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Homework Help: Change in chemical potential

  1. Nov 18, 2005 #1
    Consider a solution of particles of type A and B with the following Gibbs potential
    [tex]
    G(P,T,n_A,n_B)=n_A g_A(P,T) + n_B g_B(P,T)+ (1/2)\lambda_{AA}n_A^2/n + (1/2)\lambda_{BB}n_B^2/n + \lambda_{AB}n_A n_B/n + n_A RT \ln(x_A) + n_B RT \ln(x_B)
    [/tex]
    where the [itex]n_i[/itex]'s are the number of moles with [itex]x_i=n_i/n[/itex] and [itex]g_i[/itex] are the molar Gibbs potential of each type of particle [itex]i=A,B[/itex]. Also [itex]n_A + n_B = n[/itex] and the [itex]\lambda_{ij}[/itex] are positive constants.
    a) If we add [itex]\Delta n_B[/itex] moles of B keeping pressure and temperature constant, calculate the change in in the chemical potential of A.
    The chemical potential of A is
    [tex]
    \mu_A = \left ( \frac{\partial G}{\partial n_A} \right )_{P,T,n_A} = g_A + \lambda_{AA} n_A/n + \lambda_{AB}n_B/n + RT(1 + \ln(x_A))
    [/tex]
    so changing [itex]n_B[/itex] to [itex]n_B+\Delta n_B[/itex] only changes [itex]\mu_A[/itex] by an amount [itex]\lambda_{AB}\Delta n_B/n[/itex].
    Is this right or I'm getting the whole thing wrong?

    Edited:

    I found the trick [itex]n[/itex] as actually a depence in [itex]n_A[/itex] so you need to take account of this when you differentiate [itex]G[/itex] with respect to [itex]n_A[/itex]
     
    Last edited: Nov 19, 2005
  2. jcsd
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