Question 1
An insulated rigid chamber is partitioned into two equal halves by a rigid diathermal wall. One mole of ideal gas at <math>298K</math> and <math>10 atm</math> pressure initially occupies one of the halves. A very small hole is made in the wall which allows the gas to very slowly flow from one half into the other one until it equilibrates.
<p>
</p>
a) Calculate the final temperature of the gas
<p>
</p>
There is no heat flow on the overall system (gas + empty space) nor is any work performed on or by it, hence <math>dU = 0</math>. The following is true of an ideal gas.
<center>
<br>
<math>dU = nC_v dT</math>
<br>
<math>dU = 0</math>
<br>
<math>dT = 0</math>
<br>
<math>T_f = T_o = 298 K</math>
<br>
</center>
b) Calculate the entropy change of the gas
<p>
</p>
Entropy is a state function and we have characterized the initial and final states of the gas as <math>(T_o, V)</math> and <math>(T_o, 2V)</math>. To calculate the entropy change of the gas, all that is needed to be done is to construct a reversible isothermal expansion from <math>V</math> to <math>2V</math>. Combining the fist and second laws give the following equation.
<center>
<br>
<math>TdS = nC_v dT + pdV</math>
<br>
<math>TdS = pdV</math>
<br>
</center>
Substitute the ideal gas equation of state.
<center>
<br>
<math>dS = nR \frac
</math>
<br>
</center>
There is one mole, and value of the change of state is below.
<center>
<br>
<math>\Delta S_
= R \ln (2)</math>
<br>
</center>
c) Calculate the entropy change of the universe (system + environment)
<p>
</p>
The chamber is insulated, so the following equation is true.
<center>
<br>
<math>dS_
= \frac
= 0</math>
<br>
</center>
The entropy change of the universe reduces to following relation.
<center>
<br>
<math>\Delta S_
= \Delta S_
= R \ln (2)</math>
<br>
</center>
A common mistake was to treat part a as a reversible adiabatic expansion, but while the expansion may proceed slowly, it remains highly irreversible due to the large pressure gradient across the small hole.
<p>
</p>
d) Instead of making a small hole in the partition, I remove the partition quickly. What is now the entropy change of the gas after it has equilibrated?
Define the system as the gas initially in an occupied half.
<center>
<br>
<math>dQ = 0</math>
<br>
<math>dW = 0</math>
<br>
<math>\Delta S_
= \Delta S_
= R \ln(2)</math>
<br>
</center>
Question 2
A very thin film is epitaxially deposited on a very thick substrate. Calculate the effective heat capacity of the film under the epitaxial constraint in terms of a subset of the following properties:
<center>
<br>
<math>\mbox
</math>
<br>
<math>(C_p, C_v, E, \alpha_v, \alpha_l, \beta_T)</math>
<br>
<math>\mbox
</math>
<br>
<math>(C_p, C_v, E, \alpha_v, \alpha_l, \beta_T)</math>
<br>
</center>
Make sure to distinguish properties of the substrate from those of the film for the final answer. Note that <math>\alpha_V = 3 \alpha_L</math> and Young���s modulus (at constant <math>T</math>) is defined as:
<center>
<br>
<math>E_T = L \left ( \frac
\right )_T</math>
<br>
</center>
The epitaxy constraint implies that the in-plane lattice parameter of the film is constraint to be the same as the one of the substrate. You can assume that that the <math>z</math>-direction (normal to the film) expands freely in the same way the film material would expand if there is no constraint. You may neglect Poisson strains (which would arise due to the stress in the <math>xx</math> and <math>yy</math> directions) and treat the elastic properties of the film as isotropic.
<p>
</p>
Option 1
<p>
</p>
We are interested in a heat capacity under a strange constraint. Start with <math>dS</math>.
<center>
<br>
<math>dS = \left ( \frac
\right )_V dT + \left ( \frac
\right )_T dV</math>
<br>
</center>
The relation below follows.
<center>
<br>
<math>T \left ( \frac
\right )_
= T \left ( \frac
\right )_
+ T \left ( \frac
\right )_
\left ( \frac
\right )_
</math>
<br>
<math>C_
= C_v + T \left ( \frac
\right )_T \left ( \frac
\right )_
</math>
<br>
</center>
The partial derivative <math>\left ( \frac
\right )_T</math> of the unconstrained film is below.
