In a non-rigid diatomic molecule with an addi | vedclass

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Question

(C) For a non-rigid diatomic molecule,the degrees of freedom $(f)$ are calculated as follows:Translational degrees of freedom $= 3$Rotational degrees

Vedclass · Accepted Answer

$49 C_{v}^2=81 C_{P}^2$

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(C) For a non-rigid diatomic molecule,the degrees of freedom $(f)$ are calculated as follows:Translational degrees of freedom $= 3$Rotational degrees of freedom $= 2$Vibrational degrees of freedom $= 2$ (one for kinetic energy and one for potential energy).Total degrees of freedom $(f) = 3 + 2 + 2 = 7$.The ratio of specific heats is given by $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$.Substituting $f = 7$,we get $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$.Therefore,$\frac{C_p}{C_v} = \frac{9}{7}$,which implies $7 C_p = 9 C_v$.Squaring both sides,we get $49 C_p^2 = 81 C_v^2$.

In a non-rigid diatomic molecule with an additional vibrational mode,what is the relationship between $C_v$ and $C_p$?

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