# Degrees of freedom of a three dimensional polyatomic molecules

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?

• this question has been answered about two days back...appears to be duplicate..the wording is different. Feb 28, 2018 at 13:15
• Any link of the question? Feb 28, 2018 at 13:20
• searching for it.... Feb 28, 2018 at 13:23
• Hint: In vibrational mode - a vibration involves both kinetic and potential energy terms which are squares of velocity and coordinate thereby on the average it may contribute two degrees of freedom.. Feb 28, 2018 at 13:27
• If that is the case then how come diatomic molecules have 6 degrees of freedom ( 3 translational, 2 rotational and 1 vibrational) and not 7 degrees of freedom ( 3 translation, 2 rotational and 2 due to vibrations) Mar 1, 2018 at 8:52

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 $$8$$.