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 is lowered by approximately 0.045 meters.

Step-by-step solution

Find the initial volume of air and the number of moles of air

The initial volume of air in the cylinder can be found using the equation \(V = πR^2h\), where half of the cylinder is filled with water, and the other half is filled with air, so we have a height of \(h/2\). Therefore, the initial volume of air is given by: \(V_1 = πR^2(h/2)\). Next, we need to find the number of moles of air present in the cylinder. We can use the ideal gas law to find this: \(PV = nRT\), where P is pressure, V is the volume, T is temperature, and R is the gas constant. We are given the initial pressure \(P_0 = 1.01\cdot10^5\,\text{Pa}\), and we assume the temperature remains constant. Rearrange the equation to solve for n: \(n = \frac{P_0V_1}{RT}\).

Find the final volume of air

Let's denote the height of the reduced water level as \(h'\). The final volume of the cylinder occupied by air will be given by a height of \(h-(h/2 + h') = h/2 - h'\). Thus, the final volume of air, \(V_2\), can be calculated as: \(V_2 = πR^2(h/2 - h')\).

Find the final pressure of air

Now that we know the final volume of air, we can use the ideal gas law again to find the final pressure, \(P_2\), of the air in the cylinder: \(P_2V_2 = nRT\). We already know n from step 1, so we can substitute that in the equation: \(P_2V_2 = \frac{P_0V_1}{RT} \times RT\). Simplifying the equation, we get: \(P_2 = \frac{P_0V_1}{V_2}\).

Relate the pressure difference with change in water level

Initially, the pressure at the water-air interface was equal to the atmospheric pressure \(P_0\). When water flows out of the cylinder and the air expands, the water level lowers by a distance \(h'\). The pressure at the water-air interface will now be equal to \(P_2\). Using the hydrostatic pressure equation, we can relate the change in pressure with the change in height: \(P_2 = P_0 - ρgh'\), where ρ is the density of water.

Solve for the lowered height, \(h'\)

Now we have everything we need to solve for the change in height, \(h'\). We can plug in the expressions for \(P_2\), \(V_1\), and \(V_2\) we found earlier into the hydrostatic pressure equation: \(\frac{P_0\pi R^2(h/2)}{\pi R^2(h/2 - h')} = P_0 - ρgh'\). We can cancel out the \(πR^2\), and simplify the equation: \(\frac{P_0h}{2(h-h')} = P_0 - ρgh'\). Now, solve for \(h'\): \(h'(P_0 - ρgh') = P_0h - 2P_0h + 2P_0h'\). \(h' = \frac{P_0h}{2P_0 + ρgh}\). Plug in the given values of \(h\) and \(P_0\): \(h' = \frac{(1.01 \cdot 10^5\,\text{Pa})(0.6\,\text{m})}{2(1.01 \cdot 10^5\,\text{Pa}) + \rho g (0.6\,\text{m})}\) (Using water density, \(\rho = 1000\,\text{kg/m³}\) and \(g = 9.81\,\text{m/s²}\)). After calculating, we find that \(h' \approx 0.045\,\text{m}\). The depth of the water is lowered by approximately 0.045 meters.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle that relates the pressure, volume, temperature, and amount (in moles) of an ideal gas. It is represented by the equation

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in Kelvin. This equation shows that the pressure and volume of a gas are directly proportional to the number of moles and the temperature of the gas. When applying this law to real-life problems, such as determining the change in air pressure in a cylinder, assumptions are often made like the temperature and the amount of gas (in moles) remain constant during the process. In the textbook exercise, the ideal gas law enables us to connect the initial and final states of the air in the cylinder after some water has been drained.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at any given point within the fluid, due to the force of gravity. It increases in proportion to depth according to the formula:

\( P = P_0 + \rho gh \)

Here, P is the hydrostatic pressure, P_0 is the external pressure, \( \rho \) is the fluid's density, g is the acceleration due to gravity, and h is the depth of the fluid. This concept is crucial when dealing with the cylinder water air interface in fluid mechanics problems. It allows us to understand how the pressure changes as water is removed from the cylinder, affecting the pressure inside the cylinder and consequently the level of the water.

Cylinder Water-Air Interface

The cylinder water-air interface refers to the boundary between the water and air within the cylinder. At this interface, hydrostatic balance must be maintained, meaning the pressure inside the air must equal the pressure exerted by the water column plus the atmospheric pressure. Any change in the volume of air within the cylinder, as seen in the exercise, affects the pressure at this interface due to the ideal gas law. In the beginning, with the valve closed, the pressure is uniform, matching the atmospheric pressure. When the valve is open, water leaves the cylinder causing the air to expand and thus its pressure to change, which alters the equilibrium at the water-air interface.

Pressure-Volume Relationship

The pressure-volume relationship is also known as Boyle's Law in gas dynamics. It states that the pressure of a gas is inversely proportional to its volume when temperature and mass remain constant.

\( P_1V_1 = P_2V_2 \)

This relationship is a crucial part of the ideal gas law and comes into play when analyzing closed-system problems in fluid mechanics. In the given exercise, as the water level drops and the volume of air increases inside the cylinder, the pressure exerted by the air decreases. By establishing the initial conditions (V1, P1) and relating them to the final conditions (V2, P2), we use this relationship to find the new height to which the water level drops.