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Homework Help: Calculating pressure in high altitude?

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the change in pressure on a special-op soldier who must scuba dive at a depth of 20 m in seawater one day and parachute at an altitude of 7.6 km the next day? Assume that the average air density within the altitude
    range is 0.87 kg/m3.

    2. Relevant equations
    p=p0 + ρgh

    3. The attempt at a solution
    I understand how to calculate pressure using the above formula for something at the surface, but how do I apply it to a high altitude? In the solution manual, they just ignore atmospheric pressure, so I am confused. At that high altitude, doesn't it mean the person is not affected by the usual atmospheric pressure of 1 atm?
  2. jcsd
  3. Jan 19, 2015 #2


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    Pressure is less at the top of the sea than at the bottom of the sea ...
    we live on the bottom of a "sea of air" that's only about 8km deep.
    Pressure is zero at the "top" of the atmosphere, the edge of space.
  4. Jan 19, 2015 #3

    I doubt that the solution manual ignored the atmospheric pressure at the surface of the earth. Please show us how the manual solution got the pressure at 7.6 km.

  5. Jan 19, 2015 #4
    This is from the solution manual:

    The gauge pressure at a depth of 20 m in seawater is
    p1swgd = 2.00 x 105 Pa
    On the other hand, the gauge pressure at an altitude of 7.6 km is
    p2airgh=6.48 x 104 Pa
    Therefore, the change in pressure is
    Δp = p1 - p2 = 1.4 x 105 Pa
  6. Jan 19, 2015 #5
    The result for 7.6 km from the solution manual is incorrect. They have the sign wrong. It should be p2= - ρairgh= - 6.48 x 104 Pa. The absolute pressure at 7.6 should be 1. x 105 - 6.48 x 104 Pa. This is where the atmospheric pressure at the surface comes in.

    Of course, this error makes the final answer incorrect also.

  7. Jan 19, 2015 #6
    Isn't 105 only applies to things at roughly sea level? Wouldn't something at 7.6 km up not be subjected to this pressure?
  8. Jan 19, 2015 #7
    Sure. It''s less. That's what I calculated.

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