The maximum lung capacity of an average adult is about 4-6 L. A scuba diver takes a breath of about 2.1 L of air at a depth of 30 m where the pressure is 405 kPa. If the diver holds his/her breath while rising to the surface, where the pressure is 102 kPa, what is the volume of air in the lungs? What is the likely result?
PV is constant, so
102V = 405*2.1
>Splat< -- step back, please!
To find the volume of air in the lungs while the scuba diver rises to the surface, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional if the temperature remains constant.
First, we need to convert the pressure values from kilopascals (kPa) to atmospheres (atm) since Boyle's Law is typically expressed in terms of atm.
Pressure at 30 m depth: 405 kPa = 3.99 atm
Pressure at the surface: 102 kPa = 1.01 atm
Now, we can set up the equation using Boyle's Law:
P1 * V1 = P2 * V2
Where:
P1 = Initial pressure (3.99 atm)
V1 = Initial volume (2.1 L)
P2 = Final pressure (1.01 atm)
V2 = Final volume (the value we want to find)
Rearranging the equation, we have:
V2 = (P1 * V1) / P2
Substituting the given values:
V2 = (3.99 atm * 2.1 L) / 1.01 atm
V2 ≈ 8.36 L
Therefore, the volume of air in the lungs while rising to the surface is approximately 8.36 L.
As for the likely result, holding one's breath while ascending to the surface can lead to a condition called lung overexpansion injury or pulmonary barotrauma. The pressure decreases as the diver ascends, and if the air in the lungs is not able to escape, it can expand and cause damage to the lung tissues. This can result in pneumothorax (collapsed lung) or arterial gas embolism (gas bubbles in the bloodstream), which can be severe and life-threatening.
It is crucial for scuba divers to exhale continuously and never hold their breath during ascent to prevent lung overexpansion injuries.