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If the average kinetic energy of a gas molecule at $27^{\circ} \mathrm{C}$ is $3.3 \times 10^{-20} \mathrm{~J}$, then the average kinetic energy of the gas molecules at $127^{\circ} \mathrm{C}$ is
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The correct answer is:
$4.4 \times 10^{-20} \mathrm{~J}$
Average kinetic energy of a gas is given by
$\begin{aligned} & E=\frac{1}{2} K T \\ & \frac{E_2}{E_1}=\frac{T_2}{T_1} \Rightarrow E_2=\frac{T_2 E_1}{T_1} \\ & =\frac{(127+273) \times 3.3 \times 10^{-20}}{(27+273)} \\ & =\frac{400 \times 3.3 \times 10^{-20}}{300}=4.4 \times 10^{-20} J\end{aligned}$
$\begin{aligned} & E=\frac{1}{2} K T \\ & \frac{E_2}{E_1}=\frac{T_2}{T_1} \Rightarrow E_2=\frac{T_2 E_1}{T_1} \\ & =\frac{(127+273) \times 3.3 \times 10^{-20}}{(27+273)} \\ & =\frac{400 \times 3.3 \times 10^{-20}}{300}=4.4 \times 10^{-20} J\end{aligned}$
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