Problem 106 - Chemistry The Molecular Nature of Matter

Question

A bubble of air escaping from a diver's mask rises from a depth of \(100 \mathrm{ft}\) to the surface where the pressure is \(1.00 \mathrm{~atm} .\) Initially, the bubble has a volume of \(10.0 \mathrm{~mL}\). Assuming none of the air dissolves in the water, how many times larger is the bubble just as it reaches the surface? Use your answer to explain why scuba divers constantly exhale as they slowly rise from a deep dive. (The density of seawater is approximately \(1.025 \mathrm{~g} \mathrm{~mL}^{-1}\); the density of mercury is \(13.6 \mathrm{~g} \mathrm{~mL}^{-1}\).)

Step-by-Step Solution

Answer

The bubble is approximately 4 times larger just it reaches the surface, demonstrating why scuba divers must exhale while ascending to prevent lung overexpansion due to decreasing pressure.

Step 1: Calculate the pressure at depth

First, determine the pressure exerted on the bubble at the depth of 100 ft. Pressure at depth in a fluid is given by the formula: \( P = P_0 + \rho gh \), where \( P_0 \) is the atmospheric pressure at the surface, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth. Convert the depth from feet to meters by multiplying by 0.3048 meters/foot. Use \( 1.00 \) atm for atmospheric pressure, \( 1.025 \text{ g/mL} \) for the density of seawater, \( 9.8 \text{ m/s}^2 \) for \( g \), and \( 100 \) ft for \( h \). Remember to convert the density of seawater to SI units: \( 1 \text{ g/mL} = 1000 \text{ kg/m}^3 \).

Step 2: Convert the calculated pressure to atm

The pressure calculated in Step 1 will be in pascals since standard SI units were used. To convert the pressure to atmospheres, use the conversion ratio \( 1 \text{ atm} = 101325 \text{ Pascals} \).

Step 3: Apply the ideal gas law

Knowing that the air in the bubble behaves as an ideal gas and that no air dissolves in the water, we can use Boyle's Law, which is a part of the ideal gas law for constant temperature: \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume. Use the initial volume of \( 10.0 \text{ mL} \), the initial pressure calculated in step 2, and the final pressure of \( 1.00 \text{ atm} \) at the surface.

Step 4: Solve for the final volume

Rearrange the formula from Step 3 to solve for the final volume \( V_2 \): \( V_2 = \frac{P_1V_1}{P_2} \). Calculate \( V_2 \) to determine how many times larger the bubble becomes as it reaches the surface.

Step 5: Explain why scuba divers exhale while rising

With the calculated enlargement of the bubble, explain that scuba divers exhale to prevent the expanding air in their lungs from causing injury. As the divers rise, the pressure decreases, and the volume of air in their lungs expands according to Boyle's Law, risking overexpansion and lung injury if they hold their breath.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental principle used to understand the behavior of gases under various conditions. It is expressed as the equation: \( PV = nRT \), where \( P \) denotes the pressure of the gas, \( V \) its volume, \( n \) the number of moles, \( R \) the ideal gas constant, and \( T \) the temperature in kelvin.When examining the behavior of a scuba diver's gas bubbles under the sea, it is essential to consider that the underwater environment presents conditions where temperature is relatively stable. In such cases, the relationship between pressure and volume takes center stage, which leads us to Boyle's Law. However, the ideal gas law provides the foundation for understanding this relationship. Essentially, if the temperature and moles of gas remain constant—as they tend to for an air bubble escaping a diver's mask—the ideal gas law simplifies to Boyle's Law, highlighting the inverse relationship between pressure and volume of a gas.

Pressure-Volume Relationship

In the context of gases, the pressure-volume relationship is described by Boyle's Law, which states that for a fixed amount of gas at a constant temperature, the volume of the gas is inversely proportional to its pressure. This is expressed mathematically by the formula: \( P_1V_1 = P_2V_2 \). When interpreting this law, it is crucial to understand that as pressure increases, the volume will decrease, and vice versa, provided the temperature does not change.

Boyle's Law in Action

A practical example of this concept can be seen in scuba diving. An air bubble released from a diver's equipment at depth will expand as it rises and experiences less pressure. This expansion reflects the inverse relationship between pressure and volume. In the exercise provided, the bubble starts at a high pressure deep underwater and expands as it ascends to the surface where the pressure is lower. Therefore, the original volume of the bubble increases in direct correspondence to the decreasing pressure it faces on its ascent.

Underwater Pressure Calculations

Calculating underwater pressure is critical for understanding Boyle's Law in a diving context. The pressure at a certain depth can be found with the formula: \( P = P_0 + \rho gh \), where \( P_0 \) is the atmospheric pressure at the surface, \( \rho \) represents the fluid's density, \( g \) acceleration due to gravity, and \( h \) the depth. This formula is pivotal in understanding how pressure changes with depth, which affects how divers and marine animals experience their environment.

Application in Scuba Diving

For a scuba diver, the increase in pressure with depth means that air in any cavities, such as the lungs or a mask, decreases in volume. Conversely, as the diver ascends, the reduction in pressure causes the volume of air in these cavities to increase. Therefore, accurate underwater pressure calculations ensure divers adjust their buoyancy and control their ascent to avoid the risks associated with rapid changes in pressure, such as the bends or lung over-expansion. This scientific understanding directly influences diving practices and safety measures, further exemplifying the importance of accurate pressure-volume knowledge.