A scuba diver has an air tank with a volume of 0.010 m^3. The air in the tank is initially at a pressure of 1.0x10^7 Pa. Assume that the diver breathes 0.400 L/s of air. Find how long the tank will last at a depth of each of the following.
(a) 1.0 m
min
(b) 10.0 m
Please someone help me on this I don't know how to do this. thank you.
min
To solve this problem, we need to use the principles of Boyle's Law and the Ideal Gas Law.
Boyle's Law states that at a constant temperature, the pressure of a gas is inversely proportional to its volume. Mathematically, this can be expressed as:
P1 * V1 = P2 * V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
The Ideal Gas Law states that the pressure of a gas is directly proportional to its temperature and the number of moles of gas, and inversely proportional to its volume. Mathematically, it can be expressed as:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.
In this problem, we are given the initial volume V1 of the air tank, the initial pressure P1, and the rate at which the diver breathes air. We need to determine how long the tank will last at different depths.
(a) At a depth of 1.0 m:
To determine the final pressure P2, we can use the equation:
P2 = P1 + ρgh
where ρ is the density of the fluid (water) and g is the acceleration due to gravity. Substituting the given values:
P2 = 1.0x10^7 Pa + (1000 kg/m^3) * (9.8 m/s^2) * (1.0 m)
P2 = 1.0x10^7 Pa + 98000 Pa
P2 = 1.0098x10^7 Pa
Next, we can use Boyle's Law to find the final volume V2:
P1 * V1 = P2 * V2
(1.0x10^7 Pa) * (0.010 m^3) = (1.0098x10^7 Pa) * V2
Solving for V2:
V2 = (1.0x10^7 Pa * 0.010 m^3) / (1.0098x10^7 Pa)
V2 = 9.898x10^-3 m^3
Since the diver breathes air at a rate of 0.400 L/s, or 4.00x10^-4 m^3/s, we can calculate how long the tank will last:
Time = V2 / breathing rate
Time = (9.898x10^-3 m^3) / (4.00x10^-4 m^3/s)
Time ≈ 24.7 s
So, the tank will last approximately 24.7 seconds at a depth of 1.0 m.
(b) At a depth of 10.0 m:
Using the same process as in part (a), we can calculate the final pressure P2:
P2 = 1.0x10^7 Pa + (1000 kg/m^3) * (9.8 m/s^2) * (10.0 m)
P2 = 1.0x10^7 Pa + 980000 Pa
P2 = 1.098x10^7 Pa
Using Boyle's Law again, we can find the final volume V2:
V2 = (1.0x10^7 Pa * 0.010 m^3) / (1.098x10^7 Pa)
V2 ≈ 9.108x10^-3 m^3
Calculating the time the tank will last:
Time = V2 / breathing rate
Time = (9.108x10^-3 m^3) / (4.00x10^-4 m^3/s)
Time ≈ 22.8 s
So, the tank will last approximately 22.8 seconds at a depth of 10.0 m.