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While studying for my Statistical thermodynamics test, I encountered this graph
Source: https://www.physicsforums.com/threads/variation-of-specific-heat-with-temperature.399514/
I know this isn't the best graph you'll ever see, but the "bump" was present on several other graphs as well. So just to clarify my question once more; I'm looking for a physically intuitive explanation of the "bump", which occurs at temperatures, where rotational degrees of freedom become relevant.
Thank you!
The bump can be observed through an explicit calculation.
If $\it l$ is the angular momentum quantum number of a molecule, then the rotational energy levels are
$$ \varepsilon_\text{rot} = \frac{\hbar^2}{2I}\it l(\it l + 1) = \frac{k\theta_\text{r}}{2} \it l(\it l + 1) \,, \quad \textrm{ where }\quad \theta_\text{r} \equiv \frac{\hbar^2}{Ik} \,,$$
and $I$ is the moment of inertia of the molecule.
Since each $\it l$ is $(2\it l + 1)$-fold degenerate, the partition function over each rotational mode reads
$$ Z_\text{rot} = \sum_{l=0}^\infty (2\it l+1)\exp\left( - \frac{\theta_\text{r}}{2T} \it l(\it l + 1) \right) = \begin{cases} 1 + 3 e^{-\theta_r/T} + 5e^{-3\theta_r/T} + \mathcal O\left( e^{-6\theta_r/T} \right) &\text{for } T \ll \theta_r \,, \\ 2\frac{T}{\theta_r} + \frac13 + \frac1{30}\frac{\theta_r}{T} + \mathcal O\left( \left( \frac{\theta_r}{T} \right)^2 \right) &\text{for } T \gg \theta_r \,. \end{cases} $$
Using this, we can calculate the contribution to the internal energy per rotational degree of freedom.
$$ E_\text{rot} = NkT^2\frac{\partial}{\partial T}\ Z_\text{rot} = \begin{cases} 3Nk\ \theta_r\left( e^{-\theta_r/T} - 3e^{-2\theta_r/T} + \dots \right) &\text{for } T \ll \theta_r \,, \\ NkT\left(1 - \frac{\theta_r}{6T} - \frac{1}{180}\left(\frac{\theta_r}{T}\right)^2 + \dots \right) &\text{for } T \gg \theta_r \,. \end{cases} $$
Therefore, the contribution to heat capacity at constant volume by each rotational mode is $$ C_V^\text{rot} = \frac{\partial }{\partial T}E_\text{rot} = Nk \begin{cases} 3 \left(\frac{\theta_r}T\right)^2 e^{-\theta_r/T} \left( 1 - 6 e^{-\theta_r/T} + \dots \right) &\text{for } T \ll \theta_r \,, \\ 1 + \frac{1}{180}\left(\frac{\theta_r}{T}\right)^2 + \dots &\text{for } T \gg \theta_r \,. \end{cases} $$
The above function has a maximum of $1.1\ Nk$ at about the temperature $0.81\ \theta_r/2$. As the temperature is increased way above $\theta_r/2$, it settles down to $1\ Nk$ and we recover the classical flat result.
Graph is wrong. Translation has three degrees of freedom, specific heat 3/2 R. Rotation (excited at low temperature) adds two degrees of freedom, so specific heat becomes 5/2 R. Vibration (excited only at temperatures of thousands of degrees) adds another two degrees of freedom (one kinetic, from relative motion, and one potential, from the interatomic potential) for a total of 7/2 R.
