when we mix 2 components of ideal gas. for example we mix the flow from pure A and the flow from pure B(A and B is ideal gas). if both temperature and pressure are constant, then entropy change will be expressed as;
Sfinal-Sinitial=-Rc[yA *ln(yA)+yB*ln(yB)] (S:total entropy per mole Rc:gas constant, yi: mole fraction of component i(i=A or B)
I can't understand why this expression doesn't apply to the situation that we mix 2 equal flows. such like mixing A and A, or mixing B and B.
For example, Let's suppose that we mix the pure A flow(let it called flow 1) and the pure A flow(let it called flow 2). Although 2 flows are constituted of equal material. it seems to me that the entropy change will be expressed such as:
Sfinal-Sinitial=-Rc[y1 *ln(y1)+y2*ln(y2)],(In this condition yi is defined as mole fraction of A which came from flow i(i=1 or 2))
But my textbook illustrate entropy change equals 0, naturally
Two A molecules can not be distinguished. Their exchange does not produce a new microstate.
Every real material absorbs and emits radiation. This applies, in particular, to gases. The classical theory of ideal gases ignores the interaction of the gas particles with thermal radiation. Thus, whenever this interaction is an essential part of the phenomenon, the classical theory is bound to lead to unrealistic conclusions.
If interaction with thermal radiation is taken into account, an ideal gas performs work even when it expands in a vacuum, provided that the vacuum is filled by thermal radiation (which is always the case when the vacuum is bounded by material at a temperature different than absolute zero). This produces temperature changes both in the free expansion of the ideal gas and in its adiabatic mixing with another ideal gas [A. Paglietti, Continuum Mech. Therm. 24, 201-210 (2012)].
It is true that mixing of two identical gases which are in the same initial state before mixing does not produce any entropy change. This is simply because the process does not produce any change in the state of two gases before and after they are joined together –whether the gases are ideal or not. But then the question arises that even the smallest difference beween the two gases will produce, upon mixing, the finite change of entropy that you refer to in your formulae.
This inconsistency of the classical theory is sometimes referred to as Gibbs’ paradox. The paradox can be resolved if the gas interaction with thermal radiation is taken into account. In this case a small difference in the property of the two ideal gases produces a proportional small change (as opposed to a given finite change) in their entropy after mixing.
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
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