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 $I=2.1\times 10^{-39}gc{m}^2$.

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