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<h2>Solutions of Savchenko Physics Textbook</h2>
<p class="author">
Aliaksandr Melnichenka <br/>
October 2023
</p>
</header>
− <main>
− <article>
− <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
−
− <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.
− </p>
− <center>
− <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
+ <h3 id="back-link"><a href="../">$\leftarrow$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>
− $P_1~V_1 = nRT \;(1)$
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="5.5.16" width="250" />
+ <figcaption>
+ For problem 5.5.16
+ </figcaption>
+ </figure>
</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>
− <p class="TxSolutions">
+
+ <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>
− <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 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>
− <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>
− </p>
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