91Ó°ÊÓ

During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine \(Squalus\) sank at a point where the depth of water was 73.0 m. The temperature was 27.0\(^\circ\)C at the surface and 7.0\(^\circ\)C at the bottom. The density of seawater is 1030 kg/m\(^3\). (a) A diving bell was used to rescue 33 trapped crewmen from the \(Squalus\). The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

Step by step solution

01

Understand the problem

The exercise involves calculating the water height inside a diving bell at a depth of 73.0 m, considering the effects of temperature and pressure. Furthermore, it includes determining the gauge pressure required to remove water from the diving bell.

02

Use the Ideal Gas Law

Since the diving bell is open at the bottom, water and air reach a pressure equilibrium. The amount of air in the bell initially can be described by the Ideal Gas Law, where \( PV = nRT \). We know the initial conditions at the surface level. Translated to the depth, it modifies to \( P_2V_2 = P_1V_1 \).

03

Calculate initial conditions

At the surface temperature is 27.0°C or 300.15 K and at the bottom 7.0°C or 280.15 K. Pressure at the surface is atmospheric pressure, \( P_1 = 101325 \,\text{Pa} \), and volume \( V_1 \) corresponds to the bell height (2.3 m) initially with no water inside.

04

Calculate the pressure on the bottom

The pressure at the bottom, \( P_2 \), is calculated by the hydrostatic pressure equation, \( P_2 = P_0 + \rho gh \), where \( P_0 = 101325 \, \text{Pa} \), \( \rho = 1030 \, \text{kg/m}^3 \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 73.0 \, \text{m} \).

05

Determine the final conditions

Using \( P_2V_2/T_2 = P_1V_1/T_1 \), solve for the height of the water in the bell. Since \( V_2 = A\times(H - h_w) \) where \( A \) is the area of the bell, equalize the initial and final volumes and solve for the unknown height of water, \( h_w \).

06

Calculate gauge pressure for expulsion

To expel the water completely, the air pressure inside should equal external pressure at the bottom: \( P_\text{required} = P_2 - P_0 \). This is the definition of the gauge pressure needed.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid in equilibrium due to the force of gravity.
In our exercise, understanding hydrostatic pressure is crucial because it determines how water enters the diving bell.
The pressure applied by the water at any depth can be calculated using the equation:

• \[ P = P_0 + \rho gh \]

where:

• \( P_0 \) is the atmospheric pressure at the surface (101325 Pa)
• \( \rho \) is the density of seawater (1030 kg/m³)
• \( g \) is the acceleration due to gravity (9.81 m/s²)
• \( h \) is the depth at which the object is submerged (73.0 m)

This equation allows us to understand how the pressure at the bottom of the ocean influences the height of the water inside a diving bell.
Since the diving bell is open at the bottom, the water pressure pushes it upward inside the bell. The hydrostatic pressure difference must be overcome to keep the water out of the bell when rescuing the crew.

Ideal Gas Law

The Ideal Gas Law, written as \( PV = nRT \), connects pressure, volume, and temperature of a gas'.
At the water's surface, the interior of the diving bell contains a certain amount of air, which we must consider when the bell is submerged.
This equation can adapt to changing conditions as:

• \[ P_1V_1/T_1 = P_2V_2/T_2 \]

Where:

• \( P_1, V_1, T_1 \) are the initial pressure, volume, and temperature at the surface
• \( P_2, V_2, T_2 \) are the conditions at the bottom of the ocean

As the bell descends:- Air contracts due to increased pressure and decreased temperature.
- These changes reduce the air volume, implying a rise in water inside the bell.
To solve the exercise, we use the Ideal Gas Law to calculate how much water enters the bell due to changes in environmental conditions.

Temperature Gradients

Temperature gradients, the rate of temperature change with respect to distance, affect gas behavior significantly.
When moving from the warm ocean surface at 27.0°C to the colder deep waters at 7.0°C, the volume of air inside the diving bell changes.
Understanding this means recognizing two primary principles:

• Air at higher temperatures expands while at lower temperatures, it contracts.
• The Ideal Gas Law illustrates how temperature changes affect pressure and volume of gases (discussed above).

In the exercise, this temperature difference leads to air volume reduction inside the bell as it reaches the ocean's colder depths.
Consequently, knowing temperature gradients enables us to anticipate these changes and thus calculate any resultant pressure and volume adjustments.
This knowledge is key to understanding the behavior of the substance inside the diving bell.

Pressure Equilibrium

Achieving pressure equilibrium involves balancing different pressures acting upon a submerged object.
For the diving bell, when it reaches the ocean bottom, the pressure outside due to water must equal the pressure inside to prevent water entry.
To maintain pressure equilibrium:

• The pressure inside the diving bell (air pressure) must match the external water pressure.
• Any difference results in water entering the bell, which can be calculated using hydrostatic pressure principles.
• Compressed air needs to be supplied to the bell to offset external water pressure, calculated using gauge pressure.

This concept helps determine the gauge pressure needed for the exercise. By knowing the pressure at the bottom (\( P_2 \)) and subtracting the atmospheric pressure (\( P_0 \)), we learn the pressure required to expel water from the bell.
Successfully maintaining equilibrium allows for a safe and effective rescue through precision pressure control.