Chemistry - Ideal Gas Question with Solution | TestHub

$\overline{\mathrm{KE}}=\frac{3}{2}\mathrm{kT}=\frac{3}{2}\times \frac{8.313}{6.023\times 10^{23}}\times 298=6.17\times 10^{-21}\mathrm{J}$ (Average Kinetic energy $\overline{\mathrm{KE}}=\frac{3}{2}\mathrm{kT}=\frac{3}{2}\frac{\mathrm{R}}{\mathrm{N}}\mathrm{T})$