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Clausius clapeyron problem - i have spent 3 hours on this problem

  1. Nov 9, 2013 #1
    spent 3 hours on this, getting nowhere

    find the change in transition temperature, given: P1,P2,T1,ΔH,density (mixed so gives d1 of metal 1 and d2 of metal 2,atomic weight

    Clasius-Clapeyron-Equation.gif

    Tried solving for T1 then for ΔT simply saying ΔT = (T1 - T2) , but its clearly not correct and i havent made use of the densities or the atomic number the question gives.

    So then i approximated the problem fromt he original clausius clapeyron eqn to

    ΔP/ΔT = ΔH / (TΔV)

    but then to find ΔT, which i assume is the change in transition temperature, i need ΔV

    so i did density = m/v , PiVi = nRTi to give density = PM / RTi where M is the molar volume

    but the density would change with the pressure, so how am i supposed to use the initial density, and there's two densities not one for the compound??

    I guess what i'm asking is how do i find ΔV given the above data? why have i been given the two densities of the compound when the densities are going to change with pressure?


    Ive spent 3 hours on this question, looked for similar examples on the internet, there are NONE. General questions ask for T2 or P2, none ask for the change in transition temperature.

    Please help, erghhhhh, i better get on with my maths now
     
  2. jcsd
  3. Nov 9, 2013 #2
    You used the ideal gas law, but metal solids don't obey the ideal gas law. That's why you were given the densities. The density is the inverse of the specific volume, and, for metal, the density has negligible change with temperature and pressure (at least as far as your application is concerned).
     
  4. Nov 10, 2013 #3
    i see, thanks for the reply, i understand density = 1 / volume per unit mass, but i still need to find the change in volume (however small) somehow from the 2 densities.
     
  5. Nov 10, 2013 #4
    So what are the two densities? Why can't you do it?
     
  6. Nov 11, 2013 #5
    hi, because the volume before would not be the same as the volume after, even though there is a negligable change in volume i need that change to give me delta V, in order to plug it into the clausius clapeyron eqn to give delta t.


    and what do i do with the two densities, take an average of the two?

    and why has it given the atomic weight?
     
    Last edited: Nov 11, 2013
  7. Nov 16, 2013 #6
    just to update this, the question wanted the MOLAR VOLUME
     
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