Question 1
It is now possible to attach the endpoints of a single polymer of length <math>l</math> and impose a constant force <math>f</math> on the ends of the polymer. Experimentally, it has been found that the elongation <math>(l-l_o)</math> of a particular polymer (where <math>l_o</math> is the equilibrium length of the polymer at <math>f=0</math>) is proportional to the applied force on the ends, i.e. <math>(l-l_o)= \alpha f</math> Assume that a polymer satisfying this relation is held at constant force <math>f</math>.
<p>
</p>
a) Write an expression for the appropriate partition function for this polymer at constant temperature <math>T</math> and force <math>f</math> (leave the expression as a sum over states).
<p>
</p>
Use a prescription provided in class.
<center>
<br>
<math>dU = d \overline E</math>
<br>
<math>dU = TdS + fdl</math>
<br>
<math>U = \overline E</math>
<br>
<math>U = TS + fl</math>
<br>
<math>\frac
= \beta \overline E - \beta f l</math>
<br>
</center>
This is at constant <math>T</math> and <math>f</math>, so there is need to perform a Legendre transform. The sum over states includes a sum over all allowed lengths.
<center>
<br>
<math>\frac
- \beta \overline E + \beta f l = - \beta \phi</math>
<br>
<math>\wedge = \sum_
e^{ - \beta ( e_
- f l_
)}</math>
<br>
</center>
b) Derive an expression for <math>\overline
- \overline l^2</math> of the polymer at constant <math>T</math> and <math>f</math> and express your result in terms of parameters given in this problem.
<center>
<br>
<math>\overline l = \frac{\sum_
l_
e^{-\beta (E_
- f l_
)}}
</math>
<br>
<math>\mbox
</math>
<br>
<math>\overline l \wedge = \sum_
l_
e^{- \beta (E_
- f l_
)}</math>
<br>
<math>\mbox
</math>
<br>
<math>\frac
= \wedge \frac
+ \overline l \frac
</math>
<br>
<math>\frac
= \wedge \frac
+ \overline l_
\sum \beta l_
e^{-\beta (E_
- fl_{\eta}})</math>
<br>
<math>\frac
\left ( \sum_
l_
e^{- \beta (E_
- f l_
)} \right ) = \beta \sum_
l_
2 e{-\beta(E_{\eta - f l_
)</math>
<br>
<math>\mbox
</math>
<br>
<math>\frac
+ \overline l_
\frac{\sum \beta l_
e^{-\beta (E_
- fl_{\eta}})}
Unknown macro: {wedge}= \beta \frac{\sum_
l_
2 e{-\beta(E_
- f l_
)}}
</math>
<br>
<math>\frac
+ \beta \overline l^2 = \beta \overline
</math>
<br>
<math>\overline
- \overline l^2 = kT \frac
</math>
<br>
<math>\overline
- \overline l^2 = kT \alpha</math>
<br>
</center>
Question 2
Conceptual questions, be brief in your answers, write relevant equations if it improves clarity.
<p>
</p>
a) Give a statistical mechanical interpretation of the second law of thermodynamics for a
spontaneous change of state in an isolated system.
<p>
</p>
In an isolated system, which corresponds to constant <math>(E, N, V)</math>, it is true from the second law that a spontaneous process occurs if final state is of larger entropy than initial statem, expressed below.
<center>
<br>
<math>\Delta S = S_f - S_i</math>
<br>
</center>
From satistical mechanics, it is known that <math>S = k \ln \Omega</math>, where <math>\Omega</math> is the degeneracy at the fixed energy <math>E</math> of the system. Relations below follow.
<center>
<br>
<math>\Delta S = S_f - S_i</math>
<br>
<math>\Delta S = k \ln \Omega_f - k \ln \Omega_i</math>
<br>
<math>\Delta S = k \ln \left ( \frac
\right ) > 0</math>
<br>
<math>\Omega_f > \Omega_i</math>
<br>
</center>
In a spontaneous process, the system evolves to a state with maximal degeneracy (maximal number of microstates). This can occur if some internal constraint in the sysmte is removed, such as by adding a catalyst.
<p>
</p>
b) Give a statistical mechanical interpretation of the third law of thermodynamics.
<p>
</p>
The third law states that entropy of a system goes to zero as <math>T \right 0</math>. From statistical mechanics the following is known.
<center>
<br>
<math>\mbox
\nu \mbox
</math>
<br>
<math>P_
= \frac
</math>
<br>
<math>\Omega_0 = \mbox
</math>
<br>
<math>\mbox
\nu \mbox
</math>
<br>
<math>P_
= 0</math>
<br>
<math>S = \sum_
P_
\ln P_
</math>
<br>
<math>S = k \ln \Omega_
</math>
<br>
</center>
The value of <math>\Omega_o</math> is typically one or two and rarely exceeds <math>N \approx10^
</math>. Therefore, the following is true.
<center>
<br>
<math>S \approx k \ln 10{^23}</math>
<br>
<math>S \approx 23K</math>
<br>
<math> \frac
\approx \frac
{10^{23}} </math>
<br>
<math>\frac
\right 0</math>
<br>
</center>
Question 3
<math>M</math> arsenic atoms are adsorbed on a surface with <math>M</math> sites. Each arsenic atom vibrates in a parabolic potential on each site. The arsenic atoms do not interact with each other when adsorbed. The characteristic vibrational frequency for each adsorbed arsenic atom is <math>\nu</math>. (The adsorbed atoms vibrate independently of each other).
<p>
</p>
a) Solve for an expression of the partition function of the system of <math>M</math> arsenic atoms in terms of <math>\nu</math> (for fixed temperature).
