Question 1

A system with constant volume <math>V</math> and temperature <math>T</math> contains two non-interacting particles. The solutions to the single particle Schr��dinger equation indicate that there are two single particle quantum states <math>1</math> and <math>2</math> with energy <math>\epsilon_1</math> and <math>\epsilon_2</math> respectively.

<p>
</p>

a) Assume that the two particles in the system are distinguishable. How many microstates are available to the system. Write down the energy for each distinct microstate?

<p>
</p>

<center>

<table cellpadding=10>

<tr>

<td>
<center>
<math>\mbox

Unknown macro: {Particle 1}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

</tr>
<tr>

<td>
<center>
<math>\mbox

Unknown macro: {Particle 2}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

</tr>

<tr>

<td>
<center>
<math>\mbox

Unknown macro: {Energy of Microstate}

</math>
</center>
</td>

<td>
<center>
<math>2 \epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1 + \epsilon_2</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2 + \epsilon_1</math>
</center>
</td>

<td>
<center>
<math>2 \epsilon_2</math>
</center>
</td>

</tr>

</table>

</center>

b) Write an expression for the average energy of the system as a function of <math>T</math>. Assume now that the two particles of the system are indistinguishable.

<p>
</p>

The average energy is found with an expression below.

<center>

<br>

<math>\overline E = kT^2 \left ( \frac

Unknown macro: {partial ln Q}
Unknown macro: {partial T}

\right )</math>

<br>

</center>

A partition function is written below.

<center>

<br>

<math>Q = q^N</math>

<br>

<math>Q = \left ( e^{- \beta \epsilon_1} + e^{- \beta \epsilon_2} \right )</math>

<br>

<math>\overline E = 2kT^2 \frac

Unknown macro: {partial}

\ln \left ( e^{- \beta \epsilon_1} + e^{- \beta \epsilon_2} \right )</math>

<br>

<math>\overline E = 2kT^2 \frac{ -\epsilon_1 \left ( \frac{-1}

Unknown macro: {kT^2}

\right ) e^{-\beta \epsilon_1} - \epsilon_2 \left ( \frac{-1}

\right ) e^{\beta \epsilon_2} }{ e^{ \beta \epsilon_1} + e^{- \beta \epsilon_2} }</math>

<br>

<math>\overline E = 2 \frac{\epsilon_1 e^{- \beta \epsilon_1} + \epsilon_2 e^{- \beta \epsilon_2}}{e^{- \beta \epsilon_1} + e^{- \beta \epsilon_2}}</math>

<br>

</center>

Note that this is just twice the energy of the average energy of one particle.

<p>
</p>

c) If the particles are bosons, how many microstates are available to the system. Write down the energy for each distinct microstate.

<p>
</p>

Particles are indistinguishable and are bosons (more than one can be in the same particles state). Now there are three possible microstates.

<center>

<table cellpadding=10>

<tr>

<td>
<center>
<math>\mbox

Unknown macro: {One of the particles}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

</tr>
<tr>

<td>
<center>
<math>\mbox

Unknown macro: {The other particle}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

</tr>

<tr>

<td>
<center>
<math>\mbox

</math>
</center>
</td>

<td>
<center>
<math>2 \epsilon_1</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1 + \epsilon_2</math>
</center>
</td>

<td>
<center>
<math>2 \epsilon_2</math>
</center>
</td>

</tr>

</table>

</center>

d) If the particles are fermions, how many microstates are available to the system. Write down the energy for each distinct microstate.

<p>
</p>

Particles are still indistinguishable and are now fermions (no more than one can be in the same particle state). Now there is only one possible microstate.

<center>

<table cellpadding=10>

<tr>

<td>
<center>
<math>\mbox

Unknown macro: {One of the particles}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1</math>
</center>
</td>

</tr>
<tr>

<td>
<center>
<math>\mbox

Unknown macro: {The other particle}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_2</math>
</center>
</td>

</tr>

<tr>

<td>
<center>
<math>\mbox

Unknown macro: {Energy of Microstate}

</math>
</center>
</td>

<td>
<center>
<math>\epsilon_1 + \epsilon_2</math>
</center>
</td>

</tr>

</table>

</center>

Question 2

A magnetic solid containing <math>N</math> atoms is placed in a magnetic field <math>H</math> at constant temperature <math>T</math> and pressure <math>P</math>. Each atom in the solid has a magnetic moment <math>m_o</math> which can independently align itself parallel or anti-parallel with the magnetic field <math>H</math>.

<p>
</p>

a) Write an expression for the characteristic potential of this system in terms of appropriate Legendre transforms of the internal energy.

