Question 1
a) Consider a two-dimensional harmonic oscillator. At what temperature is it equally probably for the system to have energy <math>2 \hbar \omega</math>* as* <math>4 \hbar \omega</math>?
<p>
</p>
Probability of the system to be in any state is <math>P_j</math>, and the probability ot be in a state with energy <math>E_j</math> is <math>P_
</math>. Set two probabilities to be equal to one another.
<center>
<br>
<math>P_j = \frac{e^{-\beta E_j}}
</math>
<br>
<math>P_
= \frac{\Omega e^{-\beta E_j}}
</math>
<br>
<math>\frac{2 e^{-\beta 2 \hbar \omega}}
= \frac{4 e^{-\beta 4 \hbar \omega}}
</math>
<br>
<math>T = \frac
</math>
<br>
</center>
b) Consider a completely disordered binary solid at high temperature. If the system is slowly cooled to zero temperature, what is the maximum change in configurational entropy?
<center>
<br>
<math>\frac
= k_B \ln 2</math>
<br>
</center>
c) Two ideal gases with equal number of particles are kept in two separate equal volumes at equal temperatures. We allow the two gases to mix by removing the partition between them. What is the increase/decrease in entropy?
<p>
</p>
There are several ways to solve this problem. If the entropy is calculated through classical thermodynamic theory (expansion of two gases), the entropy increases because the particles are perceived as distinguishable.
<center>
<br>
<math>\Delta S = 2 N k_B \ln 2</math>
<br>
</center>
If the particles are different, which should have been stated in the problem and was generally assumed), the entropy increases is simply the entropy of mixing, which is the same as in b).
<center>
<br>
<math>Q \prop V^N</math>
<br>
<math>\Delta S = k_B \ln (2V)^
- 2 k_B \ln V^N</math>
<br>
<math>\Delta S = 2 N k_B \ln 2</math>
<br>
</center>
Below is the partition function of particles that are the same and indistinguishable.
<center>
<br>
<math>Q = \frac
</math>
<br>
<math>Q = \frac
</math>
<br>
<math>\Delta S = k_B \ln \frac
- 2 k_B \ln \frac
</math>
<br>
<math>\Delta S = 0</math>
<br>
</center>
The last expression is true under the assumption that Stirling's formula holds. The process actually creates a very small amount of entropy as the removal of the partition is irreversible.
Question 2
An oxide with stoichiometry <math>AO_
</math> undergoes a polymorphic transition from a low temperature alpha <math>(\alpha)</math> phase to a high-temperature beta <math>(\beta)</math> phase at <math>500^
C</math> when under normal atmospheric conditions (<math>pO_2 = 0.2 atm)</math>. Calculate how much the transition temperature changes if the oxygen partial pressure is doubled.
<p>
</p>
<math>\Delta H_
= 8 kJ/mol</math>, <math>\delta^
= 0.02</math> , <math>\delta^
= 0.01</math>
The oxygen partial pressure sets the chemical potential of oxygen. Relate changes in the chemical potential of oxygen at the transition to changes in the transition temperature.
<center>
<br>
<math>\frac
= \frac
</math>
<br>
<math>\frac
= \frac
</math>
<br>
</center>
Use an ideal gas expression of <math>d \mu_0</math>.
<center>
<br>
<math>RTd\ln(p) = d \mu_0</math>
<br>
<math>RTd\ln(p) = - \frac
dT</math>
<br>
<math>R \ln \left ( \frac
\right ) = R \ln (2)</math>
<br>
<math>R \ln \left ( \frac
\right ) = \frac
\left ( \frac
- \frac
\right )</math>
<br>
<math>R \ln \left ( \frac
\right ) = \frac
\left ( \frac
- \frac
\right )</math>
<br>
<math>R \ln \left ( \frac
\right ) = \frac
\left ( \frac
- \frac
\right )</math>
<br>
<math>T_f = 769</math>
<br>
<math>\Delta T \approx -4 K </math>
<br>
</center>
Question 3
Consider a monolayer of <math>Li</math> atoms on a bulk surface. There are two possible structural arrangements of the <math>Li</math> atoms - one close-packed low-temperature surface phase <math>(\alpha)</math> and one more open surface structure <math>(\beta)</math>. The <math>Li</math> atoms are bonded to the bulk with the same very stiff bond, regardless of the <math>Li</math> surface structure. The bonds between the <math>Li</math> atoms in the surface layer can be considered the same in both directions. However, those <math>Li-Li</math> bonds differ in strength depending on the surface structure.
