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 tank will last at different depths, we need to calculate the rate at which the air is consumed by the diver and then divide the initial volume of the tank by that consumption rate.
First, let's convert the volume of the tank from cubic meters to liters:
0.010 m^3 = 0.010 x 1000 L = 10 L
Now, let's calculate the consumption rate of the diver using the given information: the diver breathes 0.400 L/s.
(a) Depth of 1.0 m:
To calculate the consumption rate at this depth, we need to consider the increase in pressure due to the water column.
At a depth of 1.0 m, the pressure is the sum of the atmospheric pressure (1 atm = 1.01 x 10^5 Pa) and the pressure due to the water column (ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth).
ρ = 1000 kg/m^3 (density of water)
g = 9.8 m/s^2
h = 1.0 m
Pressure due to water column = ρgh = 1000 x 9.8 x 1.0 = 9800 Pa
Total pressure at a depth of 1.0 m = atmospheric pressure + pressure due to water column
= 1.01 x 10^5 + 9800 = 1.108 x 10^5 Pa
Now, let's find the new consumption rate:
Using Boyle's Law, we know that the volume of gas is inversely proportional to pressure when temperature remains constant.
Initial pressure (P1) = 1.0 x 10^7 Pa
Initial volume (V1) = 10 L
Final pressure (P2) = 1.108 x 10^5 Pa
Final volume (V2) = ?
Using Boyle's Law, we can write P1V1 = P2V2 and solve for V2:
V2 = (P1V1) / P2 = (1.0 x 10^7 x 10) / (1.108 x 10^5) = 90090.09 L
Now, let's calculate the consumption rate at this depth:
Consumption rate = 0.400 L/s
Time the tank will last = V2 / consumption rate = 90090.09 / 0.400 = 225,225.23 s
But we want the time in minutes, so let's convert seconds to minutes:
225,225.23 s = 225,225.23 / 60 min = 3753.75 min
Therefore, the tank will last approximately 3753.75 minutes at a depth of 1.0 m.
(b) Depth of 10.0 m:
Using the same steps as above, we can calculate the new consumption rate and the time the tank will last at this depth.
Pressure due to water column = ρgh = 1000 x 9.8 x 10.0 = 98000 Pa
Total pressure at a depth of 10.0 m = atmospheric pressure + pressure due to water column
= 1.01 x 10^5 + 98000 = 2.018 x 10^5 Pa
New final pressure (P2) = 2.018 x 10^5 Pa
V2 = (P1V1) / P2 = (1.0 x 10^7 x 10) / (2.018 x 10^5) = 49505.96 L
Time the tank will last = V2 / consumption rate = 49505.96 / 0.400 = 123,764.90 s
Converting seconds to minutes:
123,764.90 s = 123,764.90 / 60 min = 2062.75 min
Therefore, the tank will last approximately 2062.75 minutes at a depth of 10.0 m.