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Luisito edited Jun 14, 2024, 2:47 PM

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			<title>Savchenko Solutions</title>
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			<header>
			<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
			</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>
		@@ -70,7 +70,9 @@
			<center>
			$\frac{V_1}{V_2}=\frac{T}{T_0} \;(5)$
			</center>
		−	Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$
		+	<p class="TxSolutions">
		+	Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$
		+	</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>
			<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>
			</article>
			</main>

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	</head>		</head>

	<body id="top">		<body id="top">
	<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>

	<main>		<main>
	<article>		<article>
	<h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>

	<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.
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<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>
	</figure>		</figure>
	</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>
	@@ -70,7 +70,9 @@
	<center>		<center>
	$\frac{V_1}{V_2}=\frac{T}{T_0} \;(5)$		$\frac{V_1}{V_2}=\frac{T}{T_0} \;(5)$
	</center>		</center>
	Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$		<p class="TxSolutions">
			Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$
			</p>
	<center>		<center>
	$\boxed{T = T_0~\frac{2V_0+S(l+2x)}{2V_0+S(l-2x)}}$		$\boxed{T = T_0~\frac{2V_0+S(l+2x)}{2V_0+S(l-2x)}}$
	</center>		</center>
	<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>
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