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
min
Please someone help me out with this. Thank you!
To find out how long the air tank will last at different depths, we need to calculate the total volume of air that the diver breathes per minute and then divide the volume of air in the tank by the rate of air consumption.
Given:
Volume of air tank (V) = 0.010 m^3
Initial pressure (P_1) = 1.0x10^7 Pa
Rate of air consumption (R) = 0.400 L/s
Depth (d) = 1.0 m and 10.0 m
Step 1: Convert the rate of air consumption to m^3/min
1 L = 0.001 m^3
0.400 L/s = 0.400 * (1/60) m^3/min
R = 0.400/60 m^3/min
Step 2: Calculate the final pressure (P_2) at each depth using Boyle's law.
Boyle's law states that P_1 * V_1 = P_2 * V_2, where V_1 is the initial volume and V_2 is the final volume.
V_2 = V_1 + (V_1 / g) * d, where g is the acceleration due to gravity (9.81 m/s^2)
P_2 = (P_1 * V_1) / V_2
For depth of 1.0 m,
V_2 = 0.010 m^3 + (0.010 m^3 / 9.81 m/s^2) * 1.0 m
V_2 = 0.020 m^3
P_2 = (1.0x10^7 Pa * 0.010 m^3) / 0.020 m^3
P_2 = 5.0x10^6 Pa
For depth of 10.0 m,
V_2 = 0.010 m^3 + (0.010 m^3 / 9.81 m/s^2) * 10.0 m
V_2 = 0.110 m^3
P_2 = (1.0x10^7 Pa * 0.010 m^3) / 0.110 m^3
P_2 = 9.1x10^5 Pa
Step 3: Calculate the time (T) the tank will last at each depth using the formula:
T = V / (R * P_2)
For depth of 1.0 m,
T = 0.010 m^3 / (0.400/60 m^3/min * 5.0x10^6 Pa)
T ≈ 72 min
For depth of 10.0 m,
T = 0.010 m^3 / (0.400/60 m^3/min * 9.1x10^5 Pa)
T ≈ 624 min
Therefore, the air tank will last approximately 72 minutes at a depth of 1.0 m and 624 minutes at a depth of 10.0 m.
To calculate how long the tank will last at different depths, we need to consider that the pressure increases with depth. The formula relating pressure to depth is given by:
P = P0 + ρgh
Where:
P is the pressure at depth,
P0 is the initial pressure,
ρ is the density of water,
g is the acceleration due to gravity, and
h is the depth.
The density of water is approximately 1000 kg/m^3, and the acceleration due to gravity is approximately 9.8 m/s^2.
Let's calculate the pressure at each depth and then determine the time the tank will last:
(a) At a depth of 1.0 m:
P1 = P0 + ρgh
= 1.0x10^7 Pa + (1000 kg/m^3)(9.8 m/s^2)(1.0 m)
= 1.0x10^7 Pa + 9800 Pa
= 1.0x10^7 Pa + 9.8x10^3 Pa
= 1.0098x10^7 Pa
To find how long the tank will last at this pressure, we need to calculate the amount of air used per second:
0.400 L/s = 0.0004 m^3/s
To find the amount of time, we divide the volume of the tank by the amount of air used per second:
Time1 = (0.010 m^3) / (0.0004 m^3/s)
= 25 s = 0.417 min
Therefore, the tank will last approximately 0.417 minutes at a depth of 1.0 m.
(b) At a depth of 10.0 m:
P2 = P0 + ρgh
= 1.0x10^7 Pa + (1000 kg/m^3)(9.8 m/s^2)(10.0 m)
= 1.0x10^7 Pa + 98000 Pa
= 1.0x10^7 Pa + 9.8x10^4 Pa
= 1.098x10^7 Pa
Again, we calculate the amount of air used per second:
0.400 L/s = 0.0004 m^3/s
To find the amount of time, we divide the volume of the tank by the amount of air used per second:
Time2 = (0.010 m^3) / (0.0004 m^3/s)
= 25 s = 0.417 min
Therefore, the tank will last approximately 0.417 minutes at a depth of 10.0 m.