The temperature of a gas is -78^{circ} C and | vedclass

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(B) The average translational kinetic energy of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant a

Vedclass · Accepted Answer

$117$

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(B) The average translational kinetic energy of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.From the relation $K \propto T$,we can see that the kinetic energy is directly proportional to the absolute temperature.Given initial temperature $T_i = -78^{\circ} C$. Converting this to Kelvin: $T_i = 273 + (-78) = 195 \ K$.We want the new kinetic energy $K' = 2K$. Since $K \propto T$,if the kinetic energy doubles,the absolute temperature must also double.Therefore,the new absolute temperature $T_f = 2 	imes T_i = 2 	imes 195 \ K = 390 \ K$.Converting the final temperature back to Celsius: $T_f(^{\circ} C) = 390 - 273 = 117^{\circ} C$.

The temperature of a gas is $-78^{\circ} C$ and the average translational kinetic energy of its molecules is $K$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is: (in $^{\circ} C$)

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