A cubical box with porous walls containing an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The root mean square speed $({v_{rms}})$ of gas molecules is given by the formula: ${v_{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal g

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) The root mean square speed $({v_{rms}})$ of gas molecules is given by the formula: ${v_{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.Since the temperature $T$ is constant,we have ${v_{rms}} \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the ${v_{rms}}$ of ${O_2}$ molecules to that of ${H_2}$ molecules is:$\frac{{v_{rms, O_2}}}{{v_{rms, H_2}}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$Given that the molar mass of ${H_2}$ is $M_{H_2} = 2 	ext{ g/mol}$ and the molar mass of ${O_2}$ is $M_{O_2} = 32 	ext{ g/mol}$.Substituting these values:$\frac{{v_{rms, O_2}}}{{v_{rms, H_2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.

Similar Questions

If a gas is compressed isothermally,then the r.m.s. velocity of its molecules

At a given temperature,the ratio of $r.m.s.$ velocities of a hydrogen molecule and a helium atom will be:

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-20\,^{\circ} C$? (Atomic mass of $Ar = 39.9 \; u$,of $He = 4.0 \; u$)

At constant temperature,the vapour density of two gases are in the ratio of $6 : 9$. Compare their $v_{rms}$.

At what temperature does the $r.m.s.$ speed of air molecules become double that at $N.T.P.$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module