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)

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