Molar heat capacity of CO2 is too high. Why?

In summary, at very high temperatures, CO2 should have a specific heat capacity of 15/2 R due to its 3 translational, 2 rotational, and 4 vibrational degrees of freedom. However, experimental values are slightly higher, as shown by a figure found online. It was suggested that this could be due to an additional rotational degree of freedom gained when the molecule vibrates in the bending mode. However, this explanation may not be accurate as molecules cannot gain degrees of freedom. The approximation of treating vibrations as harmonic oscillators becomes less accurate at higher temperatures due to the increasing anharmonic character.
  • #1
Philip Koeck
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At very high temperatures CO2 should have Cp = 15/2 R, since there are 3 translational, 2 rotational and 4 vibrational degrees of freedom.
Experimental values are a bit higher than that, at least according to a figure I found on the internet.
Is that correct? And what is the explanation?
A student suggested that when the molecule vibrates in the bending mode it gets an additional rotational degree of freedom, since it is bent most of the time.
Is that a sensible explanation?
 
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  • #2
Philip Koeck said:
Experimental values are a bit higher than that, at least according to a figure I found on the internet.
Is that correct?
See
https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Units=SI&Mask=1&Type=JANAFG&Plot=on#JANAFG

At 3000 K, it is pretty much 15/2 R. It goes to higher values at higher temperatures.

Philip Koeck said:
A student suggested that when the molecule vibrates in the bending mode it gets an additional rotational degree of freedom, since it is bent most of the time.
Is that a sensible explanation?
A molecule cannot gain degrees of freedom. A linear triatomic has 2 rotational degrees of freedom, but gains one has an additional vibrational d.o.f. because bending is degenerate. If the molecule is bent, then you would loose a vibrational d.o.f. and gain a rotational one, which would actually reduce Cp, since the vibrational mode contributes 2 quadratic d.o.f. to the 1 rotational d.o.f. (see the Cp values for water).

Treating vibrations has harmonic oscillators is an approximation. The anharmonic character increases as vibrational excitation goes up, hence the approximation ##C_p = (f/2 + 1) R## is less good at higher T.

[Edit: Changed the language a bit to make it more consistent.]
 
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