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Why do scuba divers need to exhale air when they ascend to the surface of the water?

Short Answer

Expert verified
Scuba divers need to exhale while ascending to prevent lung overexpansion and potential injury due to decreasing water pressure and increasing air volume in the lungs.

Step by step solution

01

Understand Boyle's Law

Boyle's Law states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. The formula is expressed as \[ P_1 V_1 = P_2 V_2 \].
02

Analyze Pressure Changes Underwater

As a diver ascends, the water pressure decreases. According to Boyle's Law, if the pressure decreases, the volume of the air in the diver's lungs will increase, assuming a constant amount of air and temperature.
03

Risks of Not Exhaling

If the diver does not exhale while ascending, the expanding air could overinflate the lungs, leading to pulmonary barotrauma, which is damage to lung tissue due to overexpansion.
04

Exhaling to Prevent Injury

To prevent lung overexpansion and potential injury, it is crucial for divers to exhale continuously while ascending, allowing the excess air to escape and ensuring the lung volume does not exceed safe limits.

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Key Concepts

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

pressure-volume relationship
Boyle’s Law is a fundamental principle that describes the inverse relationship between the pressure and volume of a gas. It can be expressed with the formula \( P_1 V_1 = P_2 V_2 \). This law is important because it helps us understand how changes in pressure affect the volume of gases.
When a diver is underwater, the pressure around them is high due to the weight of the water. This high pressure compresses the air in their lungs, making it occupy a smaller volume. As the diver ascends to the surface, the water pressure decreases. According to Boyle’s Law, if pressure decreases, the volume of the gas will increase so long as the amount of gas and temperature stay constant.
For example, if a diver's lung has a volume of 5 liters at a lower depth with high pressure, the volume might expand to 10 liters as they ascend where the pressure is lower. Understanding this relationship is vital for divers to avoid dangerous situations.
scuba diving safety
Scuba diving is an exciting activity, but it requires knowledge about underwater physics to ensure safety. As divers go deeper underwater, the surrounding water pressure increases. Therefore, they must breathe air from their tanks at higher pressures to balance the pressure difference.
On ascending, this pressure decreases, which can pose risks if not managed properly. One critical safety practice is to exhale while ascending. This helps prevent the lungs from overexpanding due to the increase in air volume caused by decreasing pressure.
Safe scuba diving also involves regular training and the use of proper equipment. Divers should be aware of their depth and ascent rate to avoid decompression sickness and other pressure-related injuries. Continuous exhalation during ascent is a simple yet lifesaving practice that helps maintain lung health and overall safety.
pulmonary barotrauma
Pulmonary barotrauma refers to damage to the lungs that occurs due to the overexpansion of air. In scuba diving, this can happen if a diver ascends without exhaling, allowing the air in their lungs to expand uncontrollably as external pressure drops.
When the volume of air in the lungs increases too much, it can cause alveoli (tiny air sacs in the lungs) to rupture. This can lead to serious complications, such as air leaking into the chest cavity or bloodstream, potentially leading to life-threatening conditions like arterial gas embolism.
To prevent pulmonary barotrauma, divers must continuously exhale while ascending. This practice ensures that excess air can escape from the lungs, preventing dangerous pressure buildups. Understanding and respecting Boyle’s Law, as well as following safe diving practices, are essential for preventing this condition.

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Most popular questions from this chapter

An accident to the head can affect the ability of a person to ventilate (breathe in and out). a. What would happen to the partial pressures of oxygen and carbon dioxide in the blood if a person cannot properly ventilate? b. When a person who cannot breathe properly is placed on a ventilator, an air mixture is delivered at pressures that are alternately above the air pressure in the person's lung, and then below. How will this move oxygen gas into the lungs, and carbon dioxide out?

A sample of methane \(\left(\mathrm{CH}_{4}\right)\) has a volume of \(25 \mathrm{~mL}\) at a pressure of \(0.80 \mathrm{~atm} .\) What is the volume, in milliliters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas? a. \(0.40 \mathrm{~atm}\) b. \(2.00 \mathrm{~atm}\) c. \(2500 \mathrm{mmHg}\)

Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled \(24{ }^{\circ} \mathrm{C}\) at a pressure of \(745 \mathrm{mmHg}\). A balloon with a volume of \(425 \mathrm{~mL}\) drifts into the airlock where the temperature is \(-95^{\circ} \mathrm{C}\) and the pressure is \(0.115 \mathrm{~atm}\). What is the new volume of the balloon \((n\) remains constant)? Assume that the balloon is very elastic.

A sample of helium gas has a volume of \(6.50 \mathrm{~L}\) at a pressure of \(845 \mathrm{mmHg}\) and a temperature of \(25^{\circ} \mathrm{C}\). What is the pressure of the gas, in atmospheres, when the volume and temperature of the gas sample are changed to the following, if the amount of gas is constant? a. \(1850 \mathrm{~mL}\) and \(325 \mathrm{~K}\) b. \(2.25 \mathrm{~L}\) and \(12^{\circ} \mathrm{C}\) c. \(12.8 \mathrm{~L}\) and \(47^{\circ} \mathrm{C}\)

A gas has a volume of \(4.00 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\). What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if \(n\) and \(P\) do not change? a. \(1.50 \mathrm{~L}\) b. \(1200 \mathrm{~mL}\) c. \(250 \mathrm{~L}\) d. \(50.0 \mathrm{~mL}\)

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