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\begin{document} 
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\heading{E2042: Pressure of an ideal gas in a box with gravitation} 
\auname{Yair Vardi and Geva Arwas}


{\bf The problem:}

\Dn

An ideal classical gas of ${N}$ particles of mass
$m$ is in a container of height ${L}$ which is in a
gravitational field of a constant acceleration ${g}$. The gas is
in uniform temperature ${T}$. 

\Dn

\begin{itemize}
\item [(a)]
Find the dependence ${P\left( h \right)}$ of the pressure on the height ${h}$.

\item [(b)]
Find the partition function and the internal energy. Examine the limits
${mgL \ll k_{B}T}$ and ${mgL \gg k_{B}T}$.

\item [(c)]
Consider an adiabatic atmosphere, i.e.
the atmosphere has been formed by a constant entropy process in
which ${T,\,\mu}$, are not equilibrated, but ${Pn^{-\gamma}}$   =
const., equilibrium is maintained within a layer at height $h$. Find $T(h)$ and $ n(h)$
in terms of the density $n_0$ and temperature $T_0$ at $h=0$.
\end{itemize}

\Dn
{\bf Baruch's  A19.}

\Dn\Dn


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{\bf The solution:}

\Dn

\begin{itemize}
\item [(a)]
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For a layer of height $dh$ and area $A$ in mechanical equilibrium:
\begin{equation} \label{mechanical equilibrium}
A \left[ P(h) - P(h+dh) \right] = n(h) mg \cdot A dh 
\end{equation}

leading to the differential equation:
\begin{equation} \label{mechanical equilibrium differential}
dP = -n mg \cdot dh
\end{equation}

Since that for an ideal gas $P = nT$, we can write:
\begin{equation} 
\frac{dP}{P} = -\beta mg \cdot dh \qquad \Rightarrow \qquad P(h) = P_0 \, \eexp{-\beta mgh}
\end{equation}

In order to find $P_0$, we integrate over $n(h)$ to get the total number of particles:
\begin{equation} 
A\int_0^L n(h) dh = A \beta P_0 \int_0^L \eexp{-\beta mgh} dh = \frac{A P_0}{mg} \left( 1-\eexp{-\beta mgL} \right) = N
\end{equation}

from which we get $P_0 = \frac{Nmg}{A} \left( 1-\eexp{-\beta mgL} \right)^{-1}$.


\item [(b)]
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The one particle partition function is:
\begin{equation} 
Z_1 = \frac{1}{\left( 2 \pi \right)^3}
\int d^3p \, d^3x \, \eexp{-\beta \left( \frac{p^2}{2m} + mgh \right) } 
= \frac{A}{\lambda_T^3} \int_0^L \eexp{-\beta mgh } dh
= \frac{A}{\lambda_T^3} \frac{1-\eexp{-\beta mgL }}{\beta mg}
\end{equation}

The $N$-particles partition function is simply $\frac{1}{N!} Z_1^N$.%$\left( Z_1 \right)^N /N!$. 
The internal energy is given by:
\begin{equation} 
E = -\pd{\ln Z}{\beta} = \frac{N}{\beta} \left( \frac{5}{2} - \frac{\beta mgL}{\eexp{\beta mgL }-1} \right)
\end{equation}

In the high tempratures limit $\beta mgL \ll 1$ the internal energy behaves as a $3D$ ideal gas:
\begin{equation} 
E \approx \frac{N}{\beta} \left( \frac{5}{2} - \frac{\beta mgL}{1+\beta mgL-1} \right) = \frac{3}{2}NT
\end{equation}

as expected. Thus it is not surprising that we get uniform density:
\begin{equation} 
n(h) = \beta P(h) = \frac{N}{V} \frac{\beta mgL \eexp{-\beta mgh }}{1-\eexp{-\beta mgL }}
\approx \frac{N}{V}
\end{equation}
In the low temperature limit $\beta mgL \gg 1$:
\begin{equation} 
E \approx \frac{N}{\beta} \left( \frac{5}{2} - \beta mgL \, \eexp{-\beta mgL } \right) \approx \frac{5}{2}NT
\end{equation}

%This limit have no meaning since it is {\bf not} an ideal gas anymore.
and unless $h=0$, the density goes to zero:
\begin{equation} 
n(h) \approx \frac{N}{V} \beta mgL \eexp{-\beta mgh } \, \rightarrow \, \frac{N}{A} \delta(h)
\end{equation}
meaning that all particles are at the bottom of the container.

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\item [(c)]


We now have an adiabatic process where $P=Cn^\gamma$ with $\gamma > 1$. Within each layer equilibrium is maintained, so the usual state-equation of the ideal gas holds. For $h=0$ we have:
\begin{equation} \label{C}
P_0 = Cn_0^\gamma = n_0 T_0 \qquad \Rightarrow \qquad C=T_0 n_0^{1-\gamma}
\end{equation}
Using $dP=\gamma C n^{\gamma-1}dn$ , equation \eqref{mechanical equilibrium differential} now has the form:
\begin{equation} 
\gamma Cn^{\gamma-1}dn = -nmg\cdot dh \qquad \Rightarrow \qquad
\int_{n_0}^{n}{n^{\gamma-2}dn} = -\frac{mgh}{\gamma C}
\end{equation}
We obtain:
\begin{equation} 
n^{\gamma-1} = n^{\gamma-1}_0-(\gamma-1)\frac{mgh}{\gamma C}=n_0^{\gamma-1}\left(1-\frac{\gamma-1}{\gamma} \cdot \frac{mgh}{n_0^{\gamma-1}C} \right)=n_0^{\gamma-1}\left(1-\frac{\gamma-1}{\gamma} \cdot \frac{mgh}{T_0} \right)
\end{equation}
where we have used \eqref{C} in the last equality. We can now write:
\begin{equation} 
n(h) =n_0\left(1-\frac{\gamma-1}{\gamma} \cdot \frac{mgh}{T_0} \right)^{\frac{1}{\gamma-1}}
\end{equation}
and the temperature is:
\begin{equation} 
T(h) =Pn^{-1}=Cn^{\gamma-1}=T_0-\left(\frac{\gamma-1}{\gamma}\right)mgh
\end{equation}
We can see that both the temperature and the density decrease as we elevate and they vanish when we reach $h=\frac{\gamma T_0}{(\gamma-1)mg}$. 
\end{itemize}

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\end{document} 
