JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
The root mean square speed of molecules of nitrogen gas at \(27^{\circ} C\) is approximately\(.......m/s\)(Given mass of a nitrogen molecule \(=4.6 \times 10^{-26}\,kg\) and take Boltzmann constant \(k _{ B }=1.4 \times 10^{-23}\,JK ^{-1}\) )
- A \(523\)
- B \(1260\)
- C \(91\)
- D \(27.4\)
Answer & Solution
Correct Answer
(A) \(523\)
Step-by-step Solution
Detailed explanation
\(V _{ rms }=\sqrt{\frac{3 k _{ B } T }{ m }}=\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}=523\,m / s\)
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