NEET-XII-Physics
24: Kinetic Theory of Gases
- #14The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 × 10-19 J). Calculate the temperature of the air. Boltzmann constant k = 1.38 × 10-23 J K-1.Ans : We know from kinetic theory of gases that the average translational energy per molecule is `` \frac{3}{2}kT``.
Now,
Eavg= 0.040 eV = `` 0.040\times 1.6\times {10}^{-19}=6.4\times {10}^{-21}J``
`` 6.40\times {10}^{-21}=\frac{3}{2}\times 1.38\times {10}^{-23}\times T``
`` \Rightarrow T=\frac{2}{3}\times \frac{6.40\times {10}^{-21}}{1.38\times {10}^{-23}}=309.2K``