To find the volume of the bubble at the surface, we can use Boyle's Law which states that the pressure and volume of a gas are inversely proportional at constant temperature. We can set up the equation:
P1V1 = P2V2
Where:
P1 = pressure at 200m depth
V1 = volume at 200m depth
P2 = pressure at the surface
V2 = volume at the surface
First, we need to find the pressure at 200m depth using the formula:
P = P0 + ρgh
Where:
P = pressure at depth
P0 = atmospheric pressure at the surface (1x105 Pa)
ρ = density of water (1000 kg/m3)
g = acceleration due to gravity (9.8 m/s2)
h = depth (200 m)
P = 1x105 + 1000 x 9.8 x 200
P = 1x105 + 1960000
P = 2060000 Pa
Now, we can calculate the volume of the bubble at the surface:
P1V1 = P2V2
2060000 x 1.5 = 1x105 x V2
V2 = (2060000 x 1.5) / 1x105
V2 = 30.9 cm3
Therefore, the volume of the bubble at the surface is 30.9 cm3.
A submarine releases an air bubble with a volume of 1.5 cm3 from 200 m below the surface of the ocean. How large is the bubble when it reaches the surface? Assume the temperature changes from 22° C to 24° C between 200 m deep and the surface. (The density of water is 1000 kg/m3 and atmospheric pressure is 1x105 Pa.)
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