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(D) The $rms$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3kT}{M}}$,where $M$ is the mass of a single molecule.For gas $A$,the mass of a mole

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(D) The $rms$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3kT}{M}}$,where $M$ is the mass of a single molecule.For gas $A$,the mass of a molecule is $m$. Thus,$v_{rms, A}^2 = \frac{3kT}{m}$.The mean square velocity is $\langle v^2 \rangle = v_{rms}^2$. The $x$-component of the mean square velocity is $\langle v_x^2 \rangle = \frac{1}{3} \langle v^2 \rangle = \frac{1}{3} \frac{3kT}{m} = \frac{kT}{m}$. So,$w^2 = \frac{kT}{m}$.For gas $B$,the mass of a molecule is $2m$. Thus,$v^2 = v_{rms, B}^2 = \frac{3kT}{2m}$.Now,the ratio $\frac{w^2}{v^2} = \frac{kT/m}{3kT/2m} = \frac{1}{3/2} = \frac{2}{3} \approx 0.67$.

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