# Specific capacity of a diatomic molecule at low temperatures

My question is: Why at low temperatures does a diatomic molecule result in the same specific heat capacity as a monoatomic at a constant volume?

My understanding is that at very high temperature there are two additional degrees of freedom from the vibration between the atoms, this gives the specific heat capacity of $$C_v=\frac 7 2 R$$ at room temperature. Vibrational degree of freedom does not occur, so we are only left with translational and rotational.

But a very low temperature it seems that rotational motion does not occur. But why?

• For the same reason that the vibrational degrees of freedom aren't active at low temperature: there is an energy gap between the ground rotational state and first excited rotational state, so as long as $k_{\textrm{B}}T \ll$ that energy gap, the rotational degrees of freedom are "frozen out". Commented Mar 25, 2019 at 20:49
• Also, at room temperature, the correct equation is Cv=2.5R. Commented Mar 25, 2019 at 22:28

First, let's address the vibrational states since these are rarely relevant to the diatomic heat capacity. The vibrational states of a diatomic molecule, in the quantum harmonic oscillator approximation, are given by $$\begin{gather*} E_n \approx \hbar \omega \left( n + \frac{1}{2} \right) \end{gather*}$$ where $$\omega$$ is the fundamental frequency of the diatomic given by $$\omega = \sqrt{ \frac{k}{\mu}}$$ and $$k$$ is the spring constant and $$\mu$$ is the diatomic reduced mass. To get a sense of scale, applying this model to carbon monoxide gives $$\hbar \omega \approx 2169.8$$ cm$$^{-1}$$. By contrast, at a temperature of 298 K the thermal energy is $$k_BT \approx 207$$ cm$$^{-1}$$. Thus, the Boltzmann population of the vibrational levels expressed by, $$\begin{gather*} p(E_n) \propto e^{-\frac{\hbar \omega (n+\frac{1}{2})}{k_BT}} \end{gather*}$$ will be extremely biased towards only the lowest vibrational state $$n=0$$. In other words, there is no real vibrational degree of freedom at these temperatures for many diatomic molecules because there is insufficient thermal energy to actually excite them with any meaningful probability.
In an analogous fashion, the rotational motion can be modeled using the rigid rotor model, in which the rotational states labeled by $$J$$ are given by, $$\begin{gather*} E(J) \approx BJ(J+1) \end{gather*}$$ where $$B$$ is the rotational constant. For carbon monoxide, the rotational constant $$B = 1.923$$ cm$$^{-1}$$ and the degeneracy of states is $$2J+1$$. This gives a Boltzmann distribution of, $$\begin{gather*} p(E_J) \propto (2J+1)e^{-\frac{BJ(J+1)}{k_BT}} \end{gather*}$$ Since the thermal energy is still about $$k_BT \approx 207$$cm$$^{-1}$$ at 298 K, there will be appreciable populations of molecules with many different rotational states. Together with the absence of the vibrational degrees of freedom, this gives only 5 degrees of freedom (3 translational, 2 rotational) rather than the expected 6 for a diatomic molecule. Using the rule that $$C_V \approx \frac{D}{2}nR$$ where $$D$$ is the number of accessible degrees of freedom, the heat capacities are expressed in the ideal limit as, \begin{align*} C_V &\approx \frac{5}{2} nR \\ C_P &\approx C_V + nR = \frac{7}{2} nR \end{align*} We have also shown, however, that at extremely low temperatures it is also possible to freeze out the rotational degrees of freedom in addition to the vibrational degrees of freedom. This will just not happen at any reasonable everyday temperature.