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(C) The average kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2}kT$,where $T$ is the absolute temperature in Kelvin and $k$ is the Bo

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(C) The average kinetic energy $E$ of a gas molecule is given by $E = \frac{3}{2}kT$,where $T$ is the absolute temperature in Kelvin and $k$ is the Boltzmann constant.Since $T = t + 273$,where $t$ is the temperature in degrees Celsius,we can write:$E = \frac{3}{2}k(t + 273)$$E = \frac{3}{2}kt + \frac{3}{2}k(273)$This equation is in the form of a linear equation $y = mx + c$,where $y = E$,$x = t$,$m = \frac{3}{2}k$ (slope),and $c = \frac{3}{2}k(273)$ (y-intercept).Since the y-intercept $c$ is positive,the graph is a straight line that does not pass through the origin but intersects the y-axis at a positive value.Therefore,the correct graph is the one where the line starts from a positive value on the y-axis.

Graph of kinetic energy of a gas molecule versus temperature $t^{\circ}C$.

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