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1 October, 02:35

A balloon's internal pressure is 1.25 atm, and its volume is 2.50 L. If the balloon is taken to the bottom of the ocean (pressure = 95.0 atm), what would be its new volume? Assume that the temperature does not change. (Round to the nearest hundredth place) L

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  1. 1 October, 02:51
    0
    0.03 L

    Step-by-step explanation:

    Since all other properties constant, we can use Boyle's Law.

    p₁V₁ = p₂V₂ Divide each side by p₂

    V₂ = V₁ * p₁/p₂

    dа ta:

    p₁ = 1.25 atm; V₁ = 2.50 L

    p₂ = 95.0 atm; V₂ = ?

    Calculations:

    V₂ = 1.25 * (2.50/95.0)

    V₂ = 1.25 * 0.026 32

    V₂ ≈ 0.03 L
  2. 1 October, 03:01
    0
    Where temperature is constant P1V1 = P2V2 where P = pressure and V = volume.

    so 1.25 * 2.5 = 95 * V2

    V2 = (1.25 * 2.5) / 95

    = 0.03 L to the nearest hundredth ...
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