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(B) The average kinetic energy $(K_{avg})$ of an ideal gas molecule is directly proportional to its absolute temperature $(T)$ in Kelvin,given by the

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$2 E$

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(B) The average kinetic energy $(K_{avg})$ of an ideal gas molecule is directly proportional to its absolute temperature $(T)$ in Kelvin,given by the formula: $K_{avg} = \frac{3}{2} k_B T$.Given:Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.Initial kinetic energy = $E$.Final temperature $T_2 = 327^{\circ} C = 327 + 273 = 600 \ K$.Since $K_{avg} \propto T$,we have:$\frac{K_2}{K_1} = \frac{T_2}{T_1}$$\frac{K_2}{E} = \frac{600 \ K}{300 \ K}$$\frac{K_2}{E} = 2$$K_2 = 2 E$.Therefore,the final average kinetic energy will be $2 E$.

The initial average kinetic energy of the molecules was $E$,when a gas sample is at $27^{\circ} C$. When the gas is heated to $327^{\circ} C$,then the final average kinetic energy will be

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