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\begin{document}
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\exnumber{6020}
\heading{A divided box with a hole in one side}
\auname{Yotam Sherf}
{\bf The problem:}
\Dn
A cylinder of length ${L}$ and cross section ${A}$ is divided into
two compartments by a piston. The piston has mass ${M}$ and it is
free to move without friction. Its distance from the left basis of
the cylinder is denoted by ${x}$. In the left side of the piston
there is an ideal Bose gas of ${N_{a}}$ particles with mass
${\mathsf{m}_{a}}$. In the right side of the piston there is an
ideal Bose gas of ${N_{b}}$ particles with mass ${\mathsf{m}_{b}}$.
The temperature of the system is ${T}$.
Assume that the left gas can be treated within the framework of the Boltzmann approximation.
Assume that the right gas is in condensation. In items (3-5) consider separately two cases: \\
(a) A small hole is drilled in the left wall of the box.
(b) A small hole is drilled in the right wall of the box. \\
The area of the hole is ${\delta A}$.
\Dn
\Dn
\begin {itemize}
\item[(1)]
Find the equilibrium position of the piston.
\item[(2)]
What is the frequency of small oscillations of the piston.
\item[(3)]
What is the velocity distribution ${F(v)}$ of the emitted particles?
\item[(4)]
What is the flux (particles per unit time) of the emitted particles?
\item[(5)]
Is the piston going to move? If yes write an expression for its velocity.
\end {itemize}
In item (3) use normalization that makes sense for the calculation in item (4).
In item (5) assume that the process is quasi-static,
such that at any moment the system is at equilibrium.
Express your answers using ${L, A, \delta A, N_{a}, N_{b}, \mathsf{m}_{a},\mathsf{m}_{b}, T, M}$.
\[\int_{0}^{\infty}\frac{xdx}{e^{x}-1}=\frac{\pi^2}{6}\] \Dn
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{\bf The solution:}
\Dn
(1)
\Dn
Equilibrium condition
\begin{equation}P_a=P_b\end{equation}
Where for Boltzmann $P_a=\dfrac{N_aT}{Ax}$ and for the condense gas $Li_{5/2}(1)=\zeta(5/2) $ and\\ $ P_b=\zeta\left(\frac{5}2\right) \left(\frac{m_b}{2\pi}\right)^\frac{3}2T^\frac{5}2$.
\Dn
Define $x_0$ as the equilibrium position we find from (1).
\begin{equation}
x_0=\dfrac{N_a}{\zeta(5/2)A}\bigg(\dfrac{2\pi}{m_bT}\bigg)^{3/2}
\end{equation}
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\Dn
\Dn
(2)
\\\\
Looking at the total force on the piston and expanding it around $x_0$ for small oscillations. \\
\begin{equation}
F(x_0+dx)=F(x_0)+\underbrace{\dfrac{dF(x)}{dx}}_{-Mw^2}\bigg|_{x_0}dx=A\bigg(P(x_0)+\dfrac{dP(x)}{dx}\bigg|_{x_0}dx\bigg)
\end{equation}
We find
\begin{equation}w^2=-\dfrac{A}{M}P'(x_0)=\dfrac{N_aT}{Mx_0^2}~~~\Rightarrow~~~ w=\zeta(5/2)\dfrac{A}{\sqrt{N_aM}}\bigg(\dfrac{m_b}{2\pi}\bigg)^{3/2}T^2\end{equation}
\\
(3)
\\\\
(a)
In the Boltzmann case we have the known velocity distribution
\begin{equation}
n(v)=\bigg(\dfrac{m_a}{2\pi T}\bigg)^{3/2}4\pi v^2e^{-\frac{\beta mv_a^2}{2}}
\end{equation}
And for $N_a$ particles we have
\begin{equation}
F_a(v)=N_an(v)=N_a\bigg(\dfrac{m_a}{2\pi T}\bigg)^{3/2}4\pi v^2e^{-\frac{\beta m_av^2}{2}}
\end{equation}\\
(b)condense Bose gas.
\begin{equation}
F_b(v)dv=\dfrac{1}{(2\pi)^3}\bigg(\int d^3x\bigg)m_B^3d^3v\frac{1}{e^{-\frac{\beta m_Bv^2}{2}}-1}=V\bigg(\dfrac{m_b}{2\pi}\bigg)^34\pi v^2\frac{1}{e^{-\frac{\beta m_bv^2}{2}}-1}dv
\end{equation}
Where $V=A(L-x)$.\\
(4)\\
By definition $J(v)=\rho(v) v$, and $\rho(v)=\dfrac{F(v)}{V}$, so the total flux.\\
(a) Boltzmann gas.
\begin{equation}\begin{split}
&J_a=4\pi\dfrac{N_a}{4V}\bigg(\dfrac{m_a}{2\pi T}\bigg)^{3/2}\int_{0}^{\infty}v^3e^{-\frac{\beta m_Bv^2}{2}}dv=\dfrac{2N_a\pi}{Ax_0}\bigg(\dfrac{m_a}{\pi T}\bigg)^{3/2}\bigg(\dfrac{2T}{m_a}\bigg)^2\int_{0}^{\infty}xe^{-x}dx=\\&\zeta(5/2)\bigg(\dfrac{T^}{2\pi}\bigg)^2\sqrt{\dfrac{m_b^3}{m_a}}\underbrace{\int_{0}^{\infty}xe^{-x}dx}_{1}=\zeta(5/2)\bigg(\dfrac{T^}{2\pi}\bigg)^2\sqrt{\dfrac{m_b^3}{m_a}}
\end{split} \end{equation}
And the flux per unit time $ I_a=\delta AJ_a$\\
(b) Bose gas.
\begin{equation}
I_b=(\delta A)2\pi\bigg(\dfrac{m_b}{2\pi}\bigg)^3\bigg(\dfrac{T}{m_b}\bigg)^2\underbrace{\int_{0}^{\infty}\dfrac{x}{e^{-x}-1}dx}_{\pi^2/6}=(\delta A)\dfrac{m_b}{6}\bigg(\dfrac{T}{2}\bigg)^2
\end{equation}
with no dimensional dependence.
\\
\\
\\
(5)
\\
(a) In the Boltzmann case using the quasi-stationary assumption we can use equilibrium \\condition (1) in each moment, so $x_0 \rightarrow x_0(t)$.
\begin{equation}
x_0(t)=\dfrac{2\pi}{A\zeta(5/2)}\bigg(\dfrac{2\pi}{m_bT}\bigg)^{3/2}N_a(t)
\end{equation}
where ~$\dfrac{dN_a(t)}{dt}=I_a$, the piston velocity would be ($\dot{x}_0(t)=v_p$).
\begin{equation}
v_p=\dfrac{\delta A}{A}\sqrt{\dfrac{2\pi T}{m_a}}
\end{equation}
\\
(b) In the condense case we have $P_b=\zeta\left(\frac{5}2\right) \left(\frac{m_b}{2\pi}\right)^\frac{3}2T^\frac{5}2$ with no dependence on $N_b$, so the pressure will remain the same and the equilibrium condition (1) still holds .
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