Two boxes are at the same temperature. The fi | vedclass

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

Question

(C) The root mean square speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{m_1}}$.The average speed of a gas is given by $v_{avg} = \sqrt{\frac{8

Vedclass · Accepted Answer

$0.52$

Vedclass · Answer

(C) The root mean square speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{m_1}}$.The average speed of a gas is given by $v_{avg} = \sqrt{\frac{8RT}{\pi m_2}}$.Given that $v_1 = 1.5 v_2$,we substitute the expressions:$\sqrt{\frac{3RT}{m_1}} = 1.5 	imes \sqrt{\frac{8RT}{\pi m_2}}$.Squaring both sides,we get:$\frac{3RT}{m_1} = (1.5)^2 	imes \frac{8RT}{\pi m_2}$.$\frac{3}{m_1} = 2.25 	imes \frac{8}{\pi m_2}$.Rearranging for the ratio $\frac{m_1}{m_2}$:$\frac{m_1}{m_2} = \frac{3 \pi}{2.25 	imes 8} = \frac{3 	imes 3.14159}{18} = \frac{9.42477}{18} \approx 0.52$.

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

Similar Questions

The temperature at which the $r.m.s.$ speed of hydrogen molecules is equal to the escape velocity on the Earth's surface will be ...... $K$.

According to the kinetic theory of gases,the $r.m.s.$ velocity of gas molecules is directly proportional to:

At what temperature is the $R$.$M$.$S$. velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} C$ (in $K$)? (Molecular weight of hydrogen $= 2$,molecular weight of oxygen $= 32$)

The gas having an average speed four times that of $SO_2$ (molecular mass $64$) is

At what temperature in $K$ will the $r.m.s.$ speed of a hydrogen molecule be equal to the escape velocity from the Earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module