Question

A diatomic gas consisting of rigid molecules is at a temperature of $87{}^{\circ }C$. If the moment of inertia of the rotating diatomic rigid molecule is $2.76\times 10^{-39}gc{m}^2$ then the nns angular speed of the molecule is (Boltzmann constant = $1.38\times 10^{-23}J{K}^{-1}$)

• A. $6\times 10^{12}rad{s}^{-1}$
• B. $3\times 10^{12}rad{s}^{-1}$
• C. $6\times 10^{13}rad{s}^{-1}$
• D. $3\times 10^{13}rad{s}^{-1}$

Answer 1

Answer: (A)

Moment of inertia of diatomic molecule,
$I=\frac{2}{3}M{r}^2$
So, radius, $r=\sqrt{\frac{3I}{2M}}$ ...(i)
rms speed of molecule,
$v=\sqrt{\frac{3kT}{M}}$
$rms$ angular speed,
$\omega =\frac{v}{r}=\frac{\sqrt{3kT/M}}{\sqrt{3I/2M}}=\sqrt{\frac{2kT}{I}}$
$=\sqrt{\frac{2\times 1.38\times 10^{-23}\times (87+273)}{2.76\times 10^{-46}}}$
$=\sqrt{360\times 10^{23}}=\sqrt{36\times 10^{24}}$
$=6\times 10^{12}rad/s$