The average kinetic energy of a molecule of a | vedclass

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Question

(B) The average kinetic energy $(KE_{avg})$ of a single molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k_B T$,where $k_B$ i

Vedclass · Accepted Answer

$6.13 	imes 10^{-21} \, J$

Vedclass · Answer

(B) The average kinetic energy $(KE_{avg})$ of a single molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant $(1.38 	imes 10^{-23} \, J \, K^{-1})$ and $T$ is the temperature in Kelvin.Given temperature $T = 25 + 273.15 = 298.15 \, K$.Substituting the values: $KE_{avg} = \frac{3}{2} 	imes (1.38 	imes 10^{-23} \, J \, K^{-1}) 	imes 298.15 \, K$.$KE_{avg} = 1.5 	imes 1.38 	imes 298.15 	imes 10^{-23} \, J$.$KE_{avg} \approx 6.17 	imes 10^{-21} \, J$.Rounding to the nearest provided option,the correct value is $6.13 	imes 10^{-21} \, J$.

The average kinetic energy of a molecule of an ideal gas at $25 \, ^\circ C$ is .........

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