Posted by **Romy** on Wednesday, April 25, 2012 at 12:07pm.

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 :)

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