<center>
<br>
<math>\left ( \frac
\right )_T = \left ( \frac
\right )_v</math>
<br>
<math>\left ( \frac
\right )_T = - \frac{\left ( \frac
\right )_p }{ \left ( \frac
\right )_T }</math>
<br>
<math>\left ( \frac
\right )_T = \frac{\alpha_V^{(f)}}{\beta_T^
</math>
<br>
</center>
The only constrained partial derivative is <math>\left ( \frac
\right )_
</math> which is found by expanding <math>dV</math>:
<center>
<br>
<math>dV = A_
dl_
+ A_
dl_
+ A_
dl_z</math>
<br>
<math>\left ( \frac
\right )_
= A_
\left ( \frac
\right )_
+ A_
\left ( \frac
\right )_
+ A_
\left ( \frac
\right )_
</math>
<br>
<math>\left ( \frac
\right )_
= V \left ( 2 \alpha_L^
+ \alpha_L^
</math>
<br>
</center>
Combine everything.
<center>
<br>
<math>C_
= C_v + \frac{TV \alpha_V^
}{\beta_T^
} \left ( 2 \alpha_L^
+ \alpha_L^
\right )</math>
<br>
</center>
Option 2
There is interest in the heat capacity under a strange constraint. There is need to expand <math>dS</math> in an appropriate set of thermodynamic coordinates for the film. Neglect Poisson effects and consider the entropy of the film as a function of coordinates (<math>T</math>, <math>l_x</math>, <math>l_y</math>, <math>l_z</math>) where <math>l_i</math> is the length of the film in the <math>i^
</math> direction. Expand the first order differential <math>dS</math> of the film.
<center>
<br>
<math>dS = \left ( \frac
\right )_
dT + \left ( \frac
\right )_
dl_x + \left ( \frac
\right )_
dl_y + \left ( \frac
\right )_
dl_z</math>
<br>
</center>
The first term on the right hand side can be related to a constant volume heat capacity of the film. An epitaxy constraint implies that when the substrate is held at constant pressure <math>dl_x^
= dl_x^
</math> and <math>dl_z^
= dl_x^
</math>, while <math>dl_z^
</math> is allowed to expand freely. Express <math>dS</math> of the film.
<center>
<br>
<math>T \left ( \frac
\right )_
= T \left ( \frac
\right )_
+ T \left [ \left ( \frac
\right )_
\left ( \frac
\right )_
+ \left ( \frac
\right )_
\left ( \frac
\right )_
+ \left ( \frac
\right )_
\left ( \frac
\right )_
</math>
<br>
</center>
Terms such as <math>\left ( \frac
\right )_
</math> can be related to the linear expansion coefficients of the film and substrate
<center>
<br>
<math>C_
= C_v^
+ T \left [l_x \alpha_L^
\left ( \frac
_
+ l_y \alpha_L^
\left ( \frac
_
+ l_z \alpha_L^
\left ( \frac
_
\right]</math>
<br>
</center>
Use Maxwell relations to derive terms like <math>\left ( \frac
\right )_
</math>. The first-order differential of a Hemholtz-like potential (setting shear stress components to zero) is below.
<center>
<br>
<math>dF = -SdT + V \sigma_x d \epsilon_x + V \sigma_y d \epsilon_y + V \sigma_z d \epsilon_z</math>
<br>
</center>
Put everything explicitly in terms of the relevant coordinates. For instance, this means <math>V \sigm_x d \epsilon_x = A_
\sigma_x dl_x</math>. A Maxwell relation of interest is below.
<center>
<br>
<math>- \left ( \frac
\right )_
= \left ( \frac{\partial (A_
\sigma_i)}
\right )_
</math>
<br>
<math>\left ( \frac{\partial (A_
\sigma_i)}
\right )_
= A_
= \left ( \frac
\right )_
</math>
</center>
The partial above can be related to a Young's modulus and linear expansion coefficient. The derivation below is true of each of the three different terms, all of which are properties of the film only. The partial derivative is taken without the epitaxy constraint.
<center>
<br>
<math>\left ( \frac{\partial (A_
\sigma_i)}
\right )_
= A_
= -A_
\frac{\left (\frac
\right )_{\sigma_i, l_j, l_k}}{\frac
\right )_{T, l_j, l_k}}</math>
<br>
<math>\left ( \frac{\partial (A_
\sigma_i)}
\right )_
= -A_
\alpha_L^
E_T^
</math>
<br>
<math>\left ( \frac
\right )_
= -A_
\alpha_L^
E_T^
</math>
</center>
Combine everything.
<center>
<br>
<math>C_
= C_v^
+ TV \left [2 \alpha_L^
\alpha_L^
E_T^
+ (\alpha_L^
2 E_T
\right]</math>
<br>
<math>C_
= C_v^
+ TV \alpha_L^
E_T^
\left [2 (\alpha_L^
+ \alpha_L^
\right]</math>
<br>
</center>
The previous answer is recovered noting the relationships below. These follow when Poisson effects are ignored.
