If a scuba diver fills his lungs to full capacity of 5.5LL when 8.2mm below the surface, to what volume would his lungs expand if he quickly rose to the surface? Assume he dives in the sea, thus the water is salt.

Assume diver is 8.2 m below surface (and not 8.2mm), and at the surface, P=101.3 kPa.

Pressure below surface
=ρgh
=1000 kg/m² * 9.8 * 8.2 + 101.3
= 80.36+101.3 kPa
= 181.66 kPa

Assuming temperature below water and at the surface are identical, use Boyles law:
P1V1=P2V2
where
P1=101.3 kPa
V1=?
P2=181.66 kPa
V2=5.5 L
Solve for V1 to get
V1=P2V2/P1
=181.66*5.5/101.3
= 9.863 L
= 9.9 L approx.

Well, since you're asking about scuba diving, let's dive into the answer, shall we?

When a scuba diver goes deeper underwater, the pressure increases due to the weight of the water above. For every 10 meters, the pressure increases by approximately 1 atmosphere (or 101.325 kilopascals). So, if our scuba diver is 8.2mm below the surface, we can consider it negligible and say he's basically at the surface.

Now, when our diver fills his lungs to a capacity of 5.5LL (liters), that's quite impressive! But if he were to quickly rise to the surface, the pressure would decrease rapidly. As the pressure decreases, the gases in his lungs, particularly nitrogen, which is the majority of the gas we breathe, would expand.

To calculate the volume his lungs would expand to, we need to take into account Boyle's Law, which states that at a constant temperature, the pressure and volume of a gas are inversely proportional. So, as the pressure decreases, the volume increases.

However, since you mentioned the water being salt, it's important to note that the increased salinity affects the solubility of gases in water, such as nitrogen. This means the diver would need to ascend at a controlled rate to avoid decompression sickness, also known as the bends. So, in reality, the lungs wouldn't expand dramatically like a balloon.

But just to give you an approximate answer, let's assume the pressure drops to surface level instantaneously. Without factoring in general physiological limitations, his lungs might expand by roughly 2-3 times their original volume.

Remember, safety first! So, it's always best for our scuba diver friend to ascend gradually and follow proper procedures. Happy diving!

When a scuba diver descends, the pressure surrounding them increases, causing their lungs to compress and decrease in volume. Conversely, when the diver ascends, the pressure decreases, allowing the lungs to expand. To compute the volume to which the lungs would expand when the diver rises to the surface, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at a constant temperature. Boyle's Law equation is:

P1 x V1 = P2 x V2

Where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume

In this case, the initial pressure is the pressure exerted by the water at a depth of 8.2mm below the surface, and the final pressure is the atmospheric pressure at the surface. Let's assume the atmospheric pressure is 101.3 kPa.

To find the initial pressure, we can use the formula:

P1 = atmospheric pressure + pressure due to the depth

The pressure due to depth can be calculated using the formula:

Pressure = density x gravity x depth

In the sea, the density of saltwater is approximately 1025 kg/m³, and the acceleration due to gravity is around 9.8 m/s².

Let's calculate the initial pressure:

Pressure due to depth = density x gravity x depth
= 1025 kg/m³ x 9.8 m/s² x 0.0082 m
≈ 84.8932 kPa

P1 = atmospheric pressure + pressure due to depth
= 101.3 kPa + 84.8932 kPa
≈ 186.1932 kPa

Now, we can plug the values into Boyle's Law to find the final volume:

P1 x V1 = P2 x V2
186.1932 kPa x 5.5 L = 101.3 kPa x V2

V2 = (186.1932 kPa x 5.5 L) / 101.3 kPa
≈ 10.0759 L

Therefore, if the scuba diver quickly rises to the surface, their lungs would expand to approximately 10.08 L (rounded to two decimal places) in volume.

To calculate the change in lung volume as the scuba diver rises to the surface, we need to understand the relationship between pressure and volume in a gas. Boyle's Law states that for a given amount of gas at a constant temperature, the pressure and volume are inversely proportional. In other words, as pressure increases, volume decreases, and vice versa.

In this case, we know that the scuba diver fills his lungs to a volume of 5.5 liters (LL) when 8.2 millimeters (mm) below the surface. As the diver rises to the surface, the pressure decreases, causing an increase in lung volume.

To determine the new volume of the diver's lungs at the surface, we need to know the change in pressure between the starting depth (8.2 mm below the surface) and the surface itself. The change in pressure can be calculated using the hydrostatic pressure formula:

ΔP = ρgh

Where:
ΔP is the change in pressure (in Pascals or Pa)
ρ is the density of the fluid (in kilograms per cubic meter or kg/m^3)
g is the acceleration due to gravity (in meters per second squared or m/s^2)
h is the depth (in meters or m)

For seawater, the density ρ is approximately 1025 kilograms per cubic meter (kg/m^3), and the acceleration due to gravity g is approximately 9.8 meters per second squared (m/s^2). We can use these values to calculate the change in pressure as the diver rises to the surface.

Let's calculate the change in pressure using the given depth of 8.2 mm:

h = 8.2 mm = 8.2 / 1000 = 0.0082 m (converting millimeters to meters)
ρ = 1025 kg/m^3 (density of seawater)
g = 9.8 m/s^2 (acceleration due to gravity)

ΔP = ρgh = (1025 kg/m^3) * (9.8 m/s^2) * (0.0082 m)
ΔP ≈ 84.89 Pa (rounded to two decimal places)

Now that we have the change in pressure, we can calculate the new lung volume using Boyle's Law. Remember that pressure and volume are inversely proportional:

P1V1 = P2V2

Where:
P1 is the initial pressure (at 8.2 mm below the surface)
V1 is the initial volume (5.5 LL)
P2 is the final pressure (at the surface, atmospheric pressure)
V2 is the final volume (what we need to calculate)

The initial pressure, P1, is equal to the atmospheric pressure at the surface. Although the question doesn't specify the atmospheric pressure, we can assume it to be approximately 101,325 Pascals (Pa) at sea level.

P1 = P2 = 101,325 Pa (approximate atmospheric pressure)

Using Boyle's Law and rearranging the equation, we can solve for V2:

V2 = (P1 * V1) / P2
V2 = (101,325 Pa * 5.5 LL) / 101,325 Pa
V2 = (5.5 LL) = 5.5 liters

Therefore, the scuba diver's lungs would expand to a volume of 5.5 liters at the surface if he quickly rose from a depth of 8.2 mm.