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$^\circ$C

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