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!
physics someone please helpp - bobpursley, Wednesday, April 25, 2012 at 12:12pm
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.
physics someone please helpp - Elena, Wednesday, April 25, 2012 at 3:33pm
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.
physics to Elena please reply - Romy, Wednesday, April 25, 2012 at 6:36pm
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 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.