Question
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.)
Answers
GPT 3.5
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.
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.