<p>
</p>

Thermodynamic boundary conditions are constant <math>N</math>, <math>T</math>, <math>P</math>, and <math>H</math>. Below is a characteristic potential.

<center>

<br>

<math>\wedge (T, P, N, H) = E(S, V, N, M) - TS + PV - MH</math>

<br>

<math>d \wedge = - SdT + VdP + \mu d N - M dH</math>

<br>

</center>

b) Write down a formal expression for the partition function of this system and indicate which microstates you are summing over.

<p>
</p>

Find out what goes in the box by Legendre Transforming energy. Microstates being summed over correspond to all possible energies, volumes, and magnetizations.

<center>

<br>

<math>-\beta \wedge (T, P, N, H) = \frac

Unknown macro: {S(E, V, N, M)}
Unknown macro: {k_B}

- \beta E - \beta PV + \beta HM</math>

<br>

<math>\Gamma = \sum_

Unknown macro: {nu}

e^{-\beta E_

- \beta PV_

Unknown macro: {nu}

+ \beta H M_

</math>

<br>

</center>

c) Derive a general expression for the fluctuations of the magnetization <math>M</math> of the solid (i.e. <math>\overline

Unknown macro: {M^2}
  • \overline M^2</math>)

<p>
</p>

Consider the fluctuation of magnetization.

<center>

<br>

<math>\sigma_

Unknown macro: {M}

^2 = \overline

- \overline M^2</math>

<br>

<math>\overline M = \frac{ \sum_

Unknown macro: {nu}

M_

e^{-\beta E_

Unknown macro: {nu}
  • \beta PV_

+ \beta H M_

Unknown macro: {nu}

}}

Unknown macro: { Gamma }

</math>

<br>

</center>

Follow three step procedure

<center>

<br>

<math>\mbox

Unknown macro: {Step 1}

</math>

<br>

<math>\overline M \Gamma = \sum_

M_

Unknown macro: {nu}

e^{-\beta ( E_

- \beta PV_

Unknown macro: {nu}

+ \beta H M_

) }</math>

<br>

<math>\mbox

Unknown macro: {Step 2}

</math>

<br>

<math>\frac

Unknown macro: { partial (overline M Gamma)}
Unknown macro: { partial H }

= \frac

Unknown macro: {partial}
Unknown macro: {partial H}

\left ( \sum_

Unknown macro: {nu}

M_

e^{-\beta ( E_

Unknown macro: {nu}
  • \beta PV_

+ \beta H M_

Unknown macro: {nu}

) } \right )</math>

<br>

<math>\frac

Unknown macro: { partial overline M}

\Gamma + \overline M \frac

Unknown macro: {partial Gamma}
Unknown macro: {partial H}

= \left ( \sum_

Unknown macro: {nu}

\beta M_

2 e{-\beta ( E_

Unknown macro: {nu}
  • \beta PV_

+ \beta H M_

Unknown macro: {nu}

) } \right )</math>

<br>

<math>\frac

Unknown macro: { partial overline M}
Unknown macro: { partial H }

\Gamma + \overline M \left ( \sum_

\beta M_

Unknown macro: {nu}

e^{-\beta ( E_

- \beta PV_

Unknown macro: {nu}

+ \beta H M_

) } \right ) = \left ( \sum_

Unknown macro: {nu}

\beta M_

2 e{-\beta ( E_

Unknown macro: {nu}
  • \beta PV_

+ \beta H M_

Unknown macro: {nu}

) } \right )</math>

<br>

<math>\mbox

Unknown macro: {Step 3}

</math>

<br>

<math>\frac

Unknown macro: { partial overline M}
Unknown macro: { partial H }

\Gamma + \beta \frac{ \overline M \left ( \sum_

\beta M_

Unknown macro: {nu}

e^{-\beta ( E_

- \beta PV_

Unknown macro: {nu}

+ \beta H M_

) } \right )}

Unknown macro: {Gamma}

= \frac{ \sum_

Unknown macro: {nu}

\beta M_

2 e{-\beta ( E_

Unknown macro: {nu}
  • \beta PV_

+ \beta H M_

Unknown macro: {nu}

) } }

</math>

<br>

<math>\overline

Unknown macro: {M^2}

- \overline M^2 = kT \left ( \frac

Unknown macro: {partial overline M}

\right )</math>

<br>

</center>

Question 3

a) A solid is a ferromagnet at low temperature, but undergoes a second order phase transition to the paramagnetic state at <math>T_c</math>. Sketch how you expect the heat capacity to behave around <math>T_c</math> and explain why.

