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 $2.76 \times 10^{-39} \, gcm^2$ then the nns angular speed of the molecule is (Boltzmann constant = $1.38 \times 10^{-23} \, JK^{-1}$)

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

Answer 1

Answer: (A)

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