The temperature of a gas is -68^circ C. At wh | vedclass

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(A) The average kinetic energy $(E)$ of gas molecules is directly proportional to the absolute temperature $(T)$ in Kelvin: $E \propto T$.Given the in

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$137$

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(A) The average kinetic energy $(E)$ of gas molecules is directly proportional to the absolute temperature $(T)$ in Kelvin: $E \propto T$.Given the initial temperature $T_1 = -68^\circ C$. Converting this to Kelvin: $T_1 = 273 + (-68) = 205 \ K$.We want the new kinetic energy $E_2$ to be twice the initial kinetic energy $E_1$,so $E_2 = 2E_1$.Using the relation $\frac{E_2}{E_1} = \frac{T_2}{T_1}$,we get:$2 = \frac{T_2}{205 \ K}$.Solving for $T_2$: $T_2 = 2 	imes 205 = 410 \ K$.Converting the final temperature back to Celsius: $T_2(^\circ C) = 410 - 273 = 137^\circ C$.

The temperature of a gas is $-68^\circ C$. At what temperature in $^\circ C$ will the average kinetic energy of its molecules be twice that of at $-68^\circ C$?

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