<p>
</p>

At the critical point, fluctuations of the energy (which are related to the heat capacity) become very large and C_V diverges.

<center>

Unable to render embedded object: File (Heat_capacity_versus_T.PNG) not found.

</center>

b) Which of the following thermodynamic quantities are mechanical variables.

<p>
</p>

<center>
Entropy <math>S</math> Volume <math>V</math> Chemical Potential <math>\mu</math> Temperature <math>T</math> Pressure <math>P</math>
</center>

<p>
</p>

The following thermodynamic quantities are mechanical variables:

  • Volume
  • Chemical potential
  • Pressure

c) In a crystalline solid, which phonon modes require more energy to excite, long wavelength phonon modes <math>( k \right 0 )</math> or short wave length phonon modes (e.g. <math>k \right \frac

Unknown macro: {pi}
Unknown macro: {a}

</math> for a 1-dimensional crystal with lattice parameter a)?

<p>
</p>

<center>
______long wave-length phonon modes
</center>

<p>
</p>

<center>
_____short wave length phonon modes
</center>

<p>
</p>

Short wave length phonon modes require more energy.

<p>
</p>

d) The degeneracy of the quantum mechanical energy spectrum of a particular system is found to be well characterized with <math>\Omega(E) = A \exp (B(E-E_o))</math> , where <math>E_o</math> is the energy of the ground state and <math>A</math> and <math>B</math> are constants. Write an expression for the entropy of the system at <math>0 K</math>.

<center>

<br>

<math>S = k \ln \Omega</math>

<br>

<math>S = k \ln \left ( A e^{B(E - E_o) \right )</math>

<br>

<math>S = k \left ( \ln A + B(E - E_o ) \right )</math>

<br>

<math>S(T=0) = k \ln A</math>

<br>

</center>

Question 4

<math>M</math> atoms of <math>Fe</math> are crystallized on the fcc lattice. As a first approximation, the vibrational degrees of freedom of the <math>Fe</math> atoms can be treated with the Einstein model where each iron atom has an Einstein vibrational frequency of <math>\omega_

Unknown macro: {Fe}

</math>.

<p>
</p>

a) Write the constant <math>T</math>, <math>V</math> and <math>M</math>, partition function in terms of the partition function of a one-dimensional harmonic oscillator with the same frequency <math>\omega_

</math>. The fcc crystal of <math>M Fe</math> atoms has <math>M</math> interstitial sites which can be filled with carbon atoms. The binding energy of carbon in the interstitial sites of fcc <math>Fe</math> with respect to pure carbon (in the graphite reference phase) is <math>\epsilon</math>. Carbon in the interstitial sites has an Einstein frequency of <math>\omega_C</math>.

<p>
</p>

The canonical ensemble is applied to systems with fixed <math>T</math>, <math>V</math>, and <math>M</math>. Below is the energy spectrum of an Einstein harmonic oscillator.

<center>

<br>

<math>\epsilon_n = \left (n+\frac

Unknown macro: {1}
Unknown macro: {2}

\right ) \hbar \omega_

Unknown macro: {Fe}

</math>

<br>

<math>Q_

= q

Unknown macro: {Fe}

^

Unknown macro: {3M}

</math>

<br>

<math>q_

= \sum_n e^{- \beta_n}</math>

<br>

<math>q_

Unknown macro: {Fe}

= \sum_

Unknown macro: {n=0}

^

Unknown macro: {infty}

e^{- \beta \left ( n+ \frac

Unknown macro: {2}

\right ) \hbar \omega_{Fe}}</math>

<br>

<math>q_

Unknown macro: {Fe}

= \frac{e^{-\Theta_

/2T}}{1-e^{-\Theta_

Unknown macro: {Fe}

/T}}</math>

<br>

<math>\Theta_

= \frac{\hbar \omega_{Fe}}

Unknown macro: {k}

</math>

<br>

</center>

b) Derive an expression for the partition function of the fcc crystal containing <math>M Fe</math> atoms and <math>N</math> carbon atoms at contant <math>T</math> and <math>V</math>. You may assume that the carbon atoms do not interact with each other. Use this expression for the partition function to derive an expression for the Helmholtz free energy of the solid.

