Question 1
Assume a system is at constant temperature <math>T</math>, volume <math>V</math> and number of particles <math>N</math>. Determine the relationship between the heat capacity <math>C_v</math> and the energy fluctuations <math><(E - <E>)^2></math>. You may remember from the first part of the course that <math>C_v</math> is the temperature derivative of the internal energy.
<center>
<br>
<math><(E - <E>)^2> = \overline
- \overline E^2</math>
<br>
<math>\mbox{Canonical Partition Function</math>
<br>
<math>Q = \sum_i e^{- \beta E_i}</math>
<br>
<math>\overline E = \sum_i E_i e^{- \beta E_i}</math>
<br>
</center>
Define <math>Q'</math>.
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<math>Q' = \frac
_v</math> |
<br>
<math>\overline E = - \frac
</math>
<br>
<math>\frac
= - \frac
+ \left ( \frac
\right )^2</math>
<br>
<math>\frac
= \overline
</math>
<br>
<math>\frac
= -\overline
</math>
<br>
<math>\frac
= - \overline
+ \overline E^2</math>
<br>
<math>\frac
= -k_B T^2 \frac
</math>
<br>
<math>\frac
= -k_B T^2 C_v</math>
<br>
<math><(E - <E>)^2> = k_B T^2 C_v</math>
<br>
</center>
Question 2
Consider a system of <math>N</math> independent particles with <math>N</math> very large. There are two energy levels, <math>0</math> and <math>\epsilon</math>, of each particle.
<p>
</p>
a) Find the number of microstates, <math>\Omega_M</math>, with total energy <math>E = M \epsilon</math>.
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<br>
<math>\Omega_M = \frac
</math>
<br>
</center>
b) By using Stirling's approximation, derive the entropy per particle of this system as a function of particles, <math>x_e</math>, that are in the highest energy state. Sketch the entropy and mark important points (maxima, minima, etc.) on the <math>x_e</math> axis (i.e. at what value of <math>x_e</math> do they occur?).
<center>
<br>
<math>S = k_B \ln \Omega_M</math>
<br>
<math>S = k_B \left [\ln N! - \ln M! - \ln(N-M)! \right]</math>
<br>
<math>S = k_B \left [N \ln N - N - M \ln M + M - (N-M) \ln (N-M) + (N-M) \right]</math>
<br>
<math>S = -k_B N \left [\frac
\ln \frac
+ \frac
\ln \left ( \frac
\right ) \right]</math>
<br>
<math>\underline S = \frac
</math>
<br>
<math>\underline S = -k_B \left [x_
\ln x_
+ (1-x_
) \ln (1 - x_
) \right]</math>
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</center>
There is a maximum at <math>x_
= \frac
</math> and minima at <math>x_
= 0</math> and <math>x_
= 1</math>.
<p>
</p>
c) Calculate the temperature as a function of <math>x_
</math>. Discuss the result.
<p>
</p>
Option 1
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<br>
<math>\frac
= \frac
</math>
<br>
<math>\frac
= \frac
{\epsilon \partial x_
}</math>
<br>
<math>\frac
= \frac
\ln \left ( \frac{1 - x_{\epsilon}}{x_{\epsilon}} \right )</math>
<br>
</center>
Option 2
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<br>
<math>P_
= \frac{e^{-\beta
}}{1 + e^{-\beta
}}</math>
<br>
<math>P_
= x_
</math>
<br>
<math>x_
= \frac
{1 + e^{\beta E}}</math>
<br>
<math>x_
\left ( 1 + e^
\right ) = 1</math>
<br>
<math>\beta \epsilon = \ln \left ( \frac
{x_{\epsilon}} - 1 \right )</math>
<br>
<math>\frac
= \frac
\ln \left ( \frac{1 - x_{\epsilon}}{x_{\epsilon}} \right )</math>
<br>
</center>
Note that there is negative temperature when <math>x_e > \frac
</math> indicating more particles in an upper energy level than in lower level (LASER type situation). In statistical thermodynamics consider part only when <math>T > 0</math>.
<p>
</p>
Question 3
Consider a solution of <math>A</math> and <math>B</math> molecules.
<p>
</p>
If it were chosen to write the number of ways one can arrange the <math>A</math> and <math>B</math> molecules on a model lattice by the formula <math>\Omega_N = \frac
</math> where <math>N</math> is the total number of molecules <math>N_A+N_B</math>, and express the entropy of mixing as <math>\Delta_
= k_B \ln (\Omega_N_</math>
<p>
</p>
a) What assumption have been made regarding the configurational entropy of mixing? Discuss
<p>
</p>
It is assumed that there are no interactions between molecules. There is random mixing (ideal solution). Additional assumptions are the lack of vacancies, defects, etc.
