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

  • #1
119
0

Homework Statement



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

Homework Equations





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:

Answers and Replies

  • #2
119
0
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:
  • #3
119
0
sorry, ignore this post.
 

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

  • Last Post
Replies
1
Views
17K
Replies
29
Views
4K
  • Last Post
Replies
2
Views
383
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
6K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
Top