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

[Je m'appelle]

Homework Statement

http://j.imagehost.org/0069/thermo.jpg

Homework Equations

The Attempt at a Solution

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

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

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

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

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:

\frac{d\theta}{\theta} = \frac{dr}{(1 - r)}I'm lost.

Anyone?

 

[Je m'appelle]

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):

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

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

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

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:

\frac{d\theta_i}{\theta_i} = \frac{dr_s}{(1 - r_s)}But the result should be something like:

\frac{d\theta_i}{\theta_i} = 3,73\frac{dr_s}{r_s}

 

[Je m'appelle]

sorry, ignore this post.