<center>

<br>

<math>Q = \frac

Unknown macro: {M!}
Unknown macro: {N!(M-N)!}

q_

Unknown macro: {Fe}

^

Unknown macro: {3M}

q_

Unknown macro: {c}

^

Unknown macro: {3N}

</math>

<br>

<math>q_c = \sum_n e^{-\beta \epsilon_n}</math>

<br>

<math>\epsilon_n = \left ( n = \frac

Unknown macro: {1}

\right ) \hbar \omega_c + \epsilon</math>

<br>

<math>q_c = e^{\beta \epsilon} \frac{e^{\Theta _c / 2T}}{1 - e^{-\Theta _c / T}</math>

<br>

<math>\Theta_c = \frac

Unknown macro: {hbar omega_c}
Unknown macro: {k}

</math>

<br>

<math>F = -kT \log W</math>

<br>

<math>F = -3 M kT \log q_

Unknown macro: {Fe}
  • 3 NkT \log q_c - kT \left [\log M! - \log N! - \log (M-N)! \right]</math>

<br>

<math>F = -3MkT \log q_

- 3NkT \log q_c - kT \left [M \log M - N \log N - (M-N ) \log (M-N) \right]</math>

<br>

</center>

c) Use the result from b) to calculate the chemical potential for <math>Fe</math> and for <math>C</math>. Try to express your result in terms of the carbon concentration <math>x=N/M</math>.

<center>

<br>

<math>\mu_

Unknown macro: {Fe}

= \frac

Unknown macro: {partial F}
Unknown macro: {partial M}

</math>

<br>

<math>\mu_

Unknown macro: {C}

= \frac

Unknown macro: {partial N}

</math>

<br>

<math>\mu_

= -3kT \log q_

Unknown macro: {Fe}
  • kT \left [\log M + 1 - \log (M-N) - 1 \right]</math>

<br>

<math>\mu_

= - 3kT \log q_

Unknown macro: {Fe}

+ kT \left [\log \left ( \frac

Unknown macro: {M-N}
Unknown macro: {M}

\right ) \right]</math>

<br>

<math>\mu_

= -3kT \log q_

Unknown macro: {Fe}

+ kT \log (1-x)</math>

<br>

<math>\mu_c = -3kT \log q_c - kT \left [-\log N - 1 + \log (M-N) + 1 \right]</math>

<br>

<math>\mu_c = -3kT \log q_c + kT \left [\log \left ( \frac

Unknown macro: {N}
Unknown macro: {M-N}

\right ) \right]</math>

<br>

<math>\mu_c = - 3kT \log q_c + kT \log \left ( \frac

Unknown macro: {x}
Unknown macro: {1-x}

\right )</math>

<br>

</center>

Question 5

Some researchers have stated that when <math>Li_xCoO_2</math> is annealed at high temperature it is susceptible to Li loss due to the high vapor pressure of <math>Li</math> in the material. In this problem you will estimate the vapor pressure of <math>Li</math> in the material, using commonly available data.

<p>
</p>

DATA for elemental Li:

  • boiling point: <math>1597 K</math>
  • entropy of vaporization: <math>92.5 J/mol-K</math>
  • melting point: <math>454 K</math>
  • entropy of melting: <math>6.45 J/mol-K</math>

<p>
</p>

<center>

Unable to render embedded object: File (Voltage_versus_lithium_concentration.PNG) not found.

</center>

a) Using some or all of the data below estimate the vapor pressure of <math>Li</math> above liquid <math>Li</math> at <math>1473K</math>.

<p>
</p>

Use Clausisu Clapeyron to exttrapoloate from the boiling point.

<center>

<br>

<math>\frac

Unknown macro: {d ln p}

{d\left (\frac

Unknown macro: {1}
Unknown macro: {T}

\right )} = - \frac

Unknown macro: {Delta H}
Unknown macro: {R}

</math>

<br>

<math>\ln \frac

Unknown macro: {p(1473)}
Unknown macro: {p(1597)}

= - \frac

Unknown macro: {R}

\left [\frac

Unknown macro: {1473}
  • \frac
    Unknown macro: {1}
    Unknown macro: {1597}

    \right]</math>

    <br>

    <math>\Delta H = T \Delta S</math>

    <br>

    <math>\Delta H = 1597 \cdot 92.5 j / mol \cdot K</math>

    <br>

    <math>\Delta H = 147.722 kJ/mol</math>

    <br>

    <math>p(1473) = \exp \left [- \frac

    Unknown macro: {147722}
    Unknown macro: {8.314}

    \left ( \frac

- \frac

Unknown macro: {1}
Unknown macro: {1597}

\right ) \right |]</math>

<br>

<math>p(1473) = 0.39 atm</math>

<br>

</center>

b) The data below shows the potential of a <math>Li_xCoO_2</math> electrode (referenced against a <math>Li</math> metal) as function of <math>x</math> when the two are in contact through a <math>Li+</math> ion conductor. The data is taken at <math>300 K</math>. Note that a positive potential indicates that the activity of <math>Li</math> in <math>Li_xCoO_2</math> is lower than in <math>Li</math> metal. Estimate the vapor pressure above <math>Li_

Unknown macro: {0.8}

CoO_2</math> at <math>1100 C</math> assuming that the activity of <math>Li</math> in <math>Li_

CoO_2</math> at <math>1100 C</math> against hypothetical <math>Li</math> solid is the same as at <math>300K</math>.

