# Degrees of freedom of a three dimensional polyatomic molecules

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?

• this question has been answered about two days back...appears to be duplicate..the wording is different. – drvrm Feb 28 '18 at 13:15
• Any link of the question? – Shikhar Asthana Feb 28 '18 at 13:20
• searching for it.... – drvrm Feb 28 '18 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.. – drvrm Feb 28 '18 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) – Shikhar Asthana Mar 1 '18 at 8:52

## 2 Answers

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

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.