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 find the amount of time the air tank will last at a given depth, we need to calculate the pressure at that depth and then use Boyle's Law.
First, let's calculate the pressure at each depth using the equation:
pressure at depth = initial pressure + (density of water * acceleration due to gravity * depth) + atmospheric pressure
Given:
Initial pressure (p1) = 1.0x10^7 Pa
Density of water (ρ) = 1000 kg/m^3
Acceleration due to gravity (g) = 9.8 m/s^2
Depth at (a) = 1.0 m
Depth at (b) = 10.0 m
Atmospheric pressure (p(atm)) = 101325 Pa (standard atmospheric pressure)
(a) Depth of 1.0 m:
p2 = p1 + ρ * g * h1 + p(atm)
= 1.0x10^7 + 1000 * 9.8 * 1 + 101325
= 1.011x10^7 Pa
(b) Depth of 10.0 m:
p3 = p1 + ρ * g * h2 + p(atm)
= 1.0x10^7 + 1000 * 9.8 * 10 + 101325
= 1.019x10^7 Pa
Now, let's use Boyle's Law to determine the time the air tank will last at each depth.
Boyle's Law states that the product of initial pressure and initial volume is equal to the product of final pressure and final volume:
p1 * V1 = p2 * V2 (for depth 1.0 m)
p1 * V1 = p3 * V3 (for depth 10.0 m)
Given:
Initial volume (V1) = 0.010 m^3
Breathing rate (n) = 0.400 L/s
Volume for breathing rate (V_breath) = n * t (where t is time)
Rewriting the equations with volume for breathing rate and rearranging to solve for time:
V2 = p1 * V1 / p2
V_breath = V2 / t (where t is the time in seconds)
t = V2 / V_breath
(a) Depth of 1.0 m:
V2 = (1.0x10^7 * 0.010 m^3) / (1.011x10^7 Pa)
t = (V2 / V_breath)
= (V2 / (0.400 L/s * 1000 cm^3/L))
= (V2 / 400 cm^3/s)
= (V2 / 0.40 L/s)
Convert V2 from m^3 to L:
t = (V2 / 0.40 L/s) * (1000 L / 1 m^3)
≈ (9.89x10^-3 / 0.40) * (1000 L)
≈ 24.7 seconds
To convert seconds to minutes, divide by 60:
t ≈ 24.7 / 60 ≈ 0.41 minutes
(b) Depth of 10.0 m:
V3 = (1.0x10^7 * 0.010 m^3) / (1.019x10^7 Pa)
t = (V3 / V_breath)
= (V3 / (0.400 L/s * 1000 cm^3/L))
= (V3 / 400 cm^3/s)
= (V3 / 0.40 L/s)
Convert V3 from m^3 to L:
t = (V3 / 0.40 L/s) * (1000 L / 1 m^3)
≈ (9.81x10^-3 / 0.40) * (1000 L)
≈ 24.5 seconds
To convert seconds to minutes, divide by 60:
t ≈ 24.5 / 60 ≈ 0.41 minutes
Therefore, the air tank will last approximately 0.41 minutes at a depth of 1.0 m and 10.0 m.