<p>
</p>
You want to estimate the transition temperature (which is low) between the close-packed and open surface phase in <math>Li</math>. Using your favorite first-principles method you calculate the internal energy difference between the two phases: <math>E_
(T=0) - E_
(T=0) = 150 J/mol</math> of surface atoms. Additionally you know the Einstein frequencies for the <math>Li-Li</math> intra-layer vibrations: <math>\omega_E^
= 52 THz</math> and <math>\omega_E^
= 104 THz</math>
<p>
</p>
a) Derive a low-temperature expression for the vibrational entropy of the surface layer as a function of temperature, within the Einstein model of independent oscillators.
<p>
</p>
The vibrations between the surface layer and the bulk do not contribute. Therefore, the problem reduces to a 2D problem. Make a low-T expansion of the Einstein independent oscillator model.
<center>
<br>
<math>C_V = 2 N k_B \left ( \frac
\right )2 \left ( \frac{ e{\Theta_
}}{(e^{\Theta_{E/T}} - 1)^2} \right )</math>
<br>
<math>C_V \approx 2 N k_B \left ( \frac
\right )2 e{\Theta_{E/T}}</math>
<br>
</center>
Find the entropy by integrating.
<center>
<br>
<math>\Delta S (T) = 2 N k_B \int _0^T \left ( \frac
\right )2 e{\Theta_{E/T}} dT</math>
<br>
<math>x = \frac
</math>
<br>
<math>dx = - \frac
dT</math>
<br>
<math>\Delta S (T) = -2R \Theta_E^2 \int_
^
x^3 e^{-\Theta_E x} \left ( \frac
\right )^2 dx</math>
<br>
<math>\Delta S (T) = 2 R \Theta_E^2 \left ( \left [\frac{-x e^{- \Theta_E x}}{- \Theta_E} \right]_{\infty}^
- \int_
^
\frac{e^{\Theta_E x}}{\Theta_E}dx \right )</math>
<br>
<math>\Delta S (T) = 2R \left ( e^{\Theta_{E/T}} \left ( 1 + \frac
\right ) \right )</math>
<br>
</center>
b) Use the result in a) and modify it to estimate the transition temperature between the surface phases at <math>p = 0</math>.
Consider <math>\Delta F</math> at the transition temperature <math>T</math>.
<center>
<br>
<math>\Delta F = \Delta E - T \Delta S (T)</math>
<br>
<math>\Delta F = 0 </math>
<br>
<math>T_
\Delta S (T_
) = 2 R T_
\left ( e^{-\Theta_E^
/ T_{tr}} \left ( 1 + \frac{ \Theta_E^{\beta}}{T_{tr}} \right ) - e^{
\Theta_E^
/T_{tr}} \left ( 1 + \frac{\Theta_E^{\alpha}}
\right ) \right )</math>
<br>
<math>T_
\Delta S (T_
) = 150 J/mol</math>
<br>
</center>
Trying some temperatures reveals the following.
<center>
<br>
<math>T_
\approx 100K</math>
<br>
</center>
Question 4
a) Consider a carbon monoxide (ideal) gas. Evaluate the room temperature average energy for the gas. Assume that the gas particles can be treated within Boltzmann approximation. The masses for the individual atoms in carbon monoxide are <math>m(C) = 12 u</math>, <math>m(O) = 16 u</math>.
<center>
Unable to render embedded object: File (Gas_table_--_Theta_v%2C_Theta_r%2C_bond_length%2C_mass.PNG) not found.
</center>
From vibrations of <math>CO</math>, only the zero-point part contributes at room temperature. Translation, rotation, and zero-point vibrations are included in an expression below.
<center>
<br>
<math>E = \frac
+ Nk_BT + 3 N k_B \Theta_
</math>
<br>
</center>
b) Calculate the heat capacity for carbon monoxide originating from the ground state and the first rotational excitation.
<p>
</p>
Below is an expression of the relevant rotation single partition function.
<center>
<br>
<math>q_
= 1 + 2 e^{-2 \Theta_r / T} + ...</math>
<br>
</center>
Derive by first calculating an expression of the partition function. Relate to Hemholtz free energy and differentiate to find entropy. Find heat capacity with an expression relating heat capacity and entropy.
