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[Thermodynamics] Relative fractional error, Ideal-gas-scale, Celsius scale

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    http://j.imagehost.org/0069/thermo.jpg [Broken]

    2. Relevant equations



    3. The attempt at a solution

    OK, so I tried to derive θ in respect to r as in θ(r):

    [tex]\frac{d\theta}{dr} = \frac{-100}{(r - 1)^2}[/tex]

    Then, I multiplied both sides of the equation by 'dr',

    [tex]d\theta = \frac{-100}{(r - 1)^2}dr[/tex]

    Then I divided both sides of the equation by θ in order to find the fractional error but it doesn't match with the equation on the picture:

    [tex]\frac{d\theta}{\theta} = \frac{dr}{(1 - r)}[/tex]


    I'm lost.

    Anyone?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 25, 2010 #2
    I assume θ_i = T_i? Or is there a difference?

    Is θ_i in Celsius and T_i in Kelvin?

    So I tried to derive θ_i in respect to r_s as in θ_i(r_s):

    [tex]\frac{d\theta_i}{dr_s} = \frac{-100}{(r_s - 1)^2}[/tex]

    Then, I multiplied both sides of the equation by 'dr_s',

    [tex]d\theta_i = \frac{-100}{(r_s - 1)^2}dr_s[/tex]

    Then I divided both sides of the equation by θ in order to find the fractional error but it doesn't match with the equation on the picture:

    [tex]\frac{d\theta_i}{\theta_i} = \frac{dr_s}{(1 - r_s)}[/tex]


    But the result should be something like:

    [tex]\frac{d\theta_i}{\theta_i} = 3,73\frac{dr_s}{r_s}[/tex]
     
    Last edited: Apr 26, 2010
  4. Apr 25, 2010 #3
    sorry, ignore this post.
     
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