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Homework Help: Change in atmopheric pressure with height (Thermodynamics)

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Think of earth's atmosphere as an ideal gas of molecular weight [tex]\mu[/tex] in a uniform gravitational field. Let g denote the acceleration due to gravity.

    If z denotes the height above the sea level, show that the change of atmospheric pressure p with height is given by

    dp/dz = p([tex]\left([/tex]-[tex]\mu[/tex]g/RT)

    dp = change in pressure
    dz = change in height
    [tex]\mu[/tex] = molecular weight
    R = Universal gas constant
    T = temperature

    2. Relevant equations

    1) F = mg
    2) P = F/A
    3) PV = nRT

    3. The attempt at a solution

    With the relation between force and pressure, I derived an expression of p in terms of m and g. Using the ideal gas law, i obtained another equation for p. I am lost in my attempt to derive dp/dz. Please help.

    Thank in advance,
    Last edited: Apr 4, 2009
  2. jcsd
  3. Apr 4, 2009 #2


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    Gas equation is given by pV = nRT, where is the number of mole.
    n is given by [mass of the air column (ma) on unit area]/mu.
    Hence pV = [(ma)/(mu)]RT .........(1)
    Now p = rho*g*z.....(2)
    and dp/dz = rho*g......(3)
    From equation (1) find the expression for rho and substitute in eq. (3)
  4. Apr 4, 2009 #3
    thank rl.bhat!

    but the notations are confusing.
    n = (ma)/(mu)
    so ma is the mass of air column ? Can you explain a bit further please ?
  5. Apr 5, 2009 #4


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    ma is the mas of the air column on unit area, mu is the molecular weight of air.
    So number of mole = n = ma/mu.
  6. Apr 5, 2009 #5
    figured i have to derive yet another expression. This time I have to assume an adiabatic expansion and show

    dp/p = {[tex]\gamma[/tex]/([tex]\gamma[/tex]-1)}{dT/T}


    dp/dT = {[tex]\gamma[/tex]/([tex]\gamma[/tex]-1)} {p/T}

    My idea:

    since it is an adiabatic expansion, i have
    pz([tex]\gamma[/tex]-1) / T(Z)[tex]\gamma[/tex] = constant

    I used it and tried to cook the needed expression, but i got lost.
  7. Apr 5, 2009 #6


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    For adiabatic expression pV^gamma = K
    You replace V by nR(T/p)
    So p(nRT/p)^gamma = k or
    p(T/p)^gamma = K'
    i.e. T^gamma*p^(1-gamma) = K.
    Take the differentiation and simplify.
  8. Oct 14, 2010 #7
    How would you derive the hydrostatic equation in spherical polars?

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