A diatomic gas molecule has translational,rot | vedclass

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Question

(D) For a diatomic gas molecule considering translational,rotational,and vibrational degrees of freedom:Translational degrees of freedom $(f_t)$ = $3$

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(D) For a diatomic gas molecule considering translational,rotational,and vibrational degrees of freedom:Translational degrees of freedom $(f_t)$ = $3$.Rotational degrees of freedom $(f_r)$ = $2$.Vibrational degrees of freedom $(f_v)$ = $1$ (which contributes $2$ to the total degrees of freedom as it includes both kinetic and potential energy).Total degrees of freedom $(f)$ = $f_t + f_r + 2f_v = 3 + 2 + 2(1) = 7$.However,if the question implies $1$ vibrational mode as a single degree of freedom contribution (often simplified in specific contexts),then $f = 3 + 2 + 1 = 6$.Using the formula $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$:For $f = 6$,$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$.

$A$ diatomic gas molecule has translational,rotational,and vibrational degrees of freedom. The ratio ${C_P}/{C_V}$ is

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