<p>
</p>
Note the following
- Absorbed atoms are independent and indistinguishable
- single particle energies are <math>\epsilon_n</math>
<center>
<br>
<math>\epsilon_n = (n + \frac
) h \nu</math>
<br>
</center>
Write a partition function
<center>
<br>
<math>Q = q^n</math>
<br>
<math>q = \sum_
^
e^{- \beta \epsilon_n}</math>
<br>
<math>q = \sum_
^
e^{- \beta (n+\frac
)h \nu}</math>
<br>
<math>q = \frac{e^{-\beta \frac
}}{1 - e^{-\beta h \nu}}</math>
<br>
</center>
Now assume that there are only <math>N < M</math> arsenic atoms adsorbed over the <math>M</math> surface sites. Each arsenic atom still vibrates independently in a parabolic potential with frequency <math>\nu</math>.
<p>
</p>
b) Solve for an expression of the partition function in terms of <math>\nu</math>.
<p>
</p>
When <math>N < M</math> there is configurational disorder with degeneracy <math>\Omega</math>
<center>
<br>
<math>\Omega = \frac
</math>
<br>
<math>Q = \Omega q^N</math>
<br>
<math>Q = \frac
q^N</math>
<br>
</center>
c) The system of adsorbed arsenic atoms is in equilibrium with a gas of arsenic atoms. The gas phase has a constant arsenic chemical potential <math>\mu</math>. Solve for an expression for the average number of arsenic atoms adsorbed over the <math>M</math> sites in equilibrium with the gas phase at constant temperature.
<p>
</p>
Consider the average number of particles, <math>\overline N</math>, when the surface is in equilibrium with gas of arsenic at constant <math>\mu</math>. Consider the grand canonical ensemble.
<center>
<br>
<math>\Theta = \sum_
M Q(N, V, T)e
</math>
<br>
<math>\Theta = \sum_
^M \frac
q^N e^
</math>
<br>
<math>\Theta = \sum_
^M \frac
\left ( q e^
\right )N \left (q \right )
</math>
<br>
<math>\Theta = \left ( 1 + q e^
\right )^M</math>
<br>
</center>
In the grand canonical ensemble, the following is true.
<center>
<br>
<math>\overline N = \frac
</math>
<br>
<math>\overline N = kT M \frac{ \partial \ln \left ( 1 + qe^
\right )}
</math>
<br>
</center>
An alternative way of solving 3 c is to use equivalence of ensembles and work with the canonical ensemble.
<center>
<br>
<math>\mu = -kT \frac
</math>
<br>
<math>Q = \frac
q^N</math>
<br>
</center>
Question 4
Given is a plot of the cubic lattice parameter <math>a</math> for fcc binary alloy <math>Al-Ti</math> as a function of the mole fraction <math>x</math> (at <math>360 C</math>). Estimate the partial molar volume of <math>Ti</math> in the binary <math>Al-Ti</math> alloy at <math>x=0.25</math>.
<center>
Unable to render embedded object: File (Lattice_parameter_vs_composition_Al_and_Ti.PNG) not found.
</center>
<center>
<br>
<math>\overline V = \frac
{\partial N_
}</math>
<br>
<math>\overline V = \underline V + \frac
(1 - x)</math>
<br>
<math>x = \frac{N_{Ti}}{N_
- N_{Al}}</math>
<br>
</center>
In deriving an expression of volume, consider that there are four atoms per fcc unit cell.
<center>
<br>
<math>\underline V = \frac
</math>
<br>
<math>\frac
= \frac
\frac
\left ( 1 - x \right )</math>
<br>
<math>\overline V_
= \frac
+ \frac
\frac
\left ( 1 - x \right )</math>
<math>a (x = 0.25 ) \approx 3.3</math>
<br>
<math>\frac
\approx \frac
{1 - 0)</math>
<br>
<math>\frac
\approx -1.3</math>
<br>
</center>
Substitute a final expression above in an expression derived earlier of <math>\overline V_
</math>.
Question 5
Consider an idealization of a crystal which has <math>N</math> lattice sites and the same number of interstitial positions (places between the lattice points where atoms can reside).
<p>
</p>
a) Derive the number of configurational micro-states for this system when <math>n</math> atoms have gone to interstitial sites leaving <math>n</math> vacancies behind.
<center>
<br>
<math>\Omega = \left ( \frac
\right )^2</math>
<br>
</center>
b) Assume that the vacancies and interstitial atoms do not interact. Calculate the entropy and express it in terms of the interstitial concentration <math>x=n/N</math>.
<p>
</p>
Since vacancies and interstitials don't interact energetically, every configuration is of the same energy.
<center>
<br>
<math>S = k \ln \Omega</math>
<br>
<math>S = 2 k \ln \left ( \frac
\right</math>
<br>
<math>S = 2 k \left [N \ln N - N - n \ln n + n - (N-n) \ln (N-n)+(N-n) \right]</math>
<br>
<math>S = 2k \left [ N \ln \frac
\right ) + n \ln \left ( \frac
\right )</math>
<br>
<math>S = 2 k N \left [ \ln \left ( \frac
\right ) + x \ln \left ( \frac
\right )</math>
<br>
<math>S = -2kN \left [(1 - x) \ln (1 - x) + x \ln x \right]</math>
<br>
</center>
c) Assume now that the interstitial atoms are energetically attracted to the vacant sites. Will the entropy increase, stay the same or decrease from the value determined in part b ?
<p>
</p>
Entropy decreases since the entropy of part b) is already maximal. Any time certain microstates are energetically favored or are of higher probability of occurring, entropy is lower than when all microstates are of equal probability of occurring.