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(D) The average kinetic energy $(E)$ of an ideal gas is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2}

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Four times

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(D) The average kinetic energy $(E)$ of an ideal gas 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 is $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given temperatures in Celsius: $T_1 = 27^{\circ}C$ and $T_2 = 927^{\circ}C$.Converting to Kelvin: $T_1 = 27 + 273 = 300 \ K$ and $T_2 = 927 + 273 = 1200 \ K$.Substituting the values: $\frac{E_1}{E_2} = \frac{300}{1200} = \frac{1}{4}$.Thus,$E_2 = 4 E_1$,which means the kinetic energy becomes four times the initial value.

If the temperature of a gas is increased from $27^{\circ}C$ to $927^{\circ}C$,what will be the change in its kinetic energy?

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