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Problem on limits from chemistry

  1. Mar 3, 2012 #1
    I've been doing some chemistry, and come up with a confusing problem that I can't seem to solve. It's a problem I made up while trying to understand the Gibbs Free Energy, but I will try to phrase it solely as a calculus question.

    It is known that

    [itex] \Delta G + \epsilon \Delta S < 0 [/itex] and that [itex]\epsilon[/itex] is an infinitely small positive quantity. Nothing is known about the sign of [itex] \Delta S [/itex].

    Both [itex] \Delta G [/itex] and [itex] \Delta S [/itex] are finite.

    Can I prove that [itex] \Delta G ≤ 0 [/itex] ?

    I appreciate all help on this matter. Thanks!

    BiP
     
    Last edited: Mar 4, 2012
  2. jcsd
  3. Mar 3, 2012 #2

    Office_Shredder

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    By infinitely small do you mean that [itex] \epsilon[/itex] can be made as small as we desire? If so:

    Suppose that [tex] \Delta G > 0[/tex]. Then we can find [itex] \epsilon > 0[/itex] such that [tex]\Delta G + \epsilon \Delta S = 0[/tex], namely [tex] \epsilon = -\Delta G/\Delta S[/tex]

    Well, this is positive as long as [itex] \Delta S[/itex] is negative . What if [itex] \Delta S \geq 0[/itex]? In that case, [itex] \epsilon \Delta S \geq 0 [/itex] so if [itex] \Delta G > 0[/itex] we're just adding positive numbers together and of course it can't be negative
     
  4. Mar 4, 2012 #3
    By infinitely small, I mean that [itex] \epsilon > 0[/itex] is less than any positive real number.

    BiP
     
  5. Mar 4, 2012 #4

    Office_Shredder

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    No such epsilon exists in the real numbers (how can [itex] \epsilon/2 > \epsilon[/itex])? Is this supposed to be a problem in the hyperreals or something?
     
  6. Mar 4, 2012 #5
    Hmm well I don't know to say it, all I mean is that [itex] \epsilon [/itex] is a positive infinitesimal.

    I reasoned that because of its being infinitesimal it can be neglected from the equation, yielding [itex] \Delta G < 0 [/itex]. But I don't know if that's correct.

    Also your original proof does not satisfy the main constraint of the problem, i.e. [itex] \Delta G - \epsilon \Delta S < 0 [/itex]

    But I think we're getting somewhere! Thanks!


    BiP
     
  7. Mar 4, 2012 #6

    Office_Shredder

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    Can you describe a bit the setup that led you to this equation? It will help sort out exactly what kind of requirements you really have on [itex]\epsilon[/itex] - since if you know it's smaller than all positive numbers, and non-negative for example, then [itex]\epsilon=0[/itex] necessarily.
    This differs by a minus sign from the original post
     
  8. Mar 4, 2012 #7
    Sorry I meant [itex] \Delta G + \epsilon \Delta S [/itex] from the original post. The second post is a typo.

    The setup for my problem is a little complicated thermodynamics. I don't wish to go into it, but the epsilon is basically the infinitesimal difference in temperature between two reservoirs that are basically the same temperature, except one is higher in temperature by an infinitesimal amount epsilon. Heat will transfer due to that infinitesimal difference. Though the heat transfer is also infinitesimal, it is important in the calculations, whereas the epsilon reduces to the inequality above. I would love to see this problem be solved, seeing as it has been reduced to a purely mathematical problem.

    I thank you for your time and efforts, they have made me rethink infinitesimals.

    BiP
     
  9. Mar 4, 2012 #8

    Office_Shredder

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    It sounds like you really want to assume that epsilon is an arbitrarily small positive number (i.e. small enough that only first order effects occur). In this case the solution in my previous post essentially does it for us. There are two cases

    1) [itex] \Delta S \geq 0 [/itex]. Then [itex] \epsilon \Delta S \geq 0[/itex] as well, and if [itex] \Delta G \geq 0[/itex] adding these two inequalities yields [itex] \Delta G + \epsilon \Delta S \geq 0 [/itex] which we know isn't true. So it must be [itex] \Delta G < 0 [/itex]

    2) [itex] \Delta S < 0 [/itex]. We will do a proof by contradiction: suppose that [itex] \Delta G > 0[/itex]. Then we will show that if [itex] \epsilon[/itex] is small enough (smaller than a specific number which we will calculate), that [itex] \Delta G + \epsilon \Delta S > 0 [/itex] which is a contradiction.

    If [itex] \epsilon < -\Delta G / \Delta S[/itex] (which is a positive number since by assumption, delta S is negative and delta G is positive), then [itex] \Delta G + \epsilon \Delta S > \Delta G + (-\Delta G / \Delta S) \Delta S[/itex] where the inequality is > because we're increasing the coefficient of delta S, which is a negative number by assumption. Of course, cancelling the fractions then gives us [itex] \Delta G + \epsilon \Delta S > \Delta G - \Delta G = 0 [/itex] which is a contradiction.


    Note that in this case we had to assume delta G was non-zero, otherwise epsilon would have been exactly zero. But if delta G is exactly zero there's nothing to prove.

    So we had two cases: one where delta S was non-negative, and one where delta S was negative. In both cases we proved that delta G must be negative, as long as we can assume epsilon is small enough.
     
  10. Mar 4, 2012 #9
    Great! Thanks! I guess I must've just not understood your original solution, but I get this one!

    BiP
     
  11. Mar 4, 2012 #10
    Office_Shredder, all the time in the sciences like physics and chemistry, we do calculus involving infinitesimals. This can of course be put on a firm rigorous foundation in any number of ways, including hyperreals and differential forms, but we often just use techniques without always referring to their foundations. Just like you take limits all the time without writing out the epsilon-delta definition, we can write things like dA=rdrdθ without writing out the definition of the wedge product or proving Jacobi's theorem. In thermodynamics, things like δW=-PδV is often more useful for our purposes than taking an explicit derivatives, but we don't lay out the theory of inexact differentials in our calculations.
     
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