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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<h3> Statement </h3>
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$5.5.16$
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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For problem 5.5.16
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<h3>Solution</h3>
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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 $0< x < \frac{l}{2}$.
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<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>
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