# Change in chemical potential

1. Nov 18, 2005

### goulio

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

Edited:

I found the trick $n$ as actually a depence in $n_A$ so you need to take account of this when you differentiate $G$ with respect to $n_A$

Last edited: Nov 19, 2005