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
- \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
\right )_
</math>
<br>
<math>l = kT \left ( \frac
\right )_
</math>
<br>
<math>S = - \left ( \frac
\right )_
</math>
<br>
<math>S = k \ln \wedge + kT \ln \left ( \frac
\right )_
</math>
<br>
<math>\mu = - \left ( \frac
\right )_
</math>
<br>
<math>\mu = - kT \left ( \frac
\right )_
</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
\right )_
+ kTF \left ( \frac
\right )_
</math>
<br>
<math>E = kT^2 \left ( \frac
\right )_
+ F kT \left ( \frac
\right )_
</math>
<br>
</center>
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
- \overline V^2}
Unknown macro: {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}
\right] \exp \left [\frac{-pV_j}
\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
</math>
<br>
<math>\Delta \overline
= \sum_j V_j \left [\frac{-E_j}
\right] \exp \left [\frac{-pV_j}
\right]</math>
<br>
<math>\mbox
</math>
<br>
<math>\Delta \frac{\partial \overline{V}}
+ \overline V \frac
= \frac
\left ( \sum_j V_j \left [\frac{-E_j}
\right] \exp \left [\frac{-pV_j}
\right] \right )</math>
<br>
<math>\Delta \frac
+ \overline V \left [\sum_j \left ( - \frac
\right ) \exp \left [ \frac{-E_j}
\right] \exp \left [\frac{-pV_j}
\right] \right ] = \sum_j \left ( - \frac
\right ) \exp \left [\frac{-E_j}
\right] \exp \left [\frac{-pV_j}
\right]</math>
<br>
<math>\mbox
</math>
<br>
<math>\frac
+ \overline V \left ( - \frac
\right ) = \frac
</math>
<br>
<math>\frac{\overline
- \overline V^2} = -kT \left ( \frac
\right )</math>
<br>
<math>\frac{\overline
- \overline V^2}
= \frac{-kT}
\left ( \frac
\right )</math>
<br>
<math>\frac{\overline
- \overline V^2}
= \frac
{\overline V \left ( \frac
\right )</math>
<br>
<math>\frac{\overline
- \overline V^2}
Unknown macro: {overline V^2}= \fracUnknown macro: {kT}Unknown macro: {overline V}\kappa </math>
<br>
<math>\kappa = - \frac
\left ( \frac
\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
\left ( \frac
\right )</math>
<br>
<math>\kappa = \left ( \frac{-1}
\right ) \left ( \frac{-NkT}
\right )</math>
<br>
<math>\kappa = \frac
</math>
<br>
<math>\frac{\overline
- \overline V^2}
= \frac
\left ( \frac
\right )</math>
<br>
<math>\frac{\overline
- \overline V^2}
= \frac
</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
\left ( \frac
\right ) \right \infty</math>
<br>
</center>
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
</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
e^
</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
\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
\right )_
</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
\right )</math>
<br>
<math>x = \frac
</math>
<br>
<math>\mu = - \epsilon + kT \ln \left ( \frac
\right )</math>
<br>
</center>
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
{e^
- 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_
</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_
</math>?
<p>
</p>
The term <math>P_
</math> of a totally random solution is equal to <math>2x_Ax_B = 0.5</math>. Therefore, the value of <math>P_
= 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_
</math> towards <math>0.5</math>.
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_
</math>
</center>
</td>
<td>
<center>
<math>\frac{S_{tot}}
</math>
</center>
</td>
</tr>
<tr>
<td>
(a)
</td>
<td>
<center>
<math>k \ln 1 \mbox
\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
</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 ]</math>
</center>
</td>
<td>
<center>
<math>-\frac
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
\mbox
\right k\ln4N</math>
</center>
</td>
<td>
<center>
<math>0</math>
</center>
</td>
</tr>
</table>
</center>