Degrees of freedom of a three dimensional polyatomic molecules

Question

So the exact question is

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?

saurab das · Answer

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

Yuvi Patil · Answer

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.

JIPSITA JENA · Answer

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 $8$.

Answered by JIPSITA JENA on April 16, 2021

Ria Kataria · Answer

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

Degrees of freedom of a three dimensional polyatomic molecules

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