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 $(T)$,given by the relation $E = \frac{3}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.Therefore,$E \propto T$,which implies $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given temperatures in Kelvin:$T_1 = 27 + 273 = 300\,K$$T_2 = 227 + 273 = 500\,K$Given energy $E_1 = 6.21 	imes 10^{-21}\,J$.Substituting the values into the ratio:$\frac{6.21 	imes 10^{-21}}{E_2} = \frac{300}{500}$$\frac{6.21 	imes 10^{-21}}{E_2} = \frac{3}{5}$$E_2 = \frac{6.21 	imes 10^{-21} 	imes 5}{3}$$E_2 = 2.07 	imes 10^{-21} 	imes 5 = 10.35 	imes 10^{-21}\,J$.

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

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