<center>
<br>
<math>Q = \frac{(q_
)^N}
</math>
<br>
<math>F_
= -NkT \ln q_
+ kT \ln (N!)</math>
<br>
<math>F_
= -NkT \ln \left ( 1 + 3 e^{-2\Theta_r / T} \right ) + kT \ln (N!)</math>
<br>
<math>S_
= -\left (\frac{\partial F_{rot}}
\right ) T \right )_V</math>
<br>
<math>C_
= T \left ( \frac{ \partial S_
}
\right )_V</math>
<br>
<math>C_
= 12 N k \left ( \frac
\right )2 \frac{e{-2\Theta_r / T}}{\left (1 + 3 e^{-2 \Theta_r / T} \right )^2</math>
<br>
</center>
c) Use the table to choose the gas that would make the best heat absorptive medium for <math>100 < T < 500 K</math>. Motivate your answer briefly.
<p>
</p>
<math>1 u = 1.66 \cdot 10^{-27} kg</math>, <math>\hbar = 1.054 \cdot 10^{-34}</math>
<p>
</p>
There would be highest heat capacity at <math>500 K</math> with <math>I_2</math>, which is associated with the lowest vibrational characteristic temperature. Therefore, vibrations contribute to heat absorption.
Question 5
Elemental excitium <math>(Ec)</math>* can exist in two different phases, ���alpha��� and ���beta���. Shown in the graph below are the Helmholtz free energy curves for both phases as a function of temperature.*
Data:
- <math>\Delta F_0 = 1060 J/mol</math>
- <math>\Delta F_0 = 11 meV/atom</math>
- <math>\Theta_E^
Unknown macro: {alpha}</math> ���alpha��� nanoparticles and <math>m_
= 260 K</math>
- <math>\Theta_E^
Unknown macro: {beta}= 170 K</math>
<center>
Unable to render embedded object: File (Elemental_excitium.PNG) not found.
</center>
a) A collection of <math>M</math> excitium nanoparticles has been synthesized, each containing ���<math>n</math>��� <math>Ec</math>* atoms, and you want to model the thermal behavior of this collection of particles. Your system consists of* <math>m_
Unknown macro: {beta}=M)</math> held under constant temperature, and zero pressure. Neglect surface effects, and interactions between the nanoparticles. Assume that each <math>Ec</math> atom vibrates in the same fashion as in the bulk. Write down the partition function <math>Q(T,V, m_</math> ���beta��� nanoparticles <math>(m_
Unknown macro: {alpha}+ m_
Unknown macro: {alpha}, m_, m_
Unknown macro: {beta})</math>. Also, express the Helmholtz free energy (as an approximate Gibbs free energy) for this system.
<p>
</p>Each <math>Ec</math> atom oscillates as an independent oscillator, but will vibrate differently in a "beta" nanoparticle versus an "alpha" nanoparticle. There is a promotion energy <math>\epsilon_0</math> per <math>Ec</math> atom associated with being in a "beta" nanoparticle.
<center>
<br>
<math>Q(T, V, m_
Unknown macro: {beta}!} q_) = \frac
Unknown macro: {M!}{m_
Unknown macro: {alpha}! m_
Unknown macro: {alpha}= \left ( \exp \left [\frac{\Theta_{\alpha}}^{m_{\alpha}}q_
Unknown macro: {beta}^{m_{\beta}}</math>
<br>
<math>q_
Unknown macro: {2T}\right] \right )^{-3n}</math>\right] - \left [\frac{\Theta_
Unknown macro: {alpha}}
- <math>\Theta_E^
<br>
<math>q_
= \exp(-\beta \epsilon_0 n ) \left ( \exp \left [\frac{\Theta_{\beta}}
\right] - \left [\frac{\Theta_
}
\right] \right )^{-3n}</math>
<br>
<math>F(T, V, m_
, m_
\right ) = - k_B T \ln Q</math>
<br>
<math>F(T, V, m_
, m_
) = - k_B T \left [m_
\ln ( q_
+ (M-m_
) \ln ( q_
) + M \ln (M) - m_
\ln ( m_
) - (M - m_
) \ln (M - m_
) \right]</math>
<br>
</center>
b) It is impossible to experimentally constrain the nanoparticles to ���stay��� in one phase versus the other. Using your answer from part (a), solve for the equilibrium number of alpha nanoparticles as a function of temperature.
