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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The average translational kinetic energy $E$ of an ideal gas 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 $\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{600}{300} = 2$. So,$E_2 = 2 	imes E_1 = 2 	imes 6.21 	imes 10^{-21} \ J = 12.42 	imes 10^{-21} \ J$. The $rms$ speed $V_{rms}$ is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$. Thus,$V_{rms} \propto \sqrt{T}$. Given $V_1 = 484 \ m/s$,the new speed $V_2 = V_1 	imes \sqrt{\frac{T_2}{T_1}} = 484 	imes \sqrt{2} \approx 484 	imes 1.414 = 684.37 \ m/s \approx 684 \ m/s$. Therefore,the values are $12.42 	imes 10^{-21} \ J$ and $684 \ m/s$.

At a temperature of $300 \ K$,the average translational kinetic energy and $rms$ speed of a sample of oxygen gas are $6.21 \times 10^{-21} \ J$ and $484 \ m/s$ respectively. At $600 \ K$,these values will be respectively: (Assume ideal gas behavior)

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