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(C) The mean kinetic energy $E$ of gas molecules is given by the formula:$E = \frac{f}{2} k T$where $f$ is the degrees of freedom,$k$ is the Boltzmann

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(C) The mean kinetic energy $E$ of gas molecules is given by the formula:$E = \frac{f}{2} k T$where $f$ is the degrees of freedom,$k$ is the Boltzmann constant,and $T$ is the absolute temperature in Kelvin.Since $T = t + 273$,we can write:$E = \frac{f}{2} k (t + 273) = \left( \frac{f}{2} k ight) t + \left( \frac{f}{2} 	imes 273 k ight)$Comparing this with the equation of a straight line $y = mx + c$,where $y = E$ and $x = t$:Slope $m = \frac{f}{2} k$ (which is positive)Intercept $c = \frac{f}{2} 	imes 273 k$ (which is positive)Thus,the graph of $E$ versus $t$ is a straight line with a positive slope and a positive intercept on the $E$-axis. This corresponds to the graph shown in option $C$.

The graph which represents the variation of mean kinetic energy of molecules with temperature $t^{\circ} \text{C}$ is

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