Answer
0.205
Work Step by Step
Let's take,
*The pressure of the air in the tank when the tank is at the = $P_{S}$
surface of the water
*The volume of the air in the tank when the tank is at the = $V_{S}$
surface of the water
*The pressure of the air in the tank when the tank has been = $P_{h}$
lowered a distance h below the surface of the water
*The volume of the air in the tank when the tank has been = $V_{h}$
lowered a distance h below the surface of the water
Let's apply the Boyle's law $P_{1}V_{1}=P_{2}V_{2}$ to the air in the tank.
$P_{S}V_{S}=P_{h}V_{h}=>\frac{V_{h}}{V_{S}}=\frac{P_{S}}{P_{h}}$
Let's plug known values into this equation.
$\frac{V_{h}}{V_{S}}=\frac{1.01\times10^{5}Pa}{1.01\times10^{5}Pa+1000\space kg/m^{3}(9.8\space m/s^{2})(40\space m)}=0.205$
Therefore the desired volume fraction = 0.205
