Problem 25

plumleyj 08 Jan 2007 16:31

There are a total of four degrees of freedom in a two dimensional diatomic system

$q_{vib}$ is the same as for a three-dimensional diatomic gas;

$q_{rot}$ with a degeneracy of 2 for all J except J=0, for which the degeneracy is 1;

(1)\begin{align} q_{rot} = \sum_j g_j e^{-\beta \varepsilon_j} = e^{-\beta \varepsilon_0} + 2\sum_j e^{-\beta \varepsilon_j} =1+2\sum_j e^{\frac{-\beta\hbar^2 J^2}{ 2I}} = 1 + \displaystyle \int_1^\infty e^{\frac{-\beta\hbar^2 J^2}{ 2I}} dJ \approx \left( \frac{\pi T} {{\Theta}_{rot}}\right)^{\frac {1} {2}} \end{align}

Since $q_{vib}$ is the same as a three dimensional diatomic gas and the $q_{trans}$ is $\frac {2a^2\pi m k T} {\hbar^2}$ q(T) becomes

(2)\begin{align} q(T) = \left(\frac {2a^2\pi m k T} {\hbar^2}\right)\left( \frac{\pi T} {\Theta}_{rot}\right)^{\frac {1} {2}} \left(\frac {e^{\frac {-\Theta_{vib}} {2T}}} {1-e^{\frac {-\Theta_{vib}} {T}}}\right) \end{align}

The average energy becomes

(3)\begin{align} \langle E \rangle = NkT^2\left(\frac{\partial \ln q(T)} {\partial T}\right)_V= \frac {3RT} {2} + \frac {R\Theta_{vib}} {2} + \frac {R \Theta_{vib}} {e^{\frac {-\Theta_{vib}} {T} -1}} \end{align}