At what temperature is the kinetic energy of | 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}{

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$327^{\circ}C$

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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 given by $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given:$T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.$E_2 = 2E_1$.Substituting these values into the ratio formula:$\frac{E_1}{2E_1} = \frac{300 \ K}{T_2}$$\frac{1}{2} = \frac{300 \ K}{T_2}$$T_2 = 600 \ K$.Converting back to Celsius:$T_2 = 600 - 273 = 327^{\circ}C$.

At what temperature is the kinetic energy of a gas molecule double that of its value at $27^{\circ}C$?

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