Anónimo
Anónimo preguntado en Science & MathematicsPhysics · hace 1 mes

An air bubble with a diameter of 0.010 m is released underwater at a depth of 15 m. ?

(There is an increase of 1 atm for every 10.3 m in depth in water; for a sphere, v = × 3.14 × r3. There is already 1 atm at the surface.)

How large will the air bubble’s diameter be at the surface?

2 respuestas

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  • hace 1 mes

    At a depth of 15 m: P₁ = 1 + 1×(15/10.3) = 2.46 atm, V₁ = (4/3)π(0.010/2)³ m³

    At the surface: P₂ = 1 atm, V₂ = (4/3)π(d₂/2)³

    Boyle's law: P₁V₁ = P₂V₂

    2.46 × (4/3)π(0.010/2)³ = 1 × (4/3)π(d₂/2)³

    2.46 × 0.010³ = 1 × d₂³

    d₂ = ³√(2.46 × 0.010³)

    Bubble's diameter at the surface, d₂ = 0.0135 m

  • hace 1 mes

    PV = nrT so that V is inversely proportional to P

    Hence V2/V1 = P1/P2  if P1 is the pressure at depth then V2 = V1 * P1/P2  The diameter is proportional to the cube root of the volume so R2 = R1 * (P1/P2)^(1/3) = 0.010 *((1+15/10.3)/1)^(1/3)  = 0.0135 m

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