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Q. The average translational energy and the rms speed of molecules in a sample of oxygen gas at $300\,K$ are $6.21 \times 10^{-21}\,J$ and $484\,m s ^{-1}$ respectively. The corresponding values of $600\,K$, are nearly (assuming ideal gas behaviour)

Kinetic Theory

Solution:

Using $\bar{KE} = \frac{3}{2} kT$ and $v_{rms} = \sqrt{\frac{3kT}{m}}$, we get $\frac{\bar{KE}_{600}}{\bar{KE}_{300}} = \frac{600}{300} = 2 $ $i.e.$ $\bar{KE}_{600} = 2 \times 6.21 \times 10^{-21} $ = 12.42 $\times 10^{21}$ J $\frac{v_{rms \,600}}{v_{rms\, 300}} \sqrt{\frac{600}{300}} = \sqrt{2}$ $i.e. $ $v_{rms600} = \sqrt{2} \times 484 = 684.4 \, m \, s^{-1}$