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 :) and sorry for reposting, but i really need help.
No one has answered this question yet.
t = p•V/p1•Vo,
Vo = 0.4•10^-3 m^3
p1= ρ•g•h1 + p (atm) =1000•9.8•1 + 101325 =1.11•10^5 Pa,
t = p•V/p1•Vo = 10^7•0.01/ 1.11•10^5 •0.4•10^-3 =2252 s =37.5 min.
p2= ρ•g•h2 + p (atm) =1000•9.8•10 + 101325 =1.99•10^5 Pa,
t = p•V/p2•Vo = 10^7•0.01/ 1.99•10^5 •0.4•10^-3 =1256 s =20.9 min.
To find out how long the tank will last at different depths, you need to calculate the pressures at those depths using the given information.
(a) At a depth of 1.0 m, you need to calculate the pressure at that depth. The pressure at a depth in a fluid can be calculated using the formula:
p2 = p1 + ρgh + patm
Where p2 is the pressure at the depth, p1 is the initial pressure (1.0x10^7 Pa), ρ is the density of the fluid (assume water with a density of 1000 kg/m^3), g is the acceleration due to gravity (assume 9.8 m/s^2), h is the depth (1.0 m), and patm is the atmospheric pressure, which is approximately 101325 Pa.
Plugging in the values, we get:
p2 = 1.0x10^7 + (1000)(9.8)(1.0) + 101325
p2 = 1.011x10^7 Pa
Now, using Boyle's law (assuming constant temperature):
p1•V1 = p2•V2
Here, p1 is the initial pressure, V1 is the initial volume (0.010 m^3), p2 is the pressure at the depth, and V2 is the volume of air the diver breathes per unit time.
Solving for V2, we get:
V2 = p1•V1 / p2
V2 = (1.0x10^7)(0.010) / (1.011x10^7)
V2 = 9.89x10^-3 m^3
Since the diver breathes 0.400 L/s of air, which is equivalent to 0.400x10^-3 m^3/s, we can calculate how long the tank will last by dividing the volume at the depth (9.89x10^-3 m^3) by the volume of air breathed per second (0.400x10^-3 m^3/s):
Time = V2 / V(breathed per second)
Time = 9.89x10^-3 / 0.400x10^-3 = 24.7 s
Therefore, at a depth of 1.0 m, the tank will last for approximately 24.7 seconds.
(b) The process is similar for a depth of 10.0 m.
First, calculate the pressure at that depth using the same formula as above:
p2 = p1 + ρgh + patm
p2 = 1.0x10^7 + (1000)(9.8)(10.0) + 101325
p2 = 1.019x10^7 Pa
Now, plug this pressure into the Boyle's law equation to find the volume at that depth:
V3 = p1•V1 / p2
V3 = (1.0x10^7)(0.010) / (1.019x10^7)
V3 = 9.81x10^-3 m^3
Dividing this volume by the volume breathed per second, we get:
Time = V3 / V(breathed per second)
Time = 9.81x10^-3 / 0.400x10^-3 = 24.5 s
Therefore, at a depth of 10.0 m, the tank will last for approximately 24.5 seconds.
To convert these times from seconds to minutes, you simply divide the time in seconds by 60 (since there are 60 seconds in a minute).
Time in minutes = Time in seconds / 60
For the calculations above:
(a) Time in minutes = 24.7 s / 60 = 0.41 min (approximately)
(b) Time in minutes = 24.5 s / 60 = 0.41 min (approximately)
So, at a depth of 1.0 m, the tank will last for approximately 0.41 minutes, and at a depth of 10.0 m, the tank will also last for approximately 0.41 minutes.