Question

A polyatomic molecule has 3 translational, 3 rotational degrees of freedom and 2 vibrational modes. The ratio of specific heats \(\frac{C_P}{C_V}\) is

• A. \(\frac{7}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{5}{6}\)
• D. \(\frac{5}{3}\)
• E. \(\frac{6}{5}\)

Answer 1

Answer: (E)

For a polyatomic molecule with 3 translational and 3 rotational degrees of freedom, the contribution to the specific heat is: \[ C_V = \frac{3}{2} R + \frac{3}{2} R = 3R \] Each vibrational mode contributes \(R\) to \(C_V\), and there are 2 vibrational modes: \[ C_V = 3R + 2R = 5R \] For \(C_p\), we use the relation: \[ C_p = C_V + R = 5R + R = 6R \] Thus, the ratio \(\frac{C_p}{C_V}\) is: \[ \frac{C_p}{C_V} = \frac{6R}{5R} = \frac{6}{5} \]

The correct option is (E) : \(\frac{6}{5}\)

Answer 2

For a polyatomic molecule, the total degrees of freedom = translational + rotational + vibrational. 

Given: Translational = 3, Rotational = 3, Vibrational modes = 2  

Each vibrational mode contributes 2 degrees of freedom (1 kinetic + 1 potential), so: $$ \text{Vibrational degrees of freedom} = 2 \times 2 = 4 $$ 
Total degrees of freedom: $$ f = 3 + 3 + 4 = 10 $$ 
According to the equipartition theorem, $$ C_V = \frac{f}{2} R = \frac{10}{2} R = 5R $$ 
For ideal gases: $$ C_P = C_V + R = 5R + R = 6R $$ 
So, $$ \frac{C_P}{C_V} = \frac{6R}{5R} = \frac{6}{5} $$ 
Correct answer: \( \frac{6}{5} \)