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

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

Question

(D) The mean square velocity of a gas molecule is given by $v^2 = \frac{3kT}{m}$.For gas $A$, the mean square velocity is $v_A^2 = \frac{3kT}{m}$.Sinc

Vedclass · Accepted Answer

$\sqrt{\frac{2}{3}}$

Vedclass · Answer

(D) The mean square velocity of a gas molecule is given by $v^2 = \frac{3kT}{m}$.For gas $A$, the mean square velocity is $v_A^2 = \frac{3kT}{m}$.Since $v^2 = v_x^2 + v_y^2 + v_z^2$ and due to symmetry $v_x^2 = v_y^2 = v_z^2$, we have $v_x^2 = \frac{v^2}{3}$.Thus, $V_1^2 = v_{Ax}^2 = \frac{v_A^2}{3} = \frac{3kT}{3m} = \frac{kT}{m}$.Therefore, $V_1 = \sqrt{\frac{kT}{m}}$. $(1)$For gas $B$, the mean square velocity is $V_2^2 = \frac{3kT}{2m}$.Therefore, $V_2 = \sqrt{\frac{3kT}{2m}}$. $(2)$Dividing equation $(1)$ by $(2)$, we get:$\frac{V_1}{V_2} = \frac{\sqrt{kT/m}}{\sqrt{3kT/2m}} = \sqrt{\frac{kT}{m} \cdot \frac{2m}{3kT}} = \sqrt{\frac{2}{3}}$.

Similar Questions

At what temperature $K$ will the $r.m.s.$ speed of hydrogen gas be equal to the $r.m.s.$ speed of oxygen molecules at $47^{\circ}C$?

In Maxwell's speed distribution curve,for $N_2$ gas,the average relative velocity (in $m/s$) between two molecules at $300 \, K$ is:

The $N_2$ molecule is $14$ times heavier than the $H_2$ molecule. At what temperature will the $rms$ speed of $H_2$ molecules be equal to the $rms$ speed of $N_2$ molecules at $27^{\circ}C$?

Two boxes are at the same temperature. The first box contains gas with molecular mass $m_1$ and rms speed $v_1$. The second box contains gas with molecular mass $m_2$ and average speed $v_2$. If $v_1 = 1.5 v_2$,then $\frac{m_1}{m_2}$ is

Five molecules of a gas have speeds $1, 2, 4, 8$ and $16 \, m/s$ at some instant. The root mean square velocity of the gas molecules is ..... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module