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Decrease in entropy atoms problem

  1. Aug 7, 2005 #1
    If one could observe the individual atoms making up a piece of matter and note that during a process of change their motion somehow became more orderly, than one may assume which of the following in regard to the system?
    a. increase in entropy b. decrease in entropy c. gains in thermal energy d. positive work done on

    I know the answer is b. decreases in entropy, but my question is why? This may seem like a silly question, but I just wanted to make sure I understood what's going on here. If there is a decrease in entropy, that would mean that temperature is going up, correct? I'm also not quite sure on what they mean by "orderly," the atoms would be speeding up if this was the case, but I don't see what that has to do with being "orderly." Thanks for any help!
     
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  3. Aug 7, 2005 #2

    LeonhardEuler

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    A decrease in entropy will usually mean a decrease in temperature. Remember the definition of entropy: [itex]dS=dq_{rev}/T[/itex]. This means that an infintisimal change in entropy is equal to the heat supplied in a reversible path. Since a positive change in entropy coresponds to a positive intake of heat, this most often means that the temperature will increase with an increase in entropy (although exceptions are not difficult to find), and conversely that a decrease in entropy usually coresponds to a decrease in temperature.
     
  4. Aug 7, 2005 #3
    I understand what you're saying, but for some reason I just don't understand how that works. Entropy = deltaQ/T, it just seems to make sense that if you plug in a higher number for T, entropy is going to decrease. What am I missing here? Thanks a lot for your help, sorry I'm making this so difficult! I just want to make sure I understand.
     
  5. Aug 7, 2005 #4
    Also, just to see if i'm on the right track, could someone tell me if i'm right here.

    A 1.0 kg chunk of ice at 0 C melts, absorbing 80000 cal of heat in the process. Which of the following best describes what happens to this system?
    A. increased entropy b. lost entropy c. entropy maintained constant d. work converted to energy

    It would be a. increased entropy, because as this happens, the process has an effect on its surroundings. This means the process is irreversible, and if its irreversible, the total entropy always increases.

    Is that right?

    Thanks again for your patience
     
  6. Aug 7, 2005 #5

    LeonhardEuler

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    I see what the problem is. Notice that the equation is [itex]dS=dq_{rev}/T[/itex] and not [itex]S=dq_{rev}/T[/itex]. This means that a change in entropy will be lower at higher temperature. So if the entropy is already, say, 5J/K then a higher temperature will mean this: If you add a little more heat the entropy will not increase by as much as it would at a lower T. One analogy to this is if you have a neat room and throw a sock on the floor it will increase the messiness of the room more than if you threw the same sock on the floor of a sloppy room. The sloppy room is still messier than the neat one, but the one additional sock just doesn't increase it by that much.

    At the same pressure, the same pure substance will always have a higher entropy at higher temperatures. You can see this because the third law implies that we can take the entropy of a crystaline pure substance at 0°K to be 0. This means:
    [tex]S=\int_{0}^{T_1}dS=\int_{0}^{T_1}\frac{dq_{rev}} {T}[/tex]
    So if [itex]S_1[/itex] is the entropy at [itex]T_1[/itex] and [itex]S_2[/itex] is the entropy at [itex]T_2[/itex] and [itex]T_1<T_2[/itex], then:
    [tex]S_2=\int_{0}^{T_2}\frac{dq_{rev}} {T}= \int_{0}^{T_1}\frac{dq_{rev}} {T} + \int_{T_1}^{T_2}\frac{dq_{rev}} {T}[/tex]
    [tex]=S_1 + \int_{T_1}^{T_2}\frac{dq_{rev}} {T}[/tex]
    Since the last integral is positive [itex]S_2[/itex]>[itex]S_1[/itex]
     
    Last edited: Aug 7, 2005
  7. Aug 7, 2005 #6

    LeonhardEuler

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    I didn't see this before I started my rather long previous post, so I appologize for not answering it.

    If the heat is supplied at 0°C, then the process is reversible, since for the system we have
    [tex]\Delta S_{sys}=\int \frac{dQ_{rev}}{T}[/tex]
    T is constant, so:
    [tex]\Delta S_{sys}=\frac{Q}{T}=\frac{Q}{273.15°K}[/tex]
    Whille for the surroundings, we have
    [tex]\Delta S_{surr}=\int \frac{dQ_{rev}}{T}=\frac{-Q}{T}=\frac{-Q}{273.15°K}[/tex]
    And these clearly sum to zero. However, notice that the question asked for the change in entropy in the system. This means that what happened in the surroundings is not relevant. Since the heat flowed into the system, its entropy increased.
     
  8. Aug 7, 2005 #7
    Thank you very much for clearing all this up. I'm not sure if I'm grasping a complete understanding quite yet, but this has definitely helped a ton.
     
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