Three vessels A,B,and C contain gases at the | vedclass

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

Question

(B) The average velocity of a gas molecule is given by the formula $v_{av} = \sqrt{\frac{8kT}{\pi m}}$,where $k$ is the Boltzmann constant,$T$ is the

Vedclass · Accepted Answer

$V_1$

Vedclass · Answer

(B) The average velocity of a gas molecule is given by the formula $v_{av} = \sqrt{\frac{8kT}{\pi m}}$,where $k$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of one molecule of the gas.Since the temperature $T$ is the same for all vessels,the average velocity of a specific gas depends only on its molecular mass $m$.In vessel $A$,the average velocity of $O_2$ is $V_1 = \sqrt{\frac{8kT}{\pi m_{O_2}}}$.In vessel $C$,the $O_2$ molecules are at the same temperature $T$ and have the same molecular mass $m_{O_2}$ as in vessel $A$.The presence of $N_2$ in the mixture in vessel $C$ does not affect the individual average velocity of the $O_2$ molecules.Therefore,the average velocity of $O_2$ in vessel $C$ remains $V_1$.

Similar Questions

The temperature of an ideal gas is reduced from $927^\circ C$ to $27^\circ C$. The $r.m.s.$ velocity of the molecules becomes

The temperature at which the root mean square speed of a gas will be half of its value at $0\,^{\circ}C$ is ....... $^{\circ}C$.

The root mean-square $(rms)$ speed of oxygen molecules $(O_2)$ at a certain absolute temperature is $\upsilon$. If the temperature is doubled and the oxygen gas dissociates into atomic oxygen,the $rms$ speed would be

The root mean square speed of hydrogen molecules at $300 \ K$ is $1930 \ m/s.$ Then the root mean square speed of oxygen molecules at $900 \ K$ will be ....... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module