Question 1

a) A system in the canonical ensemble has a finite number of possible energy levels <math>E_j (j = 1,...,N)</math> and each energy level has a degeneracy of <math>\Omega (E_j)</math> What is the maximum value of the entropy that can be achieved in this system?

In the canonical ensemble, entropy is maximal as <math>T \right \infty</math>.

<math>S = - k\sum_

Unknown macro: {nu}

P_

\ln P_

Unknown macro: {nu}

</math>

<math>P_

- \frac{e^{-\beta E_

Unknown macro: {nu}

}}{\sum_

e^{-\beta E_{\nu}}</math>

<math>P_

Unknown macro: {nu}

= \frac{e^{-\beta E_

}}{\sum_

Unknown macro: {j=1}

^{N \Omega (E_j ) e}{-\beta E_J}}</math>

<math>\lim_

Unknown macro: {T right infty}

= P_

Unknown macro: {nu}

\frac

Unknown macro: {1}

{\sum_

^N \Omega (E_j)}</math>

<math>S_

Unknown macro: {max}

= - k \sum_j \frac

Unknown macro: {1}

{\sum_

Unknown macro: {j=1}

^N \Omega (E_j)} \left ( \frac

{\sum_

Unknown macro: {j=1}

^N \Omega (E_j)} \right )</math>

<math>S_

= k \ln \left ( \frac

Unknown macro: {1}

{\sum_

Unknown macro: {j=1}

^N \Omega (E_j)} \right )</math>

</center>

b) At constant <math>N</math>, <math>T</math> and <math>V</math>, can the probability distribution <math>Q</math> be applied to study the thermal properties of nano-particles? (Keep the answer very short, preferably one word long).

Yes, there are no restrictions on the size of the system in the derivation of the probability distribution function <math>P_

Unknown macro: {nu}

</math>

c) A new element zipium has been discovered. Preliminary solutions of the Schr��odinger equation of this element indicate that the fcc crystal structure of zipium is lower in energy than the hcp crystal structure. Furthermore, the Einstein vibrational frequency of zipium in the fcc crystal structure is smaller than that in the hcp crystal structure. Do you expect a phase transformation between the two crystal structures as the temperature is varied?

Consider the ground state energies and vibrational frequencies.

<math>E_

Unknown macro: {fcc}

< E_

Unknown macro: {hcp}

</math>

<math>\omega_

^E < \omega_

Unknown macro: {hcp}

^E</math>

</center>

Based on this information, the fcc phase is more stable than hcp at <math>T=0</math>. The fcc phase is of higher vibrational entropy than hcp.

<math>S_

Unknown macro: {fcc}

> S_

</math>

</center>

As a result free energies of fcc and hcp phases never cross and there is not a phase transformation.

Question 2

a) At constant <math>T</math>, <math>P</math> and <math>N</math>, derive an expression for the heat capacity <math>C_p</math>* in terms of fluctuations of relevant microscopic quantities.*

Constant <math>T</math>, <math>P</math>, and <math>N</math> corresponds to the isothermal-isobaric ensemble. It is known that heat capacity at constant pressure is equal to <math>\frac

Unknown macro: {partial H}

Unknown macro: {partial T}

</math>. Consider fluctuations in enthalpy. Define the enthalpy of a microstate <math>\eta</math> as below.

<math>H_

Unknown macro: {eta}

= E_

+ pV_

Unknown macro: {eta}

</math>

</center>

Write the partition function of the ensemble.

<math>\Delta = \sum_

e^{\beta (E_

Unknown macro: {eta}

+ pV_

)</math>

<math>\Delta = \sum_

Unknown macro: {eta}

e^{\beta (H_

)}</math>

</center>

Calculate fluctuations of <math>H</math> by following a three step procedure.

Multiply both sides of the expression for the average enthalpy, <math>\overline H</math>, by the partition function.
Differentiate with respect to conjugate of <math>H</math>, which in this case is <math>\beta</math>.
Divide by <math>\Delta</math>

<math>\mbox

Unknown macro: {Step 1}

</math>

<math>\overline H \Delta = \sum_

Unknown macro: {eta}

H_

e^{-\beta (H_

Unknown macro: {eta}

)}</math>

<math>\mbox

Unknown macro: {Step 2}

</math>

<math>\frac

Unknown macro: {partial overline H}

Unknown macro: {partial beta}

\Delta + \overline H \frac

Unknown macro: {partial Delta}

= \sum_

- H_

Unknown macro: {eta}

^{2 e}{-\beta H_

}</math>

<math>\frac

Unknown macro: {partial overline H}

Unknown macro: {partial beta}

\Delta + \overline H \left ( \sum_

Unknown macro: {eta}

H_

e^{-\beta H_{\eta}} \right ) = \sum_

Unknown macro: {eta}

H_

^{2 e}{-\beta H_

Unknown macro: {eta}

}</math>

<math>\mbox

Unknown macro: {Step 3}

</math>

<math>\frac

Unknown macro: {partial beta}

= \overline H^2 - \overline

Unknown macro: {H^2}

</math>

</center>

Rewrite left-hand side of equation and relate to definition of heat capacity.

