the expression obtained in (a) in the case of an ideal gas* <center> <br> <math>pV = NkT</math> <br> <math>\kappa = - \frac{1}{V} \left ( \frac{\partial V}{\partial p} \right )</math> <br> <math>\kappa = \left ( \frac{-1}{V} \right ) \left ( \frac{-NkT}{p^2} \right )</math> <br> <math>\kappa = \frac{1}{p} </math> <br> <math>\frac{\overline{V^2} - \overline V^2}{\overline V^2} = \frac{kT}{\overline V} \left ( \frac{1}{p} \right )</math> <br> <math>\frac{\overline{V^2} - \overline V^2}{\overline V^2} = \frac{1}{N}</math> <br> </center> This is a general result of the fluctuations of an extensive variable of an ideal gase. It means the fluctuations are small when <math>N</math> is large. <p> </p> *(c) In the case of a general single component system of macroscopic size, when can fluctuations in volume become large?* <p> </p> Fluctuations can become large near a critical point where the following is true. <center> <br> <math>\kappa = - \frac{1}{V} \left ( \frac{\partial \overline V}{\partial p} \right ) \right \infty</math> <br> </center> h1. Question 3 *Consider a surface with* <math>M</math> *potential adsorption sites of which* <math>N < M</math> *are occupied by argon atoms. The argon atoms do not interact with each other on this surface, but there is an energy* <math>-\epsilon</math> *associated with each adsorbed argon atom. Hence the energy of the system (surface + adsorbed argon atoms) can be written as below.* <center> <br> <math>E = -N \epsilon</math> <br> </center> *(a) What is the degeneracy (number of microstates) for this system at fixed* <math>N</math> *and* <math>V</math> *?* <p> </p> The degeneracy is the number of ways to distribute <math>N</math> particles and <math>(M-N)</math> vacancies over <math>M</math> sites. <center> <br> <math>\Omega = \frac{M!}{N!(M-N)!}</math> <br> </center> *(b) At constant* <math>N</math>*,* <math>V</math> *and* <math>T</math>*, determine the appropriate partition function for this system. (Do not leave partition function as sum over states. Write a summed form).* <p> </p> Constant <math>N</math>, <math>V</math>, <math>T</math> is associated with a canonical ensemble. <center> <br> <math>Q = \sum_j e^{- \beta E_j}</math> <br> <math>Q = \sum_E Q(E) e^{-\beta E}</math> <br> </center> Energy, <math>E</math>, depends only on <math>N</math> and not a particular arrangement of atoms. Since <math>N</math> is fixed, there is only one energy level. <center> <br> <math>Q = \frac{M!}{N!(M-N)!} e^{\beta N \epsilon}</math> <br> </center> *(c) Obtain an expression for the chemical potential of the argon atoms on the surface.* <center> <br> <math>F = -kT \ln Q</math> <br> <math>F = -kT \left [\ln \left ( \frac{M!}{N!(M-N)!} \right ) + \beta N \epsilon \right]</math> <br> <math>F = - kT \left [\ln (M!) - N \ln N + N - (M-N) \ln (M-N) + (M-N) \right] - N \epsilon</math> <br> <math>\mu = \left ( \frac{\partial F}{\partial N} \right )_{T,V}</math> <br> <math>\mu = -kT \left [- \ln N - 1 + \ln (M-N) + 1 - 1 \right] - \epsilon</math> <br> <math>\mu = -\epsilon + kT \ln \left ( \frac{N}{M-N} \right )</math> <br> <math>x = \frac{N}{M}</math> <br> <math>\mu = - \epsilon + kT \ln \left ( \frac{x}{1-x} \right )</math> <br> </center> h1. Question 4 *Short answer questions:* <p> </p> *(a) For an ideal gas, we derived the ideal gas equation of state pV=NkT and found the energy of the gas to be*<math> E=(3/2)NkT</math>*. List all assumptions used to derive this result.