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Entropy and Enthalpy for adiabatic process

  1. May 28, 2014 #1
    1) What is the reason why dH!=0 for an adiabatic(q=0) reversible process?
    The mathematical argument is irrefutable and it is clear that it has to do with the process not being isobaric:
    ΔH=ΔU+PΔV+VΔP , ΔU=work=−PΔV
    Therefore, ΔH=VΔP and this is not 0.
    However, I do not understand it conceptually...

    2) Consider an adiabatic irreversible process from state A to B. At first I would expect to have dS=0, as there is no heat transfer, However, if we construct a reverse isothermal path that connects state B and A, then S(B→A)=-S(A→B). The entropy calculated for the isothermal path is positive, so the entropy for the irreversible process is negative, which agrees with the clausius inequality.
    My question is: conceptually, why should a reversible adiabatic process has dS=0 and an irreversible adiabatic process have a dS!=0 if there is no heat transfer?

    Thanks.
     
  2. jcsd
  3. May 28, 2014 #2
    As you showed, it's actually not equal to zero. What made you think that dH = 0 for an adiabatic reverisble process?
    dS=dq/T only for a reversible path between the initial and final equilibrium states of a system. Otherwise, dS is greater than dq/T. See my blog in my PF area for a more in-depth discussion.

    Chet
     
  4. May 28, 2014 #3

    WannabeNewton

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    What's there to understand conceptually? It's just a consequence of the thermodynamic relation ##dH = SdT + Vdp## which is itself just a change of variables ##H = E + pV## of the conservation of energy equation ##dE = \delta Q - \delta W## when restricted to a quasi-static process.

    I'm just adding on to Chet here by looking at things statistically. If the process is not reversible then by definition, for the total number of accessible microstates ##\Omega_i(E)## of the initial equilibrium state and the total number of accessible microstates ##\Omega_f(E)## of the final equilibrium state, we have ##\Omega_f > \Omega_i##; here ##E## is the average energy of the system. There clearly need not be any heat transfer in order for ##\Omega_f > \Omega_i## to hold. As a very simple example, just consider a cylinder of ##N## molecules of a certain ideal gas confined to half the volume ##V/2## of the cylinder by means of a piston initially locked in place; then ##\Omega_i(E) \propto (V/2)^N##. Imagine now that the piston is removed so quickly (instantaneously) that no work gets done by the gas and no heat gets transferred to the gas; we then wait for the gas to reach a new equilibrium. This final equilibrium state will of course be the one for which the gas occupies the entire volume of the cylinder. But then ##\Omega_f(E) \propto (V)^N## with the exact same proportionality factor since the energy of the gas didn't change at all (##\Delta E = W + Q = 0##) so ##\Omega_f > \Omega_i##.
     
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