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Question

(A) The average kinetic energy $E$ of a monoatomic gas molecule is given by the formula $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T

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Less than $1\, eV$

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(A) The average kinetic energy $E$ of a monoatomic gas molecule is given by the formula $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.Given temperature $T = 30^oC = 30 + 273 = 303\,K$.Boltzmann constant $k = 1.38 	imes 10^{-23}\,J/K$.Substituting the values: $E = \frac{3}{2} 	imes (1.38 	imes 10^{-23}) 	imes 303 = 6.27 	imes 10^{-21}\,J$.To convert this energy into electron-volts $(eV)$,divide by $1.6 	imes 10^{-19}\,J/eV$:$E = \frac{6.27 	imes 10^{-21}}{1.6 	imes 10^{-19}} \approx 0.039\,eV$.Since $0.039\,eV < 1\,eV$,the correct option is $A$.

The average kinetic energy of a helium atom at $30^oC$ is

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