<h4>Statement</h4>
<p>
$5.5.16$
The gas thermometer consists in two identical containers with volume $V_0$ each one, linked by a tube of length $l$ and cross section $S$. A mercury droplet obstructs the tube. If temperatures of gasses in the volumes are equal, the mercury lies in the center of the tube. The right volume is connected to a thermostat with temperature $T_0$. Scale the thermometer finding the dependence between gas temperature in left volume and the displacement $x$ of mercury respect to equilibrium position.
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<figure>
<img src="statement.png"
loading="lazy" alt="5.5.16" width="250" />
<figcaption>
For problem 5.5.16
</figcaption>
</figure>
</center>
<h4>Solution</h4>
<p class="TxtSolutions">
Let denote left volume as $V_1$ and right one as $V_2$. Applying state equation for each container
</p>
<center>
$P_1\~V_1 = nRT \\;(1)$
</center>
<center>
$P_0\~V_2 = nRT_0 \\;(2)$
</center>
<p class="TxtSolutions">
It is easy to see that
</p>
<center>
$V_1 = V_0+S\frac{l}{2}-Sx \\;(3)$
</center>
<center>
$V_2 = V_0+S\frac{l}{2}+Sx \\;(4)$
</center>
<p class="TxtSolutions">
If droplet is in equilibrium that means pressure forces due to gasses are equal, so $P_1 = P_0$. Then, dividing $(1)$ and $(2)$
</p>
<center>
\frac{V_1}{V_2}=\frac{T}{T_0} \\;(5)$
</center>
Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$
<center>
$\boxed{T = T_0\~\frac{2V_0+S(l+2x)}{2V_0+S(l-2x)}}$
</center>
<p class="TxtSolutions">
This is mathematically valid for $0< x <\frac{V_0}{S}+\frac{l}{2}$ but according the physical situation only for $0< x < \frac{l}{2}$.
</p>
<p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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