Question:medium
The average kinetic energy of a monoatomic gas molecule kept at temperature \(27^\circ\)C is \((\text{Boltzmann constant}=1.3\times 10^{-23}\ \text{J K}^{-1})\)
The average kinetic energy of a monoatomic gas molecule kept at temperature \(27^\circ\)C is \((\text{Boltzmann constant}=1.3\times 10^{-23}\ \text{J K}^{-1})\)
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For one molecule of a monoatomic gas:
\[
E=\frac{3}{2}kT
\]
Use kelvin, not degree Celsius.
Updated On: Apr 29, 2026
- \(5.85\times10^{-21}\) J
- \(4.12\times10^{-21}\) J
- \(3.75\times10^{-21}\) J
- \(2.85\times10^{-21}\) J
- \(7.55\times10^{-21}\) J
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The Correct Option is A
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