$\begin{aligned}
& \mathrm{V}{\mathrm{ms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \
& \mathrm{~V}{\mathrm{rms}}^2=3 \mathrm{RT} / \mathrm{M}
\end{aligned}Hencewecanconcludethat\mathrm{V}_{\mathrm{rms}}^2isdirectlyproportionaltotemperature\mathrm{y}=\mathrm{mx}\Rightarrow$ Graph will be straight line



