Question

The ratio of $r.m.s$. speed to the $r.m.s.$ angular speed of a diatomic gas at certain temperature is: (assume $m=$ mass of one molecule, $M=$ molecular mass, $I=$ moment of inertia of the molecules)

• A. $\sqrt{\frac{3}{2}}$
• B. $\sqrt{\frac{3I}{2M}}$
• C. $\sqrt{\frac{3I}{2m}}$
• D. $1$

Answer 1

The correct option is C $\sqrt{\frac{3I}{2m}}$

We have,

$K.E.=\frac{1}{2}m{V}^2$ and $K.E.=\frac{3}{2}kT$

$⟹\frac{1}{2}m{V}^2=\frac{3}{2}kT$ --(1)

If $I$ is the moment of inertia then,

$\frac{1}{2}I{\omega }^2=\frac{2}{2}kT$ (diatomic gas)

$⟹\frac{1}{2}I{\omega }^2=kT$

Substituting for kT in equation 1, we get

$⟹\frac{1}{2}m{V}^2=\frac{3}{2}\times \frac{1}{2}I{\omega }^2$

$⟹\frac{V}{\omega }=\sqrt{\frac{3I}{2m}}$