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 :)
Oh no, it seems like there may have been a miscalculation. Let's do some math and figure out the correct answer!
(a) At a depth of 1.0 m, the pressure would be:
p2 = p1 + ρ•g•h1 + p(atm) = 1.0x10^7 + 1000•9.8•1 + 101325 = 1.0203x10^7 Pa
Using Boyle's law, we can calculate the new volume of the tank:
V2 = p1•V1 / p2 = 0.010•10^7 / 1.0203•10^7 = 0.0098039 m^3
To find the time the tank will last, we divide the new volume of the tank by the breathing rate:
t = V2 / 0.400 L/s = (0.0098039 m^3) / (0.0004 m^3/s) = 24.51 seconds
Now let's convert seconds to minutes:
t = 24.51 s * (1 min / 60 s) ≈ 0.409 minutes
So, the tank will last approximately 0.409 minutes at a depth of 1.0 m.
(b) At a depth of 10.0 m, the pressure would be:
p2 = p1 + ρ•g•h2 + p(atm) = 1.0x10^7 + 1000•9.8•10 + 101325 = 1.0191x10^7 Pa
Using Boyle's law again, we can calculate the new volume of the tank:
V3 = p1•V1 / p3 = 0.010•10^7 / 1.0191•10^7 = 0.009817 m^3
Dividing the new volume by the breathing rate, we get:
t = V3 / 0.400 L/s = (0.009817 m^3) / (0.0004 m^3/s) = 24.5425 seconds
Converting to minutes:
t = 24.5425 s * (1 min / 60 s) ≈ 0.409 minutes
So, the tank will last approximately 0.409 minutes at a depth of 10.0 m.
I hope this helps! But hey, if you're diving deeper than that, maybe it's time to take up a different hobby, like clown college! 😄🤡
To find out how long the tank will last at each depth, we first need to calculate the pressures at those depths.
(a) At a depth of 1.0 m, the pressure can be calculated using the formula p2 = p1 + ρgh, where p1 is the initial pressure at the surface, ρ is the density of water, g is the acceleration due to gravity, and h is the depth.
Using values:
p1 = 1.0x10^7 Pa
ρ = 1000 kg/m^3 (density of water)
g = 9.8 m/s^2 (acceleration due to gravity)
h = 1.0 m
p2 = 1.0x10^7 + 1000 * 9.8 * 1.0 = 1.009x10^7 Pa
Next, we can use Boyle's law to calculate the volume of air consumed at this pressure. Boyle's law states that the product of pressure and volume is constant at constant temperature.
p1 * V1 = p2 * V2
V2 = (p1 * V1) / p2
V1 = 0.010 m^3 (initial volume of the tank)
V2 = (1.0x10^7 * 0.010) / (1.009x10^7) = 9.890x10^-3 m^3
To convert this to minutes, we need to divide the volume by the rate at which the diver breathes air, which is 0.400 L/s.
V2 = 9.890x10^-3 m^3 * (1000 L/m^3) = 9.890 L
t = V2 / breathing rate = 9.890 L / 0.400 L/s = 24.725 s = 0.412 min (rounded to 3 decimal places)
Therefore, the tank will last approximately 0.412 minutes at a depth of 1.0 m.
(b) At a depth of 10.0 m, we repeat the same steps:
p2 = p1 + ρgh = 1.0x10^7 + 1000 * 9.8 * 10.0 = 1.019x10^7 Pa
V3 = (p1 * V1) / p3 = (1.0x10^7 * 0.010) / (1.019x10^7) = 9.810x10^-3 m^3
t = V3 / breathing rate = 9.810x10^-3 m^3 * (1000 L/m^3) / 0.400 L/s = 24.525 s = 0.409 min (rounded to 3 decimal places)
Therefore, the tank will last approximately 0.409 minutes at a depth of 10.0 m.
To find how long the tank will last at a certain depth, you need to calculate the pressures at those depths using the equation p2 = p1 + ρgh + patm, where p2 is the pressure at the depth, p1 is the initial pressure, ρ is the density of water, g is the acceleration due to gravity, h is the depth, and patm is the atmospheric pressure.
For part (a) at a depth of 1.0 m:
p2 = 1.0x10^7 + (1000 kg/m^3)(9.8 m/s^2)(1.0 m) + 101325 Pa
= 1.011x10^7 Pa
Next, use Boyle's law, which states that the product of the initial pressure and volume is equal to the product of the final pressure and volume, assuming constant temperature. In this case, we'll assume the volume of the tank remains constant.
p1•V1 = p2•V2
(1.0x10^7 Pa)(0.010 m^3) = (1.011x10^7 Pa)(V2)
V2 = (1.0x10^7 Pa)(0.010 m^3) / (1.011x10^7 Pa)
= 9.89x10^-3 m^3
Now, calculate how long the tank will last by dividing V2 by the rate at which the diver breathes air:
t = V2 / (0.400 L/s)
= (9.89x10^-3 m^3) / (0.400x10^-3 m^3/s)
= 24.7 s
Since the answer needs to be in minutes, convert seconds to minutes:
t = 24.7 s / 60 s/min
= 0.41 min
So, the tank will last approximately 0.41 minutes at a depth of 1.0 m.
For part (b) at a depth of 10.0 m:
Proceed similarly using the provided formulas and values, and you should find that the tank will last approximately 0.40 minutes at a depth of 10.0 m.
Please note that these calculations assume idealized conditions and do not account for factors such as changes in temperature or breathing patterns. It's always important to consult relevant diving guidelines and safety protocols for accurate estimates and decision-making.