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(C) The mean kinetic energy of a molecule of an ideal gas is given by $E = \frac{3}{2} kT$,where $k$ is the Boltzmann constant and $T$ is the absolute

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(C) The mean kinetic energy of a molecule of an ideal gas is given by $E = \frac{3}{2} kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature in Kelvin.Since $T(K) = t(^{\circ}C) + 273.15$,we can write the expression as $E = \frac{3}{2} k(t + 273.15) = \frac{3}{2} kt + \frac{3}{2} k(273.15)$.This equation is of the form $y = mx + c$,where $y = E$,$x = t$,$m = \frac{3}{2}k$ (slope),and $c = \frac{3}{2}k(273.15)$ (positive y-intercept).Therefore,the graph of $E$ versus $t^{\circ}C$ is a straight line with a positive slope and a positive y-intercept,which corresponds to the graph shown in option $C$.

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

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