The average translational kinetic energy of t | vedclass

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Question

(C) The average translational kinetic energy $(K_{avg})$ of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$. Given: Temperature

Vedclass · Accepted Answer

$8.28 	imes 10^{-21} \,J$

Vedclass · Answer

(C) The average translational kinetic energy $(K_{avg})$ of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$. Given: Temperature $T = 127^{\circ} C = 127 + 273 = 400 \,K$. Boltzmann constant $k_B = 1.38 	imes 10^{-23} \,J \,K^{-1}$. Substituting the values into the formula: $K_{avg} = \frac{3}{2} 	imes (1.38 	imes 10^{-23}) 	imes 400$. $K_{avg} = 1.5 	imes 1.38 	imes 400 	imes 10^{-23}$. $K_{avg} = 1.5 	imes 552 	imes 10^{-23}$. $K_{avg} = 828 	imes 10^{-23} \,J$. $K_{avg} = 8.28 	imes 10^{-21} \,J$.

The average translational kinetic energy of the oxygen molecules at a temperature of $127^{\circ} C$ is (Boltzmann constant $= 1.38 \times 10^{-23} \,J \,K^{-1}$)

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