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Homework Help: Entropy change in a irreversible adiabatic process

  1. Feb 24, 2014 #1
    One mole of an ideal monatomic gas initially at 298 k expands from 1.0 L to 10.0 L. Assume the expansion is irreversible, adiabatic, and no work is done. Calculate delta S of the gas and delta S of the surroundings.

    I know that delta dS = dq/T but q = 0 in adiabatic processes right? So does dS = 0 in all adiabatic processes?
  2. jcsd
  3. Feb 24, 2014 #2
    Hi chemkid01. Welcome to Physics Forums!!!

    dS is not equal to zero in all adiabatic processes, and dS is not equal to dq/T, except in reversible processes. Your process is not a reversible process. To figure out what the entropy change is for an irreversible process, you need to dream up a reversible path between the exact same initial equilibrium state of the system and the exact same final equilibrium state. For that path, dS = dq/T, and you can get the entropy change. For more details on this, see my Blog at my PF home page.
  4. Feb 24, 2014 #3
    Ok, so for this case, since the process is adiabatic and q = 0 and no work is done, then delta U = q + w = 0. And since it's an ideal gas, U only depends on T so T is constant as well. So could my dreamed up reversible process be a reversible isothermal expansion from 1.0 L to 10.0 L at 298 k?
  5. Feb 24, 2014 #4
    Yes. Nice analysis. That's the one I was thinking of.

  6. Feb 24, 2014 #5
    Ok, thanks for the help! One more thing, I'm not exactly sure what makes a process reversible/irreversible. For example, the question says that it is an irreversible, adiabatic expansion from 1.0 L to 10.0 L. Why can't my alternate reversible path simply be a reversible, adiabatic expansion from 1.0 L to 10.0 L. Is that possible?
  7. Feb 24, 2014 #6
    No, because you won't be able to hold the temperature constant over that reversible path. So, the initial and final equilibrium states won't be the same as for the original irreversible path.

  8. Feb 24, 2014 #7
    Hmm, ok, I don't think I quite understand reversibility. Could you explain what it is about the path being reversible that makes it so you won't be able to hold the temperature constant? Also, on a side note, since a reversible, isothermal expansion does bring you from the same initial state to the same final state as the irreversible, adiabatic expansion, does that mean that in this case, on the p-v diagram, the isotherm and the adiabat are the same line? Thanks a lot for your help by the way!
  9. Feb 25, 2014 #8
    Again, I refer you to my Blog. If a process is carried out irreversibly, the temperature and/or the pressure will vary with spatial position within the system. To go from the initial state to the final state reversibly, you need to control the temperature and the pressure at the interface with the surroundings such that the variations of temperature and pressure within the system are vanishingly small. This can be done by changing the temperature and pressure at the interface very gradually, so the system has time to respond, and such that, at any instant of time during the transition, the system is only vanishingly removed from thermodynamic equilibirum.
    In the case you are studying, no work is being done by the expanding gas, because the gas is essentially expanding against a vacuum. To carry out the process reversibly (and isothermally), you need to provide a force against the expansion, and very gradually back off on this force. This allows the system to do work against the force. To hold the temperature constant during this reversible path, you need to add heat Q. If you try to do it adiabatically, ΔU=-W, and the temperature will drop. So the reversible path between the initial and final states (which, for your problem are at the same temperature) can't be accomplished unless you add heat.

    No. The reversible path between the initial and final states is not adiabatic.

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