The specific heat, C = T(dS/dT), is the amount of energy required to raise a unit amount of a substance by 1 degree K. In this model we are only interested in the adsorbed gas, and the ‘unit amount’ is an atom; thus the units of specific heat are the same as for the entropy S, namely k/atom. The calculation proceeds similarly to the entropy calculation, by taking differences between two table entries DeltaS and DeltaT, and evaluating T(DeltaS/DeltaT). Note that DeltaS and DeltaT are evaluated at the table entries half a step above and below the temperature T. This is a good approximation on account of the mean value theorem.
Temp (K) |
Specific Heat 3.09 (k/atom) |
Specific Heat 3.19 (k/atom) |
Specific Heat 3.29 (k/atom) |
|---|---|---|---|
12.5 |
2.0217 |
1.938 |
1.8064 |
12.75 |
2.0508 |
1.953 |
1.8311 |
13 |
2.042 |
1.955 |
1.8303 |
13.25 |
2.0566 |
1.955 |
1.8535 |
13.5 |
2.0574 |
1.98 |
1.8374 |
13.75 |
2.083 |
1.978 |
1.8721 |
14 |
2.0813 |
2.001 |
1.8799 |
14.25 |
2.0918 |
1.996 |
1.873 |
14.5 |
2.1013 |
2.004 |
1.8787 |
This data table is only indicates the low T end of the data shown on the plot, but shows that the specific heat has a value ~ 2k/atom for all lattice parameters studied.
We know that the classical equipartition result for a 3-dimensional solid, valid at high T, is the Dulong-Petit law, i.e. C = 3k/atom, or k/2 per degree of freedom. At low T, these degrees of freedom get progresively frozen out, and so do not contibute to the specific heat. The models you are familiar with are the Debye (phonon) model, where the low T behaviour goes as T cubed, and the Einstein (localized) model which goes epxponentially to zero as T goes to 0. In the present model, we have used the (classical) cell model for the x-y motions parallel to the substrate, and the Einstein model for the z-motion. Can you understand why we have found C around 2k/atom in the T range studied?
Note also that the differentiation leading to these table values is starting to produce rounding errors. This could be solved by keeping more figures in the free energy F, and taking more care to use double precision arithmetic at all stages of the calculation.
Continue to the following section 2.4, or alternatively return to section 1.
