# Statistical Physics: Cubic lattice of two molecules

1. Apr 7, 2015

### Matt atkinson

1. The problem statement, all variables and given/known data
A mixture of two substances exists on a cubic lattice of N sites, each of which is occupied by either an A molecule or a B molecule. The number of A molecules is NA and the number of B molecules is NB, such that NA + NB = N. The energy of interaction is $k_BT\chi_{AA}$ between two nearest neighbour A molecules, $k_BT\chi_{BB}$ between two nearest neighbour B molecules and $k_BT\chi_{AB}$ between a neighbouring pair of an A and a B molecule. When the interaction energies are written in this form, $\chi_{ij}$ is dimensionless.

1) State,
a. what the order parameter defined as $\phi_A = NA/N$ physically represents.
b. the range of values that $\phi_A$ can take.
c. the relation between $\phi_A$ and $\phi_B$ ($\phi_B = NB/N$).

2) By determining the number of A-A, A-B and B-B nearest neighbour contacts in terms of $\phi_A$, show that, if $\chi_{AA} = \chi_{BB}$, the total energy can be written as;

$$E = E_0 + 3Nk_BT \chi \phi_A (1−\phi_A )$$
Where $E_o$ and $\chi$ are constants.

2. Relevant equations

3. The attempt at a solution
1a) The percentage of molecule A in lattice sites N.
1b) $\phi_A=0...1/2$ (due to max entropy being when there is 50% of molecule A and 50% of molecule B)
1c) $\phi_B=1-\phi_A$

2) so this is where im stuck, i found;
$N_{AA}=6N\phi_A^2$ (each site has 6 nearest neighbours, so number of A molecules x no of NN x P(NN being a A molecule)
$N_{BB}=6N(1-\phi_A)^2$
$N_{AB}=12N\phi_A(1-\phi_A)$

and then using;
$$E=N_{AA}V_{AA}+N_{BB}V_{BB}+N_{AB}V_{AB}$$
I get that
$$E=6VN+6VN\phi_A(\phi_A-1)$$
Where $V=V_{AA}=V_{BB}=V_{AB}$ as $\chi_{AA} = \chi_{BB}$.

I'm thinking i might've worked out the number of A-A, B-B and A-B interactions wrong. could someone give me a nudge in the right direction?

Last edited: Apr 7, 2015
2. Apr 8, 2015

### Matt atkinson

Figured out what i was doing wrong now! all is fine :D