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A diver rises quickly to the surface from a 5.0 m depth. If she did not exhale the gas from her lunds before rising, by what factor would her lungs expand? Assume the temperature to be constant and the pressure in the lungs to match the pressure outside the diver's body. The density of seawater is 1.03x!0^3 kg.

Respuesta :

Answer:

1.5 times

Explanation:

[tex]h[/tex] = depth of the diver initially = 5 m

[tex]\rho[/tex] = density of seawater = 1030 kg m⁻³

[tex]P_{i}[/tex] = Initial pressure at the depth

[tex]P_{f}[/tex] = final pressure after rising = 101325 Pa

Initial pressure at the depth is given as

[tex]P_{i} = P_{f} + \rho gh\\P_{i} = 101325 + (1030) (9.8) (5)\\P_{i} = 151795 Pa[/tex]

[tex]V_{i}[/tex] = Initial volume at the depth

[tex]V_{f}[/tex] = Final volume after rising

Since the temperature remains constant, we have

[tex]P_{f} V_{f} = P_{i} V_{i}\\(101325) V_{f} = (151795) V_{i}\\V_{f} = 1.5 V_{i}[/tex]

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