First slide

Rms speed

Question

Two molecules of a gas have speeds of $9 x 10^{6} {ms}^{- l}$ and $1 \times 10^{6} {ms}^{- 1}$ respectively. What is the root mean square speed of these molecules?

Easy

A

$\sqrt{21} \times 10^{6} m / s$
B

$\sqrt{41} \times 10^{6} m / s$
C

$8.5 \times 10^{6} m / s$
D

$\sqrt{17} \times 10^{6} m / s$

Solution

For n-molecules, we know that
$v_{rms} = \sqrt{\frac{{v_{1}}^{2} + {v^{2}}_{2} + {v^{2}}_{3} + . . . . + {v_{n}}^{2}}{n}}$
[ $v_{rms} = root mean square velocity$ ]
where $v_{1}, v_{2}, v_{3}, . . . v_{n}$ are individual velocities of n-molecules of the gas.
For two molecules, $v_{rms} = \sqrt{\frac{{v^{2}}_{1} + {v^{2}}_{2}}{2}}$
[ $v_{1}, v_{2}, v_{3}, . . . v_{n} are individual velocity]$
Given, $v_{1} = 9 \times 10^{6} m / s and v_{2} = 1 \times 10^{6} m / s$
$∴ v_{rms} = \sqrt{\frac{{(9 \times 10^{6})}^{2} + {(1 \times 10^{6})}^{2}}{2}} = \sqrt{41} \times 10^{6} m / s$

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