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Determining pressure at various altitudes using integration

  1. Dec 12, 2016 #1
    1. The problem statement

    Solve the following problems assuming air density is proportional to respective pressure at each height: What is the normal pressure at the atmosphere at the summit of a. Mt. McKinley, 6168m above sea level and b. Mt. Everest, 8850m above sea level c. At what elevation is the air pressure equal to one fourth of the pressure at sea level (assume the same air temperature).

    2. Relevant equations

    3. The attempt at a solution

    I started by considering a block of air with the bottom at sea level and the top at the top of each of the mountains, such that (P+dp)A+dy*A*density*g-PA=0, which simplifies to dp=-dy*density*g. I can't figure out how to go on from here though, as the density is variable.
     
  2. jcsd
  3. Dec 12, 2016 #2

    haruspex

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    Yes, and it tells you how to handle that:
     
  4. Dec 12, 2016 #3
    I just don't really understand how to do that. I've been trying to formulate an equation but I assume it involves integration which I can do in some cases but don't conceptually understand how to apply to this problem.
     
  5. Dec 12, 2016 #4
    $$\rho=kp$$ where k can be determined from the ideal gas law using the pressure and temperature at the earth's surface.
     
  6. Dec 12, 2016 #5
    I might (???) be making progress here. Assuming temperature is constant, nRT should be constant as well, thus P is inversely proportional to V, and hence directly proportional to density. Thus I can substitute in 1.29*P/P(sea level) for density, yielding the equation dp=g*(1.29P/101,325)*dy. From there I get dp/P=1.247668*10^-4. I then integrate from 0 to 6168 meters and get lnP=1.247668*10^-4*6168. Unfortunately, the answer I get is way off. Where am I going wrong here?
     
  7. Dec 12, 2016 #6

    haruspex

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    Two errors.
    You need to pay attention to signs. Does the pressure go up or down as y increases?
    You forgot the constant of integration.
     
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