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Typical question: Entropy of the universe, of surroundings

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    The evaporation enthalpy of Hg is ##59.3 kJmol^-1## at its boiling temperature ##356.6ºC##. Calculate:

    (a) the vaporization entropy of Hg at this temperature

    (b) the change in entropy of the surroundings and universe

    (c) the vaporization entropy of Hg at 400ºC.


    I was also given heat capacities for liquid and gaseous states as ##27.983 JK^-1mol^-1## and ##20.786JK^-1mol^-1##.

    2. Relevant equations

    ##dS=dQ/T##

    ##Q=\Delta_{vap} H##

    ##Delta S_{univ}=\Delta S_{system} + \Delta S_{surroundings}##
    3. The attempt at a solution

    (a)
    ##\Delta_{vap} S = \frac{\Delta_{vap} H }{T} = 59.3 / (356.6+273.15) = 0.094 kJ K^-1 mol^-1##

    (b)
    ##\Delta S_{system} = \frac{-\Delta_{vap} H}{T_{surroundings}}=\frac{-59.3}{25+273.15}= -.1988 kJ K^-1 ## Here I've assumed that only one mole is being heated, because no other information has been given. I have also assumed that the temperature of the surroundings is 25ºC.

    ##Delta S_{universe} = -.1048 kJ/K##

    (c)

    Enthalpy doesn't change with temperature (approximation). Therefore

    ##\Delta_{vap} S = \frac{59.3}{400+273.15} = 0.088093 kJ K^-1 mol^-1##


    Few comments:

    This problem violates the second law of thermodynamics because the entropy of the universe has been calculated to be negative.

    Also, I'm having doubts as to what the temperature of the surroundings is. Is it assumed to be 25ºC? Its not the same as the substance being heated? Why so?

    Would you agree with me that these questions/procedures are assuming that things are happening reversibly? Doesn't the second law of thermodynamics say that the entropy of the universe is 0 for a reversible process?

    What have I done wrong?

    Many thanks.
     
  2. jcsd
  3. May 17, 2015 #2

    Andrew Mason

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    Indeed, you have shown that the spontaneous vaporization of Hg at 356.6C will not occur in an environment at 25C. Rather, heat flow will go from the Hg to the surroundings.

    So it seems to me that the problem is with your assumption that the surroundings are at 25C.

    If the Hg is receiving heat flow at 356C and the heat flow is coming from the surroundings, what must be the temperature of the surroundings? (Hint: to find the vaporization entropy, assume that the process is reversible. For part C, assume that surroundings are at 400C)

    AM
     
  4. May 17, 2015 #3
    You mean assume that the system is at 400C (for part (C) ) ?

    I really have to be told what the temperature of the surroundings is right?
     
  5. May 17, 2015 #4

    Andrew Mason

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    For part C, assume the Hg is initially a liquid at 400C and that heat flows into the Hg from the surroundings. The issue is to determine the magnitude of the heat flow into the Hg from the surroundings (which have to be at a minimum of 400C) in order to cause vaporization. If the enthalpy of vaporization at 356.6 C is x, what will it be at 400C? - higher or lower? by how much? (hint: they gave you the specific heat of Hg liquid for a reason).

    If you assume a reversible vaporization, the surroundings just have to be at an infinitesimally higher temperature. It can't be lower. The change in entropy of the Hg + change in entropy of surroundings = 0.

    AM
     
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