Exam 2 - 02/24/03

Exam 2 CHEM 430b/530b
Statistical Methods and Thermodynamics
02/24/03 Exercise 1

(20 points) Item (1.1):Consider an ideal gas of bosons with $\mu=0$ at temperature $T=1/(\beta k)$. Show that

$$\overline{\delta_{n_k} \delta_{n_j}} =\delta_{kj} \frac{\partial \overline{n_k}}{\partial(-\beta \epsilon_k)} \Bigg)_{V,T},$$ (262)

where $\delta_{n_k}= n_k-\overline{n_k}$ and $\overline{n_k}$ is the average occupation of the one-boson energy level $k$.
(20 points) Item (1.2): Explain the minimum energy principle and show that such principle is a consequence of the maximum entropy principle.

(20 points) Item (1.3): Explain the classical limit of the quantum statistical distributions.
 

Exercise 2

Consider an ideal gas of $O_2$ molecules adsorbed on a surface of area $S$ in thermal equilibrium at temperature $T=1/(k \beta)$. Assume that each $O_2$ molecule in the gas can freely translate, vibrate and rotate but only on the 2-dimensional surface. Assume that the rotational motion of $O_2$ molecules can be described by a rigid rotor model where the rotational eigenstates have degeneracy $g(J)=2$ for all values of J except for J=0 for which g(J)=1. Assume that the rotational states have eigenvalues $E_J=\frac{\hbar^2J^2}{2I_0}$, with J=0, 1, 2, ..., where $I_0$ is the moment of inertia of the $O_2$ molecule.

(10 points) Item (2.1): Compute the rotational canonical partition function of an $O_2$ molecule as a function of its moment of inertia $I_0$ and $\beta$.
(10 points) Item (2.2): Compute the vibrational canonical partition function of an $O_2$ molecule as a function of its vibrational frequency $\omega_0$ and $\beta$.
(10 points) Item (2.3): Compute the translational canonical partition function of an $O_2$ molecule as a function of its total mass $m$, $\beta$ and the surface area $S$.
(10 points) Item (2.4): Compute the average internal energy $E$ of the $O_2$ gas as a function of $\beta$, the $O_2$ mass $m$ , the area of the surface $S$, the $O_2$ moment of inertia $I_0$ and the total number $N$ of $O_2$ molecules on the surface.
 
 
 
 

Solution:

Exercise 1: Item (1.1): Since $\delta_{n_k}= n_k-\overline{n_k}$,

$$\overline{\delta_{n_k} \delta_{n_j}} =\overline{n_k n_j} - \overline{n_k} \hspace{.1cm} \overline{n_j},$$ (263)

where

$$\overline{n_j}=\frac{1}{e^{\beta \epsilon_j}-1},$$ (264)

because $\mu=0$. Therefore,

$$\overline{n_k n_j}=\frac{1}{\Xi} \frac{ \partial^2 \sum_{n_1=0}^{......silon_2 n_2+... )} }{ \partial (\beta \epsilon_j)\partial (\beta \epsilon_k) },$$ (265)

or

$$\overline{n_k n_j}= \frac{1}{\Xi} \frac{\partial^2}{\partial (\be......ilon_k) \partial (\beta \epsilon_j)} \prod_j \frac{1}{1-e^{-\beta \epsilon_j}}.$$ (266)

Computing the first partial derivative we obtain

$$\overline{n_k n_j}= \frac{1}{\Xi} \frac{\partial}{\partial(-\beta......1-e^{-\beta \epsilon_k})^2} \prod_{l \neq k} \frac{1}{1-e^{-\beta \epsilon_l}},$$ (267)

and computing the second partial derivative we obtain

$$\begin{array}{ll}\overline{n_k n_j} & = \frac{1}{\Xi} [ \pro......q j, l\neq k } \frac{1}{1-e^{-\beta \epsilon_l}} ],\end{array}$$ (268)

where,

$$\Xi=\prod_j \frac{1}{1-e^{-\beta \epsilon_j}}.$$ (269)

