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I First Vs. Second law of thermodynamics

  1. Nov 14, 2016 #1
    Hello Pf's members

    Suppose we isolated a methane's molecule at a room with 213 kelvin.
    According to second law, methane's molecule tends to break it's bond for reaching whole system (methane's molecule and surroundings) to maximum entropy.
    On the other hand for breaking these bonds we need energy but system is isolated and no external energy can income to the room.

    Question:
    If we have unlimited time , can methane break it's bonds Eventually?
     
  2. jcsd
  3. Nov 14, 2016 #2

    mfb

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    The methane molecule can gain or lose energy by its contact to the room. The bond will break eventually, then form again (assuming the atoms don't form any other connections), break again, ...
     
  4. Nov 14, 2016 #3
    My question is how does it get energy? there is nothing in the room except a methane molecule.
    there isn't any photons. there isn't any atoms.there isn't any molecules. room is like a vacuum and room is isolated. room can not exchange any matter and energy.
    breaking bond means violate first law.
     
    Last edited: Nov 14, 2016
  5. Nov 14, 2016 #4

    dextercioby

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    One cannot speak about the entropy of a single gas molecule, hence your problem is based on a false premise.
     
  6. Nov 14, 2016 #5

    mfb

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    If the room walls cannot exchange energy with the molecule, then temperature is a meaningless concept. Temperature is a thermodynamic limit for many particles, it does not exist for single particles.
     
  7. Nov 14, 2016 #6
    A gas molecule is made of atoms. we have permission for talking about entropy of it's atoms in the room.

    Therefore, are you saying we couldn't understand a isolated vacuum room with a methane molecule inside it thermodynamically?
    If true, is there any approach in physics for understanding and predicting the future of this methane molecule?

    In addition: I still strongly believe methane molecule tends to break it's bond for reaching to maximum entropy and after that the atoms and protons decay. the decay satisfy the second law and this is the end.
     
  8. Nov 14, 2016 #7

    Bystander

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    The second law is not to be confused with "the dark side of 'the force.' "
     
  9. Nov 15, 2016 #8

    mfb

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    You can, but only as microcanonical ensemble. You cannot assign things like a temperature to it.
    Sure, the easiest description is via Newton's laws.
    You misunderstand how entropy works. It cannot get smaller. But it does not have to increase. In your example of an isolated molecule it does not change at all.

    Proton (or neutron) decay, if it happens at all, would eventually happen while the molecule is still a molecule, potentially ruining the chemistry and leading to something else, depending on which nucleon decays in which way.
     
  10. Nov 15, 2016 #9
    According to below site:
    http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/entrop.html
    Concept of entropy is that nature tends from order to disorder in isolated systems.
    Now you said me which one is more disorder?

    a) a isolated vacuum room with a methane molecules (this figure)
    ?temp_hash=ad30a7439c72ed2cb85ea3bd359964c3.png
    b) a isolated vacuum room with 5 freely atoms? (this figure)
    ?temp_hash=ad30a7439c72ed2cb85ea3bd359964c3.png
    Therefore, I think methane tends to break it's bond but it couldn't because first law is preceded to The second law.
     

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  11. Nov 15, 2016 #10

    mfb

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    Your bound molecule already has the maximal disorder for its energy. Done. Nothing will happen.
     
  12. Nov 21, 2016 #11
    I couldn't understand how you say methane has more entropy than it's components.

    In this formula 2H2+C-->CH4

    We have 3 mole of reactions and mole of production. so obviously methane has less entropy than 2 hydrogen molecules and 1 carbon and tends to break it's bond.
    If temperature be enough the breaking will happen more likely. but system is isolated and no changing in temperature could occur.
     
  13. Nov 21, 2016 #12

    mfb

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    A single molecule doesn't have a temperature. It is just flying around forever without doing anything else.
     
  14. Nov 26, 2016 #13

    DrDu

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    You can assign temperature even to a single methane molecule, but this means that you don't have full information about the state of the molecule. I.e. it could be that it is already dissociated at the very beginning.
     
  15. Nov 26, 2016 #14

    mfb

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    Not in any meaningful way.
     
  16. Nov 27, 2016 #15

    DrDu

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    You can have a methane molecule in a room with walls of a given temperature. Then the molecule will be in thermal equilibrium after a short time and you can assign it a temperature. If you now replace the walls with adiabatic ones, the temperature of the particle won't change. This kind of experiments are standard for all kinds of particles inside some particle traps. So the answer to the OPs question is: By assigning a temperature to the molecule, you already lack information on whether the molecule is intact or dissociated from the very beginning.
     
  17. Nov 28, 2016 #16

    mfb

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    OP specified no thermal contact to the walls and an intact molecule to start with.
     
  18. Nov 28, 2016 #17

    DrDu

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    All that is conserved is that there is one carbon and 4 hydrogen atoms. Whether they are bound or not is never certain at finite temperature.
     
  19. Nov 28, 2016 #18

    jbriggs444

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    Echoing what has already been said...

    You can meaningfully assign a temperature to a system of which a molecule is a part. But this does not entail that you can meaningfully and correctly assign that same temperature to all of the component parts of that system down to the level of individual molecules. In thermal equilibrium, there will be an energy distribution. Some molecules will have more energy at a given time. Some less. Even if you decide that the "temperature" of a single molecule is equivalent to its kinetic (plus vibrational plus rotational) energy, there is no guarantee that the energy of that molecule will be representative of the temperature of the larger system of which it is or had been a part.

    Once you close the system and eliminate thermal contact with the walls, the total energy of that methane molecule is fixed. If there is not enough energy to break its molecular bonds, then those bonds will not break. Without that energy, an unbound set of five atoms is not a possible state to which the system can evolve.
     
  20. Nov 28, 2016 #19

    DrDu

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    That energy of a small system representative of a canonical ensemble with a given temperature isn't fixed is well known and not at variance with the concept of temperature.
     
  21. Nov 28, 2016 #20

    jbriggs444

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    That is certainly true.

    Now then. If we had a single molecule in a room with perfectly reflective walls that neither absorb nor transmit energy, what would you mean by the "temperature" of the system?
     
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