How Does Adding Moles of B Affect the Chemical Potential of A?

[goulio]

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]

 

[Jhero]

.

Yes, you are correct. The change in the chemical potential of A when adding \Delta n_B moles of B is \lambda_{AB}\Delta n_B/n. This is because as you correctly pointed out, the number of moles n is dependent on n_A and n_B, so when differentiating G with respect to n_A, you have to take into account the change in n due to the change in n_B. This is why the change in the chemical potential of A is not simply \lambda_{AB}\Delta n_B, but rather \lambda_{AB}\Delta n_B/n.