Kvant Physics Problem 49
A U-shaped tube of constant cross-sectional area $A$ is completely filled with water of density $\rho$.
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Problem
U-shaped tube filled with water. From one arm the air has been removed; the air pressure in the other arm at temperature $t=20\celsius$ is equal to atmospheric pressure. Both ends of the tube are sealed. The difference between the water levels in the arms is 15 m. What will be the difference between the water levels in the arms if the tube is heated to $100\celsius$?
B. B. Bukhovtsev
Setup and Assumptions
A U-shaped tube of constant cross-sectional area $A$ is completely filled with water of density $\rho$. One arm is evacuated, so the gas pressure above the water column in that arm is zero in the initial state. The other arm contains a trapped quantity of air above the water column, initially at temperature $T_1 = 20^\circ\mathrm{C} = 293,\mathrm{K}$ and pressure equal to atmospheric pressure $P_{\mathrm{atm}}$.
Both ends of the tube are sealed, so the amount of trapped air remains constant. The water is treated as incompressible, and the tube walls are rigid. The gas is modeled as an ideal gas. The acceleration of gravity is $g$. The initial difference between water levels in the two arms is $\Delta h_1 = 15,\mathrm{m}$. The goal is to find the new level difference $\Delta h_2$ when the system is heated to $T_2 = 100^\circ\mathrm{C} = 373,\mathrm{K}$.
Evaporation of water, dissolution of gas, and thermal expansion of the tube are neglected.
Physical Principles
Hydrostatic equilibrium requires that pressures at the same depth in a connected fluid are equal. If $\Delta h$ is the difference between water levels in the two arms, then the pressure difference between the gas above the water in the right arm and the vacuum above the water in the left arm is balanced by the hydrostatic head of water.
This gives the relation
$$P = \rho g \Delta h,$$
where $P$ is the pressure of the trapped air.
The trapped air satisfies the ideal gas law
$$PV = nRT,$$
where $n$ is constant.
The volume of the trapped air is proportional to the height of the air column:
$$V \propto L,$$
where $L$ is the length of the air column in the right arm. In this setup, the geometry of a symmetric U-tube with equal cross sections implies that changes in water levels change the air column length proportionally to the level difference, so $V \propto \Delta h$.
Combining these relations yields a constraint between pressure, level difference, and temperature.
Derivation
Let the initial state correspond to $T_1 = 293,\mathrm{K}$ and level difference $\Delta h_1 = 15,\mathrm{m}$. Hydrostatic balance gives
$$P_1 = \rho g \Delta h_1.$$
At the same time, the gas in the right arm is at atmospheric pressure, so
$$P_1 = P_{\mathrm{atm}}.$$
Hence,
$$P_{\mathrm{atm}} = \rho g \Delta h_1.$$
After heating to $T_2 = 373,\mathrm{K}$, let the new level difference be $\Delta h_2$ and gas pressure $P_2$. Hydrostatic balance gives
$$P_2 = \rho g \Delta h_2.$$
Applying the ideal gas law between the two states,
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}.$$
Since the air volume is proportional to the level difference,
$$\frac{V_2}{V_1} = \frac{\Delta h_2}{\Delta h_1}.$$
Substituting $P_1 = \rho g \Delta h_1$ and $P_2 = \rho g \Delta h_2$ into the gas law relation yields
$$\frac{(\rho g \Delta h_1)V_1}{T_1} = \frac{(\rho g \Delta h_2)V_2}{T_2}.$$
Canceling $\rho g$ and substituting $V_2/V_1 = \Delta h_2/\Delta h_1$ gives
$$\frac{\Delta h_1}{T_1} = \frac{\Delta h_2^2}{\Delta h_1 T_2}.$$
Multiplying through,
$$\Delta h_2^2 = \Delta h_1^2 \frac{T_2}{T_1}.$$
Taking the positive root,
$$\Delta h_2 = \Delta h_1 \sqrt{\frac{T_2}{T_1}}.$$
Substituting numerical values,
$$\Delta h_2 = 15,\mathrm{m} \cdot \sqrt{\frac{373,\mathrm{K}}{293,\mathrm{K}}}.$$
Compute the ratio,
$$\frac{373}{293} \approx 1.273.$$
Thus,
$$\sqrt{1.273} \approx 1.128.$$
Therefore,
$$\Delta h_2 \approx 15,\mathrm{m} \cdot 1.128 = 16.92,\mathrm{m}.$$
Result
$$\boxed{\Delta h_2 \approx 1.69 \times 10^{1}\ \mathrm{m} \approx 16.9,\mathrm{m}}$$
Sanity Checks
The expression $\Delta h_2 = \Delta h_1 \sqrt{T_2/T_1}$ is dimensionally consistent because it relates ratios of lengths to a square root of a dimensionless temperature ratio.
The increase in temperature from $293,\mathrm{K}$ to $373,\mathrm{K}$ is about a $27%$ rise, so a moderate increase in $\Delta h$ is expected. The result shows an increase from $15,\mathrm{m}$ to about $16.9,\mathrm{m}$, which is consistent with a sublinear dependence on temperature due to the coupling of pressure and volume.
A sign error would most likely arise in the hydrostatic relation $P = \rho g \Delta h$, where reversing the definition of $\Delta h$ would incorrectly predict a decrease in level difference upon heating.