Therefore,

$$\overline{n_k n_j} = \delta_{kj} \frac{e^{-\beta \epsilon_k}}{(1-......^{-\beta \epsilon_k})} \frac{e^{-\beta \epsilon_j}}{(1-e^{-\beta \epsilon_j})},$$ (270)

and

$$\overline{\delta_{n_k} \delta_{n_j}} = \delta_{kj} \frac{e^{-\bet......\frac{\partial}{ \partial(-\beta \epsilon_k)} \frac{1}{e^{\beta \epsilon_k}-1},$$ (271)

which, according to Eq (264), gives

$$\boxed{\overline{\delta_{n_k} \delta_{n_j}} =\delta_{kj} \frac{\partial \overline{n_k}}{\partial(-\beta \epsilon_k)}.}$$ (272)

Item (1.2): See topic ``Minimum Energy Principle'' on page 55 of the lecture notes. Item (1.3): See topic ``Classical limit of Quantum Statistical Distributions'' on page 36 of the lecture notes.
 
 
 
 

Exercise 2:

Item (2.1): The rotational canonical partition function of an $O_2$ molecule is

$$q_{\text{rot}}= \sum_{J=0}^{\infty} g(J) e^{- \beta \epsilon_J}.$$ (273)

Taking the continuous limit we obtain,

$$q_{\text{rot}} \approx \lim_{\epsilon \rightarrow 0} \int_{\epsilon}^{\infty}dJ g(J) e^{-\beta \epsilon_J} = \sqrt{\frac{\pi2I_0}{\beta \hbar^2}}.$$ (274)

Item (2.2): The vibrational canonical partition function of an $O_2$ molecule is

$$q_{\text{vib}}= \sum_{\nu=0}^{\infty} e^{-\beta \epsilon_{\nu}} = \frac{e^{-\beta \hbar \omega_0 /2}}{1- e^{-\beta \hbar \omega_0}}.$$ (275)

Item (2.3): The translational canonical partition function of an $O_2$ molecule is

$$q_{\text{transl}}=\frac{S}{\pi^2} \sum_{k_x} \sum_{k_y} e^{-\beta......frac{-\beta k_x^2 \hbar^2}{2m}} \int d k_y e^{\frac{-\beta k_y^2 \hbar^2}{2m}}.$$ (276)

Therefore,

$$q_{\text{transl}} \approx \frac{S}{\pi} \frac{\pi 2 m}{\beta \hbar^2},$$ (277)

where $S=L_x \times L_y$, with $L_x$ and $L_y$ the lengths of the surface along the x and y directions, respectively. Item (2.4): The total canonical partition function of the system is

$$Q= \prod_{j=1}^{N} q_{\text{rot}} q_{\text{vib}} q_{\text{transl}}.$$ (278)

Substituting the expressions for $q_{\text{rot}}$, $q_{\text{vib}}$ and $q_{\text{transl}}$ computed in items (2.1)--(2.3) we obtain,

$$Q=\prod_{j=1}^{N} S \frac{2m}{\beta \hbar^2} (e^{\beta \hbar \ome......hbar^2}} = \prod_{j=1}^N \frac{\pi x}{\beta^{3/2} (e^{\beta y}-e^{-\beta y}) }.$$ (279)

Therefore, the average internal energy of the $O_2$ gas is

$$\overline{E}=\frac{\sum_{j=1}^{N}\partial \text{ln} Q }{\partial ......a^{3/2} y(e^{\beta y}+e^{-\beta y}))}{\beta^{3/2}(e^{\beta y}-e^{-\beta y})^2},$$ (280)

which gives

$$\overline{E} =\sum_{j=1}^{N} \frac{3/2(e^{\beta y}-e^{-\beta y}) ...... \frac{3y}{2} \frac{(e^{\beta y}+e^{-\beta y})}{(e^{\beta y}-e^{-\beta y})})*N,$$ (281)

where $x=\frac{S2m}{\hbar^2} \sqrt{\frac{2\pi I_0}{\hbar^2}}$ and $Y=\frac{\hbar \omega_0}{2}$.