[Thermo.] This problem has been bugging me for days, .

[Je m'appelle]

I've been trying to solve this for a long time now, I even posted it on this forum a long time ago but no one replied, not even a single response , I'm not sure if the folks thought I was just trying to make them solve the problem for me or if they actually did not understand it, either way I'll give it another try.

I'll be straight to the point, I'm not trying to make you solve this for me, I just need directions, any sort of enlightenment, as I have no idea what else to do.

Homework Statement

The problem statement is in the following picture:

[PLAIN]http://img15.imageshack.us/img15/3177/thermo.png

Homework Equations

OBS: I'm not sure if in the problem statement \theta_i = T_i, if it was a misstyping or something of that sort, as I have the same textbook in another language and they're written the same way, so I'll assume they are the same in this textbook also, but in any case if I'm wrong please correct me.


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

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

The Attempt at a Solution

Alright, so I basically considered T_i as a function of r_s, so I derived both sides of the following equation with respect to r_s

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

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

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

Now, I'll divide both sides of the above equation by T_i in order to get the asked ratio.

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

\frac{dT_i}{T_i} = \frac{-1}{(r_s - 1)} dr_s

Which can be rearranged as

\frac{dT_i}{T_i} = \frac{dr_s}{(1 - r_s)}

And I don't know if what I've done is correct, and if it is how do I proceed in order to get to the asked relation:

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

Please help me up, and if there is anything you didn't understand please ask.

I really need to solve this.

 

[hikaru1221]

The problem has been bugging me since yesterday, too
I'm not an expert or anything, so all I can come up with is a silly solution.

As the limit (pressure -> 0) is approached, the water vapor is approximately an ideal gas, which means P is proportional to T. Thus: r_s = \frac{T_s}{T_i}
We have: \frac{dT_i}{T_i}=\frac{dr_s}{1-r_s}=(\frac{r_s}{1-r_s})\frac{dr_s}{r_s} (1)
So now if the error is small enough (and the experiment is good enough), we can assign r_s=\frac{T_s}{T_i}=\frac{373}{273} (ice point = 0 Celsius degree; steam point = 100 Celsius degrees). So from (1):
\frac{\Delta T_i}{T_i}=3.73\frac{\Delta r_s}{r_s}