Question

A scuba diver creates a spherical bubble with a radius of 3.0 cm at a depth of 30.0 m where the total pressure (including atmospheric pressure) is 4.00 atm. Part A What is the radius of the bubble when it reaches the surface of the water? (Assume atmospheric pressure to be 1.00 atm and the temperature to be 298 K.)

Answer 1

Answer:

Radius of the bubble reaches the surface of the water = 4.68 cm

Explanation:

We know Ideal gas equation PV = nRT

Since amount of gas and temperature remains constant

so p₁v₁ = p₂v₂

Volume of Spherical bubble with a radius of 3.0 cm

⇒ v₁= $\frac{4}{3}$$\pi$$r^{3}$

⇒ v₁ = $\frac{4}{3}$ x 3.14 x $(3)^{3}$

Pressure at the depth p₁ = 4 atm

Volume of bubble when it reaches the surface of water =  $\frac{4}{3}$$\pi$$r^{3}$

Pressure p₂ = 1 atm

      p₁v₁ = p₂v₂

⇒  4 x  $\frac{4}{3}$ x 3.14 x $(3)^{3}$ =  1 x  $\frac{4}{3}$$\pi$$r^{3}$

⇒ r = 4.68 cm