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I am going through the Gibbs' paper "On the Equilibrium of Heterogeneous Substances" and in the beginning of the paper, the following is written:

If the system consist of parts between which there is supposed to be no thermal communication, it will be necessary to regard as impossible any diminution of the entropy of any of these parts, as such a change cannot take place without the passage of heat.

I tried to interpret it as since the energy is constant in each part of the system, the entropy in that part cannot increase according to the equilibrium condition given in the paper, where $(\delta \eta)_\varepsilon \leq 0$. The above state is telling me something directly opposite to the equilibrium condition.

Can you please explain this to me? I don't know much about thermodynamics, but I am trying learn it.

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  • $\begingroup$ I guess what he is saying is that the equal sign applies to such as system (which apparently starts out in thermodynamic equilibrium and stays in thermodynamic equilibrium). It's hard to answer this without more context. $\endgroup$ Commented Aug 29, 2018 at 11:27
  • $\begingroup$ It is a popular paper and it is on page 3 of the paper. I think the whole context is given in the statement. $\endgroup$ Commented Aug 29, 2018 at 12:32

1 Answer 1

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It's easier to interpret what he's saying for each part of the system using the Clausius inequality: $$\Delta S\geq \int{\frac{dq}{T_I}}$$ where dq is the differential exchange of heat with the other part of the system, and $T_I$ is the temperature at the interface between the two parts of the system (where the heat transfer is occurring). For both parts of the system (1 & 2), if dq = 0, we are left with: $$\Delta S_1\geq 0$$$$\Delta S_2\geq 0$$

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