The root mean square speed of molecules of nitrogen gas at 27 is approximately: (Given mass of a nitrogen molecule = 4.6 × 10⁻²⁶ kg and take Boltzmann constant kB = 1.4 × 10⁻²³ J K⁻¹) (a) 91 m/s (b) 1260 m/s (c) 27.4 m/s (d) 523 m/s

Answer: (d). vrms = √((3kBT)/(m)) with T = 300 K vrms = √3×1.4×10⁻²³×3004.6×10⁻²⁶ = √(2.74×10⁵) vrms ≈ 523 m/s So the answer is (d) 523 m/s.

The root mean square speed of molecules of nitrogen gas at 27 is… | JEE Main 2023 (11 Apr, Shift 2)

Q45JEE Main 2023 · 11 APR · SHIFT 2Medium

Molecular Speeds & Maxwell Distribution

The root mean square speed of molecules of nitrogen gas at $27^\circ\text{C}$ is approximately: (Given mass of a nitrogen molecule $= 4.6 \times 10^{-26}\text{ kg}$ and take Boltzmann constant $k_B = 1.4 \times 10^{-23}\text{ J K}^{-1}$ )

Show answer and solution

Answer: (d)

$v_{rms} = \sqrt{\frac{3k_BT}{m}}$ with $T = 300\text{ K}$
$v_{rms} = \sqrt{\frac{3\times1.4\times10^{-23}\times300}{4.6\times10^{-26}}} = \sqrt{2.74\times10^5}$
$v_{rms} \approx 523\text{ m/s}$
So the answer is (d) 523 m/s.

Report a mistake

Molecular Speeds & Maxwell Distribution question from JEE Main 2023 (11 Apr, Shift 2)

More questions on Molecular Speeds & Maxwell Distribution