A diver 40 m deep in 10∘C fresh water exhales a 1.0-cm-diameter bubble.

What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20∘C? Assume that the air bubble is always in thermal equilibrium with the surrounding water.

Express your answer to two significant figures and include the appropriate units.

Applying gas law:

PV = nRT = > nR = PV/T

nR = Constant at all conditions and thus;

PV/T = Constant.

Therefore.

P1V1/T1 = P2V2/T2

Where P = Pressure, V = Volume, T = Temperature in Kelvin

State 1: At 40 m deep

P1 = Atmospheric pressure + water pressure = 101325 + rho*g*h = 101325+1000*9.81*40 = 493725 Pa

V1 = Volume of the bubble = 4/3*πd^3/64 = π*1^2/4 = 0.0654 cm^3

T1 = 10+273.15 = 283.15 K

State 2: At the surface of the lake

P2 = Atmospheric pressure = 101325 Pa

T2 = 20+273.15 = 293.15 K

V2 = ?

Solving for V2

V2 = (T2*P1V1) / (T1*P2) = (293.15*493725*0.0654) / (283.15*101325) = 0.3302 cm^3

But,

V2 = 4/3*πD^3/64 = > D = cube root (3/4*V2*64) = 2.512 cm

The diameter of the bubbles as it reaches the surface is 2.5 cm rounded to two significant figures.