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Homework Help: [Thermo.] This problem has been bugging me for days, please help.

  1. Jun 23, 2010 #1
    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 :frown:, 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.


    1. The problem statement, all variables and given/known data

    The problem statement is in the following picture:

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


    2. Relevant equations

    OBS: I'm not sure if in the problem statement [tex]\theta_i = T_i [/tex], 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.


    [tex]T_i = \frac{100}{(r_s - 1)} [/tex]

    [tex]r_s = lim_{P_i \rightarrow 0} (\frac{P_s}{P_i}) [/tex]


    3. The attempt at a solution

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

    [tex]T_i = \frac{100}{(r_s - 1)} [/tex]

    [tex]\frac{dT_i}{dr_s} = \frac{-100}{(r_s - 1)^2} [/tex]

    [tex]dT_i = \frac{-100}{(r_s - 1)^2} dr_s [/tex]

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

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

    [tex]\frac{dT_i}{T_i} = \frac{-1}{(r_s - 1)} dr_s [/tex]

    Which can be rearranged as

    [tex]\frac{dT_i}{T_i} = \frac{dr_s}{(1 - r_s)}[/tex]

    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:

    [tex]\frac{dT_i}{T_i} = 3,73\frac{dr_s}{r_s}[/tex]

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

    I really need to solve this. :frown:
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 25, 2010 #2
    The problem has been bugging me since yesterday, too :biggrin:
    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: [tex]r_s = \frac{T_s}{T_i}[/tex]
    We have: [tex]\frac{dT_i}{T_i}=\frac{dr_s}{1-r_s}=(\frac{r_s}{1-r_s})\frac{dr_s}{r_s}[/tex] (1)
    So now if the error is small enough (and the experiment is good enough), we can assign [tex]r_s=\frac{T_s}{T_i}=\frac{373}{273}[/tex] (ice point = 0 Celsius degree; steam point = 100 Celsius degrees). So from (1):
    [tex]\frac{\Delta T_i}{T_i}=3.73\frac{\Delta r_s}{r_s}[/tex]
     
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