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

On any planet,the presence of an atmosphere implies ($C_{rms}$ = root mean square velocity of molecules and $V_e$ = escape velocity):

To increase the rms speed of gas molecules by $25 \%$,the percentage increase in absolute temperature of the gas is to be

The $rms$ speeds of the molecules of Hydrogen,Oxygen,and Carbon dioxide at the same temperature are $V_{H}$,$V_{O}$,and $V_{C}$ respectively. Then:

Let $\bar{v}$,${v_{rms}}$ and ${v_p}$ respectively denote the mean speed,root mean square speed,and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature $T$. The mass of a molecule is $m$. Then

The value closest to the thermal velocity of a Helium atom at room temperature $(300\,K)$ in $m/s$ is $[k_B = 1.4 \times 10^{-23}\,J/K; m_{He} = 7 \times 10^{-27}\,kg]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module