Answer:
V = 20.67 cm³
Explanation:
In this case, let's apply the Boyle's law which is:
P1V1 = P2V2
Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.
By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²
We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:
P1 = Po + dgh
Where:
Po: innitial pressure, which we can assume is 1 atm
d = density of the substance, in this case, water (1000 kg/m³)
g = gravity (9.8 m/s²)
h = distance of the bubble from the surface (115 m)
Now replacing this data in the boyle's law we have the following:
P1V1 = P2V2
V2 = P1V1/P2
V2 = (Po + dgh) * V1 / P2
Replacing the data we have:
V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5
V2 = 2,087,600 / 1.01x10^5
V2 = 20.67 cm³
