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Question

Consider a polyatomic gas which has 3 degrees of translational motion, 3 degrees of rotational motion, and k vibrational modes. Obtain a relation for Gamma.

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Solution

Polyatomic gas having 3 degrees of translational motionand 3 degrees of rotational motion and k vibrational motion, has total degrees of freedom=N =3×3-k=9-k degrees of freedomTherefore E=9-k×12RTTherefore Cv=dEdT=9-k×12Rand Cp=Cv+R= 9-k×12R+R=11-k2RTherefore γ =CpCv=11-k2R9-k2R =11-k9-k

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