<p>
</p>
b) If <math>A</math> is a solvent and <math>B</math> is a polymer with <math>n</math> segments, is the above formula still valid? Discuss. How would it be improved? Discuss. No equations necessary.
<p>
</p>
No. Each of the segments of a polymer are connected. Nor all configurations are possible in a model lattice. Only nearest neighbor sites of given segment are available to neighboring segments, and this is assumed in the Flory Huggins model.
<p>
</p>
c) Are there other contributions to the entropy of mixing? Discuss. No equations necessary
<p>
</p>
Yes.
- Vibrational entropy: this can be quite large
- Electronic entropy
- Magnetic entropy
Question 4
Below is the <math>Cd-Zn</math> phase diagram. A researcher from University Maxima claims that this phase diagram is incomplete, and that a compound with stoichiometry <math>CdZn</math> with melting point of <math>350^
C</math> exists, but that it usually is not observed because of very difficult nucleation conditions. Draw the phase diagram if the researcher were correct and this compound is present. Draw the correct phase diagram on top of the diagram shown above so that the relative location of old and new lines is clear. Explain relative shifts in lines as necessary.
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</center>
The compound is drawn as a line compound, though a compound with with off-stoichiometry can also be drawn.
<center>
</center>
Consider the lines <math>a</math>, <math>a'</math>, <math>b</math>, <math>b'</math>. These represent the two-phase equilibria between the liquid and either the <math>Cd</math> or <math>Zn</math> phase. Nothing is changed to these equilibria as one introduces the compound. This can be easily see by drawing the free energy curves of the system with and without the compound.
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</center>
When a new stable phase appears, the solubility limits in the adjoining phases decreases. Consider the lines <math>e</math> and <math>f</math> and free energy curves.
<p>
</p>
Eutectics, which are at points <math>c</math> and <math>d</math> are formed when the <math>Cd + L</math> (or <math>Zn + L</math>) two-phase region meets the liquid and compound two-phase region.
Question 5
Researchers at MIT are working hard to understand possible phases of the element Donorium (<math>Do</math>), a recently discovered element which is important because it enhances the properties of Endowium (<math>En</math>) when alloyed with it.
<p>
</p>
Below are the zero-temperature enthalpy and Einstein temperature (at <math>1 atm</math> pressure) of three crystal structures. Donorium is speculated to be stable in these phases. Donorium is not magnetic at any temperature.
<p>
</p>
What is the stable state at <math>0K</math>? Speculate on the possible phase transitions of <math>Do</math> as a function of temperature at constant <math>1 atm</math> pressure. Justify the answer (possibly with sketches of free energies) and explain why each phase transition occurs. There is no need to perform numerical calculations.
<center>
<table cellpadding="10">
<tr>
<td>
<center>
<math>\mbox
</math>
</center>
</td>
<td>
<center>
<math>\mbox
</math>
</center>
</td>
<td>
<center>
<math>\mbox
</math>
</center>
</td>
</tr>
<tr>
<td>
<center>
<math>\alpha</math>
</center>
</td>
<td>
<center>
<math>0 J/mole</math>
</center>
</td>
<td>
<center>
<math>350 K</math>
</center>
</td>
</tr>
<tr>
<td>
<center>
<math>\beta</math>
</center>
</td>
<td>
<center>
<math>5 kJ/mole</math>
</center>
</td>
<td>
<center>
<math>475 K</math>
</center>
</td>
</tr>
<tr>
<td>
<center>
<math>\gamma</math>
</center>
</td>
<td>
<center>
<math>7 kJ/mole</math>
</center>
</td>
<td>
<center>
<math>200 K</math>
</center>
</td>
</tr>
</table>
</center>
The Einstein temperature determines the magnitude of vibrational entropy. A large value of <math>\Theta_E</math> corresponds to a stiff material with low <math>S_
</math>. The relative shape of the free energy curves can be determined from <math>E(T=0)</math>, which is equal to <math>\Delta H</math>, and the entropy of each phase, <math>S = 0 \left ( \frac
\right )_p</math>.
<p>
</p>
Consider that <math>S_
> S_
> S_
</math>. The <math>\beta</math> phase never becomes stable since entropy is lower than <math>\alpha</math>. The <math>\alpha</math> phase is the most stable at low temperature since the enthalpy is lowest.
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</center>