Suppose that the ideal gas is expanding adiabatically from the state $(P_1, V_1, T_1)$ to the state $(P_2, V_2, T_2)$

I think this process can be allowed to happen reversibly or irreversibly, but then I face a puzzling contradiction regarding entropy change ($\Delta S$).

If the process occurred reversibly, $\Delta S = 0$ since there are no heat transfer and no entropy generation.

If the process occurred irreversibly, $\Delta S > 0$ since there is entropy generation by the irreversibility of the process.

However, because entropy is a state function it cannot be zero and non zero for the two fixed chosen states $(P_1, V_1, T_1)$ and $(P_2, V_2, T_2)$


2 Answers 2


If the two end states you have described are those corresponding to an adiabatic reversible process, then it will be impossible to devise and conduct an adiabatic irreversible process that passes between these same two end states. If the two end states you have described are those corresponding to an adiabatic irreversible process, then it will be impossible to devise and conduct an adiabatic reversible process between the same two end states; some heat transfer will be required.

  • $\begingroup$ I got it. Could you explain a bit more where does the impossibility in your explanation come? Or is it impossible to happen because the contradiction I mentioned in my question? $\endgroup$
    – Royalblue
    May 29, 2018 at 2:30
  • $\begingroup$ In the irreversible process, there will be entropy generated, so the final entropy will be greater than the initial entropy. So if there is a reversible process between the same two end states, you will have to have $$\Delta S=\int{q_{rev}/T}$$, and $q_{rev}$ will have to be non-zero. This means that the reversible process will have to be non-adiabatic. $\endgroup$ May 29, 2018 at 2:43
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    $\begingroup$ Ah I seem to get your point. No matter how I conduct the experiment I wont be able to obtain two end states that have the same pv^gamma value as long as the process is carried out irreversibly. $\endgroup$
    – Royalblue
    May 29, 2018 at 2:52

Not really. The fact that you are taking the gas along an adiabatic process already implies that the process is reversible (since it should be at thermal equilibrium with a reservoir during the entire path). If, however, your only requirement is that the initial and final states follow $p_1V_1^\gamma=p_2V_2^\gamma$, then, since entropy is a state function, the entropy of the gas is the same, but the entropy of the whole Universe could (and likely will) be larger.

  • $\begingroup$ I am intereded in the gas system (and its entropy change) not the whole universe! :) $\endgroup$
    – Royalblue
    May 29, 2018 at 2:31
  • $\begingroup$ But when you talk about reversibility, what matters is the whole Universe, not isolated parts of it. $\endgroup$ May 29, 2018 at 2:36
  • $\begingroup$ The second law states that the total entropy of the Universe must go up, not the entropy of your specific vessel of gas. $\endgroup$ May 29, 2018 at 2:45
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    $\begingroup$ the word adiabatic in thermodynamics means no transfer of heat, although in quantum mechanics we often use the word for entropy-conserving (isentropic in thermodynamics). An example of irreversible adiabatic effect is the cooling of gas by expansion through a throttle (Joule-Thomson effect). $\endgroup$
    – wcc
    Aug 16, 2018 at 6:04

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