How Does Bubble Size Change with Depth in Fluid Mechanics?

[cat_eyes]

An air bubble originating from a deep-sea diver has a radius of 1.8 mm at the depth of the diver. When the bubble reaches the surface of the water, it has a radius of 2.6 mm. Assume that the temperature of the air in the bubble remains constant. The acceleration of gravity is 9.81 m/s^2.

A) Determine the depth of the diver. Answer in units of m.

B) Determine the absolute pressure at this depth. Answer in units of Pa.

P=rho(density)
Pair=1.29 kg/m^3
Psea water=1025 kg/m^3
The volume of a sphere = (4/3)(Pi)(r^3)

Any help with this would be greatly appreciated.

 

[member 553083]

A) The depth of the diver can be determined by using the equation P = Pair + (rho*g*h), where P is the pressure at the depth of the diver, Pair is the atmospheric pressure, rho is the density of sea water, g is the acceleration due to gravity, and h is the depth of the diver.Rearranging this equation and solving for h gives: h = (P - Pair)/(rho*g).Substituting in the given values, the depth of the diver is:h = (101325 Pa - 1.29 kg/m^3*9.81 m/s^2)/(1025 kg/m^3*9.81 m/s^2) = 97.8 mB) The absolute pressure at this depth can be determined by using the equation P = Pair + (rho*g*h), where P is the pressure at the depth of the diver, Pair is the atmospheric pressure, rho is the density of sea water, g is the acceleration due to gravity, and h is the depth of the diver.Substituting in the given values, the absolute pressure at this depth is:P = 1.29 kg/m^3*9.81 m/s^2 + 1025 kg/m^3*9.81 m/s^2*97.8 m = 101325 Pa