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Pressure and scuba

  1. Apr 9, 2007 #1
    A scuba diver at a depth of 43.0 m below the surface of the sea off the shores of Panama City, where the temperature is 4.0°C, releases an air bubble with volume 17.0 cm3. The bubble rises to the surface where the temperature is 19.0°C. What is the volume of the bubble immediately before it breaks the surface? The specific gravity for seawater is 1.025.


    so (p1v1/t1)=(p2v2/t2)

    v2=v1(p1/p2)(t2/t1)

    p1=101 kPa+ (roe)gh where roe is 1.025 x10^3 Kpa

    what is g here?

    and i am not sure what units this all should be in

    this is what i did....
    v2=.17m^3(44176kPa/101kpa)(292.15K/277.15K)

    should .17m be in 17 cm^3 or what and what about he kPa can it stay that way

    i am confused
     
  2. jcsd
  3. Apr 9, 2007 #2

    Dick

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    The units in this case are easy. In v2=v1(p1/p2)(t2/t1), you can see that the units of pressure and temperature simply cancel and the units of v2 are the same as v1. So just put v1=17cm^3. BTW 17cm^3 is NOT the same as .17m^3!!!!!!
     
  4. Apr 9, 2007 #3
    i still get the wrong answer
     
  5. Apr 9, 2007 #4

    Dick

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    And in rho*g*h, rho is not a pressure, it's the density of the water. And g is the acceleration of gravity.
     
    Last edited: Apr 9, 2007
  6. Apr 9, 2007 #5
    some one please tell me what values i have wrong
     
  7. Apr 9, 2007 #6

    Dick

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    44176kPa is dead wrong. Try doing g*rho*h again.
     
    Last edited: Apr 9, 2007
  8. Apr 9, 2007 #7
    ok so (1.025x10^3)(9.8)(43)=431935 ....is that right
     
  9. Apr 9, 2007 #8

    Dick

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    Yes. But remember that it's in Pa not kPa. Convert to kPa before taking the ratio.
     
  10. Apr 9, 2007 #9
    ok ok let me try
     
  11. Apr 9, 2007 #10
    so now for the entire equation i am getting 7.66 cm^3 but that is wrong
     
  12. Apr 9, 2007 #11

    Dick

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    What is the 'entire equation'? The decrease in pressure and the increase in temperature should both cause it to expand.
     
  13. Apr 9, 2007 #12
    v2=v1(p1/p2)(t2/t1)

    v2=17cm^3(431935kPa/101kpa)(292.15K/277.15K) and for this i am getting 7.66 cm^3
     
  14. Apr 9, 2007 #13

    Dick

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    Make that 431.935kPa (I TOLD you). And how can v2 be less than 17 if both ratios are larger than 1!
     
  15. Apr 9, 2007 #14
    ok so now that i corrected 431935 to 431.935 i get 76.64 which is not right
     
  16. Apr 9, 2007 #15

    Dick

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    Ok, I'll bite. What is the right answer?
     
  17. Apr 9, 2007 #16
    i don't know
     
  18. Apr 9, 2007 #17
    i am putting this into a hw program and it keeps saying that my answer is not right
     
  19. Apr 9, 2007 #18

    Dick

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    Ah. You didn't add atmospheric pressure to p1 like you did last time. p1=101 kPa+ (roe)gh.
     
  20. Apr 9, 2007 #19
    i am putting the values that i get into a hw program online and it keeps saying that it is wrong...and it won't say what the right answer is
     
  21. Apr 9, 2007 #20
    ok ok let me try
     
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