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[Thermodynamics] Calculate change in entropy of closed reversible system

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Mercury is a silvery liquid at room temperature. The freezing point is -38.9 degrees celcius at atmospheric pressure and the enthalpy change when the mercury metls is 2.29 kJ/mol. Wat is the entropy change of the mercury if 50.0 g of mercury freezes at these conditions? The molarmass of mercury is 200.59 g/mol. Assume the process is reversible

    2. Relevant equations
    Q=m(h_2-h_1) (enthalpy equation)
    Q=mT(s_2-h_1) (entropy equation)


    3. The attempt at a solution

    first I calculated what the energy change is per kg with what the enthalpy change is. h2-h1 is 2.29 kJ/mol and since there are 4.0118 mols I got 9.187 J/kg.

    So then I used Q/m = T(delta S)

    I rewrote Q/m to 459.35 J (since there are 50 grams of the substance) and divided that by T which is 234.1. The answer I get is 1.96 J/K whereas the answer is -2.44 J/K.

    Not sure what I'm doing wrong.
     
    Last edited: May 19, 2012
  2. jcsd
  3. May 19, 2012 #2

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    Hi Ortix! :smile:

    Perhaps you calculated the number of moles the wrong way around?
    If 1 mol is 200.59 g, how many moles is 50.0 g?
     
  4. May 19, 2012 #3
    Hey Serena,

    Very stupid mistake indeed. It would be 50/200.59=0.249264669 but this would result in a much smaller solution and still not negative (however, it would be negative entropy since the mercury is being frozed, in other words, energy is being taken out)

    I quickly worked it out on my laptop calculator (in bed on my laptop) and my answer is: 0.0488 J/K which is obviously wrong.
     
  5. May 19, 2012 #4

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    Yup, that's why it's negative.

    Did you take the 2.29 kJ/mol into account?
    Or else what did you calculate?
     
  6. May 20, 2012 #5
    I got the answer! Pretty simple calculation, but could you perhaps tell me what "m" exactly is in the equations and what the units are?
    Q=m(h_2-h_1) (enthalpy equation)
    Q=mT(s_2-h_1) (entropy equation)

    Because I ended up with the right answer being in the form Q/mT when it should be Q/T. I think I'm doing something wrong in the conversion of mol to gram.

    EDIT:
    I think I got it. I equated both equations and let the m's drop out. Since enthalpy is given as specific enthalpy per mol, i just multiplied it by the amount of moles present in the substance and got the total enthalpy change. Divide that by T and that is the answer!

    Can anyone tell me if my reasoning is correct? :)
     
    Last edited: May 20, 2012
  7. May 20, 2012 #6

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    Yep. That's it. :)

    m is the mass in kilograms.

    And for the record, your 2nd equation should read: Q=mT(s_2-s_1) (entropy equation)
     
  8. May 20, 2012 #7
    Oh yeah, that was a typo :) Thanks for putting me on the right track! :D
     
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