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!
calculate the pressures at those depths.
Then, use the Boyle's law (constant temp).
Ptank*volumetank=pressuredepth*n*.4Liters
volume of the tank in liters is 10liters.
solve for n. On the pressure at depth, be certain to add atmospheric pressure which is on top of the water.
p1•V1 = p2•V2
p2 = p1+ ρ•g•h1 + p(atm) = 10^7 +1000•9.8•1 + 101325 = 1.011•10^7 Pa,
V2 = p1•V1/p2 = 0.01•10^7/1.011•10^7 =9.89•10^-3 m^3
t = V2/Vo = 9.89•10^-3/0.4•10^-3 =24.7 s.
p1•V1 = p3•V3,
p2 = p1+ ρ•g•h2 + p(atm) = 10^7 +1000•9.8•10 + 101325 = 1.019•10^7 Pa,
V3 = p1•V1/p3 = 0.01•10^7/1.019•10^7 =9.81•10^-3 m^3
t = V3/Vo = 9.81•10^-3/0.4•10^-3 =24.5 s.
thanks Elena I really appreciate your help, but the answer has to be in minute and I got .42 min for 24.7 s, and .41 min for 24.5 s, but its wrong I don;t know why can you please help.
Thanks :)
To calculate how long the air tank will last at a certain depth, we need to determine the rate at which the diver uses air and the total amount of air in the tank.
First, let's convert the volume of the tank to liters:
0.010 m^3 * (1000 L/1 m^3) = 10 L
The rate at which the diver breathes air is given as 0.400 L/s.
(a) To calculate how long the tank will last at a depth of 1.0 m, we need to determine the change in pressure.
The change in pressure with depth can be calculated using the formula:
∆P = ρgh
Where:
∆P is the change in pressure,
ρ is the density of water (1000 kg/m^3),
g is the acceleration due to gravity (9.8 m/s^2), and
h is the depth in meters.
∆P = (1000 kg/m^3) * (9.8 m/s^2) * (1.0 m) = 9800 Pa
The final pressure in the tank can be calculated by adding the change in pressure to the initial pressure:
Final pressure = 1.0 x 10^7 Pa + 9800 Pa
Now, we can determine the volume of air consumed at this depth:
Volume consumed = (0.400 L/s) * t
Setting the initial volume minus the volume consumed equal to zero, we can solve for time (t):
10 L - (0.400 L/s) * t = 0
Solving for t:
t = 10 L / (0.400 L/s) = 25 s
To convert the time to minutes, divide by 60:
t = 25 s / 60 = 0.417 min
Therefore, the air tank will last approximately 0.417 minutes (or 25 seconds) at a depth of 1.0 m.
(b) To calculate how long the tank will last at a depth of 10.0 m, follow the same steps as in part (a), but with a different depth:
∆P = (1000 kg/m^3) * (9.8 m/s^2) * (10.0 m) = 98000 Pa
Final pressure = 1.0 x 10^7 Pa + 98000 Pa
t = 10 L / (0.400 L/s) = 25 s
t = 25 s / 60 = 0.417 min
Therefore, the air tank will also last approximately 0.417 minutes (or 25 seconds) at a depth of 10.0 m.