<p>
</p>
The equilibrium condition is given by optimizing the free energy over the number of alpha particles.
<center>
<br>
<math>\left ( \frac
{\partial m_
\right )_
= \mu_
</math>
<br>
<math>\left ( \frac
{\partial m_
\right )_
= 0</math>
<br>
<math>0 = \ln (q_
) - \ln (q_
) - \ln ( m_
) + \ln (M - m_
)</math>
<br>
<math>\ln \left ( \frac{q_{\alpha}}{q_{\beta}} \right )= \ln \left \frac{M - m_{\alpha}}{ m_{\alpha}}</math>
<br>
<math>\ln \left ( \frac{q_{\alpha}}{q_{\beta}} \right ) = \ln \left ( \frac
{x_
- 1 \right )</math>
<br>
<math>x_
= \frac{ m_
}
</math>
<br>
<math>x_
= \left [1 + \frac{q_{\beta}}{q_
\right]^{-1}</math>
<br>
</center>
Note that when <math>q_
> q_
</math>, <math>m_
> m_
</math>. When <math>q_
< q_
</math>, <math>m_
< m_
</math>.
<p>
</p>
Consider an alternative soltion. The system fluctuates over all the numbers of "alpha" particles.
<center>
<br>
<math>\Gamma (T, V) = \sum_{m_{\alpha}} Q(T, V, m_
, M)</math>
<br>
<math>\Gamma (T, V) = (q_
+ q_
)^M</math>
<br>
</center>
The equilibrium numebr of alpha particles, <math>\overline m_
</math>, is derived below.
<center>
<br>
<math>\overline m_
= \sum_{m_{\alpha}} m_
P(m_
)</math>
<br>
<math>\overline m_
= \sum_{m_{\alpha}} \left ( \frac{ Q(T, V, m_
, M)}
\right )</math>
<br>
<math>\overline m_
= q_
\left ( \frac
{\partial q_{\alpha}} \right ) </math>
<br>
<math>\overline m_
= q_
\left ( \frac
{\partial q_{\alpha}} \right ) </math>
<br>
<math>\overline m_
= M \left ( 1 + \frac{q_{\beta}}{q_{\alpha}} \right )^{-1}</math>
<br>
</center>
c) Provide a rough sketch of the Helmholtz free energy (as a function of temperature) of the nanoparticle system on the graph given in the problem statement.
<p>
</p>
To derive the equilibrium energy as a function of temperature, plug <math>m_
^
= \overline m_
</math> into an expression from part (a). Consider two limits, <math>q_
\gg q_
</math> and
<math>q_
\ll q_
</math>. At low temperature when <math>q_
\ll q_
</math>, the number of alpha nanoparticles is essentially <math>M</math> and the free energy of the nanoparticle system follows the "alpha" curve. At high temperatures, when <math>q_
\ll q_
</math>, the number of alpha particles is zero and the free energy follows the "beta" curve. At intermediate temperatures, the nanoparticle system smoothly transitions between the two curves. An example is shown below.
<center>
Unable to render embedded object: File (Elemental_excitium_-_nanoparticle_system.PNG) not found.
</center>
d) Derive a formula for the fluctuations of <math>m_
</math>, the number of nanoparticles in the <math>\alpha</math> phase.
<p>
</p>
Use a three step trick reviewed in lecture to relate fluctuations in the number of alpha nanoparticles to a derivative of the average number of nanoparticles with respect to its conjugate. It is known that this is derivative is evaluated at <math>\mu_
= 0</math>
<center>
<br>
<math>\sigma_{m_{\alpha}}^2 = \left ( \overline {m_
^2} - \overline m_
^2 \right )</math>
<br>
<math>\sigma_{m_{\alpha}}^2 = k_B T \left ( \frac{\partial \overline m_{\alpha}}{\partial ( \beta \mu_
} \right )</math>
<br>
</center>
Extra notes not part of solution:
<p>
</p>
Fluctuations in this problem can be evaluated analytically to find an expression below. Fluctuations as a function of temperature is maximum near where "alpha" and "beta" curves cross.
<center>
<br>
<math>\sigma_{m_{\alpha}}^2 = \left ( \overline {m_
^2} - \overline m_
^2 \right )</math>
<br>
<math>\sigma_{m_{\alpha}}^2 = \left ( \frac{ \overline m_
(M - \overline m_
)}
\right )</math>
<br>
</center>