* <p> </p> The following was assumed: * Boltzmann statistics * non-interacting particles * gas particles indistinguishable * mono-atomic particles, in which electronic and nuclear excitations are neglected. *(b) We derived for bosons that the average number of particles in a single particle state* <math>k</math> *is* <center> <br> <math>\overline n_k = \frac{1}{e^{\beta (\epsilon_k - \mu )} - 1}</math> <br> </center> *where* <math>\epsilon_k</math> *is the energy of the single particle state* <math>k</math>*. What value(s) does the chemical potential* <math>\mu</math> *assume when the bosons are not conserved (i.e. the bosons are for example photons or phonons).* <p> </p> The chemical potential is zero. <p> </p> *(c) Can Boltzmann statistics be applied to fermions and bosons?* <p> </p> Yes. Boltzmann statistics can be applied at high temperature, low density, and high mass. <p> </p> *d) In a particular solid solution of* <math>A</math> *and* <math>B</math> *atoms with composition* <math>50% A</math> *and* <math>50% B</math>*, the probability* <math>P_{AB}</math> *that a particular bond is* <math>A-B</math> *(as opposed to* <math>A-A</math> *or* <math>B-B</math>)*, is* <math>0.25</math>*. This state has a given configurational entropy of mixing* <math>S_0</math>*. In order to make the entropy increase, should I increase or decrease* <math>P_{AB}</math>*?* <p> </p> The term <math>P_{AB}</math> of a totally random solution is equal to <math>2x_Ax_B = 0.5</math>. Therefore, the value of <math>P_{AB} = 0.25</math> represents short-range clustering. The restriction on the number of microstates reduces the entropy. To increase <math>S</math> it is needed to increase <math>P_{AB}</math> towards <math>0.5</math>. h1. Question 5 *Below is given a description (in some case with figure) of the macroscopic state of an* <math>AB</math> *alloy that mixes on square lattice. In each case you are asked to find the following.* * *1) the total configurational entropy is as function of* <math>2N</math>*, the number of lattice sites* * *2) the entropy per lattice site in the thermodynamic limit (i.e. as* <math>2N</math> *goes to infinity).* <p> </p> *Note that there are* <math>2N</math> *lattice sites. So in the perfectly ordered alloy the number of* <math>A</math> *is* <math>N</math> *and the number of* <math>B</math> *atoms is* <math>N</math>. <p> </p> * *a) A perfectly ordered system with composition* <math>AB</math> * *b) Same system as in a) but with one excess* <math>B</math> *atom on the sublattice of* <math>A</math> *atoms.* * *c) Same system as in a) but with* <math>1%</math> *of the* <math>A</math> *atoms replaced by* <math>B</math> *atoms.* * *d) Same system as in a) but with one nearest neighbor pair of atoms exchanged as in picture below).* <center> <table cellpadding = 10> <tr> <td> </td> <td> <center> <math>S_{tot}</math> </center> </td> <td> <center> <math>\frac{S_{tot}}{2N}</math> </center> </td> </tr> <tr> <td> (a) </td> <td> <center> <math>k \ln 1 \mbox{ or } \ln 2</math> </center> </td> <td> <center> <math>0</math> </center> </td> </tr> <tr> <td> (b) </td> <td> <center> <math>k \ln N \mbox{ (} N \mbox{ ways to insert atom)}</math> </center> </td> <td> <center> <math>0</math> </center> </td> </tr> <tr> <td> (c) </td> <td> <center> <math>-Nk \left [\ln 0.01 + 0.99 \ln 0.99 \right |0.01]</math> </center> </td> <td> <center> <math>-\frac{1}{2} k \left [0.01 \ln 0.01 + 0.99 \ln 0.99 \right |]</math> </center> </td> </tr> <tr> <td> (d) </td> <td> <center> <math>\frac{4 \cdot 2N}{2} \mbox { pairs that can be exchanged} \right k\ln4N</math> </center> </td> <td> <center> <math>0</math> </center> </td> </tr> </table> </center> |