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Thermodynamic equilibrium and Entropy

  1. Mar 23, 2007 #1
    I got this question wrong on a test:

    What is the equilibrium condition for an isolated system?

    I think the answer he wanted is the system is a maximum entropy, and that makes sense to me. But I put dS =0, which I think is basically the same thing. dS would be zero iff the system is at maximum entropy.

    Or am I way off? :rofl:
     
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  3. Mar 23, 2007 #2

    G01

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    [tex]\Delta S=0[/tex] is the condition for equilibrium for a completely isolated system.

    The entropy of the system will increase according to the second law until the system hits maximum entropy, at which point
    [tex]\Delta S=0[/tex]

    Saying this should be the same as saying that the entropy is at a maximum.

    Was there any other information given with the question?
     
    Last edited: Mar 23, 2007
  4. Mar 23, 2007 #3
    The full question is this

    Write down the relationship between heat transfer dQ and entropy change dS for an infinitesimal reversible process, in which heat is supplied to a system at temperature T. How is this expression modified for an irreversible process? What is the equilibrium condition for an isolated system?


    I got the first two parts right. After looking at it again, I think I go half credit for the dS=0.
     
  5. Mar 23, 2007 #4

    G01

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    Well, I'm assuming for one of the first two parts that you stated somewhere that:

    [tex]\Delta S = \int\frac{dq}{T}[/tex]

    For [tex]\Delta S[/tex] to be zero, dq would also have to be constant at 0. Maybe your professor wanted you to state the heat transfer would also have to be zero. You may want to discuss this problem with him.

    I'm no expert though, maybe someone else here has a little more insight, but I agree with you.
     
  6. Mar 23, 2007 #5
    I did write down that integral for clausius' theorem.


    The first two parts...

    "Write down the relationship between heat transfer dQ and entropy change dS for an infinitesimal reversible process, in which heat is supplied to a system at temperature T." dQ=TdS

    "How is this expression modified for an irreversible process?" dQ<TdS

    "What is the equilibrium condition for an isolated system?" I put dS=0. My classmate put "The system is at maximum equilibrium" and got it right.

    Im not going to argue the half point with him, as that will do me no good. I am going to ask him if dS=0 is an equilibrium condition.
     
  7. Mar 23, 2007 #6

    Andrew Mason

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    Not so. In order to determine the total change in entropy when heat flows, you have to look at the system and surroundings, which includes both reservoirs.

    [tex]\Delta S = \Delta S_H + \Delta S_C = \int dS_H + \int dS_C = \int dQ_H/T_H + \int dQ_C/T_C[/tex]

    So for [itex]\Delta S = 0[/itex]:

    [tex]\int dQ_H/T_H + \int dQ_C/T_C = 0[/tex]

    [tex]\int dQ_H/T_H = - \int dQ_C/T_C[/tex]

    AM
     
  8. Mar 23, 2007 #7

    Andrew Mason

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    Since the temperature difference is infinitesimal, [itex]T_H = T_C[/itex] and [itex]Q_H = -Q_C[/itex], [itex]dS_H = dQ_H/T_H[/itex] and [itex]dS_C = dQ_C/T_C = -Q_H/T_H[/itex], so

    [itex]dS = dS_H + dS_C = dQ_H/T_H - dQ_H/T_H = 0[/itex]

    If the temperature difference is non-zero, [itex]T_H > T_C[/itex] and [itex]Q_H = -Q_C[/itex], [itex]dS_H = dQ_H/T_H[/itex] and [itex]dS_C = dQ_C/T_C > -Q_H/T_H[/itex], so

    [itex]dS = dS_C + dS_H = dQ_C/T_C - dQ_H/T_H > 0[/itex]

    So these are the only two possibilities for a system moving to thermal equilibrium. What is the condition for thermal equilibrium then, in terms of dS?

    AM
     
    Last edited: Mar 23, 2007
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