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While studying for my Statistical thermodynamics test, I encountered this graph 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!

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  • $\begingroup$ I've never seen a figure depicting such a bump before, but then I've seen only a few plots of data and a lot of schematics that were presumably draw to show the plateaus. Do you have an example of a data plot showing such a bump. I can think of a possible reason for such a bump to appear in a practical experiment that does not correspond to some fundamental physics but to difficulties actually running the experiment, but if that was the case I'd expect a similar bump at the transition to including vibrational modes. $\endgroup$ – dmckee --- ex-moderator kitten Feb 22 '17 at 19:01
  • $\begingroup$ Where did you get the graph, and a credit/ attribution would do no harm ;) $\endgroup$ – user146020 Feb 22 '17 at 20:07
  • $\begingroup$ @dmcke We've drawn a graph like that in class as well, but our professor said that he won't go into details about it. There was no bump at the transition to vibrational modes however. Moreover, I think the bump has a theoretical background, as can be seen here: link in the "Heat capacity at low temperature" section. $\endgroup$ – JRF Feb 22 '17 at 20:07
  • $\begingroup$ This figure rkt.chem.ox.ac.uk/tutorials/statmech/hydrogen.jpg suggests two things. That it is a real phenomena and that it is related to spin degrees of freedom (which would be why it is associated with the rotational turn-on but not with the vibrational turn-on). No time to follow up now, but it promises to be a very interesting question indeed. $\endgroup$ – dmckee --- ex-moderator kitten Feb 22 '17 at 21:41
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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.

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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.

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    $\begingroup$ That's for $C_V$. The plot is depicting $C_P$. For an ideal gas, $C_P=C_V+R$. $\endgroup$ – Chris Aug 13 '18 at 23:12

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