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| <header> | | <header> |
| <h2>Solutions of Savchenko Physics Textbook</h2> | | <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
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| <main> | | <main> |
| <article> | | <article> |
| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> |
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| <h4>Statement</h4> | | <h4>Statement</h4> |
| <p> | | <p> |
| $5.5.16$ | | $5.5.16$ |
| 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. | | 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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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" alt="5.5.16" width="250" /> | | loading="lazy" alt="5.5.16" width="250" /> |
| <figcaption> | | <figcaption> |
| For problem 5.5.16 | | For problem 5.5.16 |
| </figcaption> | | </figcaption> |
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| </center> | | </center> |
| <h4>Solution</h4> | | <h4>Solution</h4> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Let denote left volume as $V_1$ and right one as $V_2$. Applying state equation for each container | | Let denote left volume as $V_1$ and right one as $V_2$. Applying state equation for each container |
| </p> | | </p> |
| <center> | | <center> |
| $P_1~V_1 = nRT \;(1)$ | | $P_1~V_1 = nRT \;(1)$ |
| </center> | | </center> |
| <center> | | <center> |
| $P_0~V_2 = nRT_0 \;(2)$ | | $P_0~V_2 = nRT_0 \;(2)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| It is easy to see that | | It is easy to see that |
| </p> | | </p> |
| <center> | | <center> |
| $V_1 = V_0+S\frac{l}{2}-Sx \;(3)$ | | $V_1 = V_0+S\frac{l}{2}-Sx \;(3)$ |
| </center> | | </center> |
| <center> | | <center> |
| $V_2 = V_0+S\frac{l}{2}+Sx \;(4)$ | | $V_2 = V_0+S\frac{l}{2}+Sx \;(4)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <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 $(2)$ | | 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> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| This is mathematically valid for $0< x <\frac{V_0}{S}+\frac{l}{2}$ but according the physical situation only for $0< x < \frac{l}{2}$. | | 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> |
| <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> | | <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
| </article> | | </article> |
| </main> | | </main> |
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