# Why doesn't beryllium, an elemental solid, obey Dulong and Petit's law?

Beryllium has a molar heat capacity of $$\pu{17.7 J mol-1 K-1},$$ which is quite far from the $$\pu{25 J mol-1 K-1}$$ as predicted by Dulong and Petit's law. What is the reason for this discrepancy?

• Boron and graphite also deviate significantly. Now, what is Dulong and Petit's "law" based on? – Jon Custer Jun 25 at 15:06
• No elementvreally obeys the"Law" which is just a high temperature approximation for the solid state. Some elements have ambient temperatures high enough, others don't. – Oscar Lanzi Jun 25 at 15:55

First, a reminder that most scientific "laws" are only convenient mathematical approximations. They aren't guaranteed to work for every case, and they aren't guaranteed to have underlying physical meaning. As such, don't think of exceptions as "not obeying the law"; it's more correct to say that the law isn't obeying reality,

In the case of the Dulong-Petit Law:

Despite its simplicity, Dulong–Petit law offers fairly good prediction for the specific heat capacity of solids with relatively simple crystal structure at high temperatures. It fails, however, at room temperatures for light atoms bonded strongly to each other, such as in metallic beryllium, and in carbon as diamond. In the very low (cryogenic) temperature region, where the quantum mechanical nature of energy storage in all solids manifests itself with larger and larger effect, the law fails for all substances.

Not a 'Law' a such it is more like a 'rule of thumb' based on a limited set of data measure at high (i.e. room) temperatures. The heat capacity is not constant but varies with temperature and for metals reaches a limiting value of approximately $$3R$$ at high temperatures. (The limiting value is higher for molecules due to the many translational, vibrational and rotational energy levels).

The heat capacity is the rate of change of the energy with temperature, i.e. the slope of a plot of energy vs temperature. At low temperatures only the lowest few energy levels are populated, so the rate of change vs temperature is almost zero. As the temperature is increased, more energy is added, more levels become populated and $$C_V$$ rapidly increases because the rate of change of energy is large. At high temperature, very many levels are now populated and the rate of increase of energy with temperature is constant, and therefore $$C_V$$ levels off becoming almost constant.

In some metals, those with large energy gaps between levels, room temperature is insufficient for them to reach the point where the slope of internal energy increase with temperature becomes constant and so the heat capacity is smaller than $$3R$$. The value at higher temperatures (say 1000 C) should reach the limiting value.

The Dulong–Petit law applies in the classical limit, i.e. when temperature is high enough that the quantisation of energy levels (as prescribed by quantum mechanics) is not readily apparent. The larger the gaps between energy levels, the higher the temperature needs to be such that you no longer "see" these gaps. The comparison is fundamentally between thermal energy $$k_\mathrm BT$$ and the energy gap $$\Delta E$$: the Dulong–Petit law applies when $$k_\mathrm BT \gg \Delta E$$.

There are three kinds of energy gaps which are under consideration here, namely translational, rotational, and vibrational. It turns out that vibrational energy levels have the largest gaps between them, and hence offer the strongest "challenge" to the requirement for $$k_\mathrm BT \gg \Delta E$$. From the simple harmonic oscillator we know that $$\Delta E = \hbar\omega = \hbar\sqrt{k/\mu}$$, so we can identify several cases where we expect the Dulong–Petit law to break down:

• when temperature is small;
• when the force constant $$k$$ is large, i.e. very strongly bonded atoms;
• when the reduced mass $$\mu$$ is small, i.e. light atoms.