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(A) The average translational kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.Given $T_1 = 300 \, K$ and

Vedclass · Accepted Answer

$12.42 	imes 10^{-21} \, J, \, 684 \, m/s$

Vedclass · Answer

(A) The average translational kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.Given $T_1 = 300 \, K$ and $T_2 = 600 \, K$,the ratio is $\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2$.Therefore,$E_2 = 2 	imes E_1 = 2 	imes 6.21 	imes 10^{-21} \, J = 12.42 	imes 10^{-21} \, J$.The $r.m.s.$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$v_{rms} \propto \sqrt{T}$.Therefore,$\frac{(v_{rms})_2}{(v_{rms})_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.414$.$(v_{rms})_2 = 1.414 	imes 484 \, m/s \approx 684 \, m/s$.Thus,the values are $12.42 	imes 10^{-21} \, J$ and $684 \, m/s$.

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

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