The average kinetic energy of a gas at -23^{c | vedclass

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Question

(D) The average kinetic energy $(E)$ of an ideal gas molecule is given by the formula: $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$

Vedclass · Accepted Answer

$10 	imes 10^{-14} \, erg$

Vedclass · Answer

(D) The average kinetic energy $(E)$ of an ideal gas molecule is given by the formula: $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.Note that the average kinetic energy depends only on the temperature $T$ and is independent of the pressure and the nature of the gas.Given:$T_1 = -23^{\circ}C = 273 - 23 = 250 \, K$$E_1 = 5 	imes 10^{-14} \, erg$$T_2 = 227^{\circ}C = 273 + 227 = 500 \, K$Using the ratio:$\frac{E_1}{E_2} = \frac{T_1}{T_2}$$\frac{5 	imes 10^{-14}}{E_2} = \frac{250}{500}$$\frac{5 	imes 10^{-14}}{E_2} = \frac{1}{2}$$E_2 = 2 	imes 5 	imes 10^{-14} = 10 	imes 10^{-14} \, erg$.

The average kinetic energy of a gas at $-23^{\circ}C$ and $75 \, cm$ pressure is $5 \times 10^{-14} \, erg$ for $H_2$. The mean kinetic energy of $O_2$ at $227^{\circ}C$ and $150 \, cm$ pressure will be:

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