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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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 Spring '07
 AREsker
 Physical chemistry, pH

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