Such behavior can be easily represented on diagrams

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Unformatted text preview: expression for the free energy of mixing of two liquids A and B is always correct as long as A and B form a homogeneous mixture. In some cases, it is experimentally observed that the partial pressure of a component above the liquid mixture increases linearly with the mole fraction of that component in the liquid phase (i.e. PA = constant x xA). Furthermore, when the constant in the expression of the vapor pressure is exactly equal to the partial pressure above the pure liquid, then this component is said to obey Raoult’s law. For component A, Raoult’s law states that: PA = xA PA* Similarly for B, Raoult’s law states that: PB = xB PB* If both components of a liquid mixture obey Raoult’s law over the entire composition range (i.e. xA between 0 and 1), then the solution is said to be an ideal solution. For such a solution, we obviously have PA / PA* = xA and PB / PB* = xB, therefore: ΔGmix = RT [nA ln(xA) + nB ln(xB)] Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 168 Furthermore, for the ideal solution, one can see (from the above equation) that: ΔSmic = - R [nA ln(xA) + nB ln(xB)] since: $ ∂[ΔG ] ' $ ∂G ' mix ) S = −& ) ⇒ ΔSmix = −& & ∂T ) % ∂T (P % (P Furthermore, ΔHmix = ΔGmix + T ΔSmix since mixing occurs at constant € temperature. Therefore: ΔHmix = 0 which implies that for an ideal solution, the enthalpy of the components in the mixed state is equal to the enthalpy of the components in the pure state. ! ∂G $ Furthermore, since V = # & ⇒ ΔVmix = " ∂P %T ! ∂ )ΔG + $ # * mix , & # ∂P & " %T and since...
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