Q140 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Physics

In non-rigid diatomic molecule with an additional vibrational mode

Correct answer
A
$81 C_{v}^{2} = 49 C_{P}^{2}$
B
$49 C_{v}^{2} = 25 C_{P}^{2}$
C
$49 C_{v}^{2} = 81 C_{P}^{2}$
D
$25 C_{v}^{2} = 49 C_{p}^{2}$

Solution

In non-rigid diatomic molecule with an additional vibrational mode, total degree of freedom $(f) =$ $3 (rotational) + 3 (translational) + 1 (vibrational) = 7$
Therefore,
$\frac{C_{p}}{C_{v}} = 1 + \frac{2}{f} = \frac{9}{7} \Rightarrow 81 C_{v}^{2} = 49 C_{P}^{2}$

Hence, the correct answer is option: 'A'.

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