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Quantification of Entropy and the 2nd Law of Thermodynamics

  1. Aug 2, 2016 #1
    All of my information comes from my current chemistry class, I just want to know where I either may have misread, misinterpreted, or was mistold information.

    With the 2nd Law of Thermodynamics we can say $$[1]\space\space \Delta{S}_{universe} = \Delta{S}_{surroundings} + \Delta{S}_{system} > 0$$
    Then using $$[2]\space\space \Delta{S}_{system} = \frac{q_{rev}}{T}$$
    with (at constant pressure) $$[3]\space\space q = \Delta{H}$$
    and (from the 1st Law of Thermodynamics) $$[4]\space\space \Delta{H}_{system} + \Delta{H}_{surroundings} = 0$$
    we can get $$[4]\space\space \Delta{S}_{system} = \frac{\Delta{H}_{system}}{T}$$ $$[5]\space\space \Delta{S}_{surroundings} = \frac{\Delta{H}_{surroundings}}{T} = \frac{-\Delta{H}_{system}}{T}$$.
    Substituting these back into [1] would give $$[6] \space\space \Delta{S}_{universe} = \frac{-\Delta{H}_{system}}{T} + \frac{\Delta{H}_{system}}{T} > 0$$
    which simplifies to $$0 > 0 $$
    which is not true.

    I'm assuming I did something wrong with this, but cannot figure out what. Where is the issue?
  2. jcsd
  3. Aug 2, 2016 #2


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    The second law states that ##\Delta S_{universe} \geq 0##. Specifically, for reversible processes (you are assuming reversibility because you've made qrev = q), ΔSuniverse = 0.
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