A polyatomic gas has f vibrational degrees of | vedclass

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(A) For a polyatomic gas, the total degrees of freedom $n$ is the sum of translational, rotational, and vibrational degrees of freedom. For a non-line

Vedclass · Accepted Answer

$\frac{4+f}{3+f}$

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(A) For a polyatomic gas, the total degrees of freedom $n$ is the sum of translational, rotational, and vibrational degrees of freedom. For a non-linear polyatomic molecule, translational degrees of freedom = $3$ and rotational degrees of freedom = $3$. If there are $f$ vibrational degrees of freedom, each contributes $2$ to the degrees of freedom (one for kinetic and one for potential energy). However, in standard physics problems of this type, the specific heat at constant volume $C_V$ is given by $C_V = \frac{f_{total}}{2} R$. Given the structure of the options, the problem assumes $C_V = (3 + f)R$ (where $3$ represents the translational and rotational contributions and $f$ represents the vibrational contribution per mole). Using the relation $C_p = C_V + R$: $C_p = (3 + f)R + R = (4 + f)R$. Therefore, the ratio $\gamma = \frac{C_p}{C_V} = \frac{(4 + f)R}{(3 + f)R} = \frac{4 + f}{3 + f}$.

$A$ polyatomic gas has $f$ vibrational degrees of freedom, then the ratio of the specific heat at constant pressure to that at constant volume will be

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