Question

A sealed vertical cylinder of radius \(R\) and height \(h=0.60 \mathrm{~m}\) is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, \(p_{0}=1.01 \cdot 10^{5} \mathrm{~Pa}\). A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.).

Short answer

Answer: The water level in the cylinder decreases by approximately 0.0515 meters.

Step-by-step solution

Consider the initial state of the cylinder

Initially, the cylinder is filled halfway with water and the upper half with air. The air is at standard atmospheric pressure, denoted as \(p_0\).

Analyze the final state of the cylinder

When the valve is opened, water flows out, and the volume of air in the upper part of the cylinder increases. The height of the air column also increases, while the pressure of the air decreases. We will denote the final air pressure as \(p_f\) and the reduced height of the water column as \(h_f\).

Apply the ideal gas laws for the initial and final states of the cylinder

Since the temperature does not change and no air leaks into the cylinder, we can apply the ideal gas law for the initial and final states. For initial state: \(p_{0}V_{0} = nRT\) For final state: \(p_{f}V_{f} = nRT\) The number of moles (n) and the universal gas constant (R) are constant. Therefore, we can express the relation between initial and final pressures and volumes as: \(p_{0}V_{0} = p_{f}V_{f}\)

Relate the volumes of air in the initial and final states

Initially, the air occupies half of the cylinder: \(V_{0} = \frac{1}{2} \pi R^2h\) Finally, it occupies the part of the cylinder not occupied by water: \(V_{f} = \pi R^2h - \pi R^2h_{f}\) Replacing the volumes in the relationship between initial and final pressures and volumes, we get: \(p_{0}\frac{1}{2} \pi R^2h = p_{f}(\pi R^2h - \pi R^2h_{f})\)

Apply hydrostatic pressure at the liquid-gas interface in the final state

Calculate the hydrostatic pressure of water at the liquid-gas interface for the final state: \(P_{hydro} = p_{f} + \rho gh_{f}\) Since the water and air are in equilibrium in the final state, they have equal pressure: \(p_{0} = p_{f} + \rho gh_{f}\)

Find the final air pressure and height of the water column

Now, combine the expressions from steps 4 and 5 and solve for \(p_f\): \(p_{0}\frac{1}{2} \pi R^2h = (\pi R^2h - \pi R^2h_{f})(p_{0} - \rho gh_{f})\) Cancel out the \(\pi R^2\) terms and simplify: \(\frac{1}{2}h = h - h_{f} - \frac{\rho gh_{f}^2}{p_{0}}\) Rearrange the equation to solve for \(h_f\): \(h_{f}^2 - 2h_f\left(h - \frac{p_0}{2\rho g}\right) = 0\) Solve the quadratic equation for \(h_f\): \(h_{f} = h - \frac{p_0}{2\rho g}\)

Determine the water level reduction

Since \(h_f\) is the height of the reduced water column, the difference between the initial and final height is the reduction in water level: \(\Delta h = h - h_f\) Substitute the expression for \(h_f\) from step 6 and calculate the water level reduction: \(\Delta h = h - (h - \frac{p_0}{2\rho g}) = \frac{p_0}{2\rho g}\) Now plug in the given values for \(p_0 = 1.01 \cdot 10^{5} \mathrm{~Pa}\), \(\rho = 1000 \mathrm{~kg/m^3}\) (density of water), and \(g = 9.81 \mathrm{~m/s^2}\) (acceleration of gravity): \(\Delta h = \frac{1.01 \cdot 10^{5} \mathrm{~Pa}}{2(1000 \mathrm{~kg/m^3})(9.81 \mathrm{~m/s^2})} = 0.0515 \mathrm{~m}\) The depth of the water is lowered by approximately 0.0515 meters.

Key concepts

The Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics which connects pressure, volume, and temperature of a gas. It is represented by the equation
\( PV = nRT \)
where P represents the pressure of the gas, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature. In the context of the exercise, this law allows us to understand how the air in the cylinder reacts to the changing volume as water is drained.

When the valve is opened on the cylinder, the volume V available to the air increases because there is less water. Since the temperature and amount of air in the cylinder remain unchanged, the pressure P must decrease to compensate for the increase in volume, as per the Ideal Gas Law. This interplay between pressure and volume is crucial for determining the final height of the water column left in the cylinder.

Fluid Mechanics

Fluid Mechanics deals with the behavior of fluids (liquids and gases) and their interactions with external forces. In this particular problem, the fluid of interest is water, which is being affected by both gravity and the air pressure from above. Fluid mechanics principles determine how fluids move and react to different forces.

One of the fundamental equations in fluid mechanics is the hydrostatic pressure equation, represented as \( P_{hydro} = \rho gh \) where P_hydro is the hydrostatic pressure, \rho (rho) is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column. In our scenario, we use this equation to find the pressure exerted by the water column on the air above, which helps to determine the new equilibrium when the water stops flowing out of the cylinder.

The Pressure-Volume Relationship

The Pressure-Volume relationship, also known as Boyle's Law in the case of a constant temperature, describes how the pressure of a gas tends to decrease as the volume increases, and vice versa, when the temperature and the amount of gas remain constant. It is an important aspect of the Ideal Gas Law and is essential in understanding the behavior of gases in various situations, including the exercise we're analyzing.

In the sealed cylinder exercise, as the volume of air increases with the decreasing water level, Boyle's Law helps us predict that the air pressure will correspondingly decrease. This concept is applied in the solution when we equate the initial and final states' pressure-volume products. By isolating the final air pressure p_f and the final water column height h_f, we can use this relationship to solve for the resulting change in the water level. Understanding this pressure-volume relationship enables students to grasp how varying conditions affect gas behavior, which is crucial for solving problems in fluid mechanics.