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(D) The average kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.$K.E. \propto T$Therefore

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Four times

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(D) The average kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.$K.E. \propto T$Therefore,the ratio of kinetic energies is given by:$\frac{E_1}{E_2} = \frac{T_1}{T_2}$Convert the temperatures from Celsius to Kelvin:$T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$$T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$Substitute the values into the ratio:$\frac{E_1}{E_2} = \frac{300}{1200} = \frac{1}{4}$This implies:$E_2 = 4E_1$Thus,the kinetic energy becomes four times the initial value.

If the temperature of a gas increases from $27^{\circ}C$ to $927^{\circ}C$,the $K.E.$ will be

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