The root mean square speed of molecules of nitrogen gas at 27° is approximately : (Given mass of a nitrogen molecule $= 4.6 \times 10^{- 26} kg$ and take Boltzmann constant $k_{B} = 1.4 \times 10^{- 23} {JK}^{- 1}$ )

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$27.4 m / s$

$1260 m / s$

$91 m / s$

$523 m / s$

answer is D.

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Detailed Solution

R.M.S speed $= \sqrt{\frac{3 RT}{M}} = \sqrt{\frac{3 KT}{M}} = \sqrt{\frac{3 \times 1.4 \times 10^{- 23} \times 300}{4.6 \times 10^{- 26}}} ≃ 523 m / s$

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