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How do I find the entropy variation?

  1. Dec 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the ∆S per mol between liquid water at -5 ºC and ice at -5ºC at 1020hPa
    Data:
    ∆CP,m fusion = 37,3 J K-1 mol-1
    ∆fusH = 6,01 kJ mol-1
    The answer is 21.3 J/K mol

    2. Relevant equations
    Usually I solve these problems by steps when they are at P=1 atm but since its at P=1020 hPa i don't really know how to proceed.
    I tried using the following equation:
    ∆S= Q/T + nCpm ln T2/T1
    Since I am given the Cpm I interpret that the pression is constant throughout the process

    3. The attempt at a solution
    I first tried to solve it like I normally would, without taking into account that the press ion a was not 1 atm.
    I want to try and find the temperature at which water soldifies at 1020 hPa but I don't know how with the data I am given. I thought that since Qp is the enthalpy maybe I could find it using Q = n∆T specific heat of water
    but even thought I know what the specific heat of water is (41814 J/K kg) I'm not given this data so I'm unsure if this is what I have to do. 25383443_1748247851875564_819929200_o.jpg 25383256_1748250738541942_717490943_o.jpg
    This is what I first tried doing but after I realized my mistake I did this:
    25401064_1748247905208892_575902280_o.jpg
    I didn't obtain the value for entropy I should have with neither of the methods I tried. Where am I wrong?
     
  2. jcsd
  3. Dec 13, 2017 #2
    You are correct about thinking of this as a multi-step process.

    ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
    ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
    ice at 0 C and 1 atm to water at 0 C and 1 atm
    water at 0 C and 1 atm to water at 0C and 1020 hPa
    water at 0C and 1020 hPa to water at 5 C and 1020 hPa
     
  4. Dec 14, 2017 #3
    fisica1.jpg fisica2.jpg
    I tried doing it this way and it really does make more sense but I can't still obtain the 21.3 J/K I should and I really don't know why. I thought maybe it's because of the Cpm I have been given , but it turns out it doesnt really differ from the one I was given, so I thought I might change the sign since one it's fusion and the other one its for solidification. Still, I haven't obtained the result I need and I don't know why. Could anyone please help me?
     
  5. Dec 14, 2017 #4
    You did this mostly correctly in your first attempt. The first thing to recognize is that 1020 hPa is essentially 1 atm, so there is no need to account for a pressure effect. So the change in entropy in going from 1 mole of ice at -5 C to water at - 5 C is given by:
    $$\Delta S=C_i\ln(273/268)+\frac{\Delta H(0)}{273}+C_w\ln(268/273)=(C_W-C_i)\ln(268/273)+22.0$$
    We know that ##C_w-C_i=37.3\ J/mol.K##
     
  6. Dec 14, 2017 #5
    I don't quite understand how you're using logarithmic properties here, could you explain it to me please?
     
  7. Dec 14, 2017 #6
    Are you saying that you are not sure how to determine the change in entropy of a solid or liquid when it is heated or cooled from an initial temperature to a final temperature (and why this should involve natural logs of temperatures)?
     
  8. Dec 31, 2017 #7
    I do, sorry. I had to revisit the basics. Thank you for your help.
     
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