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Deriving hydrostatic pressure equation for an ideal gas

  1. Sep 15, 2016 #1
    Assume: Hydrostatic Situation, ideal gas
    Use empirical formula T = T0 - Bz


    I have rechecked my work several times and can't seem to find a mistake. The answer is supposed to be:
    p2 = p1((T0 - Bz)/T0)(g/Bz)


    dp/dz = -ρg Now substitute ideal gas equation for ρ
    dp = -(p/RT)g dz
    ∫(1/p) dp = (-g/R) ∫(1/(T0-Bz) dz After substituting T = T0 - Bz ... Integrate

    let u = T0 - Bz
    du = -B dz

    ln(p2/p1) = (g/BR) ∫(1/u) du
    ln(p2/p1) = (g/BR)(ln(u))

    ln(p2/p1) = (g/BR)(ln(T0-Bz))
    p2 = p1(T0 - Bz)(g/BR)

    Where are they getting that T0 on the bottom??
     
  2. jcsd
  3. Sep 15, 2016 #2
    You forgot to substitute the lower limit of integration in the right hand side of the equation.
     
  4. Sep 15, 2016 #3
    THANK YOU, can't believe I forgot to do that!
     
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