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2nd law of thermodynamics

  1. Dec 17, 2006 #1
    According to the 2nd law of thermodynamics, a spontaneous process will occur (or rather: is very likely to occur) if [tex] \Delta S > 0[/tex].

    A chemical process occurs if [tex] \Delta G < 0[/tex], where [tex] G = H - TS[/tex].

    H = -100 kJ
    T = 1 K
    S = -10 kJ/K
    so [tex] \Delta G = - 190 kJ[/tex]. In this example, [tex] \Delta G < 0[/tex] but [tex] \Delta S < 0[/tex].
    Doesn't this contradict the 2nd law?
  2. jcsd
  3. Dec 17, 2006 #2


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    I don't see an S_{1} and an S_{2} so how did you compute \Delta S ?

  4. Dec 17, 2006 #3
    Okay, then
    [tex] \Delta H = -100 \mbox{kJ}[/tex]
    [tex] T = 1 \mbox{K}[/tex] and
    [tex] \Delta S = -10 \mbox{kJ/K}[/tex]
    because [tex] \Delta G = \Delta H - T\Delta S[/tex] where T is constant.
  5. Dec 17, 2006 #4


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    If H changes, then that means that the system is not left to itself but is externally acted upon. So we are not dealing with a spontaneous process.
  6. Dec 17, 2006 #5
    If [tex] \Delta H < 0 [/tex], then enthalpy is released from the system to the surroundings. It would make sense, if I had to add enthalpy ([tex] \Delta H > 0, \qquad H = U + pV [/tex]) and the entropy decreased, but that's not true according to the Gibbs energy.
    Last edited: Dec 17, 2006
  7. Dec 17, 2006 #6
    If the system is isolated, a spontaneous process will increase its entropy. The Universe is an isolated system in itself, so entropy is always increasing.
    But if there is a variation in entalpy (a decrease) at constant temperature T, for sure your system is not isolated. The environment is absorbing such energy, increasing its entropy in a greater extent than the entropy lost by the system (the variation, in absolute value, of the entalpy is grater than that of the entropy, as variation og G is negative). So the universe, the system and the environment, experiments an increase in entropy. This is the result of the second law. No contradiction at all.
  8. Dec 17, 2006 #7


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    No of course not. Take water. You extract enthalpy (you cool it). It freezes, and its entropy decreases...
  9. Dec 18, 2006 #8
    I get it, thanks! [tex] \Delta S_{total} > 0[/tex], even though [tex] \Delta S_{sys} < 0[/tex], because [tex] \Delta H_{sys} < 0 [/tex] according to the Gibbs energy and the conditions for a spontaneous process.

    Is it possible to calculate the [tex] \Delta S_{surroundings} [/tex], if I know the [tex] \Delta H_{sys} [/tex]? I mean, is there any proportionality between those two variables?
  10. Dec 18, 2006 #9
    Yes, but the [tex] \Delta S_{total} > 0[/tex], because the surroundings get warmer.
    (Or did I get this wrong?)
    Last edited: Dec 18, 2006
  11. Dec 18, 2006 #10
    If my memory is still readable, I would say that [tex] \Delta S_{surroundings} [/tex] is equal to [tex] \Delta H_{sys} [/tex] if volume is constant (more generally, if there is no work performed of PV type). Then [tex] \Delta H_{sys} [/tex] is the thermal energy transferred.
    If there is work performed, then enthalpy is not equal to Q, the heat.
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