/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 A bubble is released from an oce... [FREE SOLUTION] | 91Ó°ÊÓ

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A bubble is released from an ocean vent located \(20 \mathrm{~m}\) below the ocean surface. Atmospheric pressure above the ocean surface is equal to \(101.3 \mathrm{kPa}\), and the ocean temperature down to a depth of \(20 \mathrm{~m}\) is approximately constant at \(20^{\circ} \mathrm{C}\). Estimate the ratio of the density of the air in the bubble at a depth of \(20 \mathrm{~m}\) to the density of the air in the bubble just as it reaches the ocean surface.

Short Answer

Expert verified
The density ratio is approximately 2.94.

Step by step solution

01

Understand the problem

We need to find the ratio of air density in a bubble at two different depths in the ocean: one at a depth of \(20\) meters (where the bubble is released) and one at the surface of the ocean. We'll use the principles of fluid pressure and the ideal gas law to do this.
02

Calculate the pressure at 20 meters depth

The pressure at a depth in a fluid can be calculated using the formula: \( P = P_0 + \rho g h \). Here, \( P_0 = 101.3 \text{kPa} \) (atmospheric pressure), \( \rho = 1000 \text{kg/m}^3 \) (density of seawater), \( g = 9.81 \text{m/s}^2 \) (acceleration due to gravity), and \( h = 20 \text{m} \). So, \( P = 101.3\text{kPa} + 1000 \text{kg/m}^3 \times 9.81 \text{m/s}^2 \times 20 \text{m} \).
03

Simplify the calculation for pressure at 20 meters

Calculate the additional pressure due to the water column: \( 1000 \text{kg/m}^3 \times 9.81 \text{m/s}^2 \times 20 \text{m} = 196200 \text{Pa} = 196.2 \text{kPa} \).Therefore, the total pressure at a depth of \(20 \text{m}\) is: \( P = 101.3 \text{kPa} + 196.2 \text{kPa} = 297.5 \text{kPa} \).
04

Determine the final pressure at the surface

The pressure at the surface is simply the atmospheric pressure, which is given as \( 101.3 \text{kPa} \).
05

Relate pressure to density using Ideal Gas Law

The ideal gas law states \( PV = nRT \). Since the temperature is constant, \( V \) (volume) is inversely proportional to the pressure \( P \). Therefore, density \( \rho \) (which is \( \frac{mass}{volume} \)) is directly proportional to \( P \). Thus, \( \frac{\rho_{20m}}{\rho_{surface}} = \frac{P_{20m}}{P_{surface}} \).
06

Calculate the density ratio

Using the pressures calculated: \( \frac{\rho_{20m}}{\rho_{surface}} = \frac{297.5 \text{kPa}}{101.3 \text{kPa}} \approx 2.94 \).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle in fluid mechanics, especially when discussing gases and their properties under varying conditions. This law is expressed as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume of the gas.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the constant for ideal gases.
  • \( T \) is the temperature in Kelvin.
In this problem, we are assuming the temperature is constant as the bubble rises from the depth to the surface.
This means the volume of the bubble increases as the pressure decreases. Conversely, density, defined as \( \frac{mass}{volume} \), decreases when volume increases. Thus, at constant temperature and number of gas moles, density is directly proportional to pressure.
Fluid Pressure
Fluid pressure is the force exerted by a fluid per unit area. It is an essential concept when dealing with liquids and gases.
The pressure in a fluid at rest increases with depth due to the weight of the fluid above. This is given by the formula \( P = P_0 + \rho g h \), where:
  • \( P \) is the pressure at depth.
  • \( P_0 \) is the atmospheric or initial pressure.
  • \( \rho \) is the fluid's density.
  • \( g \) is the acceleration due to gravity.
  • \( h \) is the depth of the fluid.
In our example, at a depth of 20 meters in the ocean, we calculate total pressure by considering both atmospheric pressure and pressure due to the water's weight above the bubble. This helps understand how a bubble's characteristics, like size and density, change with pressure as it travels to the surface.
Ocean Depth
As the depth of the ocean increases, the pressure exerted on objects like a bubble increases significantly. This increase is primarily due to the weight of the water column above the object. The deeper the object, the more pressure it faces.
The concept of ocean depth is vital for determining how pressure changes affect a bubble's volume as it rises. Considering that the initial depth is 20 meters, we can model how the bubble changes as it rises, based on constant fluid properties, assuming no major temperature gradients.
This understanding is crucial for appreciating changes in relative air density as pressure conditions transition from 20 meters to the surface. As the bubble makes its way to the surface, it experiences decreasing pressure and therefore an increase in volume.
Density Calculation
Density calculation is important for comparing the state of a gas at different pressures. It involves understanding how closely packed the molecules are within a given volume.
Using the ideal gas law, density becomes significant because it varies inversely with volume at constant temperature. Thus, when pressure changes, density changes proportionally.
To compare the density of air within a bubble at 20 meters versus at the surface, note that pressure increases with depth, causing the bubble to have higher density at 20 meters than at the surface. Therefore:
  • The density ratio is directly linked to the pressure ratio.
  • Given that the pressure is 297.5 kPa at 20 meters and 101.3 kPa at the surface, the density ratio is approximately 2.94.
This indicates how much denser the air in the bubble is down at 20 meters, in comparison to when it reaches the surface.

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Most popular questions from this chapter

If a body of specific gravity \(\mathrm{SG}_{1}\) is placed in a liquid of specific gravity \(\mathrm{SG}_{2}\), what fraction of the total volume of the body will be above the surface of the liquid? If an iceberg has a specific gravity of 0.95 and is floating in seawater with a specific gravity of \(1.25,\) what fraction of the iceberg is above water?

What is the pressure head (of water) corresponding to a pressure of \(810 \mathrm{kPa}\) ? What depth of mercury at \(20^{\circ} \mathrm{C}\) will be required to produce a pressure of \(810 \mathrm{kPa}\) ?

A cylindrical container has a diameter of \(0.4 \mathrm{~m}\) and contains kerosene to a depth of \(0.6 \mathrm{~m}\). The temperature of the kerosene is \(20^{\circ} \mathrm{C}\). If the container, with its long axis oriented vertically, is placed on the floor of a delivery elevator that ascends with an acceleration of \(1.5 \mathrm{~m} / \mathrm{s}^{2}\), what is the pressure in the fluid on the bottom of the container? What force does the container exert on the floor of the elevator? Assume that the mass of the container is negligible compared with that of the kerosene.

(a) At what depth below the surface of a water body will the (gauge) pressure be equal to \(200 \mathrm{kPa}\) ? (b) If a 1.55-m-tall person orients himself vertically underwater in a pool, what pressure difference does he feel between his head and his toes? Assume water at \(20^{\circ} \mathrm{C}\).

A liquid is stratified such that the specific gravity of the fluid at the surface is 0.98 and the specific gravity at a depth of \(12 \mathrm{~m}\) is equal to 1.07 . Assuming that the specific gravity varies linearly between the liquid surface and a depth of \(12 \mathrm{~m}\), determine the pressure at a depth of \(12 \mathrm{~m}\). State whether this is a gauge pressure or an absolute pressure.

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