Question

An air bubble has a volume of 2.5 cm3 when it is released at a depth of 40 m of water. What will be its volume when it reaches the surface of the water? Assume that its temperature is constant and that atmospheric pressure = 10 m of water.

Answer 1

We assume that its temperature is constant. Applying Boyle's law:

$$PV = constant$$

$$(P + Hρg)V = P V'$$

$$V'=V(1+\frac{Hρg}{P})$$

We have:

$$P=hρg$$

Thus,

 

$$V'=V(1+\frac{Hρg}{hρg})=V(1+\frac{H}{h})$$

$$V'=2.5(1+\frac{40}{10})=12.5 cm^3$$