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Suppose that one mole of an ideal gas at $P_1$ and $V_1$ adiabatically expands to $P_2$ and $V_2$, reversibly and irreversibly (two separate processes). Since $\Delta U$ is a state function, why is it wrong to say that $\Delta U_{rev} = \Delta U_{irrev}$? (Since in an adiabatic process, there is no heat transfer, we know that $\Delta U = w$)?

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marked as duplicate by Jan, bon, Todd Minehardt, Wildcat, M.A.R. ಠ_ಠ Oct 15 '15 at 19:22

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    $\begingroup$ Related: Reversible and Irreversible adiabatic expansion $\endgroup$ – Loong Oct 15 '15 at 18:09
  • $\begingroup$ I can't answer this question but I still want to nitpick (how typical of me). $U$ is the state function, not $\Delta U$. $\endgroup$ – orthocresol Oct 15 '15 at 18:10
  • $\begingroup$ Both $U$ and $\Delta U$ are state functions. $\endgroup$ – notorious Oct 15 '15 at 18:11
  • $\begingroup$ A state function is a property of a system. $U$ is a property of a system, and $\Delta U$ is, loosely speaking, a "property" of a process that said system undergoes. $\endgroup$ – orthocresol Oct 15 '15 at 18:13
  • $\begingroup$ Please do not use MathJax for question titles due to searching issues. Also welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. Visit the help center to find out more about it. $\endgroup$ – Jan Oct 15 '15 at 18:15
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I think the premise of the question itself is false: if you start with a given state (P1, V1) and undergo a reversible, adiabatic (hence isentropic) expansion, you can get to state P2, V2. But if you don't have the requirement of reversibility, I don't think it is possible to gain the same second state - there will be either internal heating/cooling, causing the expected final volume (or pressure, depending on the mechanism of expansion) to not be P2 or V2.

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