Question

An air bubble is 2.81 mm in diameter at depth 96.8 meters below the surface of a lake, temperature 1.44 degrees C. What is the bubble diameter at 3.25 meters below the surface, assuming no change in temperature, also in millimeters.

Answer 1

it is an isothermal process since  no change in temperature says the question

PV = nRT = constant  

P₁V₁ = P₂V₂

P = ρ gh

V₂ /V₁ =P₁ /P₂ =ρ gh₁ /ρ gh₂

V₂ /V₁ = h₁ /h₂ = 96.8/3.25

V₂ /V₁ = 4/3*π*r₂³ / 4/3*π*r₁³ = r₂³ /*r₁³

V₂ /V₁ = r₂³ /*r₁³ =  h₁ /h₂ = 96.8/3.25

r₂³ /*r₁³ =  96.8/3.25

r₂³ = 96.8/3.25 * r₁³

D₂³ /2³ = 96.8/3.25 * D₁³ /2³

D₂³ = 96.8/3.25 * D₁³    = 96.8/3.25*2.81³

=660.862267 mm³

D₂ = (660.862267)^1/3 =8.71037766 mm answer