N molecules of gas A,each having mass m,and 2 | vedclass

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

Question

(B) The mean square velocity of a gas molecule is given by $\langle v^2 \rangle = \frac{3kT}{m}$.For gas $A$,the mean square of the $x$-component is $

Vedclass · Accepted Answer

$2: 3$

Vedclass · Answer

(B) The mean square velocity of a gas molecule is given by $\langle v^2 angle = \frac{3kT}{m}$.For gas $A$,the mean square of the $x$-component is $\omega^2 = \langle v_x^2 angle$. Since $\langle v^2 angle = \langle v_x^2 angle + \langle v_y^2 angle + \langle v_z^2 angle$ and $\langle v_x^2 angle = \langle v_y^2 angle = \langle v_z^2 angle$,we have $\langle v^2 angle = 3\omega^2$.Thus,$3\omega^2 = \frac{3kT}{m} \implies \omega^2 = \frac{kT}{m}$...$(i)$For gas $B$,the mean square velocity is $V^2 = \frac{3kT}{2m}$...(ii)Dividing $(i)$ by (ii):$\frac{\omega^2}{V^2} = \frac{kT/m}{3kT/2m} = \frac{kT}{m} 	imes \frac{2m}{3kT} = \frac{2}{3}$.

$N$ molecules of gas $A$,each having mass $m$,and $2N$ molecules of gas $B$,each of mass $2m$,are contained in the same vessel at a constant temperature $T$. The mean square velocity of $B$ is $V^2$ and the mean square of the $x$-component of $A$ is $\omega^2$. The value of $\frac{\omega^2}{V^2}$ is

Similar Questions

When the rms velocity of a gas is denoted by $V$,which of the following relations is true? ($T=$ absolute temperature of the gas)

The curve between absolute temperature $T$ and ${v^2}_{rms}$ is

$A$ gas is at a temperature of $0^{\circ}C$. To what temperature in $^{\circ}C$ must the gas be heated so that the $rms$ speed of its molecules becomes double?

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:

The average speed $\bar{v}$ and $r.m.s.$ speed $v_{rms}$ of the molecules are related as

Vedclass Test Series

Exam Paper Generator

Online Exam Module