Question 76: Thermodynamics & KTG

Question

The average kinetic energy of a molecule in air at room temperature of 20$^\circ$C

Accepted Answer

The average molecular kinetic energy is expressed as
$\overline K  = {1 \over 2}m{v^2} = {3 \over 2}kT$
Given, $T = 20^\circ C = 20 + 273 = 293\,K$
$k = 1.38 	imes {10^{ - 23}}$ J/K = Boltzmann's constant
Substituting all the values, we get
$\overline K  = {3 \over 2} 	imes 1.38 	imes {10^{ - 23}} 	imes 293$
$ = 6.07 	imes {10^{ - 21}}$ J

Answer

$6	imes10^{-22}$ J

Answer

$7.06	imes10^{-21}$ J

Answer

$6.07	imes10^{-21}$ J

Answer

$6.70	imes10^{-21}$ J

Thermodynamics & KTG

The average kinetic energy of a molecule in air at room temperature of 20^\circC

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