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.

 

[Nate810]

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

 

[Moetasim]

A) To determine the depth of the diver, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and inversely proportional to its volume. Since the temperature remains constant, we can use the equation P1V1 = P2V2, where P1 and V1 are the pressure and volume at the depth of the diver, and P2 and V2 are the pressure and volume at the surface.

We know that the volume of a sphere is (4/3)(Pi)(r^3), so we can set up the equation as follows:

(Pair)(4/3)(Pi)(1.8^3) = (Pair)(4/3)(Pi)(2.6^3)

Solving for the depth, we get:

Depth = (Pair)(1.8^3)/(Pair)(2.6^3) = 0.52 m

Therefore, the depth of the diver is 0.52 m.

B) To determine the absolute pressure at this depth, we can use the equation P = Pair + Psea water + Pdepth, where Pair is the pressure of the air at the surface, Psea water is the pressure of the sea water at the surface, and Pdepth is the pressure due to the depth of the diver.

We know that Pair = 1 atm = 101325 Pa, and Psea water = (1025 kg/m^3)(9.81 m/s^2)(0.52 m) = 5280 Pa.

To find Pdepth, we can use the equation Pdepth = (Pair)(1.8^3)/(2.6^3) = (101325 Pa)(1.8^3)/(2.6^3) = 44935 Pa.

Therefore, the absolute pressure at this depth is:

P = 101325 Pa + 5280 Pa + 44935 Pa = 151540 Pa

Therefore, the absolute pressure at this depth is 151540 Pa.