<center>

<br>

<math>- \left ( \mu_

Unknown macro: {Li}

^

Unknown macro: {LiCOO_2}

- \mu_

^

Unknown macro: {Li(s)}

= F \Delta \Phi</math>

<br>

<math>\mu_

Unknown macro: {Li}

^

Unknown macro: {LiCOO_2}

= \mu_

^

- F \Delta \phi</math>

<br>

<math>-F \Delta \phi = RT \ln a_i</math>

<br>

<math>a_i = \exp \left ( \frac{-F \Delta \phi}{RT_{300}}</math>

<br>

<math>\mu_

Unknown macro: {Li}

^

Unknown macro: {LiCOO_2}

= \left [\mu_

^

Unknown macro: {LiCOO_2}
  • \mu_
    Unknown macro: {Li}

    ^

    Unknown macro: {Li(s)}

    \right] + \left [\mu_

    ^
    Unknown macro: {Li(s)}
    • \mu_
      Unknown macro: {Li}

      ^

      Unknown macro: {Li(L)}

      \right] + \mu_

      ^
    </math>

<br>

<math>\left [\mu_

Unknown macro: {Li}

^

- \mu_

Unknown macro: {Li}

^

Unknown macro: {Li(L)}

\right] = \frac{-T_{1373}}{T_{300}} F \Delta \phi + \Delta G_s</math>

<br>

<math>\Delta G_s = \Delta H_M \left ( 1 - \frac

Unknown macro: {T}
Unknown macro: {T_M}

\right )</math>

<br>

<math>\left [\mu_

^

Unknown macro: {LiCOO_2}
  • \mu_
    Unknown macro: {Li}

    ^

    Unknown macro: {Li(L)}

    \right] = \frac{-1373}

    Unknown macro: {300}

    \times 96400 \times 4 - 434 \times 6.45 \left ( 1 - \frac

    Unknown macro: {1373}
    Unknown macro: {454}

    \right )</math>

    <br>

    <math>\left [\mu_

    ^

- \mu_

Unknown macro: {Li}

^

Unknown macro: {Li(L)}

\right] = -1.77 MJ</math>

<br>

</center>

Consider the activity with respect to <math>Li_

Unknown macro: {Liquid}

</math>.

<center>

<br>

<math>RT \ln a_i = -1.77 MJ</math>

<br>

<math>a_i = 4.3 \times 10^{-68}</math>

<br>

<math>P_

= 0.16 \times a_i</math>

<br>

<math>P_

Unknown macro: {Li}

= 6.88 \times 10^{-69}</math>

<br>

</center>

c) Recalculate the vapor pressure above <math>Li_

Unknown macro: {0.8}

CoO_2</math> at <math>1100 C</math> now taking into account the fact that the relative partial molar entropy of <math>Li</math> in <math>Li_

CoO_2</math> with respect to the entropy of <math>Li</math> in elemental <math>Li</math> metal) equals <math>-60 J/mol \cdot K</math>, and is independent of temperature.

<p>
</p>

A solution is like b but now <math>\Delta \mu_i</math> is changing with <math>T</math>.

<center>

<br>

<math>\mu_

^

Unknown macro: {LiCOO_2}
  • \mu_
    Unknown macro: {Li}

    ^{Li(s) = \left [\mu_

    ^

- \mu_

Unknown macro: {Li}

^

Unknown macro: {Li(L=s)}

\right]_

Unknown macro: {300}

+ \frac

Unknown macro: {partial}
Unknown macro: {partial T}

\left [\mu_

^

Unknown macro: {LiCOO_2}

\right] (1373 - 300) + \Delta G_s</math>

<br>

<math>\mu_

Unknown macro: {Li}

^

- \mu_

Unknown macro: {Li}

^{Li(s) = -1.77 MJ + 64.38 kJ</math>

<br>

<math>\mu_

^

Unknown macro: {LiCOO_2}

- \mu_

Unknown macro: {Li}

^{Li(s) = -1.63 MJ</math>

<br>

<math>a_i = 3.4 \times 10^{-65}</math>

<br>

<math>p_i = 5.5 \times 10^{-66}</math>

<br>

</center>

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