# Thermo. Relative Error

1. May 3, 2010

### Je m'appelle

1. The problem statement, all variables and given/known data
In a few words, show that

$$\frac{dT_i}{T_i} = 3,73\frac{dr_s}{r_s}$$

Where,

$$r_s = \lim_{P_i\rightarrow 0}\frac{P_s}{P_i}$$

There's a picture below for more details.

2. Relevant equations

$$T_i = \frac{100}{r_s - 1}$$

3. The attempt at a solution

So I tried this,

$$T_i = \frac{100}{r_s - 1}$$

Applying 'ln' to both sides,

$$ln(T_i) = ln(\frac{100}{r_s - 1})$$

And then deriving both sides by $$\frac{d}{dr_s}$$, as $$T_i$$ is a function of $$r_s$$ (right?) I get to

$$\frac{1}{T_i}\frac{dT_i}{dr_s} = \frac{r_s - 1}{100}\frac{-100}{(r_s - 1)^2}\frac{dr_s}{dr_s}$$

$$\frac{1}{T_i}\frac{dT_i}{dr_s} = \frac{-1}{r_s - 1}$$

$$\frac{1}{T_i}\frac{dT_i}{dr_s} = \frac{1}{1 - r_s}$$

$$\frac{dT_i}{T_i} = \frac{dr_s}{1 - r_s}$$

Where

$$r_s = \lim_{P_i\rightarrow 0} (\frac{P_s}{P_i})$$

But this looks completely nonsense when comparing to what I was asked to prove in the first place.

I should have gotten to something like

$$\frac{dT_i}{T_i} = 3,73\frac{dr_s}{r_s}$$

But instead I got to

$$\frac{dT_i}{T_i} = \frac{dr_s}{1 - r_s}$$

Thoughts?

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Last edited: May 3, 2010
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