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Am I doing this right? (pressure, constant density, mount everest)

  1. Feb 15, 2006 #1
    "A student decides to compute the standard barometric reading on top of Mt. Everest (29,028ft) by assuming the density of air has the same constant density as the sea level. Try this yourself. What does the result tell you?"

    Am I doing this right?
    h = 8847.7344 m
    density of air = 1.29kg/m3
    pressure = density * g * h = 111.853 kPa

    Since the by using constant density, we see that the pressure is higher on Mt. Everest. Of course, we know this to be false. Therefore, the density of air must be less on Mt. Everest than at sea level.

    I'm not sure this is right, but please check.
  2. jcsd
  3. Feb 15, 2006 #2
    What is h here. Did you think about that. The column of air above us is important. Not the one below us. Pressure is exerted due to the exertion of force over an area. The Force is applied due to the weight of the air above us and thus the atmosphere experience a pressure due to the weight of the air above it.
    So let the height of atmosphere from the surface of earth be R. Thus the chane in height of air column is R - h, where h is the height of Mt. Everest.
  4. Feb 15, 2006 #3
    is my equation correct? p = d * g * h?
    if h is R - h, then
    p = pat sea level - density * g * h

    but density * g * h = 111.853 kPa, and then p would be negative!

    I'm confused.
  5. Feb 15, 2006 #4
    I didn't plug in the numbers, but if your pressure is negative, what does that tell you about either the assumption that density of air at that altitude is the same as sea level, or g, or both?
  6. Feb 15, 2006 #5
    Why should you think it is negative. Infact now Im confused abt what you are confused???
    R - h > 0 since R>h as R is the height of air column above Earth surface -sea level - while h is the height of Mt. Everest. dgR is 101.325KPas, while dgh is less than that.
    P = dg(R - h) just as P(at sea leavel) = dgR = 101.325KP
  7. Feb 15, 2006 #6
    sorry, this is my first time learning physics...

    if P(at sea level) = 101325 Pa and P = dgR, and d = density of air = 1.29 kg/m3[/sub], then
    R = 8015m???

    ptotal = density * g * (R - h)
    ptotal = (density * g * R) - (density * g * h)
    density = 1.29kg/3, g = 9.8m/s2[/sub], h = 8847.7344 m
    then (density * g * h) = 111853.0583 Pa, which is greater than 101325 Pa, and therefore ptotal < 0.
  8. Feb 16, 2006 #7
    I got disconnected yesterday before your reply.

    Density of air is not constant along the whole column of air at atmosphere. That is why the result is interesting. I have not looked it. Thus the density of air input should be proper enough. 1.29 is the density of air nearby Earth's surface.
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