To solve this problem, we need to use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at a constant temperature.
First, we need to calculate the pressures at the two depths given: 1.0 m and 10.0 m.
At a depth of 1.0 m, we need to consider the atmospheric pressure on top of the water. The formula to calculate the pressure at this depth is:
p2 = p1 + ρgh1 + p(atm)
where p2 is the pressure at depth, p1 is the initial pressure (1.0x10^7 Pa), ρ is the density of water (1000 kg/m^3), g is the acceleration due to gravity (9.8 m/s^2), h1 is the depth (1.0 m), and p(atm) is the atmospheric pressure (101325 Pa).
Substituting the values into the formula, we get:
p2 = 1.0x10^7 + 1000*9.8*1 + 101325 = 1.011x10^7 Pa
At a depth of 10.0 m, we use the same formula but with h2 = 10.0 m:
p3 = 1.0x10^7 + 1000*9.8*10 + 101325 = 1.019x10^7 Pa
Now, we can apply Boyle's law to calculate how long the tank will last at each depth. The formula is:
p1 * V1 = p2 * V2
where p1 is the initial pressure (1.0x10^7 Pa), V1 is the volume of the tank (0.010 m^3), p2 is the pressure at the given depth, and V2 is the final volume of air consumed.
Solving for V2, we get:
V2 = (p1 * V1) / p2
For the depth of 1.0 m, plugging in the values, we have:
V2 = (1.0x10^7 * 0.010) / 1.011x10^7 = 9.89x10^-3 m^3
Now, we calculate the time it takes to consume this volume of air given a breathing rate of 0.400 L/s:
t = V2 / (0.400 * 10^-3) = 24.7 s = 0.41 min (rounded to 2 decimal places)
For the depth of 10.0 m, we follow the same steps:
V2 = (1.0x10^7 * 0.010) / 1.019x10^7 = 9.81x10^-3 m^3
t = V2 / (0.400 * 10^-3) = 24.5 s = 0.41 min (rounded to 2 decimal places)
Therefore, the tank will last approximately 0.41 minutes at both depths, based on the given breathing rate.