A Molar heat capacity of CO2 is too high. Why?

[Philip Koeck]

At very high temperatures CO₂ should have C_p = 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?

 

[DrClaude]

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.

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.]