<math>\frac

Unknown macro: {partial overline H}

= \frac

Unknown macro: {partial overline H}

\frac

Unknown macro: {partial T}

Unknown macro: {partial beta}

</math>

<math>\frac

Unknown macro: {partial overline H}

= -kT^2 \frac

Unknown macro: {partial overline H}

</math>

<math>C_p = \frac

Unknown macro: {partial overline H}

Unknown macro: {partial T}

</math>

<math>C_p = \frac{\overline H^2 - \overline {H^2}}

Unknown macro: {kT^2}

</math>

</center>

b) From classical thermodynamics we know that <math>C_p > C_v > 0</math>. Speculate why this should be the case based on what you know from statistical mechanics (be brief and qualitative).

Heat capacities are determined by fluctuations. Heat capacity at constant volume, <math>C_v</math>, is determined by fluctuations in energy microstates. Heat capacity at constant pressure, <math>C_p</math>, is determined by fluctuations in energy and volume microstates. Therefore, in the isothermal-isobaric ensemble, the system can fluctuate over many more microstates.

Question 3

A depleted uranium substrate (at constant volume) has <math>M</math> surface sites on which zipium can adsorb. The adsorption energy per zipium atom is <math>\epsilon_0</math>. The substrate and adsorbed zipium atoms are at constant temperature and in a magnetic field <math>H</math> directed perpendicular <math>t</math> the surface. Each zipium atom has a magnetic moment <math>n \mu B</math> which can be aligned in the direction of <math>H (n = +1)</math> or in a direction opposite to <math>H (n = -1)</math>. <math>\mu B</math> is the Bohr magneton. Assume that adsorbed zipium atoms do not interact with each other energetically.

a) Derive an explicit expression for the partition function when there are <math>N (< M)</math> adsorbed zipium atoms at constant <math>T</math>, <math>V</math> and <math>H</math>.

The relevant partition function at constant <math>T</math>, <math>V</math>, <math>H</math>, and <math>N</math> is below.

<math>\wedge = \sum_

Unknown macro: {eta}

e^{- \beta (E_

- M_

Unknown macro: {eta}

H) }</math>

</center>

Note that a summation over microstates <math>\eta</math> occurs over all possible configurations of <math>N</math> atoms, over <math>m</math> sites and over all magnetic states. Macroscopic energy and magnetization are defined below.

<math>E_

= N \epsilon_0</math>

<math>M_

Unknown macro: {eta}

= \sum_

Unknown macro: {i=1}

^N n_i \mu_B</math>

</center>

The term <math>n_i</math> is the index of possible magnetic orientations of particle <math>i</math>, parallel or anti-parallel to the magnetic field, <math>H</math> and are of values <math>ni = -1, 1</math>.

The explicit partition function is written below. Note that since the particles do not interact, and there can be no relaxation, or long-range interactions between the absorbed atoms, the energy can be factored out of the summation

<math>\wedge = \frac

Unknown macro: {m!}

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

e^{-\beta N \epsilon 0} \left [\sum_

Unknown macro: {n_1 = -1}

^1 \sum_

Unknown macro: {n_2=-1}

^1 \sum_

Unknown macro: {n_3 = -1}

^1 ... \sum_

Unknown macro: {n_N=-1}

^{1 e}

Unknown macro: {left (beta sum_i^N n_i mu_B right )}

H \right |]</math>

</center>

The partition function can be further simplified by using the single-particle description.

<math>\wedge = \frac

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

e^{-\beta N \epsilon0} q^N</math>

<math>q = \left [e^

Unknown macro: {beta mu_B H}

+ e^{- \beta \mu_B H} \right]</math>

</center>

b) Assume now that the system is open and is in equilibrium with a gas phase of zipium atoms constant chemical potential <math>\mu</math>. Derive an expression for the partition function of the adsorbed atoms at constant <math>T</math>, <math>V</math>, <math>H</math> and <math>\mu</math>. (Assume that the zipium atoms of the gas phase are not in a magnetic field).

