Question

A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while quietly rising to the surface, what will the volume of the lungs be (in liters) at the surface? Assume the temperature is the same at all depths. (The density of water is 1.0 x 103 kg/m3.)

Answer 1

To solve this problem it is necessary to apply Boyle's law in which it is specified that

$P_1V_1 =P_2 V_2$

Where,

$P_1$ and $V_1$ are the initial pressure and volume values

$P_2$ and $V_2$ are the final pressure volume values

The final pressure here is the atmosphere, then

$P_2 = 101325 \approx 1*10^5Pa$

$h = 10m$

$\rho_w = 1000kg/m^3$

$V_1 = 3.0L$

Pressure at the water is given by,

$P_1 = P_2 -\rho gh$

$P_1 = 1*10^5 +1000*9.8*10 =198000Pa$

Using Boyle equation we have,

$V_2 = \frac{P_1V_1}{P_2}$

$V_2 = \frac{198000*3*10^5}{10^5}$

$V_2 = 5.9L$

Therefore the volume of the lungs at the surface is 5.9L