The average translational energy and the rms | vedclass

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Question

(D) The average translational kinetic energy of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.Since the temperature increases

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The average translational kinetic energy of a gas molecule is given by $E = \frac{3}{2} k_B T$. Thus,$E \propto T$.Since the temperature increases from $300 \ K$ to $600 \ K$ (doubled),the new average energy $E'$ will be $2 	imes 6.21 	imes 10^{-21} \ J = 12.42 	imes 10^{-21} \ J$.The root mean square (rms) speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$v_{rms} \propto \sqrt{T}$.When the temperature is doubled,the new rms speed $v'_{rms}$ will be $\sqrt{2} 	imes v_{rms} = 1.414 	imes 484 \ m/s \approx 684 \ m/s$.Therefore,the correct values at $600 \ K$ are $12.42 	imes 10^{-21} \ J$ and $684 \ m/s$.

The average translational energy and the rms speed of molecules of 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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