Degrees of freedom of a three dimensional polyatomic molecules

Question

An ideal gas consists of three dimensional polyatomic molecules. The temperature is such that only one vibrational mode is excited. If $R$ denotes the gas constant, then the specific heat at constant volume of one mole of the gas at this temperature is:

The method I used is that
$$C_v=\frac{f}{2}R$$ Where Cv = specific heat at constant volume f= degree of freedom

Given that it is a 3-D polyatomic molecules, it would have the following degrees of freedoms

a) 3 translational degree freedom

b) 3 Rotational degree of freedom

c) 1 vibrational degree of freedom ( given in question)

Total degree of freedoms are 7, hence
$$Cv=\frac{7}{2}R$$

The twist, however, is that the answer is $$Cv=4R$$

Can someone explain where the 8th degree of freedom came from? or Is it that the answer given is wrong?

Answer 1

Your formula for Cv is wrong. The formula u used is for that condition when Vibrational mode is absent. Actual in general formula is Cv= ( ft/2 + fr/2 + fv) R Now using these u will get the value as 4R.

Answer 2

The degrees of freedom for a 3D polyatomic gas molecule are $6$ at normal temperature. But as there is only one vibrational mode so the degrees of freedom become $6+2=8$. Hence $C_v=fR/2$ so $C_v=4R$. Here we add 2 because in a polyatomic gas molecule no.of vibrational mode is $3N-6$ (for non linear). So $3 \cdot 3-6=3$. But only one vibrational mode is active so degrees of freedom become $8$.

Answer 3

Cv=(3+f)R. As, f=1, Cv=4R Cp=(4+f)R

Answer 4

one vibrational mode contributes 2 degrees of freedom hence 3+3+2 = 8 then (8/2)R = 4R