A diatomic ideal gas (e.g. $\mathrm{O_2}$, $\mathrm{N_2}$) at room temperature has degrees of freedom (translational + rotational, vibration not yet excited):
A$3$
B$5$
C$6$
D$7$
Answer & Solution
Correct answer: B. $5$
Diatomic molecules have 3 translational + 2 rotational degrees of freedom at typical room temperature, totalling **5**. The third rotational mode (about the molecular axis) has negligibly small moment of inertia and isn't excited classically.
Vibrational modes (option D = 7 = 5 + 2 vibrational) require higher temperatures to activate; at room temperature they are frozen out by quantum effects.
This gives $C_v = \dfrac{5}{2} R$ for diatomic gases, and $\gamma = C_p/C_v = 7/5 = 1.4$, the value used in adiabatic problems.
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