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

1. Apr 24, 2010

### Je m'appelle

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):

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

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?

Last edited by a moderator: May 4, 2017
2. Apr 25, 2010

### 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:

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

Last edited: Apr 26, 2010
3. Apr 25, 2010

### Je m'appelle

sorry, ignore this post.