At what temperature will the root mean square | vedclass

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

Question

(B) The root mean square speed $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.For the speeds to be equal,$\sqrt{\frac{3RT_1}{M_1

Vedclass · Accepted Answer

$177$

Vedclass · Answer

(B) The root mean square speed $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.For the speeds to be equal,$\sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3RT_2}{M_2}}$,which simplifies to $\frac{T_1}{M_1} = \frac{T_2}{M_2}$.Here,$M_1$ (molar mass of $O_3$) = $48 \, g/mol$,$M_2$ (molar mass of $O_2$) = $32 \, g/mol$,and $T_2 = 27 + 273 = 300 \, K$.Substituting the values: $\frac{T_1}{48} = \frac{300}{32}$.$T_1 = \frac{300 	imes 48}{32} = 300 	imes 1.5 = 450 \, K$.Converting back to Celsius: $450 - 273 = 177 \, ^oC$.

At what temperature will the root mean square speed of ozone $(O_3)$ be equal to the root mean square speed of oxygen $(O_2)$ at $27\,^oC$?

Similar Questions

Explain the Maxwell-Boltzmann distribution law with a graph.

At $133.33 \ K$,the $RMS$ velocity of an ideal gas is $(M = 0.083 \ kg \ mol^{-1}; R = 8.3 \ J \ mol^{-1} \ K^{-1})$ (in $m \ s^{-1}$)

The root mean square speeds at $STP$ for the gases $H_2, N_2, O_2$ and $HBr$ are in the order

Which of the following gases will have the lowest root mean square speed at $25\,^{\circ}C$?

On which factors does the velocity distribution of gas molecules depend?

Vedclass Test Series

Exam Paper Generator

Online Exam Module