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(C) The kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.$K.E. \propto T$Given, initial te

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$3E_0/2$

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(C) The kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.$K.E. \propto T$Given, initial temperature $T_1 = 27^\circ$$C$ = $27 + 273 = 300$ $K$.Initial kinetic energy $K_1 = E_0$.Final temperature $T_2 = 177^\circ$$C$ = $177 + 273 = 450$ $K$.Using the relation $\frac{K_2}{K_1} = \frac{T_2}{T_1}$:$\frac{K_2}{E_0} = \frac{450}{300}$$\frac{K_2}{E_0} = \frac{3}{2}$$K_2 = \frac{3}{2} E_0$.

The kinetic energy of an ideal gas is $E_0$ at $27^\circ$$C$. When the temperature is increased to $177^\circ$$C$, then the kinetic energy will be

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