A scuba diver has his lungs filled to half capacity (2.27 liters) when 10 m belo

Question

A scuba diver has his lungs filled to half capacity (2.27 liters) when 10 m below the surface. If the diver holds his breath while quietly rising to the surface, what will the volume of the lungs be (in liters) at the surface? Assume the temperature is the same at all depths. (The density of water is 1.0 ´ 103 kg/m3.) a. 5.9   b. 4.5 c. 6.4 d. 3.9 e. 3.1
A scuba diver has his lungs filled to half capacity (2.27 liters) when 10 m below the surface. If the diver holds his breath while quietly rising to the surface, what will the volume of the lungs be (in liters) at the surface? Assume the temperature is the same at all depths. (The density of water is 1.0 ´ 103 kg/m3.) a. 5.9   b. 4.5 c. 6.4 d. 3.9 e. 3.1

Explanation / Answer

V1 = 3 Liters

d = density of water = 1 x 10^3 kg/m3

h = 10 meters

Assume that atmospheric pressure = 1 atm

Calculate the water pressure: d * g * h = (1 x 10^3) * 9.81 * 10m = 9.81 x 10^4 Pascals

convert water pressure to atms:

9.81 x 10^4 Pa x (1atm / 101325Pa) = 0.967 atm

P1 (Pressure at 10 meters) = atmospheric pressure + water pressure = 1 + 0.967 = 1.967 atm

P2 (Pressure at surface) = atmospheric pressure = 1 atm

From the ideal gas law, PV = nRT, since number of moles and temperature are held constant, we know that as P and V change when the diver rises to the surface, P1V1 will still be equal to P2V2. We can now solve for V2 to find our final volume.

P1V1 = P2V2

1.967 * 3L = 1 * V2

V2 = 5.9 L

The correct answer is A.

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