Blt pb bvt pb b vt p rt lnpb p note

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Unformatted text preview: e total change in free energy, the sum of the free energy changes for A and for B, which is: VA + VB % VA + VB ( dV dV * ΔG = ΔG A + ΔGB = −RT'n A ∫ + nB ∫ ' V V* VA VB & ) % + V + VB . + V + VB .( ΔG = −RT'n A ln- A + nB ln- A 0 0* , VA / , VB /) & % +PA . +PB .( ΔG = RT'n A ln- 0 + nB ln- 0* ,P/ , P /) & Note that the last equality arises from PA = nA RT / V and PB = nB RT / V € where V = VA + VB and P = PA + PB (Dalton’s law). The same result can also be obtained considering the chemical potential of A in the mixture with B, of B in the mixture with A and of A and B pure. Indeed, we showed earlier that the chemical potential of an ideal gas under pressure P1 is related to that of the same gas under pressure P2 by: "P % µ ( T,P1) = µ ( T,P2 ) + RT ln$ 1 ' # P2 & If we take initial pressure to be P and final pressure to be PA, then the € change in chemical potential of A during the mixing process is: "P % µ A ( T,PA ) = µ A ( T,P) + RT ln$ A ' #P& € Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 164 The same equation holds for B: "P % µ B ( T,PB ) = µ B ( T,P) + RT ln$ B ' #P& Therefore the free energy of mixing, which is defined by: € ΔmixG = G(mixed gases) - G(pure gases) is equal to: [ [ Δ mix G = n Aµ A ( T,PA ) + nBµ B ( T,PB ) − n Aµ A ( T,P) + nBµ B ( T,P) * $P ' $ P 'Δ mix G = RT,n A ln& A ) + nB ln& B )/ %P( % P (. + which is identical to the equation obtained before. € We now recall that both partial pressures, PA and PB can be written in terms of the corresponding mole fractions and the total pressure. PA = yA...
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