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.
Thermodynamics – Entropy and Reversible Paths: An In-Depth Exploration
entropyreversibilitythermodynamics
Related Solutions
- For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is $$\Delta S = \int_a^b \frac{dQ}{T}$$ For reversible path between two states, entropy of the universe (Or any isolated system) is zero. $$\Delta S + \Delta S_\text{surroundings} = 0$$ So You cannot just take any system and say that entropy change between two states for this system will be zero because it is zero for a reversible process. It is not. So when you say
Surely the total change of entropy is zero.
for reversible process of closed system, it is not true. Answer to This question might help you here.
- As for the first part of your question, I don't understand what the question is. Could you edit it to be more specific?
Also, You said the following, which is false.
The entropy changes of the system are same for both cases, reversible and irreversible processes because the first and final states are unchanged. In this situation I think the surrounding also have the same first and final states for both reversible and irreversible processes.
We don't know whether surrounding has same first and final states or not. We only know about the system's first and final states. Think about it this way: In a reversible process, system is going from state A to B, and so is surrounding. Since it is reversible, $ \Delta S_{System} = - \Delta S_{Surrounding} $. So ultimately, $ \Delta S_{Universe} = 0$.
Now for an irreversible process, we know that through this irreversible path, the System goes from A to B. We don't know about surroundings. Now, since system's states are same, $ \Delta S_{System} $ will be same as above case. For the surrounding, you say that states are same as the reversible case. But then, here also $ \Delta S_{Surrounding} $ would be same as before and again $ \Delta S_{Universe} = 0$. But we know that that is not true for irreversible process. Hence, Surrounding's states must be different. So, in irreversible process, while the system goes A to B same as before, the surrounding must go from A to some C. There is no reason to believe that it would go from A to B again.
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In calculating the entropy change for an irreversible process between two thermodynamic equilibrium states of a system, the first step in the procedure should always be to FORGET ALL ABOUT THE IRREVERSIBLE PROCESS PATH ENTIRELY. The entropy change can only be calculated for a reversible path between the same two end states.
Step 2: Identify a convenient reversible process path between the same two thermodynamic equilibrium states. Any reversible path will do, because they all give the same result for the entropy change.
Step 3: Calculate the integral of dQ/T for the reversible path you have identified. This is $\Delta S$
For the example that you gave, I like use a process path where I start out by separating the two objects entirely, and then contacting each of them with a continuous sequence of constant temperature baths running from the original temperature of the object to its final equilibrium temperature. This gives me the integral that you alluded to in your question for each body.
For an irreversible process path, the temperature of the system is not even uniform spatially throughout, so what temperature do you use in evaluating the integral of dQ/T? Cauchy noted that, if you use the temperature at the boundary where the heat transfer is occurring (say for one of the bodies), you always find that the integral you get over the irreversible path is less than the entropy change of the body. This is called the Clausius inequality:
$$\Delta S\geq \int (dQ/T)_B$$where B signifies the boundary where the heat transfer is occurring.
Best Answer
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.