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Thermodynamics, cannot reach the answer (Reichl's book related)

  1. Jun 17, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider a binary mixture of particles of types 1 and 2 whose Gibbs free energy is given by ##G=n_1 \mu _1 ^0 (P,T)+n_2 \mu _2 ^0 (P,T) +RT n_1 \ln x_1 +RTn_2 \ln x_2 +\lambda n x_1x_2##.
    Where ##n=n_1+n_2##. And ##x_1## and ##x_2## are the mole fractions of particles 1 and 2 respectively.
    The book on page 159 states that the conditions ##\mu _1 ^I=\mu _1 ^{II}## and ##\mu _2 ^I=\mu _2 ^{II}## where the upperscript denotes the phase. I have no problem to understand that those conditions must be fulfilled for equilibrium when there's a coexistance of phases.
    But now comes the part where I struggle. From the condition ##\mu _1 ^I=\mu _1 ^{II}##, I'm supposed to find that ##RT \ln x_1 ^I + \lambda (1-x_1 ^I)^2=RT \ln x_1 ^{II}+\lambda (1-x_1 ^{II})^2##.
    But I don't reach this.
    2. Relevant equations
    ##\left ( \frac{\partial g}{\partial n_1} \right ) _{T,P,n_2} =\mu _1##.
    ##x_2=1-x_1##.

    3. The attempt at a solution
    ##\mu _1 =\left ( \frac{\partial g}{\partial n_1} \right ) _{T,P,n_2}=\mu _1 ^0 (P,T)+RT \ln x_1 + \lambda x_1 (1-x_1)##.
    So if I use the condition ##\mu _1 ^I = \mu _1^{II}##, I'd get ##RT\ln x_1 ^I + \lambda x_1 ^I (1-x_1 ^I )=RT\ln x_1 ^{II} + \lambda x_1 ^{II} (1-x_1 ^{II} )##. Which differs from the equation I'm supposed to find.
    I know I simply have to derivate g=G/n with respect to ##n_1## while keeping P, T and ##n_2## constant, but I'm simply failing at that apparently.
    I've rechecked the algebra many times, I really don't see where I go wrong. My friend told me he reached the good result, so I know I've went wrong somewhere.

    Thank you for any help.
     
  2. jcsd
  3. Jun 17, 2013 #2

    TSny

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    x1 and x2 are not constants. They depend on n1 and n2.
     
  4. Jun 17, 2013 #3

    fluidistic

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    Thank you TSny, I overlooked this.
     
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