Question

If a scuba diver's lungs have a normal capacity of 4.9 L at sea level (1.0 atm), what would be the volume of her lungs if the pressure at a depth of 50 ft is 975 mmHg?

Answer 1

6.3 L

Answer 2

3.8 L

Answer 3

[AI-generated answer — Explain Bot]

To find the volume of the scuba diver's lungs at a depth of 50 ft, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature is kept constant.

Boyle's Law equation: P₁V₁ = P₂V₂

Where:
P₁ = initial pressure (at sea level) = 1.0 atm
V₁ = initial volume (at sea level) = 4.9 L
P₂ = final pressure (at 50 ft depth) = 975 mmHg
V₂ = final volume (to be determined)

First, let's convert the units:
1 atm = 760 mmHg
So, 975 mmHg = 975/760 atm

Now, we can plug in the known values into Boyle's Law equation:
(1.0 atm)(4.9 L) = (975/760 atm)(V₂)

Simplifying the equation:
4.9 L = (1.2816 atm)(V₂)

Dividing both sides by 1.2816 atm:
V₂ = 4.9 L / 1.2816 atm

Evaluating the expression:
V₂ = 3.82 L

Therefore, the volume of the scuba diver's lungs at a depth of 50 ft is approximately 3.82 L.

Answer 4

The net pressure on lungs is outside-inside. Exactly what pressure is on the inside? Normally, on scuba diving, the presssure regulator keeps the inside pressure the same as outside, so net pressure is nil, and the lungs do not collapse. One does not want lungs to collapse.