The partition function at constant <math>T</math>, <math>V</math>, <math>H</math>, and <math>\mu</math>, is expressed below. Use the binomial expansion.

<math>\Gamma = \sum_

Unknown macro: {N=0}

^{m \wedge e}

Unknown macro: {beta mu N}

</math>

<math>\Gamma = \sum_

^m \frac

Unknown macro: {m!}

\left [qe^

Unknown macro: {beta (mu - epsilon 0 }

\right]^N</math>

<math>\Gamma = \left [1 - q e^{\beta (\mu - \epsilon 0 ) \right]^m</math>

</center>

c) At fixed <math>T</math>, <math>V</math> and <math>\mu</math>, how does the number of adsorbed atoms <math>N</math> depend on the magnetic field <math>H</math>

With <math>T</math>, <math>V</math>, <math>\mu</math>, and <math>H</math> as controlling variables, the characteristic potential is defined as the Legendre Transform of the internal energy with respect to <math>T</math>, <math>\mu</math>, and <math>H</math>. Below is the potential, differential, and expression of <math>N</math>.

<math>\Theta = U - TS - \mu N - HM</math>

<math>d \Theta = -S dT - N d \mu - M dH + pdV</math>

</center>

Calculate <math>N</math> from <math>d \Theta</math>.

<math>N = - \left ( \frac

Unknown macro: {partial Phi}

Unknown macro: {partial mu}

\right )_

Unknown macro: {T, H, V}

</math>

<math>\Theta = - kT \ln \Gamma</math>

<math>\Theta = -kTm \ln \left ( 1 + q e^

Unknown macro: {beta (mu - epsilon 0 )}

\right )</math>

<math>N = - \frac

Unknown macro: {partial Theta}

</math>

<math>N = m \frac{q e^{\beta (\mu - \epsilon 0)}}{1 + qe^{\beta (\mu - \epsilon 0)}}</math>

<math>q = \left [e^

Unknown macro: {beta mu_B H}

+ e^{- \beta \mu_B H} \right]</math>

</center>

Question 4

There are <math>N = \frac

Unknown macro: {M}

Unknown macro: {2}

</math> adsorbed atoms on a substrate with <math>M</math> sites. The adsorbed atoms do not interact with each other. A new technique has been devised that is able to segregate all the adsorbed atoms to the left half of the substrate (all surface sites on the left half are occupied and all the surface sites on the right are vacant). Is the segregated system in equilibrium? If not, describe the spontaneous change the system will undergo and calculate the increase of the entropy of the universe.

Consider the entropy of the segregated state. There is only one state.

</center>

An expression of the entropy of a uniform distribution over all <math>M</math> sites is below with <math>N = \frac

Unknown macro: {2}

</math>.

<math>S = k \ln \left ( \frac

Unknown macro: {M!}

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

\right )</math>

</center>

A change in entropy with <math>x = \frac

Unknown macro: {N}

Unknown macro: {M}

</math> is below. Consider when <math>x = \frac

Unknown macro: {2}

</math>.

<math>\Delta S = -kM ( x \ln x + (1-x) \ln (1-x))</math>

</center>

Question 5

A gas contains <math>N</math> argon atoms (mass <math>m_

Unknown macro: {Ar}

</math>) and <math>M</math>* krypton (mass* <math>m_

Unknown macro: {Kr}

</math>) atoms at constant <math>T</math> and <math>V</math>. Derive an expression for the partition function of the whole system in terms of single particle partition functions and write an expression for the free energy of the system.

Use Boltzmann's approximation and note that <math>Ar</math> cannot occupy <math>Kr</math> single particle states and vice versa.

<math>Q = \frac{q_

^N}

Unknown macro: {N!}

\frac{q_

Unknown macro: {Kr}

^M}

Unknown macro: {M!}

</math>

<math>q_

Unknown macro: {Ar}

= \left ( \frac{2 \pi m_

k T}

Unknown macro: {h^2}

\right )^{\frac

Unknown macro: {3}

{2}} V</math>

<math>q_

= \left ( \frac{2 \pi m_

Unknown macro: {Kr}

k T}

Unknown macro: {h^2}

\right )^{\frac

Unknown macro: {3}

{2}} V</math>

<math>F = -kT \ln \left (\frac{q_

Unknown macro: {Ar}

^N}

\right ) - kT \ln \left ( \frac{q_

Unknown macro: {Kr}

^M}

Unknown macro: {M!}

\right )</math>

</center>

The free energy of the two component ideal gas is simply the sum of their individual energies with volume <math>M</math>.

Child pages

3.20 Exam QuestionsExam22003

Question 1

Question 2

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