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The average kinetic energy for the O₂ molecule (rigid rotator) is

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$\frac{1}{2}{k_B}T$

$\frac{3}{2}{k_B}T$

$4{k_B}T$

$\frac{5}{2}{k_B}T$

answer is B.

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Detailed Solution

According to equipartition of energy, average energy absorbed by a molecule for each degree of freedom = 1/2 K_BT.

O₂ molecule has 2 degrees of freedom for rotational motion and 3 degrees of freedom for translational motion.

So f = 3 + 2 = 5.

Therefore, total average kinetic energy possessed by O₂ molecule = fK_BT/2 = 5 K_BT/2.

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