# How can the heat capacity of diamond be so low?

I'm taking an introductory thermodynamics course, and according to my professor, the heat capacity of an ideal gas is $$\frac{3}{2}R \approx$$ 12.5. Since ideal gases don't have any attraction between molecules, every bit of energy put into the system goes towards increasing the Kinetic energy of molecules.

Since average KE is literally a definition of temperature, it should be that any substance should require at least this much energy to increase their temperature. My intuition says that a higher heat capacity simply tells us some energy is being stored as PE when heat is added.

However, the molar heat capacity of diamond is about 6.11 J/(molK), which is less than an ideal gas. Even if we accept that some bizarre quantum mechanical effects are in play, isn't there a problem in the fact that adding 6.11 Joules to 1 mole of diamond increases the total Kinetic energy of the atoms by 12.5 Joules? How is this possible?

Since average KE is literally a definition of temperature [...]

This is not a good definition of temperature. Temperature is often (but not always) a measure of the average energy (in the sense that the relationship between temperature and average energy is one-to-one), but in certain systems the relationship is more complicated.

The proper definition of temperature is

$$\frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)_{V}$$ That is, if you add a small bit of energy to the system then the system's entropy changes because that extra bit of energy can be distributed between the particles in a number of different ways. The ratio of the change in energy to the change in entropy gives the temperature. If a small parcel of energy increases the entropy dramatically, then the temperature is low; if the entropy does change very much at all, then the temperature is high.

As a simple example, consider a 3D Einstein solid. It's not difficult to show that the average internal energy per particle is given by

$$\left = \frac{3\hbar\omega}{2}\coth\left(\frac{\hbar\omega}{2kT}\right)$$

This different from what one would naively expect by applying the equipartition theorem to a system of classical oscillators (in which the energy is shared equally between kinetic and potential). In that case, we would simply find that $$\left = 3kT$$. From these relations we can compute the heat capacities of our systems, $$C = \left(\frac{\partial U}{\partial T}\right)_V$$ where $$U=N\left$$. We find that for the Einstein solid,

$$C = 3k\left(\frac{\hbar \omega}{2kT}\right)^2 \operatorname{csch}^2\left(\frac{\hbar\omega}{2kT}\right)$$

whereas $$C=3k$$ for the classical solid. My intuition says that a higher heat capacity simply tells us some energy is being stored as PE when heat is added.

This intuition isn't bad. Indeed if you restrict your attention to classical physics, you'd be right - the classical counterpart to the Einstein solid has a heat capacity of $$3R$$ rather than $$\frac{3}{2}R$$ (note that $$R=N_Ak$$, where $$N_A$$ is Avogadro's number) precisely because energy is stored as vibrational potential energy.

It's the quantum effects that are the key here - specifically the fact that the set of allowed energies for each particle is discrete. When the temperature is less than $$\hbar\omega$$ (which is the spacing between allowed energy levels), there is a dramatic departure from the classical heat capacity of $$3k$$ because a small parcel of energy changes the temperature quite substantially. If $$T/\hbar\omega$$ is sufficiently small, then the heat capacity of the Einstein solid can even drop below $$3k/2$$, the heat capacity of the classical ideal gas.

Diamond is special. The rigidity it gets from its somewhat unique crystal structure makes its effective $$\omega$$ very large. As a result, its heat capacity is unusually small even for relatively large values of $$T$$. Here's a plot of its heat capacity compared with aluminum and lead. • I might suggest that the reason is not an enormous bond strength of carbon. Rather, it is the structural complexity of the sp3 bond that gives a macroscopic rigidity to the system, thereby increasing $\omega$. Otherwise, you would be wanting us to believe that something akin to polyethylene should also have a low specific heat. Specifically, see also here – Jeffrey J Weimer Oct 1 at 1:50
• @JeffreyJWeimer Yes, you're right. Bond strength was a bad way to phrase it, I meant to refer to rigidity of the diamond structure rather than generic C-C bonds. I will edit. – J. Murray Oct 1 at 2:02
• Thanks. Nicely done BTW! – Jeffrey J Weimer Oct 1 at 2:09
• @JeffreyJWeimer Thank you! – J. Murray Oct 1 at 2:10
• Thank you very much! The answer seems to indicate that the problem lies in my definition of temperature. Its interesting to know that my assumption that average KE is the same for any substance at a given temperature is not generally true. – arandhaw Oct 3 at 6:53