The heat capacity

Let us discuss the heat capacity further here, which tells us how the energy of the system responds to a change of temperature. Every degree of freedom of the system contributes to the total energy of the system and as such to its heat capacity, but the single contributions depend on how they respond to temperature changes. Consider for example a gas of molecules. Each molecule has translational degrees of freedom, plus internal degrees of freedom such as rotations and vibrations. The electrons can also be excited from their ground state, providing additional degrees of freedom, as well as nuclear excitations. We saw in the previous section that in the classical limit, in which the separation between adjacent translational energy levels is much smaller than $k_{\rm B}T$, their contribution to the energy of the system is proportional to the temperature, given by Eq. , and therefore they contribute the constant

$$C_V^{tr} = \frac{3}{2} N k_{\rm B}$$ (4.44)

to the heat capacity. For a perfect gas, from Eq.  we also have:

$$C_P^{tr} - C_V^{tr} = N k_{\rm B}.$$ (4.45)

On the other hand, if the separation between energy levels $\Delta \epsilon \gg k_{\rm B}T$ then these degrees of freedom are not excited. To see this let us consider $\Delta \epsilon = \epsilon^{int}_1 - \epsilon^{int}_0$, where $\epsilon^{int}_0$ is the energy of the ground state of a particular internal degree of freedom and $\epsilon^{int}_1$ the energy of the first excited state. According to the Boltzmann distribution the probability of finding this degree of freedom in the first excited state $p^{int}_1$, compared to that of finding it in the ground state $p^{int}_0$, is given by:

$$\frac{p^{int}_1}{p^{int}_0} = \frac{\Omega(\epsilon^{int}_1)}{\Omega(\epsilon^{int}_0)}e^{-\beta \Delta \epsilon},$$ (4.46)

which is very low if $\Delta \epsilon \gg k_{\rm B}T$, unless the degeneracy of the first excited state $\Omega(\epsilon^{int}_1)$ is much larger than that of the ground state $\Omega(\epsilon^{int}_0)$, but for low enough temperature the Boltzmann exponential will always prevail. Under these conditions only the ground state is populated and the system cannot accumulate any additional energy as the temperature increases. As a result, the contribution of these particular degrees of freedom to the specific heat is zero and they are called dormant. They are dormant because if the temperature is increased to high enough values that $p^{int}_1/p^{int}_0$ becomes non negligible, then they can spring into action and start accumulating energy, therefore contributing to the heat capacity.

Conversely, at high enough temperature the energy accumulated in every degree of freedom becomes proportional to the temperature, and so the specific heat becomes constant. As temperature is reduced from high values the heat capacity will initially stay constant, and then drop when a corresponding degree of freedom freezes out. If the temperature drops to low enough values the condition $\Delta \epsilon \gg k_{\rm B}T$ is reached for any degree of freedom and so eventually they all freeze out, meaning that the heat capacity drops to zero. This is exactly the condition that we mentioned at the end of Sec. .

Figure: Constant volume heat capacity as function of $k_{\rm B}T$ (arbitrary units) for a one-dimensional system with three degrees of freedom $\alpha_1, \alpha_2, \alpha_3$, with energies $n \epsilon_1$, $n \epsilon_2$, $n \epsilon_3, n = 1, 2, \dots$, and with $\epsilon_1 = 0.0001$, $\epsilon_2 = 0.01$, $\epsilon_3 = 10$.