<center>
<br>
<math>\alpha_L^
= \frac{\alpha_v^{(f)}}
</math>
<br>
<math>\alpha_L^
= \frac
{\beta_v^{(f)}}</math>
<br>
</center>
If the substrate were held at constant strains <math>(\epsilon_x, \epsilon_y)</math>, then the following would be true. The only term retained in the expansion of <math>dS</math> for the film would be the <math>dl_z</math> component leading to the final relationship below.
<center>
<br>
<math>dl_z^
= dl_x^
= 0</math>
<br>
<math>dl_y^
= dl_y^
= 0</math>
<br>
<math>C_
= C_v^
+ TV ( \alpha_L^
)2 E_T
</math>
<br>
</center>
Question 3
The two figures below show the Gibbs free energy <math>G(T, p=0)</math> and entropy <math>S(T, p=0)</math> of three phases of silicon: liquid, cubic diamond, and a phase with the <math>\beta-Sn</math> structure. The volume/atom ration of the <math>\beta-Sn</math> phase <math>V_
\approx 0.75 V_
</math> and <math>V_
\approx 0.9 V_
</math> when <math>T = 1000K</math> and <math>V_
= 20 \frac
</math>.
<center>
</center>
(a) Calculate the slopes of stable and metastable p-T phase boundaries at <math>p=0</math>
<p>
</p>
A goal is to find the slopes of the phase boundaries. Use the Clapeyron equation to find the flopes of each of the three choose two phase boundaries at <math>p=0</math>.
<center>
<br>
<math>\frac
= \frac
</math>
<br>
<table>
<tr>
<td>
<center>
<math>\mbox
</math>
</center>
</td>
<td>
<center>
<math>T^*</math>
</center>
</td>
<td>
<center>
<math>\Delta S</math>
</center>
</td>
<td>
<center>
<math>\Delta V</math>
</center>
</td>
<td>
<center>
<math>\frac
</math>
</center>
</td>
</tr>
<tr>
<td>
<center>
<math>\beta \right l</math>
</center>
</td>
<td>
<center>
<math>1500K</math>
</center>
</td>
<td>
<center>
<math>2.5 k_B</math>
</center>
</td>
<td>
<center>
<math>0.15 V_o</math>
</center>
</td>
<td>
<center>
<math>16.67 \frac
</math>
</center>
</td>
</tr>
<tr>
<td>
<center>
<math>cd \right l</math>
</center>
</td>
<td>
<center>
<math>1700K</math>
</center>
</td>
<td>
<center>
<math>4.0 k_B</math>
</center>
</td>
<td>
<center>
<math>-0.1 V_o</math>
</center>
</td>
<td>
<center>
<math>-40 \frac
</math>
</center>
</td>
</tr>
<tr>
<td>
<center>
<math>cd \right \beta</math>
</center>
</td>
<td>
<center>
<math>2000K</math>
</center>
</td>
<td>
<center>
<math>1.8 k_B</math>
</center>
</td>
<td>
<center>
<math>-0.25 V_o</math>
</center>
</td>
<td>
<center>
<math>-7.2 \frac
</math>
</center>
</td>
</tr>
</table>
</center>
The ratio <math>\frac
= 0.00069 \frac
</math>
<p>
</p>
(b) Calculate any phase transformation that may occur at room temperature.
<p>
</p>
At <math>(p=0, T=298K)</math> the stable phase is "cd" and two pressure induced transitions are possible <math>(cd \right \beta)</math> or <math>(cd \right l)</math>. Linearize the phase boundaries of each of the three phase equilibria.
<center>
<br>
<math>p_
^*(T) = -7.2*0.00069(T-2000)GPa</math>
<br>
<math>p_
^*(298) = 8.46 GPa</math>
<br>
<math>p_
^*(T) = -40*0.00069(T-1700)GPa</math>
<br>
<math>p_
^*(298) = 38.7 GPa</math>
<br>
</center>
The first transition is the equilibrium transition. The second only occurs if the formation of <math>\beta</math> is completely suppressed. The <math>\beta</math> phase is lower in free energy than the liquid phase in all positive pressures at <math>298 K</math>.
<p>
</p>
(c) Sketch an approximate <math>(p, T)</math> phase diagram of Si including <math>cd</math>, <math>\beta</math>, and liquid phases.
<p>
</p>
The only phase boundary of interest is <math>p_
^*(T) = 0.0115(T-1500)GPa</math> Below are the phase boundaries in the <math>(p, T)</math> plane. Determine the lowest free energy in each region and label the overall phase boundaries.
<center>
</center>
Question 4
Thermoelectric devices can transfer energy from a low temperature side to a high temperature side like a heat pump by using the entropy carried by an electric current. Is the efficiency of such a thermoelectric device limited by the efficiency of a reverse Carnot engine, a refrigerator, for which the following is true.
<center>
<br>
<math>\eta_
= \frac
</math>
<br>
<math>\eta_
= \frac
</math>
<br>
</center>
The term <math>Q_L</math> is the heat removed on the low temperature side and <math>W</math> is the electrical work put into the thermoelectric device. Yes or No?
