Q 11-12-021NEETJEE MainEasy
The average translational kinetic energy of a gas molecule at $27^\circ$C is about ($k_B = 1.38 \times 10^{-23}$ J K$^{-1}$)
Answer: (C) $6.2 \times 10^{-21}$ J
$E = \dfrac{3}{2}k_BT = 1.5 \times 1.38 \times 10^{-23} \times 300 \approx 6.2 \times 10^{-21}$ J.
Solution by Sreeraj P, M.Sc Physics