<h3 id="back-link"><a href="/#5.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>

    <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. 
    </p>
    <center>
      <figure>
        <img src="statement.png"
          loading="lazy" alt="5.5.16" width="250" />
        <figcaption>
          For problem 5.5.16
        </figcaption>
      </figure>
    </center>

    <h3>Solution</h3>
    <p>
      Let denote left volume as $V_1$ and right one as $V_2$. Applying state equation for each container
      $$P_1\~V_1 = nRT \\;(1)$$
      $$P_0\~V_2 = nRT_0 \\;(2)$$ 
      It is easy to see that
      $$V_1 = V_0+S\frac{l}{2}-Sx \\;(3)$$
      $$V_2 = V_0+S\frac{l}{2}+Sx \\;(4)$$
      If droplet is in equilibrium that means pressure forces due to gasses are equal, so $P_1 = P_0$. Then, dividing $(1)$ and $(2)$
      $$\frac{V_1}{V_2}=\frac{T}{T_0} \\;(5)$$
      Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$
      $$\boxed{T = T_0\~\frac{2V_0+S(l+2x)}{2V_0+S(l-2x)}}$$
      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 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>
    </p>



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