A scuba diver has an air tank with a volume of \(0.010 \mathrm{m}^{3}\) The air in the tank is initially at a pressure of \(1.0 \times 10^{7} \mathrm{Pa}\) Assuming that the diver breathes \(0.500 \mathrm{L} / \mathrm{s}\) of air, find how long the tank will last at depths of (a) \(2.0 \mathrm{m}\) and (b) $20.0 \mathrm{m} .\( (Make the same assumptions as in Example \)13.6 .)$

First, we need to convert the diver's breathing rate to a more convenient unit. The given breathing rate is 0.500 L/s. To convert L to m³, we use the following conversion factor: 1 L = 0.001 m³ So, the breathing rate in cubic meters per second is: (0.500 L/s) * (0.001 m³/L) = 0.0005 m³/s

The pressure at the surface is given as \(1.0 \times 10^{7}\,\mathrm{Pa}\). When the diver descends to a certain depth, the hydrostatic pressure of the water will increase the pressure in the air tank. The absolute pressure at each depth can be calculated using the following formula: \(P_{\mathrm{abs}}=P_{\mathrm{surface}} + \rho \cdot g\cdot d\) where \(\rho\) is the density of the water, \(g\) is the acceleration due to gravity, and \(d\) is the depth in meters. At 2.0 m depth: \(P_{\mathrm{abs1}}= 1.0 \times 10^{7}\,\mathrm{Pa} + 1000\,\mathrm{kg/m³} \cdot 9.81\,\mathrm{m/s²} \cdot 2.0\,\mathrm{m} = 1.0196 \times 10^{7}\,\mathrm{Pa}\) At 20.0 m depth: \(P_{\mathrm{abs2}}= 1.0 \times 10^{7}\,\mathrm{Pa} + 1000\,\mathrm{kg/m³} \cdot 9.81\,\mathrm{m/s²} \cdot 20.0\,\mathrm{m} = 1.1962 \times 10^{7}\,\mathrm{Pa}\)

Now that we have the absolute pressure at each depth, we can use Boyle's law to find the volume of the air inside the tank. Since the temperature remains constant, the following relationship holds: \(P_1V_1 = P_2V_2\) where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the pressure and volume at a particular depth. At 2.0 m depth: \(V_{2.0} = \frac{P_1V_1}{P_{\mathrm{abs1}}} = \frac{(1.0 \times 10^{7}\,\mathrm{Pa})(0.010\,\mathrm{m^3})}{1.0196 \times 10^{7}\,\mathrm{Pa}} = 0.009813\,\mathrm{m^3}\) At 20.0 m depth: \(V_{20.0} = \frac{P_1V_1}{P_{\mathrm{abs2}}} = \frac{(1.0 \times 10^{7}\,\mathrm{Pa})(0.010\,\mathrm{m^3})}{1.1962 \times 10^{7}\,\mathrm{Pa}} = 0.008360\,\mathrm{m^3}\)

Finally, we can use the available volume of air in the tank at each depth and the breathing rate to determine how long the tank will last. At 2.0 m depth: \(t_{2.0} = \frac{V_{2.0}}{\text{breathing rate}} = \frac{0.009813\,\mathrm{m^3}}{0.0005\,\mathrm{m^3/s}} = 19.626\,\mathrm{s}\) At 20.0 m depth: \(t_{20.0} = \frac{V_{20.0}}{\text{breathing rate}} = \frac{0.008360\,\mathrm{m^3}}{0.0005\,\mathrm{m^3/s}} = 16.720\,\mathrm{s}\) So, the tank will last approximately 19.63 seconds at a depth of 2.0 meters and approximately 16.72 seconds at a depth of 20.0 meters.