The average translational kinetic energy of a | vedclass

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Question

(C) The average translational kinetic energy of a gas molecule is given by the formula $E_{avg} = \frac{3}{2} k_B T$.At $NTP$ (Normal Temperature and

Vedclass · Accepted Answer

$0.56 	imes 10^{-20} 	ext{ J}$

Vedclass · Answer

(C) The average translational kinetic energy of a gas molecule is given by the formula $E_{avg} = \frac{3}{2} k_B T$.At $NTP$ (Normal Temperature and Pressure),the temperature $T = 273 	ext{ K}$.Given $k_B = 1.38 	imes 10^{-23} 	ext{ J/K}$.Substituting the values: $E_{avg} = \frac{3}{2} 	imes 1.38 	imes 10^{-23} 	imes 273$.$E_{avg} = 1.5 	imes 1.38 	imes 273 	imes 10^{-23} 	ext{ J}$.$E_{avg} = 565.11 	imes 10^{-23} 	ext{ J} = 0.565 	imes 10^{-20} 	ext{ J}$.Rounding to the nearest option,we get $0.56 	imes 10^{-20} 	ext{ J}$.

The average translational kinetic energy of a hydrogen gas molecule at $NTP$ will be. [Boltzmann's constant $k_B = 1.38 \times 10^{-23} \text{ J/K}$]

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