<p>
</p>
Yes, the efficiency of any engine (or reverse engine) is limited by its corresponding Carnot engine counterpart
<p>
</p>
When a fluid evaporates in the environment it performs work on the environment. The work could be captured with a device like the one shown below. As fluid evaporates it travels through a turbine which can generate work. The walls of the container are diabatic so the fluid can absorb heat from the environment to compensate for the enthalpy of evaporation is loses as fluid evaporates. It seems that this construction violates the Second Law as the container absorbs heat from one temperature reservoir (the environment) and performs work (as the vapor moves through the turbine). Please state whether based on the Laws of Thermodynamics this device can function as described (yes or no). If the answer is yes, explain why this device does not violate the Second Law. If the answer is no, no explanation is need.
This device does not violate the Second Law. In terms of heat-to-work conversion the Second Law only places a restriction on cyclic processes. This is not a cyclic process. After a while the evaporator runs out.
<p>
</p>
Also note that the amount of work performed is not equivalent to the heat extracted from the environment. A large fraction of <math>Q</math> extracted from the environment is expended on the bond breading process and is not recovered.
Question 5
The figure below shows three chambers connected by semi-permeable membranes. Each chamber contains <math>A</math>, <math>B</math>, and <math>C</math> atoms. The chambers are under constant <math>T</math> and <math>P</math>. The membranes are electronically insulating, but there is an unusual property of the ions. The membrane between <math>\alpha</math> and <math>\beta</math> lets neutral <math>C</math> and <math>B</math> atoms and ionized <math>A+</math> atoms (but not neutral <math>A</math>) through. The membrane between <math>\beta</math> and <math>\gamma</math> lets neutral <math>A</math> and <math>C</math> and ionized (but not neutral B) through.
<p>
</p>
Write down the equilibrium potential difference between <math>\alpha</math> and <math>\gamma</math> in terms of the chemical potentials of <math>A</math>, <math>B</math>, and <math>C</math> in <math>\alpha</math> and <math>\gamma</math>
<p>
</p>
Option 1
<p>
</p>
First, note that <math>\alpha</math> and <math>\beta</math>, <math>\beta</math> and <math>\gamma</math> must be in equilibrium independently. Focus on the equilibrium of <math>\alpha</math> and <math>\beta</math>.
<p>
</p>
At constant <math>T</math> and <math>p</math>, the following is true.
<center>
<br>
<math>dG^
+ dG^
= 0</math>
<br>
<math>\left ( (\mu_A^
- (\mu_A^
\right ) dn_A^
+ \left ( (\mu_B^
- (\mu_B^
\right ) dn_B^
+ \left ( (\mu_C^
- (\mu_C^
\right ) dn_C^
+ \left ( (\phi
- (\phi^
\right ) dq^
= 0</math>
<br>
</center>
The species <math>A</math> transfers charge. The terms <math>dn_A</math> and <math>dq</math> are coupled.
<center>
<br>
<math>dn_A^
= z F dq^
</math>
<br>
</center>
The term <math>z</math> is the valence of <math>A</math>, and <math>F</math> is Faraday's constant. In an adjusted set of units, write the following equation and substitute into the equilibrium condition.
<center>
<br>
<math>dn_A^
= dq^
</math>
<br>
<math>\left ( \mu_A^
- \mu_A^
+ \phi^
- \phi^
\right ) dn_A^
+ \left (\mu_B^
- \mu_B^
\right ) dn_B^
+ \left ( \mu_C^
- \mu_C^
\right ) dn_C^
</math>
<br>
</center>
The relationship must hold for all independent variations, so the following relations are true.
<center>
<br>
<math>\mu_B^
= \mu_B^
</math>
<br>
<math>\mu_C^
= \mu_C^
</math>
<br>
<math>\phi^
= \mu_A^
- \mu_A^
+ \phi^
</math>
<br>
</center>
Similar conditions are derived considering the equilibrium of <math>\beta</math> and <math>\gamma</math>.
<center>
<br>
<math>\mu_A^
= \mu_A^
</math>
<br>
<math>\mu_C^
= \mu_C^
</math>
<br>
<math>\phi^
= \mu_B^
- \mu_B^
Unknown macro: {beta}+ \phi^
</math>
<br>
</center>
Equate two expressions of <math>\phi^
</math> and use two relations below.
<center>
<br>
<math>\mu_A^
= \mu_A^
</math>
<br>
<math>\mu_B^
= \mu_B^
</math>
<br>
<math>
\phi^
- \phi^
Unknown macro: {gamma}- \mu_B^
= \left ( \mu_B^
\right ) + \left ( \mu_A^
- \mu_A^
\right )</math>
<br>
</center>