The mean translational kinetic energy of a gas molecule at a temperature of 47^{\circ} {\C} is (. Boltzman constant =1.38 \times 10^{-23} {\JK}^{-1})

Correct Answer is : 414×10−4eV

AP EAPCET Medical 20 May 2025 Shift 2 Paper

Question : 100 of 160

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The mean translational kinetic energy of a gas molecule at a temperature of 47∘C is
(. Boltzman constant =1.38×10−23JK−1)

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