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
Concept. Average translational KE $= \dfrac32k_BT$.
Step 1. Substitute. $1.5 \times 1.38 \times 10^{-23} \times 300 \approx 6.2 \times 10^{-21}$ J.
Answer. $6.2 \times 10^{-21}$ J (Option C).
Solution by Sreeraj P, M.Sc Physics