Consider a sample of oxygen behaving like an | vedclass

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

Question

(C) The root mean square (rms) velocity of an ideal gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The average velocity of an ideal gas is given by

Vedclass · Accepted Answer

$\sqrt{\frac{3 \pi}{8}}$

Vedclass · Answer

(C) The root mean square (rms) velocity of an ideal gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The average velocity of an ideal gas is given by $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$.To find the ratio of rms velocity to average velocity,we divide the two expressions:$\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}}$.Simplifying the expression,the terms $R$,$T$,and $M$ cancel out:$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3RT}{M} 	imes \frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}}$.Thus,the ratio is $\sqrt{\frac{3\pi}{8}}$.

Consider a sample of oxygen behaving like an ideal gas. At $300 \, K$,the ratio of root mean square (rms) velocity to the average velocity of gas molecules would be: (Molecular weight of oxygen is $32 \, g/mol$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$)

Similar Questions

The temperature at which the r.m.s. velocity of hydrogen molecules is $4.5$ times that of an oxygen molecule at $47^{\circ}C$ is (molecular weights of hydrogen and oxygen are $2$ and $32$ respectively). (in $^{\circ}C$)

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

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

In a sample of chlorine gas at $300 \ K$,the average kinetic energy per molecule is $6.21 \times 10^{-21} \ J$ and the root mean square speed $\nu_{rms}$ is $325 \ m/s$. What will be the values of these quantities at $600 \ K$?

If the oxygen $(O_2)$ has a root mean square velocity of $C \ m/s$,then the root mean square velocity of hydrogen $(H_2)$ at the same temperature will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module