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A demonstration on the necessary positive change in the entropy

  1. Nov 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello everyone. My problem is as follows: In a spontaneous process where two bodies at different temperatures [itex]T_{1}[/itex] and [itex]T_{2}[/itex], where [itex]T_{1}>T_{2}[/itex], are put together until they reach thermal equilibrium. The number of atoms or molecules of the first is [itex]N_{1}[/itex] and [itex]N_{2}[/itex] for the second one, with [itex]N_{1} \neq N_{2}[/itex], and they have heat capacities equal to [itex]C_{V_{1}}=aN_{1}k[/itex] and [itex]C_{V_{2}}=aN_{2}k[/itex], with [itex]a[/itex] given with the appropriate units. Past some sufficiently large time, the system reaches a temperature [itex]T[/itex], provided that [itex]T_{1}>T>T_{2}[/itex], which is in function of the initial temperatures and the number of atoms or molecules of the two bodies. The problem is that i can't demonstate that the change of the entropy of the system as a whole is positive, i.e. [tex]\bigtriangleup S>0[/tex]
    2. Relevant equations
    When i compute the change of the entropy for the i-th body, i get
    [tex]\bigtriangleup S_{i}=\int_{T_{i}}^T \! \frac{1}{T} \, \mathrm{d} Q=\int_{T_{i}}^T \! \frac{aN_{i}k}{T} \, \mathrm{d} T=aN_{i}k\int_{T_{i}}^T \! \frac{1}{T} \, \mathrm{d} T=aN_{i}k\ln{\frac{T}{T_{i}}}[/tex]
    With the hypothesis that the entropy is an extensive property, then [tex]\bigtriangleup S=\bigtriangleup S_{1}+\bigtriangleup S_{2}=aN_{1}k\ln{\frac{T}{T_{1}}}+aN_{2}k\ln{\frac{T}{T_{2}}}[/tex]
    So i just have to prove that [tex]N_{1}\ln{\frac{T}{T_{1}}}+N_{2}\ln{\frac{T}{T_{2}}} > 0[/tex]

    3. The attempt at a solution
    I think that i have to use the two cases ([itex]N_{1}>N_{2}[/itex] and [itex]N_{1}<N_{2}[/itex]), and using the fact that [itex]T_{1}>T>T_{2}[/itex], to prove the inequality, but i have tried to do it in very different ways, and i get nothing, so i think there is some trick to demonstrating that, but i'm still a bit of an amateur in proving tricky inequalities.
     
  2. jcsd
  3. Nov 24, 2012 #2

    haruspex

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    Don't you need conservation of energy? Otherwise there's nothing to stop both ending arbitrarily close to T2.
     
  4. Nov 24, 2012 #3
    Actually i've used conservation of energy, and i can determine the temperature [itex]T[/itex] of equilibrium, which is [tex]T=\frac{N_{1}T_{1}+N_{2}T_{2}}{N_{1}+N_{2}}[/tex]but when i substitute that expression on the inequality, the later just complicates a little bit more. In despite of this, i think i must substitute [itex]T[/itex] in the inequality, but i get nothing again.
     
  5. Nov 24, 2012 #4

    haruspex

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    I'm sure it can be done this way, but I feel that performing the integral just makes life harder. Consider a small dQ transferred. The hotter body loses dQ/T1, the cooler gains dQ/T2. Since T1 > T2, the sum has increased.
     
  6. Nov 24, 2012 #5
    Thanks haruspex. If i understand, you say that [tex]dS=dS_{1}+dS_{2}=\frac{dQ}{T_{1}}+\frac{dQ}{T_{2}}[/tex]With the fact that [itex]T_{1}>T_{2}[/itex], therefore a small dQ transferred between them would lead a change in the first entropy which is smaller than the change of the second one, without sayin anything about [itex]C_{V_{1}}[/itex] nor [itex]C_{V_{2}}[/itex]? But maybe i should take into account a small change in the temperature [itex]T_{1}[/itex] and in [itex]T_{2}[/itex] just to get a formal demonstration, don't you think?
     
  7. Nov 24, 2012 #6

    haruspex

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    [tex]dS=dS_{1}+dS_{2}=\frac{-dQ}{T_{1}}+\frac{dQ}{T_{2}}[/tex]
    Yes, I suppose strictly you should say ΔS1 ≥ -dQ/(T1 + ΔT1) and ΔS2 ≥ dQ/(T2 + ΔT2) (ΔT1 being negative). You will probably also need to specify T1 + ΔT1 > T2 + ΔT2, i.e. the deltas are small enough that the temperatures do not cross over.
     
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