If the r- m- s- speed of a gas molecule at 27 -C is 100- 2 ms -1 the r- m- s- speed at 327 -C would be

The correct option is B 200 ms ^ 1 [ v_ rms =√( 3RT / M #160; ) #160;; ...

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Standard XII

Physics

RMS and KE Expressions

If the r. m. ...

Question

If the r. m. s. speed of a gas molecule at $27^{\circ} C$ is $100 \sqrt{2} {m s}^{- 1}$ the r. m. s. speed at $327^{\circ} C$ would be

100 {m s}^{- 1}

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200 {m s}^{- 1}

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300 {m s}^{- 1}

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400 {m s}^{- 1}

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Solution

The correct option is B $200 {m s}^{- 1}$
$\begin{matrix} v_{r m s} = \sqrt{\frac{3 R T}{M}} = 100 \sqrt{2} v_{r m s}^{'} = \sqrt{\frac{3 R (21)}{M}} = \sqrt{2} \times 100 \sqrt{2} 200 m / s H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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If the r. m. s. speed of a gas molecule at 27∘C is 100√2ms−1 the r. m. s. speed at 327∘C would be

The correct option is B 200ms−1vrms=√3RTM=100√2v′rms=√3R(21)M=√2×100√2200m/sHence,optionBiscorrectanswer.

If the r. m. s. speed of a gas molecule at $27^{\circ} C$ is $100 \sqrt{2} {m s}^{- 1}$ the r. m. s. speed at $327^{\circ} C$ would be

The correct option is B $200 {m s}^{- 1}$
$\begin{matrix} v_{r m s} = \sqrt{\frac{3 R T}{M}} = 100 \sqrt{2} v_{r m s}^{'} = \sqrt{\frac{3 R (21)}{M}} = \sqrt{2} \times 100 \sqrt{2} 200 m / s H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$