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The kinetic energy of one molecule of a gas at normal temperature and pressure will be $(k=8.31 \mathrm{~J} / \mathrm{mole} \mathrm{K})$
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Verified Answer
The correct answer is:
$3.4 \times 10^3 \mathrm{~J}$
According to kinetic theory, K.E. of 1g-mole of an ideal gas, $E=\frac{3}{2} R T$
Hence, K.E. at normal temperature $0^{\circ} \mathrm{C}$
$\begin{aligned}& =273 \mathrm{~K} \\\text { or } \quad E & =\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^3 \mathrm{~J}
\end{aligned}$
Hence, K.E. at normal temperature $0^{\circ} \mathrm{C}$
$\begin{aligned}& =273 \mathrm{~K} \\\text { or } \quad E & =\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^3 \mathrm{~J}
\end{aligned}$
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