I'm a little bit confused about calculating entropy changes along irreversible paths by integrating over a reversible path. When using the central equation I can understand the argument, entropy and all the quantities we use to calculate the entropy change are state functions that have well defined end points. So the process between these two states doesn't matter, we will get the same entropy change. I find it difficult to grasp how we can use dS=dQ/T to calculate entropy changes in irreversible processes. Since Q is path dependent I don't see how the original argument applies. We can take the free expansion as an example, if we try to use dS=dQ/T we will get zero entropy change as no heat flows into the system. But using the central equation where all quantities are state functions we get a entropy change which is larger that zero as expected.
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$\begingroup$ i suggest you read up on the concept of integrating factor: en.wikipedia.org/wiki/Integrating_factor $\endgroup$– hyportnexCommented Sep 7, 2016 at 18:50
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$\begingroup$ I know the concept of integrating factor but fail to see how it has anything to do with my question. $\endgroup$– SC_ErkannCommented Sep 7, 2016 at 20:24
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$\begingroup$ $\delta Q$ is process dependent but in a reversible process the ratio $\frac {\delta Q}{T} $ is now state dependent not just process dependent. Now if you connect any two states with an irreversible process then to calculate the entropy change between them can be had by connecting the two states via a reversible process, if there is one, and there you can use the integrating factor. A state is a state irrespective of what process was used to reach it. The integration of $\frac {\delta Q}{T} $ to calculate the entropy change is only done over a reversible path. $\endgroup$– hyportnexCommented Sep 7, 2016 at 20:33
3 Answers
As you rightly say, all the quantities in the fundamental relation are functions of state, and so the change calculated using it along any 2 paths with the same end points will be the same.
When we use the fundamental relation to compute the change for an irreversible process, what we generally do is find a different path, a reversible path with the same endpoints, where we can use our nicer equations for reversible processes, such as $dS = \frac{dQ}{T}$, and integrate along that instead. Since heat and work are path dependent quantities, the amount of heat transferred and work done will generally be different along this new path to what they were on the true/actual path, but that is fine provided we have enough information to compute our integrals and end up in the right place.
How do we know we got to the correct end state if we were integrating along a totally different path? Well, our original path will have been defined by some set of constraints; to take your example, a free expansion is defined by the condition that $dU=0$. We will also have some conditions on where along that path we stopped, e.g. the final volume $V_f$. Now the original path was a 1d line in the state space of our system and our end condition gives us where on that line we finish, so the constrains and the end condition must be enough to tell us where the end point is; it is the point where all the constrains and the end condition are satisfied. So now to find whatever other thermodynamic quantities, not already specified, that we desire, we find a reversible path where all these conditions are true somewhere and integrate to that point.
So in the example of a free expansion, we know that between our start and end point $\Delta U=0$, so we will look for a reversible path where this is true. The easiest way to do this is to use the same constraint as the true path $$ dU = 0 \quad\Rightarrow \quad dQ = -dW $$ Now, since we are requiring our new path to be reversible, we can use the formulae we have for work and heat in that case. $$ TdS = pdV $$ So in order to calculate the change in entropy, we simply integrate along this new path until we reach our end point condition where $V = V_f$. Notice that along this new path heat was transferred, but it was compensated for by work being done, so the internal energy remained the same.
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$\begingroup$ It's becoming a bit more clear for me, but I had another example in my head while writing my question which still confuses me. Say you drop an object from a height which then falls onto the ground. The first law says that $Q=0$, since $\Delta U=W=mgh$. If we want to calculate the entropy change for this object along a reversible path we would get zero since $dQ=0$, how do I know that this corresponds to the actual entropy change in this case. Is $\Delta U=mgh $ and $\Delta V=0$ the necessary constraints? This approach feels similar to the faulty reasoning I gave for the free expansion. $\endgroup$ Commented Sep 7, 2016 at 20:19
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$\begingroup$ It doesn't make really make sense to talk about the volume of your ball as a variable, instead your macroscopic variables will be $T$, $S$, $h$ and $mg$. In this case the fundamental relation reads $dU = TdS + mgdh$ your constraint is that no heat was transferred and your final conditions are that you did $\Delta W$ and $h$ has gone form $h_0$ to $h_1$. From the first law $\Delta U = \Delta W$. You then have that along a reversible path $\int_{S_0}^{S^1} TdS = \Delta U - \int_{h_0}^{h^1} mg \,dh$ $\endgroup$ Commented Sep 7, 2016 at 21:03
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$\begingroup$ The right hand side you can compute since $mg$ is a constant (this is the equivalent to equation of state) The left hand side you might be able to do something with if you know the heat capacity of the ball. What this equation is saying is any work that didn't go into increasing the height of the ball went into increasing the internal energy through friction, which is, of course, irreversible. In the happy case that the work done equals $mg\Delta h$ then no energy was lost to friction the process was reversible, and no entropy produced or heat transferred as you would expect. $\endgroup$ Commented Sep 7, 2016 at 21:03
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$\begingroup$ Okay, I think I finally get it. Thanks, this has been bugging me for a while. $\endgroup$ Commented Sep 7, 2016 at 22:08
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$\begingroup$ @BySymmetry In the fundamental relation using Euler's theorem we get (ignoring chemical potential): $$U=TS - PV$$ All variables are state functions but why do we say that $U=U(S,V)$. I mean for different $P,T$ the function will have different values for the same $S,V$. Thanks in advance. $\endgroup$ Commented Mar 19, 2020 at 17:52
For an irreversible path, the integral of dQ/T is not equal to $\Delta S$. To get the change in entropy between the two end points of an irreversible path, you need to devise a reversible path between the same two end points, and calculate the integral for that path. It doesn't matter what reversible path you choose; they will all give the same value for $\Delta S$. For more details on this, see the following link:
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
This link contains contains a simple recipe for determining the entropy change for any arbitrary irreversible process, and presents several typical examples of irreversible processes encountered in thermodynamics courses, and how to apply the recipe to determine the change in entropy for each.
There is an implicit but crucial assumption in thermodynamics (your calculations are based on it): Any irreversible process can be closed by a reversible process to become a cycle. If it is false, thermodynamics would collapse:
http://philsci-archive.pitt.edu/archive/00000313/ Jos Uffink, Bluff your Way in the Second Law of Thermodynamics, p. 39: "A more important objection, it seems to me, is that Clausius bases his conclusion that the entropy increases in a nicht umkehrbar [irreversible] process on the assumption that such a process can be closed by an umkehrbar [reversible] process to become a cycle. This is essential for the definition of the entropy difference between the initial and final states. But the assumption is far from obvious for a system more complex than an ideal gas, or for states far from equilibrium, or for processes other than the simple exchange of heat and work. Thus, the generalisation to all transformations occurring in Nature is somewhat rash."
