Ncert Exercises & Solutions

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

Hint: Use the concept of RMS speed.

Step 1: Write the formula of RMS speed.

For n-molecules, we know that,

v_{r m s} = \sqrt{\frac{v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + . . . . . . + v_{n}^{2}}{n}}

[v_{r m s} = r o o t m e a n s q u a r e v e l o c i t y]

Where

v_{1}, v_{2}, v_{3} . . . . . . . . . . . . . v_{n}

are individual velocities of n-molecules of the gas

Step 2: Find the RMS speed of the gas.

For two molecules,

v_{r m s} =

\sqrt{\frac{v_{1}^{2} + v_{2}^{2}}{2}}

[

v_{1}, v_{2}, v_{3} . . . . . . . . . . . . . . v_{n}

are individual velocities]

Given,

v_{1} = 9 \times 10^{6} m / s

v_{2} = 1 \times 10^{6} m / s

v_{r m s} = \sqrt{\frac{{(9 \times 10^{6})}^{2} + {(1 \times 10^{6})}^{2}}{}}

= \sqrt{\frac{81 \times 10^{12} + 1 \times 10^{12}}{2}}

= \sqrt{\frac{(81 + 1) \times 10^{12}}{2}}

= \sqrt{\frac{82 \times 10^{12}}{2}}

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

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