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\exnumber{4215}
\heading{Chemical equilibrium with condensation}
\auname{Anne Hausmann}
{\bf The problem:}
\Dn
N Boson molecules of type B are inserted into a box with volume $V$,
the systems temperature is $T$. Each molecule is composed from two
atoms of type $A$ with mass $m.$ The inner energy of each atom is
$\varepsilon_{A}$ and the binding energy of the molecules is $\varepsilon_{B}$.
Assume there are molecules in the condensation state and that the
atoms can be treated in the Boltzmann's approximation.
\begin{enumerate}
\item What is the condition for the condensation assumption to be valid
for the molecules?
\item How many occupied excited states of molecules are in the box?
\item What is the condition for Boltzmann's approximation to be valid for
the atoms?
\item How many atoms are in the box?
\item What is the condition on the number of atoms and molecules in order
for the system to be in condensation?
\item What is the pressure on the box's walls?
\item Who pressures more (the molecules or the atoms)?
\end{enumerate}
\ \\
{\bf The Solution:} \\
{\Large Alert: mass of B molecules is $2\mass$. } \\
\begin{enumerate}
\item The ground state for each molecule is $\varepsilon_{B}$, and
\begin{equation}
N\left(\beta,\mu_{B}\right)=V\cdot c\int_{0}^{\infty}\varepsilon^{\alpha-1}d\varepsilon\left(\frac{1}{e^{\beta\left(\varepsilon-\mu_{B}\right)}-1}\right)
\end{equation}
When $\mu_{B}\to\varepsilon_{B}^{-},$ we get:
\begin{equation}
N\left(\beta,\mu_{B}\to\varepsilon_{B}^{-}\right)=\infty
\end{equation}
That's why the condition for the condensation assumption to be valid
is $\mu_{B}\to\varepsilon_{B}$.
\item The number of occupied excited states of molecules (taking into account
that the mass of each molecule is $2m$) is:
\begin{equation}
N=V\xi\left(\frac{3}{2}\right)\left(\frac{mT}{\pi}\right)^{\frac{3}{2}}
\end{equation}
\item The Hamiltonians describing the atoms and molecules energy are:
\begin{equation}
H^{A}=\varepsilon^{A}+\frac{p^{2}}{2m}
\end{equation}
\begin{equation}
H^{B}=\varepsilon^{B}+\frac{p^{2}}{2m}
\end{equation}
Therefore, the graphs of density of states is [see Y04 solution] \\
% \includegraphics[scale=1.2]{untitled(2)}\\
The Boltzmann approximation for the atoms is
\begin{equation}
F\left(\varepsilon_{A}-\mu_{A}\right)=\frac{1}{e^{\beta\left(\varepsilon_{A}-\mu_{A}\right)}-1}\approx e^{-\beta\left(\varepsilon_{A}-\mu_{A}\right)}
\end{equation}
It holds whenever the occupation is
\begin{equation}
\beta\left(\varepsilon_{A}-\mu_{A}\right)\gg1
\end{equation}
or
\begin{equation}
\varepsilon_{A}-\mu_{A}\gg T
\end{equation}
Since the molecules and the atoms are in chemical equilibrium
\begin{equation}
\mu_{B}=2\mu_{A}
\end{equation}
While taking into account the condensation condition, one gets
\begin{equation}
\mu_{A}=\frac{\varepsilon_{B}}{2}
\end{equation}
Substituting (10) in (8) one gets:
\begin{equation}
\varepsilon_{A}-\frac{\varepsilon_{B}}{2}\gg T
\end{equation}
So the requirement for Boltzmann's approximation to be valid is:
\begin{equation}
\varepsilon\gg T
\end{equation}
Where we have defined
\begin{equation}
\varepsilon=2\varepsilon_{A}-\varepsilon_{B}
\end{equation}
\item We can treat the atoms in the Boltzmann's approximation, therefore
the number of atoms in the box is:
\begin{equation}
N_{A}=V\cdot\left(\frac{mT}{2\pi}\right)^{\frac{3}{2}}\cdot\frac{1}{2}e^{\frac{\mu_{A}-\varepsilon_{A}}{T}}
\end{equation}
(where the prefactor $\frac{1}{2}$ arrose because we are counting
molecules and not atoms)\\
Substituting (10) in (14) one gets:
\begin{equation}
N_{A}=V\cdot\left(\frac{mT}{2\pi}\right)^{\frac{3}{2}}\cdot\frac{1}{2}e^{-\frac{\varepsilon}{2T}}
\end{equation}
\item The number of molecules is given by:
\begin{equation}
N=n_{0}+V\xi\left(\frac{3}{2}\right)\left(\frac{mT}{\pi}\right)^{\frac{3}{2}}
\end{equation}
Where $n_{0}$ is the number of condensed molecules.\\
The condition for condensation is:
\begin{equation}
n_{0}>0
\end{equation}
The system will be in condensation as long as the number of molecules
is bigger than:
\begin{equation}
N>V\cdot\left(\frac{mT}{2\pi}\right)^{\frac{3}{2}}\cdot\left(2^{\frac{3}{2}}\xi\left(\frac{3}{2}\right)+\frac{1}{2}e^{-\frac{\varepsilon}{2T}}\right)
\end{equation}
\item The pressure from the molecules:
\begin{equation}
p=\xi\left(\frac{5}{2}\right)\left(\frac{m}{\pi}\right)^{\frac{3}{2}}T^{\frac{5}{2}}
\end{equation}
and the pressure from the atoms :
\begin{equation}
p=\left(\frac{m}{2\pi}\right)^{\frac{3}{2}}T^{\frac{5}{2}}e^{-\frac{\varepsilon}{2T}}
\end{equation}
Altogether the pressure on the box's walls:
\begin{equation}
p=\left(\frac{m}{2\pi}\right)^{\frac{3}{2}}T^{\frac{5}{2}}\left[2^{\frac{3}{2}}\xi\left(\frac{5}{2}\right)+e^{-\frac{\varepsilon}{2T}}\right]
\end{equation}
\item Most of the pressure arises from the condensed molecules.\end{enumerate}
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\end{document}