# Homework Help: [Statistical Mechanics] An adsorption model

1. Jul 27, 2011

### Truecrimson

1. The problem statement, all variables and given/known data

"Now consider a metal surface in which the M adhesion sites are comprised of equal populations of sites of two different types..."

2. Relevant equations

The entropy and the chemical potential of a metal surface with a single type of adhesion site are shown in P7 and P8 respectively.

3. The attempt at a solution

Use the expression $$\mu=-\Delta-kT\ln \left(\frac{M-N}{N}\right)$$ separately for each type of adhesion sites A and B, so that $$\mu_A=-\Delta_A-kT\ln \left(\frac{N-A}{A}\right)$$ and $$\mu_B=-\Delta_B-kT\ln \left(\frac{N-(N-A)}{N-A}\right)=-\Delta_B+kT\ln \left(\frac{N-A}{A}\right)$$ where A is the occupation of sites A.
Then if I assume that the two types of sites are in chemical equilibrium i.e. $$\mu_A=\mu_B$$, then I can solve for A in terms of the temperature and the difference in binding energy easily. However, I'm not sure if I can assume that, but I can't think of any other way to do this. (I thought of deriving the entropy of a metal surface with two types of sites, and work out the chemical potential from that. But there were some unclear steps. If my proposed solution doesn't work then maybe I'll post that later.) Any suggestion would be appreciated.

2. Jul 28, 2011

### ardie

use the fundumental law of thermodynamics in differential form (assuming thermal equilibrium) to deduce the temperature as a function of rate of change of enthropy and the chemical potentials.

3. Jul 28, 2011

### Truecrimson

You mean $$\left(\frac{\partial S}{\partial N}\right)_{U,V}=-\frac{\mu}{T}?$$ If so, assuming thermal equilibrium
$$T_A=T_B$$
$$\left(\frac{\partial S}{\partial A}\right)_{U_A,V_A} \mu_B=\left(\frac{\partial S}{\partial B}\right)_{U_B,V_B} \mu_A$$

$\left(\frac{\partial S}{\partial A}\right)_{U_A,V_A}$ is $k\ln \left(\frac{N-A}{A}\right)$, and similarly $\left(\frac{\partial S}{\partial B}\right)_{U_B,V_B}=k\ln \left(\frac{N-B}{B}\right)=-k\ln \left(\frac{N-A}{A}\right).$

Then
$$(-\Delta_A-kT\ln \left(\frac{N-A}{A}\right))(-k\ln \left(\frac{N-A}{A}\right))=(-\Delta_B+kT\ln \left(\frac{N-A}{A}\right))(k\ln \left(\frac{N-A}{A}\right))$$

$$\Delta_A+kT\ln \left(\frac{N-A}{A}\right)=-\Delta_B+kT\ln \left(\frac{N-A}{A}\right)$$

$$\Delta_A=-\Delta_B$$

Where was my mistake(s)? Thank you, by the way.

4. Jul 28, 2011

### ardie

the mistake was in your initial writing of the fundumental law. if you have 2 objects in your system, each of which has a different chemical potential, then the total internal energy (differential) will be altered to include both the chemical potentials and the number of the particles. the total temperature of the system however can be thought of as one entity since the objects are at thermal equilibrium. this should be your starting point:
dU = T.dS - u(a).dN(a) - u(b).dN(b)

5. Jul 28, 2011

### Truecrimson

So to use the fundamental law,
1) Calculate dU from the partition function. dU then has dN(a) and dN(b) in it.
2) Write dS=dS(a) + dS(b). This again has dN(a) and dN(b).
3) Plug everything into the equation?

And are the expression for S(a) and u(a), for example, the same as in the case of single type of adhesion sites, just changing N to N(a) (as I did in my previous reply)?

6. Jul 29, 2011

### ardie

i was given infraction points for doing peoples homework before. totally sucks -.-. so I will try to point out to the right method. using the fundumental law (the thing I wrote above), you may proceed by plugging everything else in. I believe you are ok with dividing through by differantials but be careful when doing this! there isn't only one differential to be considered. dU means the infinitesimal change in the internal energy of the system as a whole, so if you are at the thermal equilibrium, you can assume that this quantity is constant... hope this helps ^^
In essence, you are looking at an extremum of the interal energy of the system where the enthropy (of the whole system) is maximised and everything is settled in in the appropropriate site.

7. Jul 30, 2011

### Truecrimson

That surely sucks!

What I don't understand though, is that if you're saying that the total entropy (and the total energy) is at its extremum, then in addition to thermal equilibrium, the system has to attain chemical (diffusive) equilibrium as well, because net flow of particles between subsystems increase the total entropy. And that condition boils down to u(a)=u(b). (Even though my u's may be wrong.)

8. Jul 30, 2011

### ardie

well in p8, you work our the chemical potential and find out that at thermal equilibrium it is only a function of the number of adhesion sites. then in the next part you are told that the total number residing on adhesion sites is half of N (constant) so this fixed the total chemical potential for adhesion sites. in the next part you want to work out how many on average would go to one type of adhesion site, provided the constraint of the systems and that the chemical potentials are defined as follows. if the two adhesion sites had the same chemical equilibrium then their population would be the same as the particles would show no preference as to where to reside. hence their population would be exactly the same. systems may still be in thermal equilibrium without the same chemical potentials being equal, the condition of thermal diffusive equilibrium is then achieved by the two sites losing/gaining particles.

9. Jul 30, 2011

### Truecrimson

The two sites' binding energies are different though. So chemical equilibrium doesn't mean that their populations would be the same(?) or am I missing something here?

10. Jul 30, 2011

### ardie

was that in reply to something I wrote or just a comment?

11. Jul 30, 2011