Differentials, Controlling Variables, Legendre Transform to find Partition Function

It may be helpful to see how problems similar to question 2 of exam 1, 2002 http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam1/2002#Question_2 and question 1 of exam 2, 2002 http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam2/2001#Question_1 are generally solved.

Defining Thermodynamic Potentials

How is it known whether combinations of extensive and intensive quantities, such as force and length or applied field and magnetization, are added or subtracted in defining thermodynamic potentials? Examples are question 2 of exam 1, 2001 http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam1/2001#Question_2 and question 1 of exam 1, 2003 http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam1/2003#Question_1 Guidelines in how to derive potentials may be helpful

Heat Capacity

Thin Film on Thick Substrate

It may be helpful to see why the following logic is true in a solution of problem 2 of exam 1, 2005 http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam1/2005#Question_2.

<p>
</p>

We are interested in a heat capacity under a strange constraint. Start with <tex>dS</tex>.

<center>

<br>

<tex>dS = \left ( \frac

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

\right )_V dT + \left ( \frac

Unknown macro: {partial V}

\right )_T dV</tex>

<br>

</center>

The relation below follows.

<center>

<br>

<tex>T \left ( \frac

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

\right )_

Unknown macro: {epi}

= T \left ( \frac

Unknown macro: {partial T}

\right )_

Unknown macro: {V}

+ T \left ( \frac

Unknown macro: {partial S}

\right )_

Unknown macro: {T}

\left ( \frac

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

\right )_

Unknown macro: {epi}

</tex>

<br>

</center>

Bosons

Could it be explained what it means physically in a situation wherein bosons are not conserved and why <tex>\mu</tex> is zero in an equation below? I believe this is related to question 4 of exam 2, 2001 http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam2/2001#Question_4.

<center>

<br>

<tex>\overline n_k = \frac

Unknown macro: {1}

{e^

Unknown macro: {beta (epsilon_k - mu )}

- 1}</tex>

<br>

</center>

Entropy

AB alloy

Could a solution of <tex>S_

Unknown macro: {tot}

</tex> found in part c of question 5 in exam 2, 2001 be explained?http://www.onethread.org/wiki/index.php?title=3.20_Exam_Questions/Exam2/2001#Question_5 A solution is below.

<center>

<br>

<tex>S_

= -Nk \left [\ln 0.01 + 0.99 \ln 0.99 \right ]</tex>

<br>

</center>

Intercept rule

To get the partial molar volume of B or A start with the definition of
partial molar quantities

<center>

<br>

<tex>V = \overline

Unknown macro: {V}

_A N_A + \overline

_B N_B</tex>

<br>

</center>

where <tex>N_A</tex> and <tex>N_B</tex> is the number of moles of <tex>A</tex> and <tex>B</tex> respectively. This holds because <tex>V</tex> is an extensive quantity, so it is a homogeneous function of degree one. Now divide by the total number of moles:

<center>

<br>

<tex>\underline

Unknown macro: {V}

= \overline

_A x_A + \overline

Unknown macro: {V}

_B x_B (1)</tex>

<br>

</center>

<tex>\underline

Unknown macro: {X}

</tex> means "molar <tex>X</tex>" and <tex>\overline

_i</tex> is the partial molar <tex>X</tex> of component "<tex>i</tex>". using the constraint that <tex>x_A + x_B = 1</tex> and taking the derivative we have

<center>

<br>

<tex>d\underline

= (\overline

Unknown macro: {V}

_B - \overline

_A) dx_B</tex>

<br>

</center>

so

<center>

<br>

<tex>\overline

Unknown macro: {V}

_B = \overline

_A + d\underline

Unknown macro: {V}

/dx_B</tex>

<br>

</center>

substitute in (1) for <tex>\overline

_A</tex> and solve for <tex>\overline

Unknown macro: {V}

_B</tex> you get

<center>

<br>

<tex>\overline

_B = \underline

Unknown macro: {V}

+ (1-x_B) d\underline

/dx_B</tex>

<br>

</center>

which is the intercept rule.

Now, to solve the problem from 2002, E2, Q4, you just have to cast the
r.h.s. in terms of the lattice parameter

<center>

<br>

<tex>\underline

Unknown macro: {V}

= a^3/4</tex>

<br>

</center>

and

<center>

<br>

<tex>d\underline

/dx_B = d\underline

Unknown macro: {V}

/da * da/dx_B</tex>

<br>

</center>

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