I have seen many similar questions but have not found my answer. Why do we use the actual heat involved in the process to calculate entropy change of surrounding? The only answer I can think of is that we are treating the process as reversible with respect to the surroundings even if it is irreversible with respect to the system. But, this is an irreversible path for the system, and a reversible path will not involve the same heat, so now, the entropy of surrounding calculated must be different. This seems to contradict the idea that the entropy of surroundings must be a state function. So why is this reasoning incorrect?
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$\begingroup$ See my answer in the following thread: chemistry.stackexchange.com/questions/42681/… $\endgroup$– Chet MillerJan 3, 2016 at 15:05
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When you evaluate the change in entropy of an entity (in this case the surroundings) between an initial and final (equilibrium state), you separate that entity from everything else, and then you only look at how that entity has changed between the two states, disregarding the actual process path. In the case of the surroundings (treated as an infinite reservoir), it had received a certain amount of heat between the initial and final states, and its internal energy and entropy increased as a result. In the formalism of thermodynamics, changes in an ideal infinite reservoir are always regarded as taking place reversibly, irrespective of other irreversibilities in the actual process. So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir).
You may also find my response in the following link pertinent to this question: Thermodynamics entropy and temperature
answered Jan 3, 2016 at 15:22
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$\begingroup$ Yes, I understand that the process can be modelled as reversible with respect to the surroundings. But consider a reversible process between the same two states for the system, then this process is also reversible for the surroundings, but since it involves a different heat change, the entropy calculated must be different for the same process in two different ways. $\endgroup$ Jan 3, 2016 at 15:41
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$\begingroup$ As I said, to get the overall change in entropy, you need to look at each of the entities separately between their actual initial and final equilibrium states. You don't run the entire process over again reversibly for the combination of system and surroundings to get the change in entropy. You take each entity separately over a reversible path that may or may not bear any resemblance to the actual path followed by the system in the irreversible process (or to the reversible paths selected for the other entities). $\endgroup$ Jan 3, 2016 at 15:58
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$\begingroup$ So now if I treat the surrounding as one entity, any process is reversible for this entity due to it being modelled as a large reservoir.Why should I calculate the entropy change using the q for the actual process. As you said, the entropy change for any entity must depend only on the amount of heat extracted reversibly, independent of their resemblance to the actual process.What I claim is that there are infinite ways in which I can take the surrounding between the same states reversibly, involving different amount of heat extracted. $\endgroup$ Jan 3, 2016 at 16:35
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$\begingroup$ That is just not so. If the surroundings is purely an infinite reservoir, then the change in internal energy of the reservoir is just Q for the actual process (no work is done by the reservoir). Once the internal energy change is fixed (i.e., the final state is fixed) and the temperature is fixed, the entropy change can only be Q/T. $\endgroup$ Jan 3, 2016 at 16:45
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2$\begingroup$ Sure. The internal energy per unit mass is just another example of an intensive property. And we know that it takes two intensive properties (in this case internal energy and temperature) to define a thermodynamic equilibrium state of a single phase system. $\endgroup$ Jan 3, 2016 at 19:31