A diatomic gas consisting of rigid molecules | vedclass

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Question

(A) For a rigid diatomic molecule,the rotational kinetic energy is associated with two degrees of freedom. According to the equipartition theorem,the

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$6 	imes 10^{12} 	ext{ rad s}^{-1}$

Vedclass · Answer

(A) For a rigid diatomic molecule,the rotational kinetic energy is associated with two degrees of freedom. According to the equipartition theorem,the average rotational kinetic energy is given by $K.E. = 2 	imes (\frac{1}{2} k_B T) = k_B T$. Given the moment of inertia $I = 2.76 	imes 10^{-39} 	ext{ g cm}^2 = 2.76 	imes 10^{-39} 	imes 10^{-3} 	ext{ kg} 	imes (10^{-2} 	ext{ m})^2 = 2.76 	imes 10^{-46} 	ext{ kg m}^2$. Temperature $T = 87 + 273 = 360 	ext{ K}$. Equating rotational kinetic energy to $\frac{1}{2} I \omega_{rms}^2$: $k_B T = \frac{1}{2} I \omega_{rms}^2$ $\omega_{rms} = \sqrt{\frac{2 k_B T}{I}} = \sqrt{\frac{2 	imes 1.38 	imes 10^{-23} 	imes 360}{2.76 	imes 10^{-46}}} = \sqrt{\frac{993.6 	imes 10^{-23}}{2.76 	imes 10^{-46}}} = \sqrt{360 	imes 10^{23}} = \sqrt{36 	imes 10^{24}} = 6 	imes 10^{12} 	ext{ rad s}^{-1}$.

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