Q 11-12-074JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
The root mean square speed of molecules of nitrogen gas at $27^\circ\text{C}$ is approximately (Given mass of a nitrogen molecule $=4.6\times10^{-26}\ \text{kg}$ and take Boltzmann constant $k_B=1.4\times10^{-23}\ \text{J K}^{-1}$)
Answer: (D) $523\ \text{m s}^{-1}$
$$v_{rms}=\sqrt{\frac{3k_BT}{m}}=\sqrt{\frac{3\times1.4\times10^{-23}\times300}{4.6\times10^{-26}}}=\sqrt{2.74\times10^5}\approx523\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics