Physics Asked on March 26, 2021
I am studying heat and thermodynamics by zemansky. In the chapter entropy, there is a statement under the section ‘Entropy and Irreversibility’-
In the case of any process involving the isothermal transformation of work W done by a system into internal energy of a reservoir there is no entropy change of the system because the thermodynamic coordinates of the system do not change.
I don’t understand how the thermodynamic coordinates do not change. Only the temperature is constant isn’t it?
It was a fun challenge to try to decipher this without having the book, but I believe I cracked the code.
I am fairly certain that he is talking about processes in which the initial state of the system is the same (thermodynamically) as the final state, not just the temperatures.
So this would be a process where energy somehow "flows" through the system, but at the end ends up in the reservoir, with the system ending up in the exact state it started from.
Here's an example: if I wave my arms in my room (which is let's say in thermal equilibrium with the outside), I will do some work against the air molecules, which will temporarily heat the room up a little bit. But, assuming the temperature outside stays constant, all that energy I added to the room will eventually dissipate into the outside, leaving the room in the exact state it was before.
Correct answer by ReasonMeThis on March 26, 2021
The section 8.7 you are referring to starts with the sentence "Examples are those processes involving the isothermal dissipation of work through a system (which remains unchanged) into internal energy of a reservoir, such as, etc.,". Therefore it is assumed that $Delta U=Delta Q + Delta W=0$ and then $Delta Q = -Delta W$, and then all internally and irreversibly generated entropy is being "pushed out", so to speak, from the system into the reservoir whose entropy will be increased by $Delta S = -frac{Delta Q}{T}=frac{Delta W}{T}$
Answered by hyportnex on March 26, 2021
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