Subfield: Thermodynamics

A hollow insulated cylinder of height $2 \mathrm{H}$ and volume $2 \mathrm{~V}$ is closed from below by an insulating piston. The cylinder is divided into two initially identical chambers by an insulating diaphragm of mass $m$. The diaphragm rests on a circular ledge and a gasket between them provides tight contact. Both chambers are filled with gaseous helium at pressure $p$ and temperature $T$. A force is applied to the piston, so that it moves upwards slowly.

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Context question:
a. Find the volume of the lower chamber $V_{0}$ when the gas starts to leak between the chambers.
Context answer:
\boxed{$V(1+x)^{-3 / 5}$}


Question: b. Find the temperature $T_{1}$ in the upper chamber when the piston touches the diaphragm.

Solution:
Let $p_{1}, V_{1}, T_{1}$ denote the (time dependent) pressure, volume, and temperature in the upper chamber, and $p_{2}, V_{2}$, $T_{2}$ - those in the lower one. Note that $V_{1} \equiv V$ does not change.

Consider a parcel of volume $v$ below the diaphragm containing $n$ moles of helium. It is convenient to imagine it bounded by two fictitious free thin massless pistons. During slow perturbations the parcel undergoes an adiabatic process. The pressure and the temperature for the parcel are actually the pressure and the temperature for entire lower chamber $p_{2}$ and $T_{2}$. The energy conservation for the parcel is

$$
0=p_{2} \mathrm{~d} v+\mathrm{d}\left(\frac{3}{2} n_{v} R T_{2}\right)=\frac{5}{2} p_{2} \mathrm{~d} v+\frac{3}{2} v \mathrm{~d} p_{2}
$$

This gives

$$
v^{5} p_{2}^{3} =\text { const }
\tag{1}
$$

$$
T_{2}^{5} p_{2}^{-2} =\text { const }
\tag{2}
$$

The leak begins when the pressure below the diaphragm exceeds that in the upper chamber by $\Delta p \equiv$ $p_{0}-p=m g H / V=p x$, where

$$
x=\frac{m g H}{p V} \tag{(a., b., c.)}
$$



a. We may let $v=V_{2}$ before that.

$$
\begin{gathered}
V^{5} p^{3}=V_{0}^{5}(p+\Delta p)^{3}=V_{0}^{5} p^{3}(1+x)^{3} \\
V_{0}=V(1+x)^{-3 / 5}
\end{gathered}
$$

b. The energy conservation for the whole system:

$$
\begin{gathered}
0=p_{1} \mathrm{~d} V_{1}+\mathrm{d}\left(\frac{3}{2} n_{1} R T_{1}\right)+p_{2} \mathrm{~d} V_{2}+\mathrm{d}\left(\frac{3}{2} n_{2} R T_{2}\right)= \\
\frac{5}{2}\left(p_{1} \mathrm{~d} V_{1}+p_{2} \mathrm{~d} V_{2}\right)+\frac{3}{2}\left(V_{1} \mathrm{~d} p_{1}+V_{2} \mathrm{~d} p_{2}\right)= \\
\frac{5}{2} p_{2} \mathrm{~d}\left(V+V_{2}\right)+\frac{3}{2}\left(V+V_{2}\right) \mathrm{d} p_{2}
\end{gathered}
$$

since the pressure above the diaphragm remains lower than that below by the same margin $\Delta p$ during the later process and $\mathrm{d} V_{2}=\mathrm{d}\left(V+V_{2}\right)$. Similarly to (1), we get

$$
\left(V+V_{2}\right)^{5} p_{2}^{3}=\text { const }
$$

The pressure $p_{2}^{\prime}$ in the lower chamber when the piston touches the diaphragm is found from the equation

$$
\begin{gathered}
V^{5} p_{2}^{\prime 3}=\left(V+V_{0}\right)^{5} p_{0}^{3}=V^{5}\left(1+\frac{1}{(1+x)^{3 / 5}}\right)^{5} p^{3}(1+x)^{3} \\
p_{2}^{\prime}=p\left(1+(1+x)^{3 / 5}\right)^{5 / 3}
\end{gathered}
\tag{3}
$$

The pressure in the upper chamber at this moment is

$$
p_{1}^{\prime}=p_{2}^{\prime}-\Delta p=p\left(\left(1+(1+x)^{3 / 5}\right)^{5 / 3}-x\right)
$$

The temperature in the upper chamber is found from the equations of state $p V=n R T$ and $p^{\prime} V=(2 n) R T^{\prime}$

$$
T_{1}^{\prime}=\frac{T}{2}\left(\left(1+(1+x)^{3 / 5}\right)^{5 / 3}-x\right)
$$

Final answer: $\frac{T}{2}\left(\left(1+(1+x)^{3 / 5}\right)^{5 / 3}-x\right)$