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Entropy Change in Reversible vs Irreversible Gas Expansions

  1. Feb 26, 2017 #1
    • Poster has been reminded to post schoolwork-related questions in the HH forums and use the Template
    Hello! I have this GRE question:
    In process 1, a monoatomic ideal gas is heated from temperature T to temperature 2T reversibly and at constant temperature. In process 2, a monoatomic ideal gas freely expands from V to 2V. Which is the correct relationship between the change in entropy ##\Delta S_1## in process 1 and the change in entropy ##\Delta S_2## in process 2?

    So, I am not sure in the first process what do they mean by, at constant temperature, if the temperature goes from T to 2T, but leaving that aside the correct answer is: ##0 < \Delta S_2 < \Delta S_1 ##. However, another option is ##0 =\Delta S_1 < \Delta S_2##. My main question is: isn't the change in entropy always 0 in a reversible process? So, shouldn't be this answer the right answer (being also the only one with ##0 =\Delta S_1##)?

    Thank you!
  2. jcsd
  3. Feb 26, 2017 #2


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    I think it is a misprint. Probably, the question meant to to say "from volume V to volume 2V reversibly and at constant temperature" the idea being to compare the entropy change for the two processes when the volume changes are identical.
  4. Feb 26, 2017 #3
    But still the change in entropy should be 0, as the process is reversible, right?
  5. Feb 26, 2017 #4
    Not necessarily. It depends on whether you are talking about the system or about the system plus surroundings.
  6. Feb 27, 2017 #5
    Hello! I am a bit confused. When you say reversible process, don't you always talk about system plus surroundings? If you talk only about the system, then basically any process that bring the system to the initial state would be reversible, as the change in entropy of the system would be zero (but of course the change in entropy of the universe would increase). So isn't somehow pointless to define reversibility only for the system?
  7. Feb 27, 2017 #6
    No. If the system undergoes the exact same path it would have in a reversible process (i.e., a continuous sequence of thermodynamic equilibrium states), even if the surroundings did not undergo a corresponding continuous sequence of thermodynamic equilibrium states (i.e., it underwent an irreversible path), the process is still considered reversible for the system but not for the surroundings. An example of this would be where you manually (by hand) cause a gas in a cylinder/piston arrangement to undergo a reversible adiabatic compression (or expansion) by controlling the history of the force that you apply to the piston. Certainly the biological and mechanical processes taking place in your body during this process are not reversible. So the gas has undergone a reversible process, but not the surroundings. In Fundamentals of Engineering Thermodynamics by Moran et al, this is referred to as an internally reversible process.
  8. Feb 28, 2017 #7
    Hello! I think I formulated my statement in a wrong way, but this is also what I said. There are was to make just the system reversible. Shouldn't the term "reversible" refer to both the system and the surroundings (and another term for just the system - like you said)? But in any case, in the problem that I posed initially, doesn't reversible means a 0 change in entropy for the system they are talking about?
  9. Feb 28, 2017 #8
    In both process 1 (with the typo corrected to go from V to 2V, rather than T to 2T) and process 2, the change in entropy of the system is greater than zero, and is the same for both processes. It has to be the same for both processes because they both start in the same state and both end in the same state. Here are a couple of Physics Forums Insights articles I wrote which may help to supplement your understanding:



    In the second article, pay particular attention to Example 3.

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