Physics please someone help me
posted by Romy .
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
(b) 10.0 m
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).
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.