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I started a mock paper for one of my entrance exams with this fairly easy question. I rearranged the equation and got 5/2, knowing that a diatomic molecule has a Cv of 5/2, i concluded that the answer must be either A or C, but I do not know the difference between rigid diatomic and non rigid diatomic, can someone explain the difference?

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At sufficiently high temperature, a diatomic molecule may have a thermodynamic degree of freedom that corresponds to bond length oscillation. That would change the '5/2' to '6/2'.

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  • $\begingroup$ The vibratory behavior adds two quadratic terms to the Hamiltonian, so the change is from 5/2 to 7/2. This is a good reason to be careful how you talk about equipartition: it's very easy to be sloppy about about [the meaning of 'degree of freedom' ](physics.stackexchange.com/a/317605/520) and fool yourself. $\endgroup$
    – dmckee --- ex-moderator kitten
    Apr 27 '17 at 10:37
  • $\begingroup$ Just to confirm , when we increase the energy of the system it is able to get two more degrees of freedom , thereby changing from rigid diatomic (5/2) to non rigid diatomic (7/2) ... That made sense thanks, both of you! $\endgroup$
    – Harshit Pandey
    May 2 '17 at 1:17
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Rigid diatomic molecule has $C_v = 5/2$ and non rigid has $C_v = 7/2$. From this we can fairly distinguish between both. Examples of rigid molecules include $\mathrm{Cl_2}, \mathrm{Br_2}, \mathrm{N_2}, \mathrm{O_2}$ and so on. An example of a non-rigid molecule is $\mathrm{CO}$.

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  • $\begingroup$ These might be good examples, but they don't explain why some are rigid while others are not. $\endgroup$
    – Quantum Mechanic
    Aug 26 '21 at 14:00

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