...

{html}
<P> </P>{html}

 \[ n=0,1,2 \] 

{html}
<P> </P>{html}

 \[ \nu = \frac{1}{2 \pi} \sqrt{\frac{f}{m}} \] 

There is a relation between a spring constant and mass. Light masses or masses with a stiff spring vibrate rapidly. Write the mini-partition function. and simplify with the relation pertaining to a geometric series.

...

{html}
<P> </P>{html}

 \[ q_{\mbox{vib}} = \sum_{n=0}^{\infty} e^{\left ( n + \frac{1}{2} \right ) h \nu} \] 

{html}
<P> </P>{html}

 \[ q_{\mbox{vib}} = e^{\beta \frac{h \nu}{2}} \sum_{n=0}^{\infty} e^{\beta h \nu n} \] 

{html}
<P> </P>{html}

 \[ \mbox{Geometric Series} \] 

{html}
<P> </P>{html}

 \[ x < 1 \] 

{html}
<P> </P>{html}

 \[ \sum_{n = 0}^{\infty} x^n = \frac{1}{1-x} \] 

{html}
<P> </P>{html}

 \[ q_{\mbox{vib}} = \frac{e^{-\beta \frac{h \nu}{2}}}{1-e^{-\beta h \nu}} \] 

Express in terms of the characteristic vibrational temperature,

...

{html}
<P> </P>{html}

 \[ q_{\mbox{vib}} = \frac{e^{-\frac{\theta_{\nu}}{2 T}}}{1-e^{-\frac{\theta_{\nu}}{2 T}}} \] 

The characteristic termperature,

...

{html}
<P> </P>{html}

 \[ P_n = \frac{e^{-\beta \epsilon_n}}{q_{\mbox{vib}}} \pi_{n>0} \] 

A general expression of being in any state other than the ground state is below, as well as a calculation at

...

{html}
<P> </P>{html}

 \[ \pi_{n>0} = e^{-10} \] 

{html}
<P> </P>{html}

 \[ \pi_{n>0} \approx 10^{-5} \] 

The probability of a molecule being in an excited state is very small. It is hard to excite a vibrational state above the ground state.

...

{html}<p>{html}

 \[ q_{\mbox{rot}} = \sum_{\epsilon_j} = (2j + 1) e^{-j (j+1) \frac{\theta}{T} } \] 

{html}<p>{html}

 \[ \theta_r = \frac{h^2}{8 \pi^2 I k_N} \] 

Case 1

The first case involves high temperatures, or temperatures much greater than\theta_r. If the temperature is large, the exponential term goes to zero. States are dense, and it is possible to approximate with an integral.

...

{html}<p>{html}

 \[ q_{\mbox{rot}} = \frac{T}{\theta_r} \] 

Case 2

Consider when the temperature is close to\theta_r. Caclulate a few of the terms, and see that the terms decrease rapidly. Assume that the larger terms are zero. Consider the first couple terms of the summation.

...

{html}<p>{html}

 \[ F = -k_B T \ln Q \] 

{html}<p>{html}

 \[ F = -k_B T \left ( \ln q_{\mbox{trans}} + \ln q_{\mbox{rot}} + \ln q_{\mbox{vib}} - \ln N\! \right ) \] 

{html}<p>{html}

 \[ F = F_{\mbox{trans}} + F_{\mbox{rot}} + F_{\mbox{vibr}} \] 

Pressure

Calculate the pressure. Only the translational component of the partition function is proportional to volume.

...

{html}<p>{html}

 \[ E_{\mbox{rot}} = N k_B T \] 

Vibrational energy

The vibrational energy is expressed below.

...

{html}<p>{html}

 \[ E_{\mbox{rot}} = N k_B T \left ( \frac{ \frac{h \nu}{k_B T}}{e^{\frac{h \nu}{k_B T}} - 1} \right ) + \frac{N h \nu}{2} \] 

Analyze the vibrational energy with two limits.

...

{html}<p>{html}

 \[ \lim_{x \to 0} \frac {x}{e^x - 1} = 1 \] 

{html}<p>{html}

 \[ E_{\mbox{vibr}} = N k_B T + \frac {N h \nu}{2} \] 

Case 2

An expression of the vibrational energy in the case of temperature much less than

...

{html}<p>{html}

 \[ C(v, \mbox{trans}) = \frac{3 N k_B}{2} \] 

Rotational component of heat capacity

...

{html}<p>{html}

 \[ C(v, \mbox{rot}) = N k_B \] 

Vibrational component of heat capacity

...

{html}
<P> </P>{html}

 \[ C(v, \mbox{tot}) = \frac{5}{2} N k_B \] 

Case 2

An expression of the heat capacity is below in the case of the temperature greater than or equal to

...

{html}
<P> </P>{html}

 \[ C(v, \mbox{tot}) = \frac{7}{2} N k_B \] 

Comments

Define different kinds of motion. Solve the Schrodinger equation. Evaluate as well as possible. Take limits in the rotational case. Boltzman is a system approximation. Don't need to be too occupied with the math.

...

{html}
<P> </P>{html}

 \[ N \mbox{atoms} \] 

{html}
<P> </P>{html}

 \[ x = \frac{N}{M} \] 

Assume now that the energy is only a function of concentration. Fix the concentration. With

...

{html}
<P> </P>{html}

 \[ \Omega (E) = \frac{M!}{N!(M-N)!} \] 

{html}
<P> </P>{html}

 \[ F = E - TS \] 

{html}
<P> </P>{html}

 \[ F = E - k_B T \ln \Omega \] 

{html}
<P> </P>{html}

 \[ S = k_B \ln \Omega \] 

The term

 \[ S \] 

is the configurational entropy. Construct a phase diagram. Consider an electronic or vibrational shift. Configurational entropy is most important. Below is an expression simplified with Stirling's approximation. The configurational entropy can be expressed as a function of concentration.

...

{html}
<P> </P>{html}

 \[ S_{\mbox{config}} = k_B ((M - N) \ln M + N \ln M - N \ln N - (M - N) \ln (M - N) \] 

{html}
<P> </P>{html}

 \[ S_{\mbox{config}} = -k_B \left ( \frac{N} {M} \ln \frac{N}{M} + \frac{(N-M)} {M} \ln \frac {M-N}{M} \right ) \] 

{html}
<P> </P>{html}

 \[ S_{\mbox{config}} = -k_B M ( x \ln x + (1 - x) \ln (1 - x)) \] 

The expression above is of the ideal solid solution configurational entropy.

...

{html}
<P> </P>{html}

 \[ F = E_{G, ST = 0} - T (S_{\mbox{conf}} + S_{\mbox{el}} + S_{\mbox{vibr}} ) \] 

The largest impact is from the configurational entropy.