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(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$, given by the relation $E = \frac{3}

Vedclass · Accepted Answer

$4.4 	imes 10^{-20} \,J$

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(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$, given by the relation $E = \frac{3}{2} k_B T$. Therefore, the ratio of kinetic energies at two different temperatures is $\frac{E_2}{E_1} = \frac{T_2}{T_1}$. Given: $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$, $E_1 = 3.3 	imes 10^{-20} \ J$, and $T_2 = 127^{\circ} C = 127 + 273 = 400 \ K$. Substituting the values: $E_2 = E_1 	imes \frac{T_2}{T_1} = 3.3 	imes 10^{-20} 	imes \frac{400}{300}$. $E_2 = 3.3 	imes 10^{-20} 	imes \frac{4}{3} = 1.1 	imes 4 	imes 10^{-20} = 4.4 	imes 10^{-20} \ J$.

If the average kinetic energy of a gas molecule at $27^{\circ} C$ is $3.3 \times 10^{-20} \,J$, then the average kinetic energy of the gas molecules at $127^{\circ} C$ is

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