At 300 K the average velocity of a gas is 3 | vedclass

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

Question

(A) The formula for average velocity is $\mu_{av} = \sqrt{\frac{8RT}{\pi M}}$.Since $\mu_{av} \propto \sqrt{T}$, we have the ratio $\frac{\mu_{av_1}}{

Vedclass · Accepted Answer

$6 	imes 10^2$

Vedclass · Answer

(A) The formula for average velocity is $\mu_{av} = \sqrt{\frac{8RT}{\pi M}}$.Since $\mu_{av} \propto \sqrt{T}$, we have the ratio $\frac{\mu_{av_1}}{\mu_{av_2}} = \sqrt{\frac{T_1}{T_2}}$.Given $T_1 = 300 \ K$, $T_2 = 1200 \ K$, and $\mu_{av_1} = 3 	imes 10^2 \ cm / s$.Substituting the values: $\frac{3 	imes 10^2}{\mu_{av_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore, $\mu_{av_2} = 2 	imes 3 	imes 10^2 = 6 	imes 10^2 \ cm / s$.

At $300 \ K$ the average velocity of a gas is $3 \times 10^2 \ cm / s$. The average velocity (in $cm / s$) of it at $1200 \ K$ is

Similar Questions

If the $V_{rms}$ of a gas is $(30R)^{1/2}$ at $27\,^{\circ}C,$ then calculate the molar mass of the gas in $kg/mol$.

At the same temperature,the root mean square velocities of two gases are $v_1$ and $v_2$,and their molar masses are $m_1$ and $m_2$ respectively. Which of the following relations is correct?

At $T \ K$,the $RMS$ velocity of $He$ is equal to the $RMS$ velocity of $SO_2$ at $127^{\circ} C$. What is $T$ (in $K$)?

Which of the following pairs will diffuse at the same rate through a porous plug?

If the $RMS$ speed of nitrogen at a certain temperature is $3000 \ ms^{-1}$,the approximate kinetic energy of one mole of nitrogen at that temperature in $kJ$ is (assume nitrogen as ideal gas)

Vedclass Test Series

Exam Paper Generator

Online Exam Module