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Thermodynamics: Irreversible process.

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider a gas with adjustable volume V, and diathermal walls, embedded in a heat bath of constant temperature T, and fixed pressure P. The change in the entropy of the bath is given by:

    delta S_bath = delta Q_bath/T = -deltaQ_gas/T = -1/T (delta E_gas + P deltaV_gas)

    By considering the change in entropy of the combined system, establish that ¨the equilibrium of a gas at fixed T and P is characterized by the minimum of the Gibbs free energy G = E + PV - TS¨

    3. The attempt at a solution
    I tried assuming that the gas was ideal and found that delta S_gas would be NR ln (V2/V1) but I don´t see how that helps. I also tried writing down from the equation of G, that dG = dE + PdV - TdS, and saw that ifyou substitute the thermodynamic identity, dE = TdS-PdV, dG will equal zero but I dont think this is valid here.

    Any ideas?

    Thanks,
    Vince
     
  2. jcsd
  3. Feb 10, 2008 #2

    Mapes

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    Your dG equation is incorrect, it should read [itex]dG=dE+P\,dV+V\,dP-T\,dS-S\,dT=-S\,dT+V\,dP[/itex]. Does this help?
     
  4. Feb 10, 2008 #3
    But it says P and T are constant so the SdT and VdP terms would be zero.

    Any other ideas?

    Thanks anyway.
     
  5. Feb 10, 2008 #4

    Mapes

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    Right, so doesn't dG=0 imply a minimum (or maximum)? Isn't this what you're trying to show?

    Edit: Looking again, I shouldn't have said your dG equation was incorrect. I was responding to the lack of the two differential terms VdP and -SdT, not noticing that you had already cancelled them out. Sorry about that. I think that your original conclusion that dG=0 answers the question.
     
    Last edited: Feb 10, 2008
  6. Feb 10, 2008 #5
    Really? I don't seem to be using anything in the equation: delta S_bath = delta Q_bath/T = -deltaQ_gas/T = -1/T (delta E_gas + P deltaV_gas)

    which was given. And the question asks us to use the change of entropy of the entire system, which I don't know if I'm properly using here. Can I just replace all the d's with delta's?

    The thermodynamic identity dE = TdS-PdV cancelling out everything else seems a bit too trivial to solve this question, considering that the other four questions in the problem set were much harder.
     
  7. Feb 10, 2008 #6
    Somebody please?
     
  8. Feb 11, 2008 #7

    Mapes

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    You're trying to show a basic result (it's common to start a thermo problem by stating that G is minimized at constant T and P), so it's hard to know what you're "allowed" to assume.

    You've already shown that dG=0, therefore G is at an extreme (maximum or minimum) when the gas is at equilibrium. The Second Law says that S is maximized at equilibrium (dS=0). Callen's Thermodynamics sets up the problem this way:

    [tex]dS=\frac{1}{T_g}dU_g+\frac{1}{T_b}dU_b-\frac{P_b}{T_g}dV_g-\frac{P_b}{T_b}dV_b=0[/tex]

    where the subscripts are for the bath and the gas. Any energy or volume lost by the gas adds to the bath, so [itex]dU_g=-dU_b[/itex] and [itex]dV_g=-dV_b[/itex]. From this we can deduce that [itex]T_g=T_b[/itex] and [itex]P_g=P_b[/itex] at equilibrium.

    Callen also offers a nice argument by reductio ad absurdum to show why entropy maximization corresponds to energy minimization. We assume that it doesn't; that we have a system at equilibrium with maximized entropy but non-minimized energy. We can therefore remove some of this energy in the form of work (doing work doesn't change the entropy) and reinsert the extracted energy by heating the system. This does increase the entropy. Since we end with the same energy but higher entropy, we must conclude that our original assumption was wrong and the energy must have been minimized.

    With dP=0 and dT=0, it's clear that U minimization implies G minimization, and you are done.
     
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