Theoretically speaking, entropy should increase while mixing two ideal mixtures (from Second law of Thermodynamics). But when mixing happens at constant pressure, (deltaH) =(delta Q) and thus (delta S) = (delta H)/T and for ideal mixing (delta H)=0. So shouldn't (delta S)=0?
Can anyone explain this??
While mixing ideal solutions, the mixing process is not coupled with heat tinting (delta H = 0) as well as with changes in the total volume of the mixture (delta V = 0). Although delta H = 0, delta S does not need to be zero, since it is defined as follows:
delta S = - n R ( x1 lnx1 + x2 lnx2 )
Since 0 < x1, x2 < 1, lnx1 and lnx2 will both be negative and lead to a positive delta S.
There is no change in H and V for ideal solution . I.e delta H =delta V=0 .. Why entropy of mixing is not zero beacouse of Configuation entropy alsway predominate when we talikng about mixing .. Its depend on amount only ..
when you speak about mixing, configurational entropy has to be accounted and how the entropy for a solution is defined globally. However, so called thermal entropy i.e., dS=dH/T is zero..!!!
therefore, in general ideal mixing entropy is positive that comes from configurational part (probability of occupying the positions by the atoms during 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
Rule of additive pressure can be used to predict total pressure of mixtures which is also called Dalton's rule. Total pressure of mixture P = sum (y_i*P) where y_i is the mole fraction of each component. When mixing occurs, whether in constant pressure or not, pressure of each components reduced to its partial pressure. So this pressure reduction for each component cause increase in entropy.
The entropy change delta S = delta Q/T (= delta H/T at constant pressure) only for reversible processes. For irreversible processes such as mixing delta S > delta Q/T (= delta H/T at constant pressure). Thus for mixing of ideal gases or ideal solutions delta Q = 0 but delta S > 0.
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