Question

The temperature of the gas consisting of rigid diatomic molecules is T = 300K. Calculate the angular root mean square velocity of a rotating moelcule if its moment of inertia is equal to $I = 2 .1 \times 10^{-39} g\,cm^2$.

• A. $6.3 \times 10^{12}$ rad / sec
• B. $6.8 \times 10^{12}$ rad / sec
• C. $3.6 \times 10^{12}$ rad / sec
• D. $3.2 \times 10^{12}$ rad / sec

Answer 1

Answer: (A)

$6.3 \times 10^{12}$ rad / sec

Answer 2

By formula, if $I$ is the moment of inertia then $\frac{1}{\alpha} I \omega^{2}=\frac{1}{2} \times f \times K_{B} T$ where $f=$ degree of freedom $K_{B}=$ Bult mann conotant $T=$ temperature the root mean square angular velocity, $\begin{array}{r} \omega^{2}=\frac{f K_{B} T}{I} \\\ \Rightarrow \omega=\sqrt{\frac{f K_{B} T}{I}} \end{array}$ for diatanic molecule, $f=2$ $w=\sqrt{\frac{2 \times 1.3 \times 10^{-23} \times 300}{2.1 \times 10^{-39}}}=6.3 \times 10^{12} rad / sec$ The angular root mean square velocity is $6.3 \times 10^{12} rad / sec$