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The entropy change of an ideal gas is calculated as \begin{equation} \Delta S = C_V n\ln(\frac{T}{T_0}) + nR\ln{\frac{V}{V_0}} \end{equation} derived by integrating $dU = TdS - pdV$. It is only true for reversible processes, since we have assumed that the change in heat is $DQ = TdS$. This equation (according to the solution of an exercise) is applied when you have a container with two different gases separated by a wall, and then you suddenly remove the wall and let the two gases mix. The way I see it, this is an irreversible process, yet we apply an equation which we have derived from an assumption only true for reversible processes. This is a rather frequent problem of mine, I see $DQ = TdS$ applied in a lot of exercises where I would have assumed that the process is irreversible. What am I missing here?

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Entropy is a state variable. That means for every state you have a single value for entropy of that state. So the entropy difference of two states does not depend on the process that takes you from one state to the other. Therefore you can use this formula if the initial state is $(T_0,V_0)$ and final state is $(T,V)$. However the change in the Entropy of universe (system+surrounding) does depend on the process: $\Delta S_{universe}=0$ for reversible process and $\Delta S_{universe}>0$ for irreversible process.

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  • $\begingroup$ Why does the enthropy of the Universe depend on the process? The argument you said should also apply to it. $\endgroup$ May 12 '15 at 11:49
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    $\begingroup$ Imagine two processes one reversible and the other irreversible. Both of them takes the system from state A to B but in one process the state of surrounding goes from say C to D in the other process from C to D' not D. That's why the overall change of entropy of universe is different. $\endgroup$
    – richard
    May 12 '15 at 12:48
  • $\begingroup$ I'm starting to understand things. However I have problems with determining what is the system and what is the environment. Consider the Gay-Lussac experiment for example. A gas is in one half of a container held in place by a rigid and heat insulating wall. The container is also rigid and heat insulating. We then suddenly remove the wall, and let the gas expand in an irreversible process. However, if I would let the wall slide in the box like a piston very slowly and frictionless, the expansion would be reversible. Waht is the universe that is taken to different states in the two processes? $\endgroup$ May 12 '15 at 18:16
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    $\begingroup$ Ok, in the second process you need to do one more thing in order to get the same state for the gas (system) as in the first process and that is some reversible heat flow into the gas because you have taken some of its internal energy in the form of work in the expansion. That heat comes from outside (surrounding) and lowers the entropy of surrounding and in the end the entropy of universe is unchanged. $\endgroup$
    – richard
    May 13 '15 at 6:59

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