## At the depth of 10 m, the diver breathes in air from the scuba tank, filling her lungs with air whose pressure equals that of the surroundin

Question

At the depth of 10 m, the diver breathes in air from the scuba tank, filling her lungs with air whose pressure equals that of the surrounding water. How does the number of air molecules in her lungs compare with the number she had in a full breath at the surface?

A. Half as many

B. The same

C. Twice as many

D. Four times as many

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2 weeks 2021-08-30T03:28:34+00:00 1 Answers 0 views 0

The number of air molecules in her lung compared with the number she had in a full breadth at the surface is;

C. Twice as many

Explanation:

The depth at which the scuba diver fills her lungs with air, d = 10 m

Pressure of fluid = Density, ρ × Acceleration due to gravity, ‘g’ × Depth, ‘d’

Where;

ρ = 970 kg/m³

g ≈ 9.81 m/s²

∴ The pressure of the water of the 10 m water column = 970 kg/m³ × 9.81 m/s² × 10 m = 95,157 Pa

The atmospheric pressure, P₁ = 101,325 Pa

The total pressure at the 10 m depth = The atmospheric pressure + The pressure of the water of the 10 m water column

∴ The total pressure at the 10 m depth, P₂ = 95,157 Pa + 101,325 Pa = 196,482 Pa

By the combined gas law, we have;

P₁·V₁/n₁ = P₂·V₂/n₂

Where;

P₁ = The atmospheric pressure

P₂ = The pressure at 10 m depth

V₁ = V₂ = The volume of her lungs

n₁ = The number of air molecules at the surface that fills her lungs

n₂ = The number of air molecules that fills her lungs at the 10 m depth

Therefore, we have;

P₂/P₁ = n₂/n₁ × V₁/V₂

From V₁ = V₂, we have;

P₂/P₁ = n₂/n₁ × V₁/V₁ = n₂/n₁ × 1

∴ P₂/P₁ = n₂/n₁

P₂/P₁ = 196,482/101,325 ≈ 2

∴ P₂/P₁  = n₂/n₁  ≈ 2

n₂ ≈ 2 × n₁

The number of air molecules that fills her lungs at the 10 m depth, n₂ is approximately 2 times (twice) as many as the number of air molecules at the surface that fills her lungs, n₁

Therefore the correct option is twice as many molecules as she had in a full breath at the surface is required to fill her lungs at the 10 m water depth.