Question

Knowing that the atmospheric pressure at the surface of the lake is 1.013x105 N/m2, what is its depth if an air bubble doubles in volume when it rises from the bottom which is at T = 15.0°C to the top where T = 28.0 °C?

Answer 1

Using Ideal gas law:

PV = nRT

When air bubble rises from the bottom to the top then

PV/T = Constant, So

P1*V1/T1 = P2*V2/T2

P1 = Pressure at the surface of lake = 1.013*10^5 N/m^2

V1 = Volume of bubble at the surface

V2 = Volume of bubble at the bottom,

given that V1 = 2*V2

T1 = Temperature at the top = 28.0 C = 301 K

T2 = Temperature at the bottom = 15.0 C = 288 K

So,

P2 = Pressure at the bottom = P1*V1*T2/(T1*V2)

P2 = 1.013*10^5*2*V2*288/(301*V2)

P2 = 1.013*10^5*2*288/(301)

P2 = 193849.83 N/m^2

Now We also know that

Pressure at the bottom = Pressure at the top + rho*g*H

P2 = P1 + rho*g*H

rho = density of water = 1000 kg/m^3

So,

H = (193849.83 - 1.013*10^5)/(1000*9.81)

H = 9.43 m = Initial depth of the bubble

Let me know if you've any query.