The average kinetic energy of a gas molecule | vedclass

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(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$: $E \propto T$.Therefore,the ratio i

Vedclass · Accepted Answer

$10.35 	imes 10^{-21} \, J$

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(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$: $E \propto T$.Therefore,the ratio is given by: $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given:$E_1 = 6.21 	imes 10^{-21} \, J$$T_1 = 27^oC = 27 + 273 = 300 \, K$$T_2 = 227^oC = 227 + 273 = 500 \, K$Substituting the values:$\frac{6.21 	imes 10^{-21}}{E_2} = \frac{300}{500} = \frac{3}{5}$.Solving for $E_2$:$E_2 = \frac{6.21 	imes 10^{-21} 	imes 5}{3} = 2.07 	imes 5 	imes 10^{-21} \, J = 10.35 	imes 10^{-21} \, J$.

The average kinetic energy of a gas molecule at $27^oC$ is $6.21 \times 10^{-21} \, J$. Its average kinetic energy at $227^oC$ will be

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